Thermodynamics of One-Dimensional Solvable Models

Thermodynamics of One-Dimensional Solvable Models

Exactly solvable models are very important in physics. They are important not just from a theoretical point of view but also from the experimentalist's perspective, because in such cases theoretical results and experimental results

can be compared without ambiguity. This is a book about an important class of exactly solvable models in physics.

The subject area is the Bethe-ansatz approach for a number of onedimensional models, and the setting up of equations within this approach to determine the thermodynamics of these systems. It is a topic that crosses the boundaries between condensed matter physics, mathematics and field theory. The derivation and application of thermodynamic Bethe-ansatz equations for one-dimensional models are explained in detail. This technique

is indispensable for physicists studying the low-temperature properties of one-dimensional substances.

This book, written by one of the top physicists in this field, and the originator of much of the work in the field, will be of great interest to theoretical condensed matter physicists.

THERMODYNAMICS OF ONE-DIMENSIONAL SOLVABLE MODELS Minoru Takahashi Institute for Solid State Physics, University of Tokyo, Tokyo, Japan

CAMBRIDGE UNIVERSITY PRESS

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo

Cambridge University Press The Edinburgh Building, Cambridge C132 2RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521551434

© Cambridge University Press 1999

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1999 This digitally printed first paperback version 2005

A catalogue record for this publication is available from the British Library

Library of Congress Cataloguing in Publication data Takahashi, Minoru, 1944Thermodynamics of one-dimensional solvable models / Minoru Takahashi. p. cm. Includes bibliographical references and index. ISBN 0 521 55143 9 (hb) 1. Bethe-ansatz technique. 2. Statistical thermodynamics. 3. Mathematical physics. I. Title. QC20.7.b47T35 1999 530.15-dc2l

98-34997

CIP

ISBN-13 978-0-521-55143-4 hardback ISBN-10 0-521-55143-9 hardback

ISBN-13 978-0-521-01979-8 paperback ISBN-10 0-521-01979-6 paperback

Contents

page xi

Preface

Part one: Thermodynamics of non-interacting systems and ground states of interacting systems 1 1 Free energy and correlation functions of the XY model 1 1.1 The isotropic XY model 1 1.1.1 Introduction and historical overview 1 1.1.2 Energy eigenvalues of the Hamiltonian and the partition function 3 1.1.3 Correlation functions 4 1.2 The anisotropic XY model 7 1.2.1 The subspace a = 1 8

1.2.2 The subspace a = -1

9

1.2.3 The free energy 2 Systems with a delta function potential 2.1 The boson problem 2.1.1 The c = 0 case 2.1.2 The c = oo case 2.1.3 Scattering states of Bose particles with finite interaction 2.1.4 Periodic boundary conditions 2.1.5 Linear integral equation for the distribution of quasi-momenta 2.1.6 Bound states in the case c < 0 2.1.7 Elementary excitations for repulsive bosons 2.2 The fermion problem 2.2.1 The c = 0 case 2.2.2 The two-body problem 2.2.3 The three-body problem

9

2.2.4 The M = 1 and arbitrary N case 2.2.5 The arbitrary M and N case 2.2.6 Periodic boundary conditions v

10 10 10 11

12 14 15 16 18

20 20 21

22 25 25 26

Contents

vi

2.2.7 The ground state for c > 0 2.2.8 The ground state for c < 0 2.2.9 Expansion from small n j 2.2.10 A unified form of the integral equations 2.3 Bosons and fermions with arbitrary spin 3 The isotropic Heisenberg model 3.1 The ferromagnetic case 3.2 The string solution of an infinite system 3.3 The Hulthen solution for an antiferromagnet 3.4 The des Cloizeaux-Pearson mode of an antiferromagnet 3.5 The magnetic susceptibility and magnetization curve for J < 0 3.6 Wiener-Hopf type integral equation 4 The XXZ model 4.1 Symmetry of the Hamiltonian 4.2 The Bethe-ansatz wave function 4.3 The string solution for A > 1 4.4 The lowest energy state for A < -1 4.5 The magnetization curve for a field in the z-direction at A < -1 4.6 The lowest energy state for fixed M and -1 < A < 1 4.7 The magnetization curve for a field in the z-direction at

-1
4.8 Susceptibility for -1 < A < 1 4.9 The long range order of the XXZ model 4.10 Excitations from the ground state 4.10.1 A < -1 and the energy gap 4.10.2 Excitations for -1 < A < 1 4.10.3 Spin-wave like excitations for 0 < A < 1 4.10.4 The spin-wave bound state 5 XYZ and eight-vertex models 5.1 Transfer matrix of the eight-vertex model 5.1.1 Relation between the six-vertex model and the XXZ model 5.1.2 The eight-vertex model and the XYZ model 5.2 The symmetry of the XYZ model 5.3 Modulus l and modulus k 5.4 The case Jx = 0 5.5 The ground state for JZ > 0 5.6 Long range order 5.7 Elementary excitations 5.7.1 The nearly degenerate ground state 5.7.2 Spinon excitations

28 29 31

32 32 34 34 37 37 39 41 44 46 46 47 48 49 51

52 53 54 56 59 59 61 62 64 67 67 67 70 73 74 76 77 79 80 80 80

Contents

5.7.3 Spin-wave excitations 5.7.4 Spin-wave bound states 6 The Hubbard model 6.1 Symmetry of the Hamiltonian 6.1.1 Particle-hole symmetry 6.1.2 S U(2) symmetry 6.2 The Bethe-ansatz equation for the one-dimensional Hubbard model 6.2.1 The wave function for a finite system 6.2.2 Periodic boundary condition 6.2.3 Fredholm type integral equations for the ground state 6.2.4 Analytic solution for the ground state in the half-filled case 6.2.5 Spinon excitation in the half-filled case 6.2.6 Energy gap of the charge excitation 6.2.7 Susceptibility and magnetization curve of the half-filled case 6.3 11 U' expansion 6.4 Perturbation expansion in the half-filled case 6.5 Asymptotic expansion from U = 0 Part two: Finite temperature integral equations for un-nested systems 7 Repulsive delta function bosons 7.1 Uniqueness of the solution 7.2 Holes of quasi-momenta and their distribution function 7.3 Thermodynamic equilibrium 7.4 Elementary excitations 7.5 Some special limits 7.5.1 The c = oo limit

7.5.2 The c = 0+ limit 7.5.3 The T = 0+ limit 8 Thermodynamics of the XXX chain 8.1 String solution of an infinite XXX chain 8.2 String hypothesis for a long XXX chain 8.3 Thermodynamic Bethe-ansatz equations for the XXX chain 8.4 Some special cases and expansions 8.4.1 The J/T 0 case 8.4.2 High-temperature expansion or small J expansion 8.4.3 The low-temperature limit 8.4.4 The fugacity expansion 9 Thermodynamics of the XXZ model 9.1 Thermodynamic equations for the XXZ model for A > 1

vii 81

83 85

85 85 87 88 88

90 90 92 92 95 97 101

102 104

109 109

109 110 111

113 115 115 116 116 117

117 118 120 123 123 124 126 127 130

130

Contents

viii

9.2 Theory for the JAI < 1 XXZ model 9.2.1 String solution of the infinite XXZ model with JAI < 1 9.2.2 Scattering phase shift among strings 9.2.3 Bethe-ansatz equation for the XXZ model with JAI < 1 9.3 Some special limits 9.3.1 The T oc or J -* 0 limit

9.3.2 ThecaseJ >0, -10, 0 0, Jx < 0 and T -+ 0 limit 10.2.3 The JZ > 0, Jx > 0 and T -* 0 limit 10.2.4 The Jx = 0 case 11 Low-temperature thermodynamics 11.1 The XXZ model 11.1.1 The XXZ model at 2h > J(1 - A), A < 1 11.1.2 The case J(1 - A)/2 > h > 0 11.2 Roger's dilogarithm and specific heat at h = 0 11.2.1 Specific heat of the XXX antiferromagnet 11.3 The ferromagnetic chain and modified spin-wave theory 11.3.1 Numerical analysis of the thermodynamic Bethe-ansatz equation 11.3.2 Spin-wave calculation of the 1D ferromagnetic chain 11.4 The antiferromagnetic XXX model 11.5 The XYZ model at IJxI < Jy < JZ Part three: Finite temperature integral equations for nested systems

12 S = 1/2 fermions with repulsive potential in the continuum 12.1 Derivation of the thermodynamic equations 12.2 Some special limits 12.2.1 c - * 0+

12.2.2 coo 13 S = 1/2 fermions with an attractive potential 13.1 Derivation of the thermodynamic equations 13.2 Some special limits 13.2.1 c -* 013.2.2 T --> 0+

133 133 138 139 143 143 144 144 144 145 145 149 149 150 150 151 152 152 152 152 153 153 155 155 155 156 157

159 159 159 164 164 166 167 167 171 171

172

Contents

ix

14 Thermodynamics of the Hubbard model 174 14.1 Strings of the Hubbard model 174 14.2 Thermodynamic Bethe-ansatz equation for the Hubbard model 177 179 14.3 Some special limits 14.3.1 The limit U oo 179 14.3.2 The limit U 0 180 14.3.3 The limit T -* 0 182 14.3.4 The limit t -* 0 183 Part four: The quantum transfer matrix and recent developments

185

15 The transfer matrix and correlation length 15.1 The transfer matrix for the Ising chain 15.2 The transfer matrix for the classical Heisenberg model 16 The spin 112 XXZ model in a magnetic field 16.1 The diagonal-to-diagonal transfer matrix 16.2 The limit of an infinite Trotter number 16.3 Analytical solutions for special cases 16.3.1 The Ising limit 16.3.2 The XY limit 16.3.3 The T = h = 0 case 16.4 Numerical calculations of the XXZ model 17 The XYZ model with no magnetic field 17.1 The transfer matrix for the XYZ model 17.1.1 Baxter's theory for the inhomogeneous eight-vertex model 17.1.2 Transcendental equations 17.1.3 The limit of M -* oo 17.2 Special cases and numerical methods 17.2.1 The T -* 0 limit 17.2.2 The Jx = 0 case (anisotropic XY chain) 17.3 Numerical calculations 18 Recent developments and related topics 18.1 Numerical analysis of the S = 1 chain Appendix A The Young tableau and the theorem of Lieb and Mattis Appendix B The number of string solutions Appendix C The commuting transfer matrix and spectral parameter Appendix D The matrix Q(v) Appendix E Special functions

185 185 186 189 189 198 201 201

Bibliography Index

246

202 203 205 210 210 211 212 215 218 218 220 221 223 223 226 228 231

235 239

251

Preface

The subject of this book is on the borders of condensed matter physics, mathematics and field theory. It is believed that the N body problem is soluble only for N = 2, but is not soluble for systems with N >_ 3. But, considering one dimensional systems, there are many solvable models, like the XXZ model. In condensed matter theory, the essential problem is that of solving many-body interacting systems and it is rare that we encounter solvable cases. But in these rare cases we can compare the theoretical results and experimental fact in detail. This kind of work is very valuable. Actually, orbits of planets and the energy spectrum of a hydrogen atom are treated in the regime of the exactly solvable case of two-body systems. Thus, the knowledge of exactly solvable systems is important not only for theorists but also for experimentalists. This book is planned for readers who have taken an elementary course of statistical mechanics and quantum mechanics. There are also solvable two-dimensional classical systems like the six-vertex

model. The Hamiltonians of 1D quantum systems and the transfer matrices of 2D classical systems sometimes have common eigenstates. In many cases

we can write down many-body eigenfunctions of these matrices by the method of the Bethe-ansatz. The N-body wave function is represented as a linear combination of N ! plane waves with N quasi-momenta. The energy eigenvalue of the lowest energy state in the thermodynamic limit is reduced to a distribution function of the quasi-momenta. The distribution function must satisfy a linear integral equation. The energy per unit length is obtained by solving this integral equation. Elementary excitations from the ground state are expressed by the deviation of the distribution of quasi-momenta from the equilibrium. In many cases the quasi-momenta are all real numbers for the ground state. But for general eigenstates, quasi-momenta are complex numbers. Fortunately for the delta-function Boson case, quasi-momenta are always xi

xii

Preface

real. Yang and Yang introduced the thermodynamic theory for this case. The total energy is represented by the distribution function of quasi-momenta. Entropy is represented by the distribution functions of quasi-momenta and holes. By the condition of minimizing the free energy, we can derive equations for these two distribution functions. These are coupled non-linear equations. We then obtain the thermodynamic potential as a function of the chemical potential and temperature. We can calculate all thermodynamic quantities such as energy, entropy, pressure, specific heat and so on. This method can be extended to other Bethe-ansatz soluble models. Then next simplest case is the spin-1/2 Heisenberg model. In this case one cannot

obtain the correct free energy assuming that the quasi-momenta are all real. But complex quasi-momenta group together to form a string in the rapidity plane in the limit of a big system. The length of the string is an arbitrary natural number 1, 2, 3, ... We must then consider the distribution functions for strings with length 1, 2, ... This is a set of coupled integral equations which has an infinite number of unknown functions. It looks very complicated, but we can solve this set of coupled equations analytically in some special limit. The asymptotic behaviour of the distribution function can be calculated for a very long string. We can thus approximate this set

of equations by one with a finite number of unknown functions and get thermodynamic quantities numerically for a given temperature. In almost all cases, the thermodynamic Bethe-ansatz equations have an infinite number of unknown functions, as in delta-function fermions, the Kondo problem and the Hubbard model. The string assumption is indispensable for determining the thermodynamics of Bethe-ansatz soluble models.

In this book we mainly treat the delta-function quantum gas, the 1D Heisenberg model and the Hubbard model. We stress the thermodynamic properties of these models. Except in the case of a boson gas system, strings appear in the complex plane of rapidity. For general eigenstates, these strings should be considered. The length of the string is dependent on the

parameters of the system. But once we accept the string assumption we can construct the equation for the free energy of these models at a given temperature. In many cases the equation contains an infinite number of unknown functions. But, surprisingly, this set of equations can be solved in some special limits and we obtain the exact free energy of these complex models in the limit of an infinite system. We can treat the problem using numerical methods for the finite systems. Bethe-ansatz equations give very interesting results which cannot be predicted using direct numerical analysis. For example the susceptibility exponent of a 1D Heisenberg ferromagnet

was expected to be less than two for a long time. But it was established

Preface

xiii

that y is exactly two by the thermodynamic Bethe-ansatz method. This was later confirmed by the numerical Monte Carlo method. Very recently the anomalies of the susceptibility of a Heisenberg antiferromagnet was also found by this method. For these soluble models many analytical and important results are obtained. The ground state energy properties are also derived as a special case of the zero temperature limit. We can calculate low temperature properties of these models. To get analytical results we need

various mathematical tools. In the process of calculation of Bethe-ansatz equations, many mathematical tools have been invented. Physically new results are also obtained by these methods. The author is grateful to Dr. E. Williams for his critical reading of the manuscript.

Part one Thermodynamics of non-interacting systems and ground states of interacting systems 1

Free energy and correlation functions of the XY model 1.1 The isotropic XY model 1.1.1 Introduction and historical overview

As the simplest case of a solvable model we consider the following spin 1/2 Hamiltonian, N

N

ix Si+1 + si S+1

SN+1

2h

i=1

= S1

i=1

[Sk, Sh

= i5klEafySk.

Each site has two states: an up-spin state and a down-spin state. Si, and S are represented by the Pauli matrices, 1

six = 26j

U)-\1O

SY

Sj

= 26 0

1

= 26j, i

0

1

0)'

-1

For this Hamiltonian there are 2N states, and it can be transformed as follows: N

S+Si+I + Si Si+1 - 2h

Si , Sk - Sk ± iSk .

(1.4)

Lieb, Schultz and Mattis61 and Katsura47 investigated this model in detail. The operators Sk satisfy the relations {S+, Si } = 1,

[Si+, Sjj = 0 1

for

i

j.

(1.5)

Free energy and correlation functions of the XY model

2

The above commutation relations are neither fermionic nor bosonic. But if we introduce the operators k-1 Ck =

k-1

11(2S S

- 1)S+ ck = SC fl(2SItSti

i=1

- 1),

i=1

these satisfy the fermionic commutation relations {Ckt,CI}

= 6kl,

{Ck,CI} = {Ck,

}

= 0.

(1.6)

Fortunately one can find the expression for spin operators using these fermionic operators k-1

SC

k-1

St =

(1 - 2Cti C1)Ck, i=1

1(l - 2ci Cj)Ck.

(1.7)

i=1

The Hamiltonian (1.1) is transformed as N-1

2

c+1ci + Ci Ci+1 + 2 a(CiCN

+ cNC1)

- hN + 2h

i=1

where a -

N cti cl,

(1.8)

i=1

nk1(1-2ckck). The total number of down-spins M is a constant

of motion. The value of a is (-1)M. We introduce the Fourier transformation of these fermionic operators: Cq =

N

1

N k=1

exp(-igk)ck,

q = 2irn/N,

(1.9)

where n is integer (half-odd integer) for odd (even) M. These operators satisfy {Cq, Cpl

=

{Cg, Cp}

= 0,

{Cq, Cp} = bpq.

(1.10)

The Hamiltonian is the same as that of one-dimensional spinless fermions .Yf

= -hN + E(2h - J cos q)cgcq.

(1.11)

The lowest energy state at fixed M is M

T7 t

11 Crz(M+1-21)/NIO)

(1.12)

l=1

and the total energy is

- h(N - 2M) - J

M

cos(7r(M + 1 - 21)/N). l=1

(1.13)

1.1 The isotropic X Y model

3

In this way the XY model is treated by the transformation of spin operators to fermion operators. (1.7) is called the Jordan-Wigner transformation.

1.1.2 Energy eigenvalues of the Hamiltonian and the partition function

General eigenstates for fixed M are M

11 ct(2itIj/N)1O).

(1.14)

j=1

Here {I1,12, ..., IM} is a set of different integers or half-odd integers. If two

sets of integers are different, the two corresponding states are orthogonal. The total number of states with M down-spins and N - M up-spins is given by the binomial coefficient

CM N

N! (1.15)

M!(N-M)!

The total number of states represented by (1.14) is

CN+CN+...+CNN =2N.

(1.16)

Thus the states represented by (1.14) give a complete orthonormal set of eigenstates of the Hamiltonian (1.1). The partition function of this system is N

Z=z-N/2(2[[1(l+ze

Jcos(2n1/N)

ze

T

)]

l=1

1=1

N

N

T

Jcos(n(21-1)/N) T

N

)+rl (1-ze

l=1

Jcos((21-1)N)

)]),

(1.17)

l=1

where z - exp(-2h/T). The first bracket gives the odd M states and the second gives the even M states. The second term in each bracket is much smaller than the first term, and so we can neglect them in the thermodynamic limit.

In the case of the lowest energy state, the energy per site in the large N, M limit is 1

e = -h +

7CM/N

27 -,M/N

_ -h(1 - 2M/N) +

(2h - J cos q)dq

(J sin( rM/N))

(1.18)

4

Free energy and correlation functions of the X Y model

The free energy per site in the thermodynamic limit is the same as for non-interacting fermions

f = -NlnZ =

-h-TJrln[1+exp(-(2h-Jcosq)/T)]dgr.

(1.19)

The entropy per site of this system is as follows, S

_

Of

_

dq

aT27ru

2h - J cos q \ T

u(x) = ln(2 cosh x/2) - 2 tanh x/2.

(1.20)

The function u(x) is a symmetric and rapidly decreasing function of x

u(x) = u(-x),

foo u(x)dx =

3

.

(1.21)

7rT

(1.22)

Thus the low-temperature entropy is s

-

T 71 J cos-1(2h/J) 1

00

f =-=

u(x)dx =

00

3J cos-1(2h/J)

The specific heat per site is C

T

as

7rT

OT

3J cos-1(2h/J)

(1.23)

On the other hand the velocity of a low energy excitation is

vs = J cos-1(2h/J). Then the specific heat is written as

C7rT

(1.24)

3vs

1.1.3 Correlation functions The static correlation function (S, 'S,,) is called the longitudinal correlation function and (Si S,) is the transverse correlation function. These can be calculated analytically65. In the fermion representation, the correlation functions are written as follows :

(S'S.') = ((1 - 2ci cl)(1 - 2cmcm))/4 =

4

- N + (c

CICLCr),

(1.25)

M-1

(SI Sm) _ (cl f (1- 2CkCk)Cm). k=1+1

(1.26)

1.1 The isotropic XY model

5

It should be noted that the longitudinal correlation function is represented by averages of four fermion operators, but the transverse correlation function is represented by averages of many fermion operators. The value of (cl clcmcm) is given by (cl Cl) (cmcm) - (c cm) (cmcl). Here we have used Wick's theorem. Thus we have 2

(Sl Sm) = (2 - N Ulm = (cl

(1.27)

- ulmuml,

(1.28)

cm).

In principle we can decompose the thermal average of complicated operators of non-interacting fermions into products of averages of two fermion operators using Wick's theorem. Thus we can calculate (1.26) analytically. The highest order term is m-1

(_2)m-1-1(Cl

r7 (c cj )C

k=ll+l

This is decomposed as

2m-l-ldet

ul,l+1

u1+1,1+1

u1+2,1+1

...

um-1,1+1

U1,1+2

u1+1,1+2

u1+2,1+2

...

um-1,1+2

u1,1+3

u1+1,1+3

U1+2,1+3

...

um-1,1+3

ul+1,m

u1+2,m

...

um-1,m

L ul,m

J

The other terms are also written as determinants of this kind. Summing up all terms we have the transverse correlation function, 2m-1-1

(SI

L

x det

u1,1+1

u1+1,1+1 - 2

u1+2,1+1

um-1,1+1

ul,l+2

u1+1,1+2

1.11+2,1+2 - 2

um-1,1+2

U1,1+3

u1+1,1+3

ul+2,1+3

um-1,1+3

ul,m-1

ul+l,m-1

u1+2,m-1

ul,m

Ul+i,m

u1+2,m

um-l,m-1 -

1

2

um-1,m

In the limit N -4oo we have Ulm for the ground state at N = 2M 1

ulm = 27r

rc/2

-n/2

1

ik(1-m) dk 2

=

sin it(l-m)/2 7r(1-m)

for l = m, for l = m.

6

Free energy and correlation functions of the X Y model

Then we have the longitudinal correlation function

for I= m,

1

4

(SI Sm) =

(1

-

for l = m.

(1.29)

and a determinant expression of the transverse correlation function, Sxy(m

- l) = 2(Si Sm) 0_ 1

0

1

1

0

-3

0

1

0

...

1

2m-l 7r-(m-l)det

3

(1.30) 1

0

3

1

0

3

1

0

0

1

The determinant is represented by An for 1 - m = 2n and AnAn_1 for l - m = 2n -1, where An is the determinant of an n x n matrix with elements xl>i = (-1)1 1(2(i - j) + 1), 1

-3 3

2n (-l )M

...

2n-3

An - det

(1.31)

-3 L(1)"-'

2n-1

1

1

.

One can eliminate x1,j, j > 2 by putting x' . = x.. - x1,1x1j without changing the value of the determinant. The new elements are

- 1)(j - 1)(-1)`-i (2i - 1)(2j - 3)(2(i - j) + 1)' 4(i

1-

2

By this operation one gets the following recursion relation: Al = 1 , A n

((2n - 2)!!)2 - (2n-1)!!(2n-3)!!`

Then we have

- 2j)2J

n-1

(2n

i=1

(2n + 1 -

An =

) 4 n-1.

2j)2j-1

If we define Bn = (2/7C)nAn, the two-point function is given as Sxy(2n) = Bn, Sxy(2n - 1) = BnBn-1.

(1 . 32

(1.33)

1.2 The anisotropic XY model

Bn satisfies:

=

Bn+1 Bn

Using 2/it =

fh1(1-

2

n

j_111

rl -

7

-

1 )2 1-1

2i

(2j)-2), we have: 00

In B 1 = n

ln(1

- (2j)-2)

-4n1

(1.34)

^-' .

j=n+1

Then we can expect that Bn behaves as n-1/4. The two-point function Sxy(n) decays as n-1/2. On the other hand, SZZ(n) decays as n-2. Thus the correlation exponents are different for SZZ and Sxy. We find that two-point functions decay algebraically and not exponentially at zero temperature. At finite temperature these decay exponentially. 1.2 The anisotropic XY model

We consider the anisotropic case of the XY model: N

N

=J,xSjx

(1.35)

j=1

j=1

This is written in terms of S± operators as follows:

J(S SJ+1 + S 5+1) + J'(Si

5+1

j=1

N

-2hESj J Z

ix

+ Jy 2

j=1

J

+ Si S-+1)

ix - Jy 2

(1.36)

This Hamiltonian changes the number of down-spins by two. Thus space is divided by the parity of number of down-spins. By the Jordan-Wigner transformation (1.7) we have N-1 I -Ot(c cN LLLLL

-

+ CNCl) +j=1 E Cj+1Cj + N-1

J/ 2

a(cC+ 1 N CNC1) +

c1c E j+ +CjCj+11 j=1

N

a = fl(1 j=1

(1.37)

8

Free energy and correlation functions of the XY model

We can show that a2 = 1 and [,*, a] = 0. The Hamiltonian and a are simultaneously diagonalized and the eigenvalue of a is ±1. The Fourier transformation of these fermionic operators is N

E exp(-igk)ck,

Cq =

q=

(1.38)

k=1

where n is integer (half-odd integer) for a = -1(+1). dr = -hN + E(2h - J cos q)(cgcq + Cfgc_q) + J' sin q(cg cfq + c_gcq ). (1.39) q

Here Eq means the sum over 0 < q < 7r. 1.2.1 The subspace a = 1

In this Hamiltonian, particles with momentum q and -q are coupled. We apply the following transformation for fermion operators Cq and c_q, Cq = COS °grlq + sin Ogilfq,

C_q = -sin Ogrg + COs Og1l_q,

1lq = COS Oq Cq - sin OgCt q, 1l-q= + Sm OgCg + Cos 0gC_q.

(1.40)

Nq/2ir is half-odd integer. The Hamiltonian (1.36) is transformed as

,YP = -hN + E 2 sin 2 Oq(2h - J cos q) + 2 sin 0q cos OqJ' sin q q

+[(2h - J cos q) cos 20q - J' sin q sin 20q] (gggq + gfgq_q) +[(2h - J cos q) sin 20q + J' sin q Cos 20q](rlggfi q +'l_ggq).

(1.41)

The last term is removed if we put tan 20q =

J' sin q

J cos q - 2h

(1.42)

Thus the Hamiltonian becomes 1

_

e(q)(gqqq

- 2)

q

e(q) =

/(J Cos q - 2h)2+ (J' sin q)2.

(1.43)

The lowest energy state I'F) must satisfy rlq 1W) = 0. The following state satisfies this condition, IT)

= fl(cos 0q + sin Ogcgcfq)l0). q

(1.44)

1.2 The anisotropic XY model

9

As we are considering the case a = 1, the total number of quasi particles must be even. So the number of states which belong to this subspace is not 2N but 1 j CN = 2N-1 1.2.2 The subspace a = -1 The number of particles must be odd in this subspace. Then the Hamiltonian is the same form as (1.43) but qN/2ir must be an integer. The lowest energy state in this subspace is IT) = co [J(cos 0R + sin Oqc c R)I0).

(1.45)

R

The general states are given by an even number of excitations from this state. The number of states is also 2N-1. These states are orthogonal to each

other and therefore all these states together form a complete set of wave vectors.

1.2.3 The free energy

Using the results of 1.2.1 and 1.2.2, one obtains the partition function of the system,

Z =exp

(E)

2{fl(1+e-E(R)lT)+e-F(R)lT)} R

(q'))

+exp

R

2{fl(1 + eF(R')VT) - fJ(1 -

ee(R7/T)},

(1.46)

R'

R'

where qN/27r is a half-odd integer and q'N/27r is an integer. In the thermodynamic limit the second term in {...} is much smaller than the first term and we obtain the free energy per site. The free energy is given by the logarithm of the partition function

f =-N1nZ = eo -

T 27r

f

n

ln(1 + exp(- (J cos q - 2h)2 + (J' sin q)2/T ))dq. E

(1.47)

Here eo is the ground state energy per site

eo = -

1

47r

n

(J cos q

- 2h)2 + (J' sin q)2dq.

2

Systems with a delta-function potential

2.1 The boson problem

2.1.1 The c = 0 case Here we consider the system N i=1

02

a2 Ox?

+ 2c E a(xi - x;).

(2.1)

i<j

At first we consider the cases c = 0 and c = oo. We assume periodic boundary conditions. In second quantized form, the Hamiltonian (2.1) at c = 0 is written as -YP

= E k2akak, k = 2irn/L, k

[ak, aq] = [ak, at ] = 0.

[ak, at] = 6k,q,

(2.2)

The eigenstates and eigenvalues are given by 11(nk

!)-1/2(ak)nk j0),

k

k2nk.

(2.3)

k

A set of integers {nk} gives an eigenstate. The ground state of this system is no = N, nk#o = 0. The partition function of the grand canonical ensemble at chemical potential A is

8=f k

exp(-(k2

- A)nk/T) = fl(1 - exp[-(k2 - A)/T])-1.

nk=0,1,2,...

(2.4)

k

The Gibbs free energy is given by

G = -T In 7. = T E ln(1 - exp[-(k2 - A)/ T]). k

10

(2.5)

2.1 The boson problem

11

The number of particles is given by

N=

OA L exp[(k2 - A)/ T] - 1

The pressure is p

=

aL 1

L

-T aL E ln[1 - exp((-(27rn/L)2 -+- A)/T)] n=-oo

k exp((k2 -A)/ T) - 1 2k2

In the thermodynamic limit we have 1

p

27r

2k2dk

J exp(k2 - A)/T - 1 f ln(1 - exp[(A - k2)/T])dk = lim -G/L, L-oo

271

(2.8)

by partial differentiation with respect to k. 2.1.2 The c = oo case

It is difficult to treat this case via the second quantization representation. So we assume that the wave function is represented by the Bethe-ansatz wave function. We assume N quasi-momenta k1, k2, k3,..., kN, A(P) exp(i

1(xl, x2, ..., xN) = P

kpixi),

(2.9)

1

in the region x1 < x2 < X3 < ... < XN. In the other N! - 1 regions the wave function is determined by the full symmetry of the wave function. The energy eigenvalue must be given by E _ 7,1 kl . The wave function must be zero at xn = xn+l. This condition is satisfied if A(P) = e(P). This means that A(12) = 1, A(21) = -1, for N = 2, A(123) = A(231) = A(312) = 1,A(213) = A(132) = A(321) = -1,

for N = 3.

(2.10)

The periodic boundary condition is 1V(xl, x2, ..., XN) = 1V(x2, x3, ..., XN, x1 + L).

(2.11)

This condition is satisfied by exp(iki) = (-1)N-1. This means that the quasi-momenta are 27r/Lxinteger for odd N and 27r/Lx half-odd integer for even N. If two quasi-momenta are the same, the wave function vanishes,

Systems with a delta function potential

12

thus they must be always different. The ground state is given by the set of quasi-momenta {k} = {(N - 1)7r/L, (N - 3)7r/L,..., -(N - 1)7r/L}. The partition function of this system for the ground canonical ensemble is

+

e(A-((2n-1)n/L)2)/T) + fJ(1

n

-

e(A-((2n-1)7r/L)2)/T )]

n

+ [Tj(1 +

- fl(1 -

e(A-(2n,,/L)2)/T)

n

e(A-(2n7r/L)2)/T)].

(2.12)

n

In the first bracket, even order terms with respect to fugacity z = exp(A/T) survive. This is the contribution of states with even number of particles. The second bracket is the contribution of states with odd number of particles. This formula resembles that of the free fermion model but is slightly complicated because the parity of the quasi-momenta change according to the number of particles. In the thermodynamic limit the second terms in each bracket are much less than the first terms. The Gibbs free energy per unit length is

g = limes -T In E/L =

-2 fnu + exp[-(k2 - A)/T])dk.

(2.13)

Thus the boson problem at c = oo resembles that of noninteracting fermions. The momentum distribution problem of this case was investigated by Girardeau31 and Lenard58.

2.1.3 Scattering states of Bose particles with finite interaction

The applicability of the Bethe-ansatz method was first shown by Lieb and 60,59 At first we consider the scattering problem of two particles, Liniger60,

a2

a2

ax - ax+ 2c8(xi - x2). 2

2

1

2

(2.14)

We assume that the wave function is represented by a superposition of plane waves

f (xi, x2) = A(12) exp(i(kixi + k2x2)) + A(21) exp(i(k2xi + kix2)),

(2.15)

for xi < x2. By the symmetry condition of the Boson wave function we have f (xi, x2) = A(12) exp(i(kix2 + k2xi)) + A(21) exp(i(k2x2 + klxi)),

(2.16)

2.1 The boson problem

13

at xi >- x2. This wave function is continuous at x1 = x2 but its derivative is discontinuous, 02

02

X2)= (k2 + k2)f + 2ib(xi 7xi ax2]f x(A(12) - A(21))(k1 - k2) exp(i(ki + k2)xi). (x1,

- x2) (2.17)

Thus if i(ki - k2)(A(12) - A(21)) + c(A(12) + A(21)) = 0 is satisfied, the function f satisfies the eigenvalue conditionf = (ki +k2)f. This means that the amplitudes A(12) and A(21) must satisfy the following condition,

ki-k2+ic

A(12) A(21)

(2.18)

ki - k2 - is *

Next we treat the three-body problem. We assume the wave function is of the form

.f(xi,x2,x3) = +A(213)ei(k2xi+kix2+k3x3)

A(123)ei(kixi+k2x2+k3x3)

+A(132)e`(kixi+k3x2+k2x3) + A(321)e`(k3x'+k2x2+kix3) +A(231)ei(k2xi+k3x2+kix3)

+A(312)e`(k3xi+kix2+k2x3)

(2.19)

for x1 < x2 < x3. We have six conditions for six amplitudes A(ijk):

A(123)/A(213) = -(ki A(123)/A(132) = -(k2 A(213)/A(231) = -(ki A(132)/A(312) = -(k1 A(321)/A(231) = -(k3 A(321)/A(312) = -(k2

- k2 + ic)/(ki - - ic), - k3 + ic)/(k2 - - ic), - k3 + ic)/(ki - - ic), - k3 + ic)/(ki - - ic), - k2 + ic)/(k3 - - ic), - ki + ic)/(k2 - ki - ic). k2 k3

k3 k3

k2

(2.20)

These relations must be dependent, because only five independent relations are possible. Rewriting (2.20), we get A(123) = Y12A(213),A(123) = Y23A(132),A(213) = Y13A(231), A(132) = Y13A(312),A(321) = Y32A(231),A(321) = Y21A(312),

Yab = -(ka - kb + ic)l (ka - kb - ic). To get A(cba) from A(abc) we have two paths A(cba) = YabYacYbcA(abc), A(cba) = YbcYacYabA(abc).

(2.21)

14

Systems with a delta function potential

The final results become equivalent if Yab = Yba,

(2.22)

Yab Yac Ybc = Ybc Yac Yab

One can prove these relations from the definition. The boson problem is the simplest example of a Yang-Baxter relation. The answer is

-

-

A(123) = C(ki k2 + ic)(ki - k3 + ic)(k2 k3 + ic), A(213) = -C(k2 - ki + ic)(ki k3 + ic)(k2 - k3 + ic), A(132) = -C(ki k2 + ic)(ki - k3 + ic)(k3 - k2 + ic), A(321) = -C(k2 ki + ic)(k3 - ki + ic)(k3 k2 + ic), A(231) = C(k2 - ki + ic)(k3 - ki + ic)(k2 - k3 + ic), A(312) = C(ki k2 + ic)(k3 - ki + ic)(k3 - k2 + ic).

-

-

-

-

(2.23)

For the case of an N-body problem the relations (2.22) are sufficient to construct a consistent solution. For the N-body problem one must determine N! coefficients A(P) in the wave function which holds for xi < x2 < ... < XN. The answer is

f = > A(P) exp[i(kpixi + kP2x2 + ... + kPNxN)l P

A(P) = Ce(P) [J(kpj

- kPk + ic).

(2.24)

j
The energy eigenvalue is N

E=>k2j,

(2.25)

j=i

The total momentum operator is defined by N j=i

7xj

The wave function (2.24) is also an eigenstate of this operator, N

A-f =Kf, K =>kj.

(2.26)

j=1

2.1.4 Periodic boundary conditions

In the previous section we considered the N-body problem in a onedimensional space of infinite length. Here we consider the N-body problem

2.1 The boson problem

15

on a ring with length L. The wave function must satisfy { (x1,x2,...,X1,...,XN)

J

= f(xl,x2,...,x1 +L,x1+1,...,JCN)

j = 1, ..., N.

(2.27)

This condition is satisfied if exp(ikPNL) =

A(P N, P 1, P 2, ..., P (N - 1))

A(P 1, P2, P3, ..., PN)

for any permutation P. This condition gives k1 + ic), exp(ik1L) _ H (kj k1 - ki - is

j = 1, 2, ..., N.

1

(2.28)

(2.29)

These are N conditions for N numbers. Taking the logarithm of these equations, we have N k1L=27rI1->2tan-'(k1-kI)

c

1=1

(2.30)

In the limit c oo we have k1 = 2irI1/L. For the ground state (lowest energy state with fixed N) {I1}

= {(N -1)/2, (N - 3)/2,..., -(N -1)/2}, or I1 = (N + 1)/2 - j. (2.31)

This is a set of integers or half-odd integers which determines an eigenstate.

One can show that the set of integers gives a unique real solution for the k1 s. The ground state is always unique and is never degenerate with other states. Thus the integers (2.31) give the ground state. 2.1.5 Linear integral equation for the distribution of quasi-momenta

We take the limit of infinite N and L, setting the density as finite. The distribution function of ks is denoted as p(k). (2.30) becomes

k = 2n f k p(q)dq - J B 2 tan 1

(k c q) P(q)dq

Differentiating with respect to k, we have 1B

1

it

(2.32) P(q)dq. B C2 + (k - q)2 This is called the Lieb-Liniger equation. The energy and number of particles

P(k)

=

27r

per unit length are

e= f

+

i

k2p(k)dk, B

n=f

p(k)dk. B

(2.33)

Systems with a delta function potential

16

Equation (2.32) is a Fredholm type linear integral equation. If one puts p = Bx, 2 = c/B, we have 1

P (x) = 27r

! 1

+

.

1

1

22

+ (x - y)2 p ( y )d y, 1

e = B3F(2), F(2)

x2p(x)dx, 1

n = BG(2), G(2)

J p(x)dx.

(2.34)

1

Then we have e/n3

= F(A)/G3(2), c/n = 2/G(2).

(2.35)

Thus we find that energy per unit length e is expressed as e = n3u(c/n),

(2.36)

by a certain function u(x). The perturbation expansion for a 1D boson system with delta-function potential gives

u(x) = x - fx3/2 + (6

- -) x2 + 0 (x512).

(2.37)

The Bethe-ansatz solution of delta-function bosons is used to check approximate theories for Bose These results are compared in Fig. systems99,100

(2.1).

2.1.6 Bound states in the case c < 0

In the case of an attractive interaction, c < 0, we have solutions with complex quasi momenta. We consider the two-body wave function (2.15) in the infinite one-dimensional space, f (xl, x2) = exp(i(klxl + k2x2))[A(12)6(xl - x2) + A(21)0(x2 - X01 + exp(i(k2xi + klx2)) [A(21)6(xl - x2) + A(12)6(x2 - xl )].

The term exp(i(kixi + k2x2)) is rewritten as exp(i(k1 + k2)(x1 + x2)/2 + i(ki -

k2)(x1 - x2)/2). This term diverges if s(k1 + k2) * 0. Here Z(x) means the imaginary part of x. Moreover if Z(kl - k2) > 0, we should have A(12) * 0 and A(21) = 0. Then from equation (2.18) we have kl = k2 + iIcl and therefore kl = a + i1cJ/2,

k2 = a - i1cI/2,

2.1 The boson problem

17

Fig. 2.1. The function u(x) determines the zero temperature properties of a deltafunction Bose gas, the thick line is the result of numerical calculation of (2.34). Line 1 is the result of primitive perturbation. Line 2 is the result of Bogoliubov theory. Line 3 is the perturbation result (2.37) by the author99.

where a is a real number. The energy, momentum and wave function of this state is

E = 2a2 - c2/2,

K = 2a,

f (X 1, x2) = exp(iXK) exp(-cIx1 - x21/2).

(2.38)

Then we have E = K2/2 - c2/2. This means that the mass is doubled and the binding energy is -c2/2. This can be regarded as a bound state of two particles.

Next we consider the three-body problem in the infinite one-dimensional space, and we assume kl > Sk2 > sk3. In this case equation (2.19) is

normalizable if A(312) = A(231) = A(321) = 0. This is satisfied by the following two cases,

kt = k2 - ic, k3 = k2 + ic,

kt = a - is/2, k3 = a + is/2, k2=#. The former is the bound state of three particles. We call this the string of

Systems with a delta function potential

18

length three. The second is the scattering state of one bound state and one independent particle. Generally we have dispersion for an n-string E

_

K2

n(n2

12

n

kj = a

- 1)c2

- (n + 1- 2 j)c/2,

j = I,-, n.

(2.39)

In the attractive case of bosons, there is no thermodynamic limit, because the ground state energy is of the order N3. This situation is completely different from the repulsive case, where the ground state energy is of order N.

2.1.7 Elementary excitations for repulsive bosons

The ground state is characterized by the quantum numbers Ij which are successive. Two kinds of excitations exist from the ground state. One is the particle state given by the integers

{b} _ {m +

22

N 11

- 21 ..., -

N

}.

(2.40)

Here the number of particles changes from N to N + 1. The total momentum

of this state is 2irm/L. We put the solution of (2.30) for the ground state as {kl, k2, ..., kN} and for this excited state as {q, ki, k2, ..., kN} Putting AkJ - = k - ki we can write down the equation for Akj,

AkjL = -7r+ 2 tan 1q-kl-

N 2(tan

c

- tan-1

c

.

(2.41)

.i=1

As Akj is of the order of N-1, we can write this equation as follows: Akj = - L (7t

- 2 tan 1 q

c kj

-

c

c2+ (kj - k1 )2

(Ak j

- Aki)).

(2.42)

This equation becomes Akj (1 +

1

N

L

c2 +

j

( k2c

1q 1 (ir - 2 tan = 2 ) - k i) L 1

L

2c

k

c

E c2 + (kj - ki)2 Aki.

Using (2.32) this is written as 27rAk

- k i)

L(7r-2tan- 1 q cki)

(2.43)

2.1 The boson problem

+f

19

B

(2.44)

Then we can define the back flow J(k, q) = LA(k)p(k), J(k, q)

-f

B 7r

c2 + (k

- k')2

J(kl, q)dk' = 2 (7r - 2 tan-i q - k) .

(2.45)

The energy change AE(q) and momentum change AK(q) are DE(q) = q2 + fB B 2kJ(k, q)dk,

OK(q) = q + fBB J(k, q)dk, q > B.

(2.46)

We define the Green function for this type of integral equation

e1

U(k, k') -

f

cdk"

B it c2+(k-k")2

U(k', k") = 6(k

- k').

(2.47)

Using this Green function we can write p(k) and J(k, q),

f U(k, k')dk', (ii - 2 tan -1 J(k, q) = - f p(k) =

1

27r

B

B

U(k, k')

B

q c

k'

27r

) dk'.

(2.48)

Considering that U(k, k') = U(k', k) we can calculate AK(q), B

AK(q)

= q - fB p(k) ii - 2 tari i q C- kdk

= -7rn + q +

_ -itn +

f

f

B

2 tan-1 q - k p(k)dk c

B

q

dk'{ 1

+

f

B B

c2

+

(kc- k)2

p(k)dk}

q

-icn + 27r f p(k)dk.

(2.49)

0

At q = B we have AK = 0. The energy change AE is the chemical potential at zero temperature. This is called a type I excitation, and corresponds to the Bogoliubov excitation. Next we consider the hole state of elementary excitation, given by integers {

N N-1, ... N-m+1, N- m - 1, ...,- 2 + 1}.

(2.50)

jj} 2' 2 2 2 2 The number of particles is N - 1. The energy and momentum change are

Systems with a delta function potential

20

given by AK(q) = -q + fs J(k, q)dk, e a AE(q) = -q2 + 2kJ(k, q)dk, a

f

al

c

J(k, q) - fB 7E - c2+(k-k')2 J(kl, q)dkl

= 1271

(7r + 2

tan-1 k

q

,

q < B.

(2.51)

C

This excitation is a hole excitation in the fermionic distribution. Ishikawa and Takayama40 identified that this excitation is solitonic in nature. It is noteworthy that the boson system resembles a spinless fermion system. The excitations near the ground state are described by the sum of these two excitations.

2.2 The fermion problem

2.2.1 The c = 0 case In second quantization form the Hamiltonian is written as

_

k2(aklakl + aklaki.),

k = 27rn/L,

k

{ak,, aq6'}

=

ok,q&y,6',

{akc, aqc'}

=

{ak6, aq6,}

= 0.

(2.52)

The state is given by flk.(at)nka 10). nk, is the number of particles in state k, a and takes only the values 0 or 1. The total energy is given by E = E k2nk0.

The total number of particles is E nk6. The grand partition function at chemical potential A is

_ E exp[(E - AN)/ T] = 11(1 + exp(-(k2 - A)/ T)). nk,Q=°,1

(2.53)

kQ

The free energy is given by 00

G(T, A, L) = -2T E ln(1 + exp[-((27rn/L)2 - A)/ T]).

(2.54)

n=-oo

In the limit of L

co this becomes 00

G = -2TL f (27r)-i ln(1 + exp[-(k2

- A)/T])dk.

(2.55)

2.2 The fermion problem

21

The pressure is given by -aG/aL,

p = 2T L(2)-' ln(1 + exp[-(k2 - A)/T])dk.

(2.56)

00

2.2.2 The two-body problem

The wave function of particles with spin has not only coordinates for position but also spin coordinates, (2.57)

f (x151, X2S2, x353, ..., XNSN).

So we have four functions to represent an eigenstate of the two S = 1/2 particles system fTT(x1,x2),fTJ(x1,x2),fjT(xl,x2),fjj(xl,x2).

(2.58)

The fermion's wave function must be antisymmetric and we have fTT(x1,x2) _ -fTT(x2,x1),

fT1(xl,x2) = -f1T(x2,x1), f11(xl,x2) = -f11(x2,x1).

(2.59)

Thus we obtain fTT(xl,xl) = 0 and f11(x1,x1) = 0. This means that in the fully polarized case the delta-function potential has no physical effect. The wave function is represented by a determinant fTT(x1,x2) = Cdet

exp(ikixi) exp(ik2x1)

eXp(ikix2) exp(ik2x2)

(2.60)

Next we consider the case where one fermion has spin up and the other has spin down,

fl1(xl,x2) = 9(x2 - xl)([12,12]ei(klx,+k2x2) + [12,21]e`(k2x'+kix2))

+9(xl - x2)([21,12]e`(k2x'+k,x2) + [21,21]ei(k2x2+kixi))

aa2

2

a2

ax2

fT1(xl,x2) = (k1

(2.61)

+k2)fT1(x1,x2)+ei(ki+k2)xi

x [2ib(xl - x2)(k1 - k2) ([12,12] - [12,21] + [21,12] - [21,21] )

-6'(x1 - x2)([12,12] + [12,21] - [21,12] - [21,21])].

(2.62)

Systems with a delta function potential

22

To satisfy * f11(x1,x2) = EftI(x1,x2), we put

E=ki+k2, [12,12]+[12,21]-[21,12]-[21,21] = 0, 2i(ki - k2) ([12,12] - [21,12] + [12,211 - [21, 21])

+2c([12,12]+[12,21]) =0.

(2.63)

From these equations we have [12,12])

-(

([21,12])

U21 - 1

U21-1)([21,21])'

U21

[12 12]

If we write x(12)

k1-k1 ujl = kj - kl + is

[12, 21]

U21

[21:12] ) we

(2.64)

have

x(12) = [(u21 - 1)I + u21P12](21),

(2.65)

where I and P12 are identity operator and pair-exchange operator, respectively. One solution is

kl -A -ic'

x(12) _ [-(k2 -A+ ic

x(21) _

(k2-A-ic') [

kl - A + ic'

I '

c, - c/2.

(2.66)

2.2.3 The three-body problem

In this case there are 3! x 3! coefficients [Q, P]. The wave function is given by

[Q' P] exp[i(kpixQl + kP2xQ2 + kP3xQ3)],

f (xl, x2' x3)

(2.67)

P

at xQl < xQ2 < XQ3. We write P as a 3!(= 6) dimensional vector. The relation between x(123) and x(213) is c (123) = Y21 c(213)

1;(123) =

Y21

=

[123,123] [132,123] [321,123] [213,123] [231,123] [312,123]

,

u21-1

0

0 0

U21-1

U21

0 0

x(213) =

0 0

0 0 0

U21-1

U21

0

[123,2131[132,2131 [321, 213] [213, 213] [231, 213] [312, 213] U21

0 0

0

U21-1

u21

0 0

0 0 U21

0 U21 - 1 0

0 u21

0 0 0

U21-1

2.2 The fermion problem

23

The other relations are x(123) =

x(132) = Y31 x(312), x(321) = Y23 x(231),

c (321) =

x(213) = Y31 x(231).

(2.68)

The inverses of these relations are x(213) = Y12 x(123), c (132) = Y23 x(123), c (312) = Y,1x(132), x(231) = c (312) = x(231) = (2.69)

Here Yib

(U jl - 1)1 + ujlPab, ujl =

kj-

kj

k

k, + is

(2.70)

and Pab is the operator which exchanges elements a and b. Sometimes we write Pab as (ab). In general N the coefficients [Q, P] should satisfy

[Q,P] = (ur(l+1)pl-1)[Q,P(1,1+1)]+up(l+l)pl[Q(1,1+1),P(1,1+1)]. (2.71) As (1, 1 + 1) is the exchange of l and 1 + 1, P (l, 1 + 1) is a permutations as follows : 1

2

P1 P2

...

1

1+ 1

...

N

...

P(l+1)

PI

...

PN

1+ 1 Ql

...

N QN

In the same way Q(l, l + 1) is 1

Q1

2 Q2

...

1

...

Q(l + 1)

...

To get c (321) from x(123) there are two paths,

x(123) - x(213) - x(231) - x(321), x(123) - x(132) - x(312) - x(321). For the two paths to be equivalent, the Ys must satisfy 12

23

12

23

12

23

Y23113Y12 =112Y13Y23

(2.72)

The left-hand side is [(u23 - 1013 - 1)(u12 - 1) + (u13 - 1)u23u12]I + [u23(u13 - 1012 - 1) + U12(U23 - 1)(u13 - 1)]P12

+(u23 - 1)u13(u12 - 1)P23 + u23u13u12P13

+(u23 - 1)u13u12P23P12 + U23U13(U12 - 1)P12P23

(2.73)

Systems with a delta function potential

24

The right-hand side is [(u12 - 1)(u13 - 1023 - 1) + (u13 - 1)u12u23]I +[(U12 - 1)(u13 - 1)u23 + U12(U13 - 1023 - 1)]P23

+(u12 - 1)(u23 - 1)u13P12 + u12u13u23P13 +u12u13(u23 - 1)P23P12 + (u12 - 1)u13u23P12P23

(2.74)

Here we have used P12P23P12 = P23P12P23 = P13 and p22 = I. It is apparent that the coefficients of I, P13, P23P12, P12P23 are the same. We can show that u23(u13 - 1012 - 1) + u12(u23 - 1013 - 1) = u12u13(u23 - 1) from the definition

of uji. The coefficients of P(12) and P(23) coincide in the l.h.s. and r.h.s. Thus we have proved the identity (2.72). Moreover x(123) - x(213) - x(123) requires that Y a,a+l Y a,a+1 =I . ij jl

(2 . 75)

The l.h.s. is

[(uji - 1)(uij - 1) + ujiuij]I + [(uji - 1)ulj + uji(uij - 1)]Pjl.

One can show that (uji-1)(ulj-1)+ujiulj = 1 and (uji-l)uij+uji(uij-1) = 0. Thus if x(123) is given, we can construct all the P consistently.

We assume that particles 1 and 2 have spin up and particle 3 has spin down. Then f (x1, x2, x3)

= -f (x2, x1, x3) and

[123, P] = -[213, P], [321, P] = -[312,P], [132, P] = -[231, P].

(2.76)

Therefore we should determine 6 x 3 coefficients

t(I;P) = e(P) [312,P], J(2;P) = e(P)[321,P], t(3;P) = e(P)[123,P]. (2.77)

The relations among the coefficients are '(y; P) = di(y; P(l, l + 1)),

1(l; P) = (1

for

y * 1,1 + 1,

- up1,P(i+1))'(1; P(l, l + 1)) + upl,P(i+1)t(l + 1; P(l, l + 1)),

1(1+1;P) _ (1-upl,P(,+1))1(1+1;P(1,1+1))+upl,P(l+1)1(1;P(1,1+1)). (2.78)

A solution of these equations is D(1; P) = D(kP2

- A - ic')(kP3 - A - ic'),

(D(2; P) = D(kpi - A + ic')(kp3 - A - ic'), (D(3; P) = D(kpl - A + ic')(kP2 - A + ic').

(2.79)

2.2 The fermion problem

25

2.2.4 The M = 1 and arbitrary N case The solution of N -1 up-spin electrons and 1 down-spin electron was solved by

McGuire66, 67

(2.80)

c(Y ; P) = DFP (Y, A), Y-1

FP (Y, A)

N

11 (kp j

- A + ic) 11 (kP1- A

j=1

- ic ).

(2.81)

1=y+1

The case M = 2 was solved by Flicker and

Lieb26.

2.2.5 The arbitrary M and N case Later Gaudin29 and Yang117 derived the solution for a general number of down-spins. Assume that there are M down-spins in an N-particle system. The relations of I(y1, y2, ..., YM ; P) are as follows : I(Y1,Y2,...,YM;P) = (D(Yi,Y2,...,YM;P(l,1 + 1)), for ya * 1, l + 1,

D(Y1,Y2,...,YM;P) = (1 - up 1,P(1+1))I(Yl,...,ya,...,YM;P(1,1 + 1)) +uP1,P(1+1)(D(Yl,..., Ya + 1, ..., yM; P(l, l + 1)),

for

ya = l and ya+1 * 1 + 1

I(Y1,Y2,...,YM;P) = (1 - uP1,P(t+1))t(Yl,...,ya, +uPI,P(1+1)(D(Y1,...,Y«

for

ya = l + 1

- 1,...,YM;P(1,1 + 1)),

YM;P(l, l + 1))

and Ya_1 * 1, (D(Y1,Y2,...,YM;P(l,1 + 1)),

for

ya = l and Ya+i = 1 + 1.

Regarding Yi, Y2,..., yM as particle coordinates on the lattice with N sites, we assume the generalized Bethe-ansatz, (D(Y1,...,YM;P)

_

A(R)FP(AR1, Y1)FP(AR2, Y2)...FP(ARM, YM).

(2.82)

Here the R s are permutations of 1, 2,..., M, A(R)FP (ARa, Ya )FP (AR(a+l), Ya + 1)

+A(R')FP(AR(u+1),Ya)FP(ARu,Yu + 1) = A(R)FP,(ARa, Ya)FP'(AR(a+1), Ya + 1)

+A(R')FP'(AR(a+l),Ya)FP'(ARa,Yu + 1),

(2.83)

Systems with a delta function potential

26

and we put P' - P(ya, y, + 1), R' - R(a, a + 1). Thus we have A(R)(AR(a+1)

- ARa - ic) + A(R(a, a + 1))(ARa

- AR(a+1) - iC) = 0.

(2.84)

- ic).

(2.85)

This is satisfied if

A(R) = e(R) fl(ARj - ARI j
Thus we can construct a wave function which satisfies Sri' = E`l' for arbitrary set of numbers kl,..., kN and Al, ..., Am. 2.2.6 Periodic boundary conditions The periodic boundary condition requires

x(123) =

x(132) =

x(213) = P23Pi2(132) exp(ik2L),

x(231) =

x(312) = P23P12c(123) exp(ik3L),

exp(ik2L),

x(321) =

exp(ik3L). (2.86)

These six equations yield three equations for x(123): X32X12(123),

Xj1 = PjiYll.

(2.87)

Here we have used Pal = I, PjkPjl = PjIPIk. One can show that the three operators X21X31,X32X12,X13X23 commute with each other using

XijXji = I,

XjkXikXijXkjXkiXji = I,

XijXkl = Xk!Xij,

(2.88)

where 1,), k, l are all unequal. Thus the three operators can be diagonalized simultaneously. The periodic boundary condition for the wave function is 1V(x1,x2,...,.xN) = 1V(XN -L,x1,x2,...,XN-1).

(2.89)

This is satisfied by [Q1, Q2,..., QN; P 1, P2,..., PN] exp(ikPNL)

= [QN, Q1, ..., Q(N - 1); PN, P I,, P(N - 1)].

(2.90)

This is rewritten as 1(Y1, Y2,..., YM; P) exp(ikPNL)

_ 1(Y1 + 1,Y2 + 1,..., ym + 1;PN,P1,...,P(N - 1)), 1(yl, ..., YM-1, N + 1; P) (l, y1, Y2, ..., YM-1; P ).

(2.91) (2.92)

2.2 The fermion problem

Assume that yM

27

N. Using (2.81) and (2.82) we have M

exp(ikPNL) = a=1

kPN - Aa + ic' kPN - A,, - lC

(2.93)

In the case yM = N we have

M kPN- Aa+ic'

I(Y1,...,yM-1,N;P)11

kPN- Aa - iC' _ i(1,Y1 + 1,Y2 + 1,...,YM-1 + 1;PN,P1,...,P(N - 1)),

(2.94)

by substituting (2.93) into (2.91). This is satisfied if A(R) = A(RM, R1, .., R(M

- ARM - ic' - 1)) ;_1 `kP(kP;j -ARM + lC

(2.95)

fN

Substituting (2.85) we get

ft(kf-ARM+ic' j=1

M /ARM-A;-iC ARM-A.+IC

(2.96)

1

I

j=1

Thus the periodic boundary condition is satisfied by ex p(ikjL)

_ JJ a =1

Aa + ic/ l j l Qj

- Aa - iC

N7(Aa-k;+ic'l _ 1

j=1

\ na - k j - ic' /

-

,

j=1, .. ,N,

icl ,(Aa-A#+l

7`'g

=1 11=

`na - n# - is /'

(2.97)

a=

1, ...,

M. (2.98)

These are N + M coupled equations for N + M unknowns {kj} and {A,,}. Taking the logarithm of (2.97) and (2.98) we have

kjL=2irIj-2Etari k;c A M

(2.99)

a=1

2 tan-1

Aa

- kj = 2rrJa + 2

c'

j=1

M

tan-1

Act A

(2.100)

/3=1

where I j is an integer (half-odd integer) for even (odd) M, and Ja is an integer (half-odd integer) for odd (even) N - M. The total momentum is N

K = kj = j=1

L

(2.101)

j

a

Systems with a delta function potential

28

2.2.7 The ground state for c > 0 We consider the ground state in the case c -+ oo. In this limit the As are proportional to c. We denote xa = Aa/c'. The ks remain finite. Then equations (2.99), (2.100) become

kiL=2lrIi+yi- ki

1

tan-1 xa,

y1 = 2

CY2,

Y2 =

1 + xa'

a

- x#

(2.102)

- 2Y2 +O(c_2)].

(2.103)

M

xa

2N tan- xa = 27rJa + ) 2 tan-' 1

The total energy is

(2irI

E = Eke = E i i

+Y1)2 [1

To minimize this in the limit of very large c for even N and odd M, Ii and Ja should be taken as follows,

Ii = (N + 1)/2 - j, Ja = (M + 1)/2 - a.

(2.104)

In the case of even N and odd M the lowest energy state is unique. The determination of xa is the same problem as the antiferromagnetic Heisenberg chain with N sites and M down-spins in section 3.1.

We put the distribution function of ks as p(k) and that of As as 6(A). From equations (2.102) and (2.104) we have

fk k = 27E

B

p(t)dt -

r Q 2 tan 1 Q

2(A

fB tan

1

2(k

- A) a(A)dA,

(2.105)

c

- k) p(k)dk

c

= 271 f a(x)dx + fB 2 tan-1

A

- A' a(A)dA'.

(2.106)

c

Differentiating with respect to k and A we have 1B

1

P(k)

=

6(A) +

27r

B

+

B 7r(c

B

2ca(A)dA 7r(c2 + 4(k

2+ cu(A')dA' (A -

A,)2)

(2.107)

A)2)1

- 1Q

Q

2cp(k)dk

7r(c2+

4(k - A)2)

(2.108)

We put the energy, number of particles and number of down-spin particles

2.2 The fermion problem

29

per unit length as e, n, ni,

e=

k2p(k)dk,

n=

nj _

p(k)dk, Q

Q

a

a(A)dA.

(2.109)

In the case of B = oo equation (2.108) can be treated by the Fourier transformation, 6(A)

fQ 2c

c A) ) p(k)dk.

(2.110)

Q p(k)dk = n/2.

(2.111)

sech

Q

This is a special point and we have

ni =

foo r(A)dA = 2 fQ

In the limit c = 0+ we have a(A) = p(A)/2 = (21c)-1 for JAI < Q. Then e

2Q 2Q3

n=

(2.112)

2Q.

2.2.8 The ground state for c < 0

In the case that N is even and M is odd, the ground state at c = 0+, the solution of (2.93), becomes

Aa = L (M + 1 - 2a), ka = ka+M = Aa, k2M+n = -kN+1-n = L (M -1 + 2n).

(2.113)

Then at c = 0, two real ks and one A coincide. In the case c = 0- these two ks become complex conjugates of each other. A remains on the real axis. In the limit of L -* oo these two ks and A form the following bound state, ka

= Aa + 2i + O(exp(-8L)), kM+a = Aa -

-i + O(exp(-8L)).

(2.114)

Thus in the ground state 2M ks become complex and N - 2M ks are real. From equation (2.93) we have M

kj = 2iIj + 2> tan-1 [2(kj - Aa)/Icl] ; j = 2M + 1, ..., N. a=1

For ka, equation (2.93) is

ex ikaL)-

ka ka

- Aa+ic' Aa-A#+is - Aa - ic'(Aa-Al3

(2.115)

Systems with a delta function potential

30

The first term of the r.h.s. should be treated carefully. In the same way we have exp(ik«+ML)

k«

- A« + ic'7-

A«

- Ap

_ k«-A«-ic' fl*a A« - Afl - is

The product of these two equations is k«

exp(2iA«l) =

- A« + ic' k« - A« + ic'

(A« - Aj + is

k«-A«-ic'k«-A«-ic' 11(A«-A - ic

From (2.98) we have

-k«-A«-ic'k«+M-A«-ic'

ka -A« +ic'k«+M -Aa + ic'

kj

N

j=2M+1

- A«- ic'

kj-2

+ic'.

Thus

- A - ic'

(A« - A# + is l

k.-A«+ic'

`A«-Afl-ic/

kj

N

exp(2iA«l) =

11

j=2M+1

Taking the logarithm we have 2A« L = 2

iJ

«

- A# + 2 j=1 E tan-1 2(A« - kj)

A

+ 2 E tan 1 fl=1

;

Icl

ICI

a = 1, ..., M.

(2.116)

The total energy is given by M

2

E = E [ (A« + 2

+

(A«

- 2) 2] +

nE

0.

.

(2.117)

j=1

a=1

For the ground state of even N and odd M, the Ii s and the J« s are

N-2M-1 N-2M-3 2 2 M-1 M-3 M-1 2 J« = 2 2

N-2M-1

Ij =

2

(2 . 118)

(2 . 119)

,

We set the distribution function of the real ks as p(k) and that of the As as 6(A). We have equations for these functions: P(k)

=2

6(A) +

1

_ ir

/B

J

f

B

B

I

I

( )

(2.120)

B c + 4(k aA)2'

Icla(A')dA' c2 + (A A')2

-

- -

/.Q

2IcI p(k)dk

1

1

Ir

7r J Q c2 + 4(k - A)2 (2.121)

2.2 The fermion problem

31

The particle numbers and energy per unit length are

n = 2 fB o (A)dA + f p(k)dk, Q Q

nT - ni =

p(k)dk,

_Q

e=f

B B (2A2

c

Q

- 2)a(A)dA + f p(k)dk.

(2.122)

It should be noted that one A and two ks form a bound state of two fermions with binding energy c2/2. In contrast with an attractive boson system, there are no bound states of more than two fermions. The system has a thermodynamic limit. 2.2.9 Expansion from small nj We consider the small B limit of equations (2.107) and (2.108). We define the function f (n1, nj; c') as the internal energy per unit length as a function of c' = c/2. Using the small B expansion we obtain at c > 0

f(n-x,x;c')=

R33

2

-xf 2(

+7cn2)tari 1

+4x2n{tari 1

nn 12 c

n-2c'n} C/

7r

+ O(x3).

(2.123)

67 The first order term is obtained by and Yang117. the quadratic term is obtained by Suzuki87. For c < 0 we have from (2.120), (2.121) and McGuire66,

(2.122) X33

f(n-x,x;c') +4x2n{7r

_

+ tans

-x f 2(_

-n 12

1 Win)-2c'n}

+ 0(x3).

(2.124)

C/

These expansions can be written in a unified form as follows: 2

_3 -x{2(_ +7rn2)(2 C 0(x3). -4x2n{ 2 -tan 1 n }2 + f(n-x,x;c') _

3

The saturation magnetic field is given by 2

-tan £n) -2c'n} (2.125)

i

hs = (- + 7rn2) (2 -tan-' 7cn) - cn'.

(2.126)

Systems with a delta function potential

32

As a function of c' this is analytic even at c' = 0, notwithstanding that the two sets of equations seem to be quite different. We can construct a unified integral equation which analytically continues from c > 0 to c < 0. Then we can show that the internal energy f (nT, ni ; c) is analytic on the real axis of c when nT * n1.

2.2.10 A unified form of the integral equations

We can eliminate p(k) from the integral equations for (2.107)-(2.108) or (2.120)-(2.121),

f

K(A, A')a(A')dA' = I g(A),

6(A) + .

(2.127)

B

with z

K(A,A')

flxl>Q (c2 + (A - x)2)(c2 + (x - A')2)'

7T2

g(A) = 1 - I (tan-'

C'

Q+A

+tan-

c'

1

Q-A

.

(2.128)

The internal energy density f, particle density n and ni are

f=

Q3 37r

B B

+

dA

+n(A2

n=

Q

ci2 + (Q + i)2 {2Qc' - Ac log 72

71

- cr2)g(A)}a(A),

-+

A) 2

(2.129)

B

+

71

J- B

dAg(A)o (2),

(2.130)

fBB

nl =

a(A)dA.

(2.131)

As far as B < Q, the integration kernel K and inhomogeneous term analytically continue from c' > 0 to c' < 0. Then if nT ni, f also continues from c > 0 to c < 0. This means that f (n, n - x, c') is expanded by a Taylor series of c', and that series has a finite convergence radius for x n/2.

2.3 Bosons and fermions with arbitrary spin Sutherland showed that the spin 1/2 fermion theory can be generalized to

fermions with larger spins84 for the repulsive case (c > 0). He obtained integral equations for the distribution. The author investigated the case of

2.3 Bosons and fermions with arbitrary spin

33

attractive fermions with arbitrary spin92. Lai and Yang investigated the mixture of spin 1/2 fermions and spin zero bosons with the same mass

and repulsive interactions. In the attractive case of bosons, there is no thermodynamic limit.

3 The isotropic Heisenberg model

3.1 The ferromagnetic case

The Heisenberg model was the first model to be treated by the method of the Bethe-ansatz13. In the beginning of the 1930s only the ferromagnetic model was considered:

= -J

N

Sl ,

SI S1+1 + Si S +1 + Si S1+l - 2h l=1

l=1

h>0, J > 0,

Si.

SN+1

This Hamiltonian is defined on a 2N dimensional space, but the space is classified by the total SZ = T S. The ground state is the state where all spins are up and the total SZ is N/2, *'I0) = EoIO),

Eo = -JN/4 - Nh.

(3.2)

We write eigenfunctions at SZ = N/2 - M as follows: `P =

f (n1, n2,..., nM)Sn-, Snz...S;M I0),

(3.3)

where 1 < nl < n2 < ... < nM < N and 2M < N. In terms of f (n1, n2,..., nM) the eigenvalue equation is

-2 E(1-8nj+1,nj+,){f(nl,...,nj +1,nj+1,...,nM) +f (nl,..., nj, nj+l - 1, ..., nm) }

+{Eo - E + (J + 2h)M - J E and+1,n;+ }f (nl, n2,..., nM) = 0. 3

34

3.1 The ferromagnetic case

35

Next we assume that the wave function is written as follows : M

M!

f(ni,n2,...,nM) = E A(P) exp(i E kpjnj), P

(3.5)

j=1

where P is a permutation of 1, 2,..., M, 2,

1,

P

P1, P2,

...,

M

...,

PM

(3.6)

kl, k2,..., km are called quasi-momenta. As 2M < N there is at least one set of (nl, n2,..., nM) which satisfies nj + 1 < nj+1. Then we have M

E = Eo +>[J(1- cosk1) + h].

(3.7)

j=1

The situation is almost the same as for 6 function bosons, the difference is that here the space is discrete. The shift operator T is defined as follows, T f (nl, n2,..., nM)

= f (nl + 1, n2 + 1, ..., nM + 1).

This wave function is also an eigenstate of the shift operator TT = We call K is the total momentum, which is given by

(3.8) iKqJ

M

K = Ekj. j=1

Next we consider the case n j + 1 = n j+l and ni + 1 < n1+1. f should satisfy

f(nl,...,nj + l,nj + 1,nj+2,...,nM) +f(nl,..., nj, nj, nj+2,..., nM) -2f (nl,..., nj, nj+1, nj+2,

...,

nM) = 0.

(3.10)

This is satisfied if 0 = A(P)(euI i + e-kP(i+i)

-

2)e`k°(j+i)

+A(P(j, j + 1))(eikP(1+i) + e-kPi - 2)e' Pi.

There are N!(N

(3.11)

- 1)/2 equations. These equation are satisfied if we put A(P) = e(P) fJ(ei(kP,+kP;) + 1 - 2e P').

(3.12)

1<j

The periodic boundary condition requires : f (nl, n2, ..., nM-1, N + 1) = f (1, ni, n2, ..., nM-1).

(3.13)

The isotropic Heisenberg model

36

This is satisfied if the following condition stands,

exp(ikp,yfN)A(P1,P2,...,PM) = A(PM,P1,...,P(M

- 1)).

(3.14)

This is equivalent to

exp(ikjN) =

(-1)M_1 r,

exp[i(kj + k1)] + 1 - 2 exp(ikj)

lJ exp[i (kJ j + k j)] + 1 - 2 exp(ikl)'

j = L21 ..., M.

(3.15)

This is a set of complicated coupled equations for M unknowns. If we have a solution of this set, we have one eigenstate and its energy eigenvalue and total momentum. We introduce rapidity parameters xj = cot(kj/2). In terms of the new parameters

exp(ikj) = l

xj +\J

x-

ii

(3.16)

( j

and equation (3.15) becomes xj + l N

xj - x,

xi - i

+2i)

xj - xl -

l

2i(3.17)

The r.h.s. is a function of the difference of the x js. The wave function is

f(n1,n2,...,nM) = EA(P)fl( j_1

p

xP+i nj ) J

xP-i J

,

A(P) = De(P) fJ(xpj - xpl - 2i).

(3.18)

j<1

The energy and momentum are

E =Eo+E[

2J

+1 j=1 xJ

II(xj+i

+2h],

xj j=1

(3.19)

M

K -7rM - 2 E tan

1

xj

mod(27r).

(3.20)

k=1

In the lattice problem the momenta K and K + 2ir x integer are equivalent.

This point is different from the problems of chapter 2 in the continuum space.

3.2 The string solution of an infinite system

37

3.2 The string solution of an infinite system In the M = 1 case, the elementary excitation is

E - Eo = J(1 - cosK) + 2h.

(3.21)

This is the spin wave excitation of the ferromagnetic Heisenberg model. Bethe13 found that the bound state of spin waves exists in M = n subspace,

xj =a+i(n+1-2j),j=1,2,...,n.

(3.22)

The energy and momentum of this excitation are 2nJ

E = E + a2

+ n2

K=

+ 2nh,

I

In

1

in). « - in

(x

+

(3.23)

Thus the dispersion of these excitation is

E = Eo +

J(1 n

- cosK) + 2nh.

(3.24)

These string solutions play an essential role in the thermodynamics of soluble models.

3.3 The Hulthen solution for an antiferromagnet

The antiferromagnetic case is the J < 0 case of equation (3.1). In actual materials, the ferromagnetic case is rare. Typical one-dimensional magnetic substances are antiferromagnetic. From equation (3.17) we have M

tan-'

2N tan-1 xj = 27rIj + 2

xj

- xl

(3.25)

2

1=1

where Ij is an integer (half-odd integer) for odd (even) N - M. The total momentum is given by:

K =7r(1 - (-1)M)/2 -

271

One can show that for even N the lowest energy state in the subspace of fixed total Sz is given by

Ij = (M + 1 - 2j)/2, j = 1, 2,..., M.

(3.26)

In the thermodynamic limit the xjs distribute from -B to B. From equation (3.25) we have:

tan-1

x= 7

x

p(t)dt +

f

B

B

tan 1

X

_ 2

yp(y)dy.

(3.27)

The isotropic Heisenberg model

38

Differentiating with respect to x we have P(x)

_

1

1

- f B it (x - y)2 + 4 P(y)dy. 1

it x2 + 1

2

(3.28)

The energy and magnetization per site are: B

-4 12h - x2 I+I N f N = 2 - f p(x)dx. E

-h+

B

p(x)dx,

(3.29)

1

(3.30)

B

This integral equation can be solved in the case of infinite B. We define the Fourier transform of p(x) as follows: P((0) = r-00 e `wxp(x)dx.

(3.31)

Using the formula f ir-In/(x2 + n2)exp(-ixw)dx

exp(-nIwI) we can

rewrite (3.28) as follows: P(w)(1 + e 2I w) = e-10)1.

Then we have p(w) = 1/(2 cosh (o) and 00

p(x) =

4sech(2 ).

2-

(3.32)

Substituting this into (3.30) we have:

e = -IJl (ln2

- 4) = -0.4431471JI,

sZ = 0.

(3.33)

Then we find that the case B = oo is the true ground state at It = 0 and that the magnetization is zero. Thus we have an analytical result for a onedimensional antiferromagnet (Hulthen39). The first neighbour correlation is derived from this result (SizS+1) = 12(1 -41n2) = -0.14771573.

(3.34)

An analytic expression of the second neighbour correlation function is also known for this model (Takahashi'°'), (SFS+2) =

2(1

-

0.06067977.

(3.35)

3.4 The des Cloizeaux-Pearson mode of an antiferromagnet

39

3.4 The des Cloizeaux-Pearson mode of an antiferromagnet We consider excitations from Hulthen's ground state at M = N/2, where the I s are

N'-1 N'-3 2

N'-1

2

'

2

N' = N/2.

'

(3.36)

It is difficult to find a real solution of ks for Stotal = 0, except for the ground state. One possibility is the following state in Stotal = 1 subspace

N' N'

2 2 -1,...,N'2 -r+1,N'2 -r-1,

N' ..., 2

2 -s+1, N'2 -s-1,- N'

(3.37)

We put 0 < r < s < N'. In this case all M - 1 ks can be real numbers. The total momentum is ir(1-M)/2+27r(N' - r - s)/N. Then the change of total momentum is AK = 21r(r + s)/N. This state has two holes at positions Xr and xs. In the thermodynamic limit we have AK = 2ir (J

° p(x)dx + f p(x)dx)

xr

s

tan-1(e-"xr/2)

=2

+2

tan-1(erzxs/2).

(3.38)

The change in energy is not 21JI/(1 + x;) + 21JI/(1 + xs) because the other xjs move of the order of N-1. To get the change of total energy one must calculate the sum of the small shifts of the other ks. This is called the back-flow effect. From equation (3.25) we have

2N(tan 1(xj + Axe) - tan' xj)

-2

N'-1

tan_1(xi

+ Ax j - x,,, - Ax,,

m=1

2

1 -tan-1(x i -2xm 1

_ it [sign(xj - x1) + sign(xj - xs) - 1] - 2 tan-1

Xi - xN 2 (3.39)

The last term of the r.h.s. is replaced by -7r because XN' = -oo. From this we have 20x1

xJ + 1

1

N'-1

4(0x1 - Oxm)

N E (x - xm)2 + 4 M=1

J

It

N

[sign(xj-xr)+sign(xj-xs)-2]. (3.40)

If we put NAxp(x) = J(x) this equation becomes

The isotropic Heisenberg model

40

J(x) f

J(x)

2

4p(y)dy

00

p(x) J_ (x - y)2 + 4 +

x2 + 1 p(x)

4J(y)dy (x - y)2 + 4

= 7r(sign(x - Xr) + sign(x - xs) - 2).

(3.41)

Using (3.28) we have

[-6(u - xr) - 6(u - xs)]du. (3.42)

ao (x - y)2 + 4J(y)dy = rX This is the integral equation for the back flow J(x). The analytic solution of this equation is

J(x) +

00

J(x) _ 1 [-6(u - Xr) + R(u - Xr) - 8(u - xs) + R(u - xs)]du, x 00

R(x) =

! D-1)n-1x2 n=1

2n

(3.43)

+ 4n2

Thus the change of energy AE is AE = 21JI [xN 1

+1-

f

dx x2 + 1

J(x)dx] .

By partial differentiation, AE = 2IJI

_

1rJI

- R(x - xr) - R(x - xs) - 6(x - xr) + 6(x - xs)x2+1 dx +sech7r2s)

(sech7r2r

= 21J17r(p(xr)+p(xs)).

(3.44)

Putting q1 = 2 tan 1 e-"xrl2, q2 = 2 tan 1 e-"xs/2 we have

AE =

rzln II

(sin q1 + sin q2),

AK = ql + q2.

(3.45)

This spectrum is shown in Fig. (3.1). Thus Stotat = 1 states with real ks are composites of two excitations. One spinon with momentum q has energy (irIJI/2) sin q and spin 1/2. Actually the spinon number must be even. Des Cloizeaux and Pearson19 obtained analytic expressions for these excitations by putting one spinon at zero energy ql = 0 or 7r,

AE =

7T I2 III

sin K I .

This is the lowest energy state with momentum K. This is called the des Cloizeaux-Pearson mode. This means that the ground state has excitations with an infinitesimal energy gap. We find the velocity vs of this elementary excitation at low momentum,

vs = l odAE/dK = 7rIJI/2.

(3.46)

3.5 Magnetic susceptibility and magnetization curve

41

(E-EO)/J

Fig. 3.1. Spinon excitation of XXX model at T = 0 and h = 0.

According to the linear spin-wave theory (Anderson2, Kubo54) the excitation spectrum is

AE = IJil sinKI.

(3.47)

Then spin-wave theory gives a slower velocity than the exact Bethe-ansatz solution. But one should note that the coincidence of spinon lower edge and spin wave spectrum is very accidental. In a one-dimensional XXX antiferromagnet there is no Neel order even at zero temperature. 3.5 The magnetic susceptibility and magnetization curve for J < 0 In a sufficiently strong magnetic field the ground state is completely magnetized. The energy of the ground state is E0 = h), and the energy of an excited state is

E = E0 + [2h - IJI(1 - cosk)].

(3.48)

- 21JI at k = it.

Then at h < J1 the

The minimum of this energy is 2h

completely magnetized state cannot be the lowest energy state. To minimize the energy we should have ce/8B = 0 for equation (3.29). We must consider

the change B --> B + AB and the change p(x) -* p(x) + Ap(x). Then the equation for Ap(x) is B

Ap(x) +

B

a2(x

- y)Ap(y)dy

= -(a2(x - B) + a2(x + B))p(B)AB,

an(x) -

1

n

7rx2+n2

(3.49)

The isotropic Heisenberg model

42

From equation (3.29) we have the change of energy B

De = 2h 120Bp(B) + fB Ap(x)dx]

-2irIJI I2ABp(B)ai(B) + f B ai(x)Ap(x)dxI.

(3.50)

B

This is written as follows, = 4ABp(B) [h (1

I

-itIJI (ai(B) -

B B

-L

D(x)dx)

ai(x)D(x)dx)],

(3.51)

where D(x) is defined as the solution of the equation B

D(x) + f a2(x - y)D(y)dy = a2(x - B).

(3.52)

B

The expression inside the second parenthesis of the r.h.s. of (3.51) is p(B). The expression inside the first parenthesis is F(B), where F(x) is defined as the solution of the equation B

(3.53)

F(x) + f a2(x - y)F(y)dy = 1 B

The magnetic field h for a given B is h=

7rP(B)

(3.54)

J1

F(B)

The magnetization is given by (3.30). Thus one can plot the magnetization curve as a function of magnetic field by changing B from 0 to oo

(Griffiths32).

See Fig. (3.2a).

In a small magnetic field the energy becomes minimum at large B. The equation (3.28) can be transformed as follows

- y)p(y)dy.

(3.55)

P(x) = POW + f l>B R(x - y)p(y)dy, Po(x) = 4sech(2 ).

(3.56)

P(x) + f 00a2(x - y)p(y)dy = ai(x) + Jyl>B a2(x

po(x) is the solution at B = oo or h = 0. Substituting this into (3.30) we have

sZ = 2 -

f

p(y)dy + f

vl>B

p(y)dy

=2

f'>B p(y)dy.

(3.57)

3.5 Magnetic susceptibility and magnetization curve

43

Here we have used f po(x)dx = f R(x)dx = 1/2. The energy change is fB

e - eo = -2hsz + 27rIJI If ai(x)po(x)dx = -2hsz + 21rIJI

f

ai(x)p(x)dx] B

CO

po(x)p(x)dx.

(3.58)

xI >B

Putting p(x) = 2 exp(7rB/2)p(x + B), equation (3.56) becomes

p(x) = e nx/2 + f(R(x - y) + R(x + y + 2B))p(y)dy,

(3.59)

and sz = e-nB12a(B),

a(B)

e - eo = -2hsz + IrJIe ,,Bb(B),

= (1/2)1"0 p(x)dx,

b(B) = 7r

J

f enx/2p(x)dx.

(3.60)

The functions a(B) and b(B) behave for B >> 1 as follows:

a(B)=ao[1+

1

+O(InB)] ,

21rB

b(B)=bo[i+O(B-2)].

B2 ))J

(3.61)

In the thermodynamic equilibrium ae/aB = 0. Thus we have (bo/ao)IJje-nB12[1

h= 2sz

+

(27rB)-1]-i,

2a2

x = h = boIJI (1 + 2a2 boJJI [1

(27rB)-1)2 +

O(ln B/B2)

InIInhl

1

+ 21n(boIJI/(aoh)) + O

(T1

n h)2

)] .

(3.62)

There is a logarithmic singularity in the magnetic field as a function of the magnetic field. Z has an infinite slope. See Fig. (3.2b-c). Griffiths 32 estimated

numerically that ao = 0.48394, bo = 1.15573 and predicted that 2a2

-4

2.

bo

(3.63)

Yang and Yang120 proved this analytically using the Wiener-Hopf factorization as shown in the next section. Thus x is related to the excitation velocity obtained in the previous section:

X=

2 7rvs

.

(3.64)

The isotropic Heisenberg model

44 (a)

(b)

0.25

h/J 0.2

0.4

0.8

0.6

h/J 1

(c)

szJ/h 0.24r

0.22

0.1

0

Fig. 3.2. Magnetization curve (a) and sZ/h - h curve (b), (c) for the XXX chain. Near zero field Griffiths found a logarithmic anomaly in the sZ/h - h curve. It has infinite slope near h = 0.

3.6 Wiener-Hopf type integral equation We should consider the solution of the following integral equation, 100

R(x - y)p(y)dy = e

p(x) -

nx/2.

(3.65)

We define (3.66)

P(w) = fr p(x)e`wxdx. The l.h.s. of (3.65) is

E

dw -iwx e

00 27c

I

exp(-21wI) + 1

p (w ).

(3 . 67)

3.6 Wiener-Hopf type integral equation

45

Here we define lo-)l - sign(91((o))co. The denominator has zeros at w = (n + 1/2) ti and a branch cut on the imaginary axis. We factorize the denominator exp(-21(ol) + 1 = G+(w)G_(w),

(3.68)

where G+(w) and G_(w) are analytic and non-zero in the upper and lower half-plane respectively, plus the real axis, G+(w) = G_(-(9),

G+(oo) = 1.

(3.69)

The analytic expression of G+(w) has a discontinuity or branch cut on the negative imaginary axis only. Using (E.43) and (E.46) (see Appendix E) we find the following function satisfies this condition, G+(w)

i'(z

+ 2)zZe Z ,z = -w/n.

(3.70)

The solution of equation (3.65) is

NO

= G+(w)G_(-iri/2) =

2/ezze

7r/2-iw

(z+2)r(z+z)

(3.71)

By the definition we have 2/(ire), bo = np(ici/2) =n/e.

ao = p(0)/2 =

(3.72)

Using these values we have (3.63). In the detailed calculation we have Z =

4 J7C2

[l +

1

lnln(ho/h)

21n(ho/h)

4(ln(ho/h))2

ho = J 7r3/(2e) = 2.388155328J.

1

+ o C (ln(ho/h))2)1 (3.73)

The same kind of anomalies are also found in the Hubbard model and other itinerant electron models50

4

The XXZ model

4.1 Symmetry of the Hamiltonian In this section we consider the following Hamiltonian: N

i(J, A, h) = -J

Sf ,

Si Si+1 + Si Sl+1 + ASl Sf 1 - 2h l=1

1=1

h > 0, SN+1 = S1-

This Hamiltonian contains a new parameter A. The case A = 1 is the XXX model treated in the previous section. The case A = 0 is the XY model treated in chapter 1. The limit of very large A is the Ising model. The generalization of Bethe's method to A * 1 was done by Orbach77 and Walker114 Yang and Yang investigated the ground state of this model in detail' 111-122 Bonner and Fisher16 investigated this model using the exact diagonalization method up to N = 12. In this Hamiltonian the magnetic field is applied in the z-direction. For a magnetic field in another direction, the exact solution is not known. Let us consider the following unitary transformation,

r(J, A, h) = U1 r(J, A, -h) U1 1,

U1 - 1125 = U1

(4.2)

l=1

By this unitary transformation Stotal changes its sign and thus we can treat the N > M > N/2 case. In the case of even N we can show that (-J, -0, h) = U2,XP(J, A, h) U21,

U2 = [J 2S1 = U.

(4.3)

l=even

By this unitary transformation Sf, Sly, Sf change to -Six, -S11, Sl at l = even. 46

4.2 The Bethe-ansatz wave function

47

4.2 The Bethe-ansatz wave function We consider the state where all spins are up and the total SZ is N/2, X10) = EoIO),

Eo = -JON/4 - Nh.

(4.4)

We assume the wave function of equation (3.3). Corresponding to (3.4) we have 0

J = - 2 E(1 j

- bnj+l,nj+l) {f (nl,..., nj + 1, nj+1,..., nM)

+.f (n1, ..., n j, n j+1 - 1, ..., nm) }

+{Eo - E + (JA + 2h)M - JA

bnj+1,nj+,

}f(ni, n2,..., nM).

j (4.5)

The wave function of the type of equation (3.5) can be an eigenstate with M

E = Eo +>[J(A- coskj) + h], j=1

0 = A(P)(eil i + e ikP(j+')

- 2A)e`kP(j+n

1))(e"P(j)

+A(P(j,j +

+

2A)e`kPj,

(4.7)

A(P) = e(P) fl(ei(kPl+kpj) + 1 - 2Ae`k"t ).

(4.8)

1<j

The periodic boundary condition is as follows,

exp(ikjN) _

(-1)M-1

exp[i(kj + k1)] + 1 - 2A exp(ikj)

H ,=,j exp[i(kj + k1)] + 1 - 2A exp(iki)'

j = 1, 2,..., M.

(4.9)

The transformation (3.16) is not convenient because the phase factor exp[i(kj + k1)] + 1 - 2Aexp(ikj) exp[i(kj + k1)] + 1 - 2A exp(iki)'

cannot be written as a function of x j - xi except the case A = 1. If we put exp(ikj) =

sin(xj + i) sin (xj - i)

(4.10)

in place of (3.16), the phase factor becomes cos (xj + xi)(cosh 0 - A) + (A cos ±2 (xj - xi + 2i) - cos ±(xj - xi)) 2 cos (x j + xi)(cosh 0 - A) + (A cos ±2 (xi - x j + 2i) - cos ±2 (xi - x j))

The XXZ model

48

Then if we put cosh 0 - A = 0, this phase factor becomes a function of x j - xl and independent of x j + xt, sin 0 (x j - xi + 2i)/ sin 0 (xj 2

2

- xi - 2i).

Then for A > 1, (4.9) becomes 1sin 0 (x j + i) N

sin O (x j

'sin (xj -

2(

,,j

- xl + 2i)

xj-xl - 2i)' ¢>0.

cosh-1 A,

(4.11)

ForA< -l we put exp(ikj) = -

sin 0 (x j + i)

sin 2 (x j

-

(4.12)

i)

Equation (4.9) becomes sin O (x j + i) N

= 11

(sin±(xj-i)1 2

l

sin O (x j - x j + 2i) sin 02 (xj -

xi - 2i)'

0 = cosh-1(-A),

4) > 0.

(4.13)

In the case -1 < A < 1 we put expik j

_

sinh i(xj + i) sinh 2 (x j

- i)

(4.14)

Equation (4.9) becomes 1sinh 2 (x j + i) N sinh 2 (x j - i) )

_ 11 sinh 2 (x j - xi + 2i) sinh 2 (x j - xl - 2i)' 1

y = cos-1(-A), t > y > 0.

(4.15)

4.3 The string solution for A > 1 By the transformation (4.10), the wave function and eigenvalue are f(ni,n2,...,nM)

_ E e(P) fl sin - (xp j - xpl + 2i) P

j
M

sin 2

j=1 sin 2((xPjJ + i)) ni (4.16)

M

E = Eo +

.=1

(2h +

J sinh2 1 cosh 0 - cos Ox j /

K = y 2 cot-1

tan(Ox j/2) (4.17) tanh(4)/2)

4.4 The lowest energy state for A < -1

49

From the normalizability of the wave function, the following string solutions are possible for complex xjs for the N = oo case,

xj =a+(M+1-2j)i,

(4.18)

where a is a real number, -it < a < it. The total momentum and energy are given by

K = 2 cot-1

tan(oa/2) tanh(M4/2)

E = Eo +

J sinh 0 sinh MO + 2Mh.

cosh MO - cos 4a

(4.19)

From these we have :

E = Eo + 2Mh + J sinh

r

cosh MO - cos K l J sinh MO

(4.20)

This excitation energy gives (3.24) in the limit A --* 1. In the limit of large A, the energy is JA + 2Mh. This state is M successive down spins in the sea of up spins. The lowest energy state for A > 1 with M down spins is the M string state given by (4.20) with zero total momentum. Thus, E

=-J40 -(N-2M)h+Jsinh0tanh

2

(4.21)

+ 7r].

(4.22)

4.4 The lowest energy state for A < -1 For A < -1 equation (4.13) yields M

N61 (x1, 4)) = 2irI j +

62(x; - x1, 4)),

tan-1(tan(Ox/2))

4i) = 2

tank no/2

+27, [Ox

27r

Example of this function are shown in Fig. (4.1). Here we take the branch

of tan -I x as -7r/2 < tan -I x < 7r/2. xj moves in the region -Q < xj < Q, Q - it/4). The distribution function of the xjs at the lowest energy state should satisfy B

Oi(x, 4)) = 21T j p(t)dt + fB 02(x - y, 4))p(y)dy x

Differentiating with respect to x we have B

al(x,4)) = p(x) + fB a2(x

an(x' ) -

27r ax 6n(x,

)

- y,4))p(y)dy,

- 27r cosh no

cos Ox

(4.23)

The XXZ model

50

Fig. 4.1. The function 0,(x, 0) at 0 = 0.4. The solid line is for n = 1, the dashed line is for n = 2 and the dotted chain line is for n = 3.

Here B varies from 0 to Q. The magnetization and energy are B

1

sZ = -

LB

p(x)dx,

e = - JO 4 - h + JBB (2h - 2mJ sinh

al (x)) p(x)dx.

(4.24)

At B = Q equation (4.23) can be solved by Fourier series, fQ

P(n) = J Q p(x) exp(-in4x)dx,

exp(-Inlfi) = P(n)(1 +exp(-21nl4))).

(4.25)

Thus we have the distribution function for the ground state po(x), 1

Po(n)

Pox) =

2 cosh no' 1

dn(Q , u),

E exp(in4x)P(n) =

2Q n=-oo

27C Q

(4.26)

4.5 The magnetization curve at A < -1

51

where dn(x, u) is the Jacobian elliptic function and the modulus u is determined by K( 1 - u2) = K(u)

=

1

is

Q

dp

K(u) =

0

(4.27)

1- u2 sin2 p

The magnetization is zero and the energy density is

e=-

JA 4

00

-1

- J sinh 0 1-2 + 2 E

1

e2n¢

n=1

+

1].

(4.28)

4.5 The magnetization curve for a field in the z-direction at A < -1 The energy per site, which is given by (4.41) and (4.24) at a given magnetic field, must be minimized. To minimize the energy we should have ae/aB = 0 for equation (4.41). We must consider the change of the region B -+ B + AB and the change p(x) -* p(x) + Ap(x). The equation for Ap is B

Ap(x) + fB a2(x - y)Ap(y)dy = -(a2(x - B) + a2(x + B))p(B)AB. (4.29) Thus the change of energy becomes B

Ae = 2h I2ABp(B) + fB Ap(x)dxl 271 sinh OJ

B

[2ABai(B)p(B) + LB ai(x)Ap(x)dx].

(4.30)

This is written as follows Ae

4ABp(B) =

h(1

-

7L sinh OJ (al(B)

fB

-B

D(x)dx)

J - LB ai(x)D(x)dx), B

(4.31)

where D(x) is defined as the solution of the equation D(x) +

B

a2(x - y)D(y)dy = a2(x - B).

(4.32)

The second parentheses of the r.h.s. in (4.31) is equal to mp(B). The magnetic field h for given B is Jir sinh 0 p(B) 1

0

F(B)'

(4.33)

The XXZ model

52

where F(x) is defined as the solution of the equation B

F(x) + fB a2(x - y)F(y)dy = 1.

(4.34)

The magnetization is given by (4.24). At B = 0 we have F(B) = 1, J(sinh q5/4)p(B) = J(1 - A)/(2ir). Then the corresponding magnetic field is h = J(1 - A)/2. Above this magnetic field all spins are up and sZ = 1/2. At B = Q we have F(x) = 1/2 and (4.26). Then the corresponding magnetic field is

h, - J sinh OK(u')u'

(4.35)

At zero temperature and h < h,, the state is not magnetized. The phase diagram and magnetization curve are given in Figs. (4.2) and (4.3).

4.6 The lowest energy state for fixed M and -1 < A < 1 Assume that J > 0. In the case of M down spins, the number of states is C. The off-diagonal matrix elements of the Hamiltonian (4.1) are -J or 0. Then the lowest energy state wave function is always nodeless and non-degenerate. We take the logarithm of (4.9), M

kjN = 2irIj -

O(k j, ki),

A sin 2, (x - y)

O(x' y) =_ 2 tan-1

cos i (x +Y) - A cos i (x - y)

(4.36)

where Ij are different integers (half-odd integers) for odd (even) M. Set

I=M

2

1

- J, j = 1, ..., M.

(4.37)

At A = 0 this gives k1 = ic(M + 1 - 2j)/N and the corresponding energy coincides with (1.13). At -co < A < 1 equations (4.36) with (4.37) have unique real solutions of {kj}. Thus the lowest energy state at fixed M at 0 < M < N/2 is certainly expressed by Bethe-ansatz. Using the transformation U2 in (4.3) we find that the lowest energy state at J < 0, N = even, A > -1 is expressed by the Bethe ansatz. The analysis for XXX antiferromagnets with J < 0, A = 1 in §3.3 is justified. At -1 < A < 1 equation (4.37) is written as follows

2N tan-'

[tanh(yxj/2)] [tanh y(xj - xi)/21 = 2icIj + 2 tan' tan(y/2) tan y

(4.38)

4.7 The magnetization curve for afield in the z-direction at -1 < A <

53

1

Writing the distribution functi on of the xjs as p(x), we have 2 tan-1 B

[tanh(y x/2) = 2 ir tan(y /2)

2 tari

1

]

tank y(x

f

x

p (t)dt +

- y)/2 p y )dy. ]

tan y

B

(

(4 . 39)

By differentiation

a(x, l) = p(x) + fB a(x - y, 2)p(y)dy, B

a ( x,n )

y sin ny

1

4 . 40)

21i cosh yx - cos ny '

The energy density and magnetization density are B

1

sz = 2 - f B p(x)dx,

-h+

- 2irJ sine Y a x, 1)] p x )dx.

( ( (4 41) = -J At B = oo the integral4equation can be solved analytically with the use of a

e

B

[2h

.

Y

Fourier transformation, sinh(po - 1)w sinh pow

-

p(w) (1 +

sinh(po - 2)w sinh pow

PO =71T

(4.42)

Thus we have

4sech(2 ). 2c shw' p(x) =

p(w)

(4.43)

Using (4.41) we have the ground state energy per site at sZ = 0, e = -

JA 4

- J

sin y y

f,,

sinh(po - 1)w dw. cosh w sinh pow

(4.44)

4.7 The magnetization curve for a field in the z-direction at -1 < A < 1 The energy per site, which is given by (4.41) at a given magnetic field, must be

minimized. To minimize the energy we should have ae/aB = 0 for equation (4.41). We must consider the change of the region B -+ B + AB and the change of p(x) -* p(x) + Ap(x). Then the equation for Ap is B

Ap(x) + fB a(x - y, 2)Ap(y)dy = -(a(x - B, 2) + a(x + B, 2))p(B)AB. (4.45)

The XXZ model

54

Thus the change of energy becomes B

De = 2h [21\Bp( B) + LB Ap(x)dx]

2nJ sin y

y [2AB a (B , 1) p (B)

B

+ L a( x, 1)Ap(x )d x ] .

4. 46)

B

This is written as follows : De

h(1

4ABp(B) 7TJ ysin

-

y (a (B , 1)

JB

B

-

D( x )dx )

B

LB

a( x, l)D( x )d x ) ,

(4 . 47)

where D(x) is defined as the solution of the equation

D(x) + J a(x - y, 2)D(y)dy = a(x - B, 2).

-BB

(4.48)

The second parenthesis of the r.h.s. of (4.47) is equal to np(B). The magnetic field h for a given B is

h = Jm sin y p(B) y F(B)' where F(x) is defined as the solution of the equation F(x) + fB a(x - y, 2)F(y)dy = 1.

(4.49)

(4.50)

B

The magnetization is given by (4.41). At B = 0 we have F(B) = 1, J sin yp(B)/y = J(1 - A)/(2ir). Then the corresponding magnetic field is h = J(1 - A)/2. Above this magnetic field all spins are up and sZ = 1/2. 4.8 Susceptibility for -1 < A < 1 In the case -1 < A < 1, in the limit B = oo, the corresponding magnetic field is zero, because F(x) = 1/2, p(oo) = 0. The system is magnetized by a weak magnetic field. We can calculate the magnetic susceptibility. In the small magnetic field the energy becomes minimal for large B. The equation (4.40) can be transformed as follows,

P(x) + f 00 a(x - y, 2)p(y)dy = a(x, 1) +

lyl>B

a(x - y, 2)p(y)dy.

(4.51)

By Fourier transformation we have

P(x) = Po(x) + fyI>B R(x - y, y)P(y)dy,

Po(x) = 4 sech (2 ), (4.52)

4.8 Susceptibility for -1 < A < 1

R(x, y) -

1

27r

r

e i(OX

55

sinh(po - 2)w 2 cosh co sinh(po - 1)w

dw.

(4.53)

The integrand of (4.53) has poles at pop11 i and 7r(n + 2 )i. Then for large x, O(e-2x)

=

R(x, y)

+ O(e °p 'x) 77

po(x) is the solution at B = oo or h = 0. Substituting this into (4.41) we have

sz = 1

-

00

0

p(y)dy +

p(y)dy = iyi>B

it

/'

2(it - y) .l lyi>B

Here we have used f po(x)dx = 1/2, f R(x, y)dx = (7r energy change is

e - eo = -hsz +

2irJ sin y Y

B

LB

- 2y)/2(7r - y).

00

[f- 00 a(x,1)po(x)dx sin y

= -hs+

a(x,1)p(x)dx]

Y

p(y)dy.

(4.54)

The

-

po(x)p(x)dx. JIxj>B

Putting p(x) = 2 exp(7rB/2)p(x + B) we have

p(x) = e 7cx/2 + f(R(x - y, y) + R(x + y + 2B, Y))p(y)dy, sz =

it

7E -y

exp(-irB/2)a(B),

e - eo = -hsz + (J

sin 2y

y ) exp(-7rB)b(B),

a(B) = (1/2) or p(x)dx, 0

b(B) = it f e

2p(x)dx.

(4.56)

The functions a(B) and b(B) behave for B >> y as follows: n')

a(B) = ao[1 + O(eP

+ O(e B")],

2Bn

b(B) = bo [1 + O(e) + O(e B")]. We put p(x) = p1(x) + P2(x) + P3(x) + ...

p1(x) -

jR(x - y, Y)p1(y)dy = e 7rx/2

00

po(x) -

J

00

R(x - y, Y)p.i(y)dy =

J

R(x + y + 2B, Y)p.i-1(y)dy

(4.57)

The XXZ model

56

The integration kernel is factorized as follows : sinh(po(o)

1 - R(w, y) = 2 cosh(es) sinh((po - 1)(o)

-i ) (PO 2(1- po

G+(ro)G+(-w)'

z

1)po-i

F(Poz + 1)

F(z + z)T((Po - 1)z + 1)

Po

Using this function we can obtain ao and bo, ao = it-iG+(0)G+(i2),

bo

=

(G+(0)/7r)2

=

bo = G+(i2),

2(

73

y)

The higher order corrections are O(exp(-7rB)) and O(exp(-27tB/(po - 1))), because R(x+y+2B, y) decays in this way. In the thermodynamic equilibrium ae/ 7B should be zero. Thus, h = bo J(7r - y) sin y

hZ

ao

7ry

=x

ZJ(7[

4 Jir(7r

y) sin y

2Bir

[1 + O(e per) + O(e-B")l ,

b [I + O(h) + O(hn YY )] 2

y)2 sin Y o

[I + O(ho) + O(hn vv )].

(4.58)

There is an algebraic singularity as a function of the magnetic field. X has an infinite slope for -1 < A < -0.8. 4.9 The long range order of the XXZ model At A > 1 and h = 0 the ground state is doubly degenerate : I TTT ... T),

I III ... I).

Thus the two-point function of the ground state is always 1/4.

(SfSjZ)=4. In the limit -A >> 1 the system is equivalent to the Ising antiferromagnet. For the case of N = even, the ground states in the limit A -+ -oo are the following two wave functions

I Tit

...

1),

1 iTl

...

T).

4.9 The long range order of the XXZ model

57

Fig. 4.2. Phase diagram of the XXZ model in the magnetic field. In regions A and C the ground state has an energy gap. In region B the system becomes gapless. Point

E is the ferromagnetic XXX point. Point D corresponds to the antiferromagnetic XXX point.

The energy is -NJIA /4. Then in the Ising limit the two-point correlation function of the ground state is 4

For finite A. A < -1, the two-point correlation function of the ground state is not constant but remains finite. The two levels degenerate in the limit A = oo become slightly different, though this difference is very small. These two states are called nearly degenerate. The third lowest energy states are at -NJ I AI /4 + J I AI. Then the third states have a considerable energy gap from the nearly degenerate ground state.

The calculation of the two-point function at the ground state is very difficult. Direct calculation from the wave function was not successful. But it is expected that the two-point function in the z-direction will approach to a finite value, i

lim 4(-1)`-j(S j00

Po P.

The XXZ model

58

m

h/J Fig. 4.3. Magnetization curve of the XXZ model at zero temperature for a magnetic

field in the z-direction. We take A = -3,-2,-l,-0.5,0,0.7. At A < -1 the system is not magnetized at h < h,

As will be shown in the next chapter, the ground state of the XXZ model relates to the thermodynamics of the six-vertex model. Especially, the XXZ

ground state at A > 1 is related to the F model below the transition temperature. Po is the spontaneous order of this model. Baxter10 obtained that (o0 1-t2n PO

= CfJ 1 + ten)

2 ,

t = exp(-O).

(4.59)

1

The result is very simple and coincides with the perturbation expansion from the Ising limit. This is transformed in various forms, 00

Po = ri tanh2 no = n=1

2(u')1/2K(u')

=

21r

eXp(-(n + n=0

The last formula converges rapidly for small 0.

1)27r2/0) 2.

(4.60)

4.10 Excitations from the ground state

59

4.10 Excitations from the ground state 4.10.1 A < -1 and the energy gap

In the case of even N, the true ground state at h = 0 has M = N/2 and quantum numbers I1 given by (4.37),

N/2 - 1 N/2 - 3

N/2 - 1

2

2

2

(4.61)

This is an Ising-like antiferromagnet. We have a nearly degenerate state with this ground state,

I

- N12+1 N/2 - 1 - 2 2

N/2 - 3 2

(4.62)

The former has total momentum zero, the latter has total momentum 7r. The latter has slightly higher energy than the former but the difference is o(1/N).

In this state k1 is always it. In the Ising limit (A -> -oo), these two states correspond to the following wave functions: I

TITS....) ± I IT1T ...).

Next we consider the two-holes state, with M = N/2 - 1 N' N' N' N'

If = 2 , 2 - 1, ..., 2 - r + 1, 2 - r - 1, N'

..., 2

N' -s+1, N'2 -s-1,-2,

(4.63)

with 0 < r < s < N'. The total momentum is K = 27r(r + s)/N + it. From equation (4.22) we have

N01 (x', 4) = 27rh + E 02(x' - xi,

(4.64)

t4r,s

Subtracting (4.22) from (4.61) we have N'-1

2ira2(x1- xl)(Ax1- Axi)

27cNa1(x1)Ax11=1

_ lc(sign(x1- Xr) + sign(x1 - xs) - 1) - 02(x1- XN', 4) (4.65)

Putting J(x1) = NAx1p(x1) yields

J(x) + fQQ a2(x - y)J(y)dy = I (sign(x - Xr) + sign(x - xs)) -1-

tan-' (tanh 0 tan(x/2)). 7C

(4.66)

The XXZ model

60

The Fourier transform of this yields 21ni¢) J(n)

(1 + e

[2(-1)n - einx, - einxs - (-1)n(1 - e ZIn11)]

in

for n * 0

-xr - xs - 27r

for n = 0. (4.67)

The excitation energy DE becomes:

-2h +

27rJ nh

f

Q

1ai(xN') -

0

-

_ -2h + J sink [tanh

dxai(x)J(x)dxl Q

n

2 1 + e 21n1o [-einx, - einxs + I

no

(-1)n(1 + e 21n1,)]] _ -2h + J sinh ¢ [1 + 0

= -2h + 27rJS1

_ -2h +

einx, + einxs

2 cosh no

[PO(Xr) + Po(xs)]

J sinh OK (u)

[dn (K (U)xr , u) + do (K 71

7E

(u)xs

7E

(4.68)

Here the modulus u and the elliptic integrals K(u) and K' = K(u') are determined from the parameter 0 by (4.27). The momentum is given by

qr =

27rr

N =

= 2it

fQ

po(x)dx = sin-1 sn(K(u)x u) Q

J

it

xr

-2 -

sin-i

sn (K

xr

u) .

(4.69)

Thus we have

DE = -2h +

JK(u' inh

[

1 - u2 cos2 qr +

1 - U2 cos2 qs],

AK =qr+qs-it.

(4.70)

This excitation has Stotal = 1. By the unitary transformation U1 we have the following excitation at Stotal -1:

-

AE = 2h +

JK(u') sinh

0

[

1 - u2 cos2 qr + 1- u2 cos2 qs].

(4.71)

4.10 Excitations from the ground state

61

(E-EO)/J

(E-EO) /J

(E-EO)/J

1

2

4

3

-E 5

6

Fig. 4.4. Spinon excitation of the XXZ model at A = -1.5, A = -2 and A = -3. We put h = 0.

The spinon spectrum in a zero field is shown in Fig. (4.4). The lowest energy

point of (4.71) at h=0isatq,.=q3=0, AE = 2JK(u')u'sinh

(4.72)

0

The energy gap goes to zero as A

1+. The excitation spectrum were

investigated by des Cloizeaux and Gaudin20, Johnson, Krinsky and McCoy44

Ishimura and Shiba41 calculated the dynamical correlation function of the finite chains.

4.10.2 Excitations for -1 < A < 1

As will be shown in part 2, the XXZ chain has many kinds of string solutions which are dependent on the anisotropy parameter A = - cos y. For the ground state, the xjs are on the real axis. For equation (4.39) the ground state is given by the set of half integers

-{ N/22 - 1 N/22 - 3 ,...,

{Ij } -

,

N/2 - 1 2

}.

(4.73)

The XXZ model

62

The simplest excitation is the hole excitation. Taking the quantum number Ij as (4.63) we have the following energy and momentum change:

AE = 27rJ,siny

J(x) +

[a(x, 1)

-f

dxa(x,1)J(x)dx],

a(x - y, 2)J(y)dy 7r

=

2(sign(x - xr) + sign(x - xs)) - 1 -

y

00

AK = 2i LIxr r po(x)dx +

f po(x)dx] - r.

(4.74)

s

The energy expression is given by

AE =

2irJ sin y

(4.75)

[Po(xr) + Po(xs)]. Y

At zero field, the ground state p(x) is given by (4.43). Thus,

AE =

J2r sin y 2 [sin qr + sin qs],

AK = qr + qs -7r.

(4.76)

Y

The hole state has almost the same spectrum as the XXX antiferromagnet except for the velocity coefficient. The spinon velocity is given by VS =

J7r sin y

(4.77)

2

Y

This velocity goes to zero as A approaches to 1. In this limit the dispersion becomes completely flat. This dispersion also resembles (4.70). The main difference between the excitations of the Ising-like XXZ model and the XYlike XXZ model is that the former has a finite energy gap but the latter is gapless.

4.10.3 Spin-wave like excitations for 0 < A < 1 We assume that there are N/2 - 1 particles on the real axis and 1 particle on the Zx = 7r/y axis, with y at it > y >7r/2. Then the Bethe-ansatz equation becomes N'-1

-N9(y, po - 1) = 27rJ -

9(y - x1, po - 2),

(4.78)

1=1

N'-1

NO(xj,1) = 2irh - O(xj - y, po - 2) +

O(x; - x1, 2). 1=1

(4.79)

4.10 Excitations from the ground state

63

Here y = fix, J is an integer and I1 are integers (half-odd integers) for odd (even) N' - 1 and tanh yx/2 \ O(x, a) - 2 tan 1(

ltanay/2)

We assume a symmetric distribution of the Ij,

N'-2 N'-4 2

2

'

N'-2 '

2

Then the total momentum is

K = O(y,po - 1) - f 0'(x,1)J(x)dx.

(4.80)

The energy change is e

27tJ sin

a

fa', f

x 1 J x dx .

1

(4.81)

00

Y

Here a(x, a) _ (27,)-i

00(x, a)

8x

and J(x) is the back flow for this excitation,

J(x) + f a(x - x', 2)J(x')dx' =

1

- poI +

27r

O(x - y, 2 - po).

(4.82)

The solution of this equation is

J(x) _ -

+

1

2(po - 1)

xY

dt L,

Jo

dw cosh(po - 1)w eiwt 2ir cosh w

(4.83)

Then we have e(Y)

_

y

27rJ sin y

= 4 1sech

1

d

27r dy

K(y) =

o0

dwe iwy

cosh(2 - po)w cosh w

2 (y - i(2 - po)) + sech 2 (y + i(2 - po))].

(4.84)

Integrating with respect to y yields

K = 2[tan t ez(y-i(2-P°)) + tan-' K = sin(po - 1)ir/2 tan 2 sinh try/2

ePy+i(2-po))],

(4.85)

We can eliminate y and obtain

e(K) = trJ sin y y

sin 2 VI1 + cot2

(PO

2 1)7r sine

K.

2

(4.86)

The XXZ model

64

At low momentum this excitation has the same velocity as a spinon excitation. The spinon excitation has periodicity 7E. But this excitation has periodicity 27r. As A goes to 1, y -+ it and po - 1. Then this excitation becomes

e(K) = 2J sin2 2 = J(1 - cosK).

(4.87)

This is the spin wave excitation from the ferromagnetic ground state of the XXX ferromagnet. It should be noted that this excitation exists only for

0 < A < 1. If we calculate energy and momentum for -1 < A < 0, we find that the bare energy and momentum are cancelled by the back flow at the ground state. At finite temperature or at finite magnetic field this cancellation is not complete and we can calculate the excitation spectrum. 4.10.4 The spin-wave bound state

A length n string can exist in the ground state of XXZ model, if (po - 1)n < 1. In the other region the bare momentum and energy cancelled with the back flow. The positions of rapidities are x; = y + [(-1)npo + (n + 1 - 2j)]i.

(4.88)

The bare momentum and energy of this state are 0(y, n(po - 1)), J sin y/yO'(y, n(po - 1)).

(4.89)

The back flow density should satisfy

J(x) + f a(x - x', y)J(x')dx' = -n/po + (27r)-1O(x - y), O(x) = 8(x, po - (n -1)(po

- 1)) + 9(x, po - (n + 1)(po - 1)). (4.90)

The solution of this equation is

J(X) = J(

+ 2(po - 1) n

JY dt

- 1w e0".

Edw cosh n(po cosh w 27r

(

4.91

)

Thus we have y

e(Y)27rJsiny

= 4 [sech 2 (y

_

1

d

27r dy

f

iwy

cosh(1 - n(po - 1))w cosh o-)

- i(1 - n(po - 1))) + sech 2 (y + i(1 - n(po - 1)))]

.

(4.92)

4.10 Excitations from the ground state

A=0.9

A=0.8

(E-E0)/J

A=0.6

(E-EO) /J

(H-EO) /J

A=

-2

-1

1

2

A=0.3

-1

2

A = 0.001

(E-EO)/J

(E-EO)/J

1

A=-0.5 (E-EO)/J

-2

65

0.6

2

-2

1

2

A = -0.001 (E-EO)/J

2

A=-0.8 (E-EO)/J

A = -0.999 (E-EO) /J

4.5. Elementary excitations for various values of A from the zero field ground state. If the system is magnetized or the temperature is not zero, the other excitations have finite momentum region also. We put A = 0.9,0.8,0.6,0.3,0.001,-0.001,-0.5,-0.8,-0.999. Fig.

The XXZ model

66

We can eliminate y and get

e(K) _

7rJ sin y 1

{

cots

n(p0

Y

2

1)7r

sing 2

(4.93)

One should note that this excitation exists for cos(lr/(n + 1)) < A < 1. In the limit A -* 1 we have y -* 7r, p0 -* 1 and obtain

e(K) = J(1 - cosk)/n.

(4.94)

Thus we have the spinon excitation, spin-wave excitation and bound state excitation for the ground state of the XXZ model for JAI < 1. All these excitations are gapless and have a common low-momentum group velocity. These excitations are shown in Figs. (4.5). In the case of the XYZ model there are corresponding excitations, but these are all massive and the system has the energy gap. The selection rule for the length and parity of strings will be discussed in a later section.

5 XYZ and eight-vertex models

5.1 Transfer matrix of the eight-vertex model 5.1.1 Relation between the six-vertex model and the XXZ model Here we should consider the partition function of a two-dimensional classical model. Consider a square lattice with N x M sites and six kinds of vertices

as shown in Fig. (5.1) The Boltzmann weights for a vertex are assumed to be w1, w2, W3, W4, W5, W6. The Boltzmann weight of one configuration is W=W NIW N2WN3w4 4w5 Sw6 6. The partition function is given by the sum of Boltzmann weights of all configurations, Z = E Wconf.

(5.1)

conf

This partition function is the trace of the product of M transfer matrices, Z = TrTM.

(5.2)

Hereafter we put w1 = w2 = a, W3 = W4 = b and W5 = W6 = c for simplicity. The transfer matrix T is a 2N x 2N matrix given by T(61, 62, ..., 6N I61, 62, ..., 6N)

=

... T1

T2

R(61, 61)TN,TlR(62, 62)T],T2...R(6N, 6N)TN_1,TN

(5.3)

TN

This matrix element can be written as the trace of products of 2 x 2 matrices, T (al, 62, ..., 6N I6'1, 6'2, ..., 6'N) = TrR(61, a )R(62, a')...R(6N, 6N).

(5.4)

Here we have four kinds of matrices,

R(+-I-) = (a

R(-+) =

R(+-) =

0

0) ,

C

0I'

0

0) b '

0

0)' R(--) = (0 67

) a

(5.5)

XYZ and eight-vertex models

68 W1

W2

W4

W3

W5

W6

Fig. 5.1. Vertices and their Boltzman weights of the six-vertex model.

The problem of the thermodynamics of an infinite M system is to calculate the largest eigenvalue of the transfer matrix T, 2N

Z=

M. l=1

The free energy per site is

f

T 1nZ nil 00NM

Here T is the temperature. One trivial eigenstate of this matrix is T,..., T) with eigenvalue aN + bN. The next simplest solution is the state with one down spin. Two transfer matrices T(a, b, c) and T(a', b', c') commute if I

a2+b2-c2 2ab

- ai2+b'2-c'2 2a'b'

(5.8)

To ensure the commutation, a, b and c are parametrized as follows,

a = p sin(v + i),

b = p sin(v - rl),

c = p sin 2r1.

(5.9)

p and v might be different but cos 2i must be common for a, b, c and a', b', c' to commute T(a, b, c) and T(a', b', c'). Then these transfer matrices T(v) have the common set of eigenvectors. One can construct another series of matrices Q(v) which satisfy T(v)Q(v) = 4(v - ii)Q(v + 2i) + 4(v + ii)Q(v + 2r), [T(v), Q(v')] = [Q(v), Q(v')] = 0,

0(v) = (p sin(v))N.

(5.10)

(5.11)

As T and Q always commute, they are diagonalized by a certain linear transformation which is independent of the parameter v. As all elements of T(v) and Q(v) are entire functions of v on the complex plane, diagonal elements of T and Q must be so as well. Thus we have an equation for one diagonal element in the expression: T(v)Q(v) = 0(v - i)Q(v + 2i) + 0(v + q)Q(v + 2rl).

(5.12)

5.1 Transfer matrix of the eight-vertex model

69

Moreover they must be periodic functions with period 21r. Assume that Q(v) has n zeros in the region -7r < Rv < 7r, n

Q(vj) = 0, j = 1, 2, ..., n,

Q(v) = C fl sin(v

- vi).

(5.13)

i=1

Putting v = vj in (5.12) we have sin(v j +,i) N

sin(vj - rl)

-

n sin(v1- vi + 2i1) 1_11

sin(vj - vi - 2rl)

j = 1, 2, ..., n.

(5.14)

The corresponding eigenvalue is given by T(v) = pN [sinN(v - q) n j=1

+ sinN(v + f)

sin(v - vi. + 2q) sin(v vj)

-

sin(v - v j - 2rl) sin(v - vj j=1

(5.15)

The R matrix is represented as follows : R(6l, 6i)T,_1,T1 =

+ (5.16)

Let us consider the transfer matrix at v = rl. Here we have a = c, b = 0 and Tj. In this case R(a1, al)TI-1,T! = a861,T1-1 (61,62,...,UNIT(11)1c',a...... ' ) =a N6616266263...6QNSi

(5.17)

This means that the transfer matrix is merely the right-ward shift operator times aN. Then the total momentum K of an eigenstate is given by

expiK = a-1T(rl) _, sin(v j + rl) j=1

(5.18)

sin(vj - rl)

Now let us consider T'(rl)T-1(q), (a

I a')

dv T(v)

V=17

_

pN(sin

2;)N-1

[cos 2q

61-bd cSQi-ad-ad_1+a61+1+),,

W1 T- (q) I a") _ (p sin 2q )-N fj bul,Qlf i i+1 i

(5.19) (5.20)

XYZ and eight-vertex models

70

a

b

c

Fig. 5.2. Vertices and Boltzmann weights of the eight-vertex model.

Thus we have (al dvT(v)

f

v=nT-'(vl)la") = (sin217)-1

il,l+1

l

x [cos

2

(sin

2n)_1

[cos 2i(1

+i

i+i) + i 6i+i + 6i 6 +

(5.21)

This is just the Hamiltonian of the XXZ model plus a constant. Thus the energy eigenvalue of the Hamiltonian

_ -J

N

Sl Sl+1 + Sf Sl+i + ASl SIZt

(5.22)

1=1

is given by

E _ -J2 sin2n dvd InT(v) Ill=n JN4 4

JNA 4

n

- J sin 2n j=1

s in

cos 2

11

2

11

- s n2 i

(5.23) v

If we put 11 = i0/2,vj = 4x/2, equation (5.14) coincides with (4.11). The energy eigenvalue also coincides with (4.6) at h = 0.

5.1.2 The eight-vertex model and the XYZ model

The exact solution of the eight-vertex model was found by Baxter8'9. In eight-vertex model vertices shown in Fig. (5.2) are possible. Here we have four kinds of matrices,

R(++) = ( 0 0b)' R(+-) =

R(-+) =

d

0), R(--) =

(0

0)'

\0

al

(5.24)

5.1 Transfer matrix of the eight-vertex model

71

In this case two transfer matrices, T(a, b, c, d) and T(a', b', c', d'), commute with each other if

a2+b2-c2-d2 2(ab

- cd)

a'2+b'2-c'2-d'2 2(a'b'

- c'd')

ab/cd = a'b'/c'd'.

(5 . 25)

(5.26)

Details are in Appendix C. This condition is parametrized by Jacobian elliptic theta functions,

a = po(2q)o(v - q)H(v + q), b = pO(2q)H(v - rl)O(v + rl), c = pH(2q)O(v - q)o(v + q), d = pH(21)H(v - rl)H(v +'1).

(5.27)

Here H(x) and O(x) are 00

H(x)

2 E(-1)n+1gn(n-1)+1/a sin(2n

- 1)ZK

n=1 00

O(x)

1 + 2 E(-1)ngZ cos K , q = exp(- K ).

(5.28)

n=1

These functions have the following properties and are doubly quasi-periodic on the complex plane,

H(x) _ -H(x + 2K) = -iq1/4ei7rx/2'O(x + 1K') _ -geirzx/'H(x + 2iK')' O(x) = O(x + 2K) = -iq1/4einx/2KH(x + iK') _ -ge"`x/x O(x + 2iK').

(5.29)

The modulus k and parameter rl are determined by k

1+l' 2

S1. (

1)_ 1

l

A2-A2 A2_Ax'

AZ - Ax + AZ - AY _ ( VA

2

Ax +Ay

AZ - a2 + b2 - c2 - d2, Ay = 2(ab + cd), Ax = 2(ab - cd).

(5.30)

One can construct the matrix Q(v) which satisfies T(v)Q(v) = 4(v + zj)Q(v - 2q) + 0(v - i)Q(v + 2q), 0(v) = (p®(0)H(v)0(v))N' [T(v), Q(v)] = [Q(v), Q(v')] = 0.

(5.31)

XYZ and eight-vertex models

72

Then T(v) and Q(v) are diagonalized at the same time. Details are in Appendix D. An eigenvalue of Q(v) is factorized as follows, N/2

Q(v) = exp(_

iV2K7lv

) fj h(v - wj), h(x) = H(x)O(x).

(5.32)

j=1

The sum of the wjs should satisfy N/2

1: wj = -K(v" + N/2 + even integer) + ivK'/2.

(5.33)

j=1

Here v is an even (odd) integer if N/2 minus the number of down spins is even (odd). v" is an even (odd) integer if the eigenstate is symmetric (antisymmetric) for reversing up-spin and down-spin. We obtain transcendental equations for the wjs, 0

= O(wj + n)Q(wj

- 2q) + O(wj - q)Q(wj + 2q),

or

N/2

(h(wi + q) N = -e 2iv,r,1/K

11

h(wl-r1))

h(wl - wj + 2r1)

h(wi-wj-271')

5.34)

The corresponding eigenvalue of T(v) is given by T(v) = 4(v

+ ,7)ei7rvn/K

+O(v - 71)e

"zvn/K

11 1-1

h(v - wl - 211) h(v - wi)

1 h(v - wl + 2q) 11-11

h(v - wi)

(5.35)

At v = n we have b = d = 0, a = c and therefore (a1, 62, ..., 6N I T (71) j a1, a2, ..., 6N )

_ (pe(0)h(2i))N6C,6266;63...6QN671

(5.36)

The logarithmic derivative is d (61

T(v)

dv(S

v=n

T-1(n)Ia")

_

jl ba, a1, 1

i#1,1+1

X [P1(bai+ba/+l+ba +bal+1+ + bal-bal+l-bad

+P2(ba1+bd+1+a

+ bat-ba,+1-6U +6U1+1+)

+P3(bai+bd+1-bad +ba1+1- + bal-b6i+1-S6i +ba,+1+)

+P4(ba1+bd+1-bQi -bat+1+ + ba,-bai+l+ba, +ba1+1-)]

(5.37)

5.2 The symmetry of the XYZ model

73

where p1, P2, P3, P4 are the logarithmic derivatives of a, b, c, d,

H'(21) H'(0)0(21l) H'(0)H(2q) O'(2_) P1 = H(2q) , P2 = H(2rl)O(0) , P3 = ®(2i) , Pa = ®(2i)®(0)

(5.38)

In Pauli matrix representation (5.37) becomes [P1 + P3 2

+ P1 - P3 z,z + P2 - P4 6v y + 2 t+l 2 t t+l

P2 + P4

1

2

6Xa,+,]. x l

(5.39)

Moreover from (5.38) we have

h(2q) cn(2rl)dn(2q) P1+P3 = h(2rl), P1-P3 = P2±P4 = sn)d Then the eigenvalue of the XYZ Hamiltonian [Sfsi+1

+

sn(2q)+ksn(2rl).

1- ksn22rl y y

1+ ksn22rl x x cn2rldn2q S1 Sl+l + cn2ndn2q S1 S1+1],

(5.40)

(5.41)

is given by E

_

Jz

d

sn2rl

2 cn2jdn2ij dv

In T(v)

IV--q

1h'(wi

Jz sn2q 2 cn2gdn211 -1 h(wj - ,j)

h'(w.i+11) h(wj + q)

+ h(2,j) (5.42)

From (5.36) we find the shift operator is given by S

SS

66162V 6263...VQN61

_

1

y

(5.43)

4i(2Yl)T (il)

Using (5.35) we find that the total momentum K is given by eiK

= ei7rvglK(k)

N12 1=1

h(wl + 11) h(wl

-)

(5.44)

Thus one can calculate the energy of the XYZ model in the same way as the XXZ Hamiltonian. But the solution of the XYZ model with a non-zero magnetic field is not known.

5.2 The symmetry of the XYZ model Here we consider the symmetry of the following Hamiltonian, N

=-

JXSi Si+l + Jysf Sl+l + JZSf 1=1

Sf+l.

(5.45)

XYZ and eight-vertex models

74

We assume N is even. By the transformation

U2rU2-1,

U2 = 11 2S1, l=even

we have (Jx, Jy, JZ) (-Jx, -JY, Jz ). In the same way we have (Jx, Jy, JZ) - (Jx, -JY, -Jz) and Y'(JX, Jy, JZ) (Jx, -Jy, -Jz ). Specifically the energy spectrum of this Hamiltonian is unchanged on reversing the signs of two Js. It is evident that the spectrum is unchanged under exchanging Jas. Then it is sufficient to treat only the case Jz >- Jy > JJxJ > 0. Comparing (5.41) and (5.45) we can determine k and tj, 1_

1 + l'

k

2- 2

l=

JZz2

-

(5.46)

JY2x

j2z- J2 x+

1Jy-Jx sn2rl=-kJy+Jx 2

J2 _ J2 z

Jy+Jx

y

2

(5.47)

In the case of the Ising-like XXZ model (JAI > 1), Jz = JO and Jy = Jx = J. Then we have k = 0, 2rl = i cosh-1 A. In the case of the XY-like XXZ model (IAI < 1), we have k = 1, 2rl = i cos-1 A. Thus the zero field XXZ model is included as a special case of the XYZ model.

5.3 Modulus l and modulus k In previous sections we used modulus k. But one can use 1, defined in (5.46). We have the following relation : Kk

= 2K;, Kk = K'

1Ki, Kk = (1 + 1)KI.

1

(5.48)

2

The function h(x) = H(x)O(x) has zeros at x = 2mKk + inK'. On the other hand H1, (x) has zeros at x = 2mK1 + 2inKl. Then Hlp(1+l) has common zeros

with h(x). One can show that h(x)

(1+ll `2x

l'KI' HI'

PI

l

exp (-KIKI(1

2ix

+ l)2) H1(l + l

(5.49)

Next we consider the function sn(x)/(cn(x)dn(x)). This function has poles at (2m 1)K + inK' with residue (-1)m. On the other hand the function sn(x)

-

5.3 Modulus l and modulus k

75

has poles at 2mK + i(2n + 1)K' with residue (-1)'n/k. Thus we have sn(x,k) cn(x, k)dn(x, k)

2ix

1 + l'

1) -

2ix

1 + l'

2i

1 - sn2(x,1), dn(x, 1) =

Using cn(x, 1) = cn

_ I+ l sn

(5.50)

lJ .

1 -12sn2(x, 1) we have

1 + ksn2(x, k) 2ix do cn(x, k)dn(x, k) 1+l

1)

-

1 - ksn2(x, k) cn(x, k)dn(x, k)

(5.51)

We define the following parameters: iij

xj

KI + 12+

M+

1),

Q = K'/C.

(5.52)

Then equations (5.41) and (5.34) become JZ [Sl Sl+1 + dn2'S, S +1 -

'Ye

Hl(i (xl + i)) N (Hi(ir(xi i)) =

-

Sl+l],

- xj + 2i)) -e-21riv'/po 11 Hl(it,(xl Hi(ir(xl - xj - 2i))'

(5.53)

(5.54)

N/2

xa = K'v' + iKIv.

(5.55)

a=1

Substituting (5.49) into (5.44), we have the total momentum 7zv' P0

N/2

L

Hi(i(x« + i)) HI(i(xa i))

(5.56)

-

Equation (5.42) is N/2 E_- JZsn(2l;) 2 E g(xj),

j=1 H'(i/T(x i))

-

H'(iiT(x + i))

g(x) - H(ii; (x - i))

H(i?(x + i))

(5.57)

H'(2C)

(5.58)

The function g(x) has poles at x = +i+(2mK' +2inKl)/C with residue :Fir-1 and is doubly periodic. On the other hand the function b(x)

2snCcnCdnC = sn2c sn2(il;x)

-

(5.59)

is also such a function. Then g(x) - b(x) is a doubly periodic function with no singularity. By the Liouville theorem this must be a constant as a function

XYZ and eight-vertex models

76

of x. Putting x = 0 this constant is again a function of l;,

- H'(2i) -2

2H(_)

H(20

H(l;)

(5.60)

snt

This has periodicity 2K, i2K' and poles at 0, K, iK', K + 1K' with residues 1 , _2, 2, -1, respectively. On the other hand the poles and residues of

-

cnCdnC

snl;

are 0, 1K' and 1, -1. The poles and residues of snt; dn (ln(cn ))' cnC

are K, iK' and 1, -1. The poles and residues of _12 snl; cnC dnl;

are iK', K + 1K' and -1, 1. Then we find that the following quantity is a constant:

(5.60) +

snCdnC-

1

sn

2

(5.61)

dn I'

cn

This must be an odd function of C and therefore must be zero. Then we have finally

_ g(x)

2snCcnCdn

sn2

= b(x) +

1

- sn2(iCx) 2 sn; - cn n = b(x) + snl;

sn2C

l2sn;cnC

1

dnC

cnC

I sn2C

+ b(Q)). - 1(b(0) 2

(5.62)

Then (5.57) becomes N4z

E

11-

s

2

(b(0) + b(Q))]

-

Jz

2 2

b(x1).

(5.63)

l=i

5.4 The case J,, = 0 In this case we have cn2l4 = 0 and 2C = K1. The r.h.s. of (5.54) becomes

2i)) = -1. Then equation (5.54) becomes very simple,

(Hill 2 (xa + l))\N Hl(i (xa - i))

2

- -(-1)

(5.64)

5.5 The ground state for Jz > 0

77

This is equivalent to N K ( l'tn(2 (1- ixa, l)))

_ -(-1)" .

(5.65)

Thus we have (l1)-1/2e iqa,

tn(2 (1 -ixa)) =

qa ° N(1 + v" + Na),

(5.66)

where Ia is some integer. This yields

sn(iKxa/2) =

VF + e2iga

- sign(cos qa) 2

l'+ e 2iga

(5.67)

1 + l' cos qa

The r.h.s. is always real or pure imaginary. This means that xa must be on the real axis or on ixa = 2. Substituting (5.67) into (5.59) we have g(xa) = sign(cos qa) (1 - 11)2 + 41' cos2 qa + 1 - l'.

(5.68)

Substituting this into (5.63) we have the lowest energy, or the ground state energy, N/2

E=-2

(1 - 11)2 + 41' cos2 N (N/2 + 1 - 2j).

(5.69)

.i=1

For JZ > 0, the ground state xas are on the real axis and for JZ < 0 they are on the poi axis.

5.5 The ground state for JZ > 0 Taking the logarithm of equation (5.54) we have N/2

N6(xa,1) = 2irIa -

E O(xa - x,, 2), fl=1

PO

B(x, n) =-

1

i

In (-Hi(.

ni))

(x -

2 (( P 0 ni))) ++27r

1

x+Q 2Q

PO

n < 2po.

(5.70)

The function O(x, n) has the following properties,

6(x, n) = -6(-x, n), 6(jQ, n) = j7r (po - n) PO

The ground state should be characterized by the following set of integers, v'

= 0, Ia=

N/2 - 1 N/2 - 3 2

,

2

N/2

2

-

1

(5.71)

XYZ and eight-vertex models

78

Thus the distribution function p(x) of xas on the real axis for -Q < x < Q should satisfy

a(x, l) = p(x) + f Q a(x - y, 2)p(y)dy,

(5.72)

Q

a(x'

n)

1

d

0°

27r dx B(x' n)

J=-co

sin pa

1

2po cosh ,(x-2

)

PO

- cos -"' PO

(5.73)

where po - K1/l;. The energy (5.63) and momentum in (5.56) are rewritten as follows,

E = -NJ z [1

- 7rsn2t; (a(0,1) + a(Q, 1))] - Jzirn2l; 1/2 a(xl,1), 1=1

(5.74) N/2

K = E 9(x«,1) +'rv

(5.75)

.

PO

a=1

The Fourier series 1(m) for a function f (x) in the region [Q, -Q] is defined as follows, QQ

f (x) exp(-i7rmx/Q)dx,

!(M)= J

12Q E I (M) exp(i7rmx/Q). 00

f (x) =

(5.76)

m=-co

The functions a(x, n) are transformed as follows, sinh(irm(po - n)/Q) sinh(irmpo/Q)

'

for m

0

a(m, n) =

(5.77)

Po - n

for m = 0.

Po

The equation (5.72) becomes a(n,1) = p(n) + a(n, 2)p(n).

(5.78)

Thus we obtain 1

7(m) = p 2 cosh(7rm/Q)' 00

p(x) = E 4 sech (2 (x - 2jQ) = J=_00

dn(K (u')x, u).

(5.79)

5.6 Long range order

79

Here the modulus u is determined by

K(u')_ 1 K(u)

Q'

The energy per site is given by

[1 -

NJZ 7rsn2C

s2 (a(0,1) + a(Q,1))]

4

1

p(m)a(n,1)

2Q m=-co 1

-

isn2 . r

4

sinh(po 1)in Lcosh(in) sinh('rnpo)

2C

- sinh 2(po - 1)in sinh 2poin

'

(5.80)

where r - 7E/Q.

5.6 Long range order

In the case JZ > Jy > IJJ >- 0, the two-point function of the ground state of the system as described by (5.45) is lim lim (Sl Sl+n) =

n-*co N-+oc

1 p02.

4

(5.81)

Po is called the spontaneous polarization. This is equivalent to the order parameter of the eight-vertex model below the transition point. For the case Jy = J, the system is the ferromagnetic Ising-like XXZ model. In this case it is evident that Po = 1. At Jy = -Jx this is equivalent to the antiferromagnetic Ising-like XXZ model, discussed in §4.9. Baxter and Kelland12 conjectured that 00

1 +q Po-ll(1-qn n

1 - t2n 2 x 1+t2n)

n= 1

t

exp(-2K(l'))'

q

ex

p(-

irK(k') 2irK(l) ). K(k) ) = ex p(_

K(l')

(5.82)

Later, Jimbo, Miwa and Nakayashiki43 derived this formula using the corner transfer matrix.

XYZ and eight-vertex models

80

5.7 Elementary excitations 5.7.1 The nearly degenerate ground state For the case N = even, it is possible to consider the state with the following quantum numbers : V'

= 1, I1 =

N'+1 N'-1 2

2

,

N'-3

... ,

2

,

For this state the total momentum is t. The true ground state has zero total momentum. In this case all zeros shift slightly to the right, NAxja(xj,1) = 1

- PO- + 1

a(xj - xl, 2)(Ax1 - Ax1).

(5.83)

I

In the N - co limit we introduce the flow-function J(x) = NAxjp(x). We have the following equation for J(x):

J(x) + fQ a(x - y, 2)J(y)dy =1-

1.

(5.84)

PO

Q

This has a very simple solution: J(x) = 1/2. We have energy and momentum change as follows:

AE _

-Jsn 12C

Q Q

a'(x,1)J(x)dx = 0,

K = 27r f a(x,1)J(x)dx +

7r

= it.

(5.85)

PO

Actually, for the finite system this state has a slightly higher energy than the true ground state. But the energy difference decreases rapidly as N becomes large. This state is unique and has no dispersion, but it is important that the ground state is actually degenerate. 5.7.2 Spinon excitations For the ground state all zeros are on the real axis. We consider one simple excitation where we have one string with length 2 and several holes in the distribution of zeros on the real axis. From equation (5.74) we have

11 a (x,1)J(x)dx

AE

-a(xr,1) - a(xs,1) + a(xo, 2)], Q

Q

K = 271 1J po(x)dx + f po(x)dx] - it. x,

(5.86)

s

(5.87)

5.7 Elementary excitations

81

Here xr and xs are the positions of two holes on the real axis. xo is the real part of the length 2 string. The equation for the back flow J(x) is

J(x) + Q a(x - y, 2)J(y)dy = -9(x - xr, 2) - 9(x - x, 2) Q

+012(X - xo),

(5.88)

012(x) = 9(x,1) + 9(x, 3).

(5.89)

We can calculate J(x) analytically using the Fourier transform. The terms in (5.86) which contain xo cancel each other and we get

AE =

JZ7rsnl22;

[P0(xr)

+ Po(xs)],

C

K = am(K'xr) + am(K'xs).

(5.90)

Here am(x) = sin-1 sn(x). Then this excitation is represented by two spinons

OE =

J,K(u')sni2C

K

[

2C

1 - u2 cos2 qr +

1 - u2 cos2 qs] (5.91)

=qr+qs-ir.

The energy gap of one spinon is Ospinon =

Jz sn122K (u')u'

(5.92)

The energy gap from the ground state to the excited state is 2Aspinon, because

the spinon number is even for a ring. One should note the low-temperature specific heat behaves as exp(-Ospinon/T) and not as exp(-20splnon/T), as will be shown later. This spinon excitation exists for any set of parameters

of J, I 5 Jy < J. 5.7.3 Spin-wave excitations

We consider the case where J. > 0 and 1 < po(- K(1')/() < 2. Assume that N' 1 zeros are on the real axis and that one zero is on the poi axis. Then,

-

AE =

Jgr 12

[- f

a'(x,1)J(x)dx - a(-Q,1) + a(y, po - 1)I, Q

K = 9(y, po - 1) - 9(-Q, 1) -

Q Q

9'(x,1)J(x)dx.

Here

a(x, a) -

(27r)-189(x, a)

ax

(5.93)

XYZ and eight-vertex models

82

and J(x) must satisfy

J(x)+Q a(x-x', 2)J(x')dx' = 27t -[-n +O(x+Q, 2)+9(x-y, 2-po)]. (5.94) fQ The solution of this equation is

J(x) =

Q+x

dteiimt m

Q

X -Y

+

dteiimt

sinh(po - 2)Tm 2 cosh tm sinh(po - 1)im

cosh(po - 1)im

(5.95)

cosh im

JO

Then we get

AE =

JK(uu')sn12l;

[dn(K(u')(y - i(po - 2)))

+dn(K(u')(y + i(po

-

(5.96)

2))) 1

K = it - am(K(u')(y - i(po - 2))) -am(K(u')(y + i(po - 2))).

(5.97)

Taking the cosine of (5.97) we have

cosK = -

1 - 2sn2z + sn4z(2u2sn2a1 - u4sn4a1) ,

(1 - u2sn2zsn2al)2

x - K(u)y, al - K(u)i(2 - po).

(5.98)

This is a quadratic equation for sn2z. The solution is 1 + cos K

sn2z =

(5.99)

2 - u2sn2a1(1 - cos K)

By the addition theorem of elliptic functions (5.96) is transformed,

4dnaldnz

AE = 2C

(5.100)

1 - u2sn2zsn2al

Substituting (5.99) we have the spin-wave dispersion

AE = Jzsnl K(u') /(1 + ctnu,(f) sin2

2) (1 - dnu,(/l) cost

2

),

(5.101)

where fi = K(u')(po - 1). This state corresponds to (4.86) of the XXZ model. This excitation has a minimum at K = 0. The energy gap for one spin wave is

Aspin-wave -

JZsn12l;K(u')usnu

(K(u' )(po

- 1)).

(5.102)

5.7 Elementary excitations

83

This excitation exists only at Jx > 0 and not at Jx < 0. This energy gap is competitive with the spinon energy gap (5.92). The condition that the spin-wave gap is lower than the spinon gap is snu,(K(u')(po 1)) < 1/2. In

-

region Y of figure (11.2) the spin-wave gap is dominant. 5.7.4 Spin-wave bound states

Assume that 1 < po < 1 + 1/n. Consider the condition that N' - n zeros are on the real axis and n zeros form a string, xi = y + i{po + (po - 1)(n + 1 - 21)} + deviation, l = 1, 2,..., n.

(5.103)

The energy and momentum difference from the ground state are

AE _

JZirsni2

L-

Q

a(x,1)J(x)dx

Q

-na(-Q, 1) + a(y, n(po

- 1))]

,

K = O(y, n(po - 1)) - n9(-Q, 1) - f Q O'(x,1)J(x)dx. (5.104)

J(x) satisfies J(x) + fQ Q a(x

- x', 2)J(x')dx' _

2 [-nn + nO(x + Q, 2) + 0(x - y)], O(x) = O(x,po - (n -1)(po

- 1)) + O(x,po - (n + 1)(po - 1)). (5.105)

The solution of this equation is

J(x ) = n

+

i

Q+x

dte`imt

Q f, in X -Y

J

dte`Tmt

-

sinh(po 2)im 2 cosh im s i n h(po - 1 )im

cosh n(po - 1)im

5 . 106

cosh TM

O

(

)

Substituting this into (5.104) we get

AE = JK(u')snl2t; 2C

Idn(K(

u ')(y

+ i(1 - n (po

+dn(K(u')(y - i(1- n(po - 1))))

K = n - am(K(u')(y + i(1 - n(po - 1)))) -am(K(u')(y i(1 - n(po - 1)))).

-

- 1)))) (5.107)

(5.108)

XYZ and eight-vertex models

84 {jy, ix)-

(E-E0(/J. 3.5r

(E-E0)/Jz 3.5 (jy, ix)= (0.99, 0.6}

0.81 0.6)

Fig. 5.3. Elementary excitations for various values of jy = Jy/JZ and jx = J/JZ from the ground state.

Eliminating y we get the dispersion relation of this excitation at p0 < 1 + 1 In,

DE _

Jz sn12CK (u')

(1 + ctnu,(nf) sin2

2

) (1- dnu,(nf) cost

).

(5.109)

2K=0 This excitation corresponds to (4.93). The minimum energy is at Jzsn12CK(u')u'snu,(nK(u')(pO

- 1)).

(5.110)

This is always bigger than Aspin_wave defined in (5.102). This excitation cannot be dominant at low temperature. These spinon, spinwave and spinwave bound state excitations are shown in Fig. (5.3)

6

The Hubbard model

6.1 Symmetry of the Hamiltonian 6.1.1 Particle-hole symmetry In the problem of electrons with a narrow band, the Hubbard model is the most famous and has been well investigated by many authors. It relates to the metal-insulator transition (or Mott transition). In the half-filled state with strong repulsion, the system becomes insulating and the effective

Hamiltonian is represented by the XXX antiferromagnet. The Hubbard Hamiltonian is N

.(t, U,A, h) _ -t E E (c c + (iJ)

c1ciTc

v

cil

i=1

NQ

NQ

-A j:(c TciT + cc) - h j:(c TciT - c jci1). i=1

(6.1)

i=1

and ccr are creation and annihilation operators of an electron at site j. (ij) means that sites i and j are nearest neighbours. Na is the number of atoms. These operators satisfy Here

{Civ,CJ(7'}

- CT 'Cjo'I - 0,

!CT'CjQi}

(6.2)

Each atomic site has four states, namely the vacant state 10), the up-spin Thus the state the down-spin state cj110), and the full state total number of states is 4Na. The following quantities commute with the Hamiltonian and are therefore conserved quantities,

Nt -

N1

ctit c.it,

-

cj1c.i1

Now we assume that the lattice is divided into two sub-lattices. The nearest

neighbour atoms of one atom of the 2 sub-lattice belong always to the 4 85

The Hubbard model

86

sub-lattice. We call such a lattice a bipartite lattice. In the case of a bipartite lattice there are many symmetries of this Hamiltonian. Let us rewrite the field operators as follows, at = at

c

ail = ciT,

,

= -cil, aji = -cal, aiT = -cal,

ail =

cT

for i E ,, for j E A

,

a.iT =

(6.3)

By this transformation we have Na

-t

atia

aj, +

a

1=1

Na

Na

+UNa - (U - A) j:(a4ai1 + a jaij) + h >(a TaiT - atlail). i=1

(6.4)

i=1

Thus the Hamiltonian and particle number are changed as follows, K(t, U, A, h) _+ UNa + $ (t, U, U - A, -h),

NT+Na - NT,

NI-, Na - Nj.

(6.5)

It should be noted that the more than half-filled case (N > Na) is transformed to the less than half-filled case (N < Na). If we exchange particles and holes only for spin-up electrons,

ail = c+, aiT = -cat, aft = -c)tT, ail = ciT,

for i c- d, for j E -4,

(6.6)

we have Na

-t

aTaitcjcij (if)

-(A + h)Na + (h +

i=1

2

) E(aTait + c jciJ) + (A -

E(ait ail - cjcil), i=1

i=1

and therefore

1(t,U,A,h)

-(A+h)Na+Ye (t,-U,-hNT 4 Na - NT,

U,

2 -A),

Ni

Nj.

(6.7)

The attractive Hubbard model (U < 0) can be obtained by this transformation from the repulsive Hubbard model (U > 0).

6.1 Symmetry of the Hamiltonian

87

6.1.2 SU(2) symmetry The total magnetization operators are defined as follows, -'to ot

2

Stot ° 2

v =-

ci,c.il +

Stot

t

t

1

2i

.

cll ci?

ciT cil

cjTciT - Cj1c.il

Among these operators and the Hamiltonian (6.1), the following commutation relations exist, IStov St t] = istot, ,

Stot] = 0,

LStvot, Stot]

1_-Y' Stot]

= iStot,

[-'tot, -'tot] = iStt,

= -2hiStt,

[

,

= 2hiStot.

St t]

(6.9)

Defining

cj cil, Stot - -'tot - iSt t =

Sit - Stot + iSt t =

c.ilciT

(6.10)

.i

1

we have LStt, Slog = 2Stot,

[Stot, St t]

= ±St ,

[

,

Sttl = -F 2hStt.

(6.11)

For an eigenstate of the Hamiltonian and Stot, IW) with eigenvalue E and 1, St tIW) and StotIW) are also eigenstates with eigenvalues E :F 2h and l + 1,

if they are not zero. Thus we can generate a class of states from the state with highest 1. Assume that IW) is an eigenstate of and Stot, and that

Si t I W) = 0. We can show that (W l(Stot)n(Stot )"lip)

I(Sit)n-1(Stot)n-1IW).

= n(2l + 1 - n)(W

(6.12)

This series of states must be finite. Then 21 + 1 should be an integer and therefore l must be an integer or half-odd integer. We have 21 + 1 states,

[fl k(2l + 1- k)]

112(Stot)n1W), n

= 0,1, ..., 2l.

(6.13)

k=1

-

The eigenvalues of Stot for these states are 1, 1 1, 1 - 2,..., -l. This is the S U(2) symmetry of the spin degrees of freedom. This type of symmetry is observed in the XXX model and delta function fermion theory. But the Hubbard model has another hidden S U(2) symmetry. By the transformation (6.6) Stot, Si t, Stot are changed as follows, '1Z _

t1+=(-1)ja.iTa.il,

r1

=(-1)jati1atil. (6.14)

The Hubbard model

88

These operators satisfy the relations [q+,

] = 2rjz,

[q', q-±] = ±11±

,

P4" q±] = +(U - 2A)rl±.

(6.15)

Thus we can generate a series of states by successively applying the operator to an eigenstate of the Hamiltonian. Therefore the Hubbard model with nearest neighbour interactions on a bipartite lattice has a S U(2) x S U(2)

symmetry.118 Then the eigenstates of the Hubbard Hamiltonian (6.1) are characterized by q, riz, S, SZ. For an eigenstate of Hamiltonian lw) which satisfies

l'Iw),St tlw) = 0,AZtot = llw),

Iw> =

(6.16)

we can obtain the following states: )" (S )n [ 11=1 k(21 + 1- k) flk,-1 k'(21' + 1 - k')] n = 0,1, ..., 21, n' = 0,1, ..., 21'. (q

In,n )

Iw),

(6.17)

The eigenvalues for the operators, Sz and q' are E + 2nh + (U - 2A)n', l - n and l' - n', respectively. 6.2 The Bethe-ansatz equation for the one-dimensional Hubbard model 6.2.1 The wave function for a finite system

For the one-dimensional Hubbard model with nearest neighbour hopping, Lieb and Wu62 found that this is solvable by the Bethe-ansatz method, just after the discovery of the solution for spin 1/2 delta-function fermions by Gaudin29 and Yang117. In the first quantization representation, the eigenvalue equation for the Hamiltonian (6.1) is

-t1: E f(x1,x2,...,xj+S,xj+1,...,xN) j=1 s=±1

8jl-AN-h(N-2M)-E)f(x1,x2,...,XN)=0.

+(U j
(6.18)

Here N is the total number of electrons and M is the number of down-spin electrons. We assume 2M < N < Na. The other cases are obtained from the symmetry relations in §6.1.1. x j is the coordinate on the lattice of the j-th electron and takes integer values 1, 2,..., Na. We assume that the 1, 2,..., M-th

6.2 The Bethe-ansatz equation for a ID Hubbard model

89

electrons have down spins and that the others have up spins. Applying the Bethe-ansatz we assume N

IQ, P] exp(iEkPjxQj),

f(x1,x2,...,xN) _

j=1 XQ1 < .Q2 < XQ3 < ... < XQN. P

for

(6.19)

Here Ps and Qs are permutations of the integers 1,2,..., N. The energy eigenvalue and total momentum are given by N

E = -2t

N

cos kj - AN - h(N - 2M), K= j=1

kj.

(6.20)

j=1

As we treat spin 1/2 fermions on the lattice we only consider two-body scattering by these particles. From the continuity condition on the boundary of the regions we get [Q, P] _ (uPj,P(j+1) - 1) [Q, P'] + upj,P(j+1) [Q', P'],

for j = 1,2,...,N- 1,

(6.21)

where unm

sin k - sin km sink, sink,, + i U/(2t)'

Q1

-

= Q(j j + 1),

P' = P (j, j + 1). (6.22)

We regard the coefficients [Q, P ] as an elements of an N ! x N ! matrix. Then the above equation can be rewritten as follows:

P = ' Pj,P(j+l) P',

Yklb

= (ukl - 1) + uklPab.

(6.23)

Here cP is the column vector for a permutation P. There are N! x (N - 1) relations between N! vectors. But by the Yang-Baxter relations Y0abYjiab

-

_ I,

ab

be

ab

-

Yjk 1 ik Yij -

be

ab

be

if 1 ik Yjk ,

(6.24)

these are mutually consistent with each other. The coefficients [Q, P] must be antisymmetric for the exchange of spin-up electrons and for the exchange of spin-down electrons. Then we can write [Q, P] = e(Q1)e(Q2)(D(Y1, Y2, ..., YM ; fl,

1 <_ Y1 < Y2 < ... < YM <- N. (6.25)

Here Q1 is the ordering of the first M electrons and Q2 is the ordering of the of the M + 1-st to N-th electrons. y1, y2, ..., yM are the coordinates of down

The Hubbard model

90

spins in the lattice with length N. In the same way in §2.2 we have b(Y1,Y2, ,YM; P) A(R)FP (Y1, AR1)FP (Y2, AR2)...FP (YM, ARM),

(6.26)

R

Y-1

N

FP (y,A)=fJ(sin kpj-A+iU') fi (sin kP1-A-iU'), j=1

U'

1=y+1

U

=4t,

(6.27)

A(R) = e(R) fl(ARj - ARl - 2iU').

(6.28)

j<1

6.2.2 Periodic boundary condition

The periodic boundary condition is exp(ik jNa)

M sinkj-A#+iU'

=

sinkj - A# - iU'

N A«-sinkj+iU'

M A«-Ap+2iU'

111A«-sinkj-iU'=1A«-Af-2iU'

(6.29)

(6.30)

Taking the logarithm of these equations we have

Nak j = 27rI j - 2ME

tari1

sin k j - A# Ul

(6.31)

/3=1

N 2 tan- 1 1:

j=1

A«

in k j= Ut

2irJ« +

M

2 tan 1 A 2 U'

(6.32)

/3=1

Here the I js are integers (half-odd integers) for even (odd) M and the J«s are integers (half-odd integers) for odd (even) N - M. Using these equations and (6.20) we have the total momentum:

K = Na [> I j + > J«] .

(6.33)

6.2.3 Fredholm type integral equations for the ground state

For the ground state in the case U > 0, I j and J« are successive numbers centered around the origin. For the lowest energy state at even N and odd

6.2 The Bethe-ansatz equation for a 1D Hubbard model

91

M they are

Ij= N-1 2, N-3 2

M-3 Ja= M-1 2 2

2,

N-i

M-1 2

.

(6.34)

Putting N/Na and M/Na as constants, we take the limit Na - oo. We put the distribution functions of ks and As as p(k) and a(A). From equations (6.31) and (6.32) we get

-f

k = 27r f (k)

2 tan

fQ

B

1

sink - A U,

a(A)dA,

(6.35)

B

sink p(k)dk

A

1

2 tan-

U

Q

= 2icg(A) +

f

B

2 tan 1

A

' -1Aa(A')dA'.

(6.36)

B

Here f (kj) = I j/Na, g(Aa) = Ja/Na and therefore f'(k) = p(k), g(A) = 6(A). ±Q are the upper and lower bounds of the distribution of ks, and +B are bounds on the As. Differentiating with respect to k and A we get p(k) =

11

+ cos k B

f

B

a1(sink

- A)a(A)dA, fQ

6(A) + f a2(A - A')a(A')dA' = J Q a1(A B

(6.37)

- sin k)p(k)dk, (6.38)

an(x) =

7r (nU')2

(6.39)

+x2

Q and B are determined by

f

Q Q

p(k)dk = N/Na = n = nT + ni,

f B a(A)dA = M/Na = nl.

(6.40) (6.41)

B

From equation (6.20) we have the energy per site e

= f Q (-2t cos k - A - h)p(k)dk + 2h Q

f

B

a(A)dA.

(6.42)

B

Using these equations we can determine the energy at t > 0, U > 0, N < Na, 2M < N. The energy in other regions can be obtained by the symmetry relations given in the previous subsection.

In the Q = n case we can show that n = 1, by substituting (6.37) into (6.40). This means that the system is in the half-filled state at Q = it. In the

The Hubbard model

92

B = oo case equation (6.38) is transformed as follows:

6(A) _ LQ

(7r(A2sink)) p(k)dk.

4U'sech

(6.43)

Substituting this into (6.41) we have nl = nT = n/2. 6.2.4 Analytic solution for the ground state in the half-filled case Coupled integral equations (6.37) and (6.38) can be solved analytically at Q = ir, B = oo. Substituting (6.37) into (6.43) we have

a(A) = 1

I sech

7E(A - sink) dk

4U' Substituting this into (6.37), 271

I n

p(k) = _

2U'

1 + cos k f

1

27r

cosk

1

27r

i

+

n

f

=

R

1 f°°

Jo((o)eiwndo)

2 cosh U'co

27r

.

(6.44)

sink - sink' l dk'

U' \ U' )27r Jo(w) cos(aw sink) dug. 1 + exp(2U'co)

(6.45)

Substituting these into (6.42) we get:

e - -A - 4t

00 Jo

Jo(w)Ji(w)dw w[1 +exp(2U'w)]

(6.46)

6.2.5 Spinon excitation in the half-filled case

Consider the case of Na even and Na/2 odd. We assume that the external magnetic field h is zero. The excitation spectrum is represented analytically 78,91

In this case the half-filled ground state is characterized by the integers :

Na1 Na3 2

Ij

,

Ja =

Na1 2

2

Na/2 - 1 Na/2 - 3 2

2

Na/2 2

-

1

(6.47)

At first we consider the excitation where the number of electrons is the same and the number of down spins decreases from Na/2 to Na/2 - 1. Here we have

Il=

Na Na-2 2

2

...,

Na-2 2

-1,...,Na+1-r,Na -1-r,...,Na Ja= Na,Na +1-s, 4 4 4 4 4

4a-1s,...,- 4a 0
(6.48)

6.2 The Bethe-ansatz equation for a ID Hubbard model

93

There are two holes in the distribution of A. The situation is almost the same as with the calculation of the XXX model. The total momentum of this state is

K = Na (r + s) = 2ir (f a(A)dA + j a(A)dA).

(6.49)

Using (6.31) and (6.32) we have the change of each quasi-momentum as follows : tan-1

AkjNa =7r + 2

sin k1 - ANa/2

U'

NQ/2-1

-2ir

al(sin k1

- Aa)(cos k1Ak1- AAa),

(6.50)

a=1

(AA, - cosklAkl)a1(A, - sinkl) _ IT [sign(Aa - Ar) + sign(Aa - As)] NQ/2-1

a2(Aa - Af)(AAa - AAfl).

+27t

(6.51)

13=1

From (6.50) we have 27rAk1Na

(2 + cos k1 f a1(sin k1- A)a(A)dA)

= 2irNa f al (sin k1 - A)AAa(A)dA.

(6.52)

The substitution of (6.37) yields a1(sin kj - A)AAa(A)dA.

p(k1)Ak1

(6.53)

From (6.51) we have AAa{1: a1(Aa - sink1) -

a2(Aa - A13)} #

1

1

+ E a2(Aa - Aa)AA13 = 2 (sign(Aa - A,) + sign(Aa - As))

+

cos k1Ak1a1(Aa - sin k1).

(6.54)

Using (6.38) and (6.53) we obtain

AAaa(Aa) + I a2(Aa - A')u(A')AA'dA'

=

1

2Na

[sign(Aa - Ar) + sign(Aa - As)].

(6.55)

The Hubbard model

94

Putting NaAAa(A) = J(A) we have an equation for J(A):

J(A) +

f a2(A - A')J(A')dA' = 2 [sign(A - Ar) + sign(A -

AS)].

(6.56)

Thus the energy difference is

AE = 2tNa

f

sin kp(k)Akdk

f dk sink foo dAa1(sink - A)J(A) =2t f Cdksink f dA41 = 2t

n

00

x 2 [sign(A - Ar) + sign(A

e(A) = 2U'

f

it

dk cost k sech [

- AS)] = e(Ar) + e(AS), 7t (A - sink) 1 r2U'

(6.57)

7r

Using (6.49) we have

K = q(Ar) + q(As),

sech(7r(A2sink)

J

q(A) = 2zc fA

dx fit

dk 1 (7r(x - sink) sech 27r 4U' 2U' (6.58)

Then the excitation spectrum e(q) is defined by the parameter A. The lower edge of the spinon spectrum is shown in Fig. (6.1). At A >> 1, U', e = 4ti1(2U,) exp(

_ An 2U'),

q=

Ir

exp(

An 2U').

Thus the group velocity of this excitation is I1(2tir/U) vs = 2tIo(2t7r/U)

(6.59)

The excitation spectrum in the sector SZ = 1 and N = Na is represented by two spinon excitations. At U = 0+ we can put (2U')-1sech(7rx/2U') = 5(x). Then we have K = cos-1 Ar + cos 1 AS = qi + q2, and

AE = 2t( 1 - AY

1 - As) = 2tI sin ql

+ 2tI sin q21.

At U/t >> 1 the distribution of A becomes very broad on the real axis. On the other hand sink is restricted to [-1, 1]. Then equation (6.57) becomes:

K = ql + q2,

AE =

27rt 2

[I sin q1I + I sin q2I].

(6.60)

This mode corresponds to des Cloizeaux and Pearson's result as shown in

6.2 The Bethe-ansatz equation for a 1D Hubbard model

95

(E-EO) /t

Fig. 6.1. Lower edge of the spinon excitation of the half-filled Hubbard model. We put U/t = 0.4,2.0,4.0,8.0.

(3.45) if we put -J = 4t2/U. In fact the half-filled Hubbard model at U >> t is represented by the antiferromagnetic Heisenberg model. 6.2.6 Energy gap of the charge excitation

Next we consider the case where N = Na - 1 and M = Na/2 - 1. The excitation spectrum for charge also has an analytical solution70. The halffilled ground state is characterized by (6.47). This excitation is

Ij =Na,N'a-1,...,Na-r+1,N'a-r-1,...,-Na+1, Ja =

NQ-1 NQ-1 2

,

2

- 1, ...,

NQ-1 2

Na-1

-s+

-S- 1, ..., -

2

NQ-1 2

(6.61)

where N'a - Na/2. We have one hole in the I j and one in the J. The former is called a holon and the latter is a spinon, which appeared in the previous subsection. The total momentum of this state is

K=Na ( r+s

2

Na 4

)

(6.62)

The equations for the shift of the ks and As are

AkjNa = it sign(kj - kr) + 2 tan

1 sin k j - ANa/2

U'

Na/2-1

-2i E a1(sinkj-Aa)(coskjAkj-AA,,), a=1

(6.63)

96

The Hubbard model

- sink) = it sign(Aa - A,.)

(AA, - cos kjAkj)al (A,

+2 tan-

1

Aa - sin kNQ U/

NQ/2-1

+27r E a2(Aa - Af)(AA, - AA#).

(6.64)

fl=1

Putting D(k) = NaAkp(k), J(A) = NaAAa(A), we have D(k) = 2 {1 + sign(k

- kr)} +

c

al(sink

- A)J(A)dA.

(6.65)

Substituting this into (6.64) we have:

J(A) + E a2(A - A')J(A')dA' = 2 sign(A - As) +

I tan-1 A -sink, Ut

(6.66)

7E

The energy difference is

AE = A + 2t cos kNa + 2t L sin kD(k)dk n

= A + 2t cos kr + 2t f dk sin k f G dAal(sink - A)J(A) = e(AS) + A + 2t cos kr + 2t f

U'

n

=A+2t{coskr

f O

sink

n kr si n k ,

7E

+2J1(w)cosco(sinkr)dco} (o(e2U'w +

) dk (6.67)

Using (6.62) we have the momentum change,

AK = q(AS) + 27r f p(k)dk r

- 2.

(6.68)

This holon excitation spectrum is shown in Fig. (6.2). The minimum point of AE is kr = ±7r, A, = ±oo. In the half-filled state the chemical potential should be taken as A = U/2. Thus the energy gap of a hole excitation in the half-filled state is AE=2t{U'-1+200

JO

Jl(w)

w(e2U w + + 1) dco}

(6.69)

at AK = ±z. This gap is shown in Fig. (6.3) as a function of U/t. This quantity is always positive at U > 0. To make a particle we also need the same energy. Then we can conclude that the half-filled state is always an insulator for positive U. It has been believed that the half-filled Hubbard

6.2 The Bethe-ansatz equation for a 1D Hubbard model

97

model is metallic at U < U, and insulating at U > U, and that there is metal-insulator transition at some finite U = U, But Lieb and Wu62 found that U, = 0 for the 1D Hubbard model. 6.2.7 Susceptibility and magnetization curve of the half-filled case

We consider the case Q = 7r of the coupled integral equations (6.37) and (6.38). Substituting (6.37) into (6.40) we have n = 1. The energy per site and equation for a(A) are

e = -A - h +

f

B

dAa(A) [2h -

2t f

B

7r

cos2 kai(A

- sin k)dk],

7r

(6.70)

a2(A - A')F(A')dA' = f7r ai(A - sink)

a(A) +

dk 2-.

B

(6.71)

We change B to B + AB. The change of energy is given as Ae = 2ABa(B) [2h

- 2t if

cos2 kai (B

- sin k)dk]

+ f B Aa(A) [2h - 2t £ cos2 ka1(A- sink)dk] dA.

(6.72)

Ao (A) is the change in 6(A) and satisfies

Aa(A) + fB a2(A - A')Aa(A')dA' B

_ -Aa(B){a2(B - A) + a2(-B - A)}.

(6.73)

Then the change in the energy is written as follows, Ae = 2ABa(B){2hF(B) - 2tG(B)},

(6.74)

F(A) + f a2(A - A')F(A')dA' = 1,

(6.75)

BB

G(A) + f a2(A - A')G(A')dA' = f BB

kai(A it 7r

- sin k)dk. 6.76)

Thus for a given B the corresponding magnetic field and magnetization are:

h= tF(B)

m = nT -ni = 1- 2 f B a(A)dA.

(6.77)

At B = 0 we have h = 2t[.,/t- + U12 - U'], m = 1. Above this magnetic field the system is completely magnetized. The magnetization curve is obtained

The Hubbard model

98 (a)

(E-EO+U/2)/t

K

(b) (E-EO+U/2)/t

K

(c)

(E-EO+U/2)/t

-3

-2

-1

K 2

3

Fig. 6.2. The holon excitation of the half-filled Hubbard model at U/t = 2.0,4.0,8.0.

6.2 The Bethe-ansatz equation for a 1D Hubbard model

99

(E-EO+U/2)/t 17.5 15

12.5

Fig. 6.3. Energy gap for the charge excitation as a function of U.

by numerical calculations of these equations. At B = oo the magnetization is zero. To get the magnetic susceptibility for weak magnetic fields we must consider the behaviour of these equations at B >> 1, U. Equation(6.71) is transformed as follows:

6(A) -

f

u7

{R(A u' ) + R(A u'

I6(A')dA' = 6o(A),

)

(6.78)

(6.79)

6o(A) 60

is the solution at B = oo. The energy (6.70) is as follows: m=2

"0

6(A)dA,

(6.80)

B

e = eo - hm + 4t

Go(A)a(A)dA,

(6.81)

JB'

Go(A) =

n f7r

dk

`7r(A

cos2 k sech

sin

(6.82)

Go(A) is the solution of (6.76) at B = oo. The magnetization curve for the

The Hubbard model

100

half-filled Hubbard model was investigated by the author90. 6o(A) and Go(A) behave as follows: "

6o(A)

1 2U'

exp

1

=

2U'

exp(- 7r(A-sink) ldk

) /'' - exp ( (- 2U 'jA J_ 7' 27 l

Go(A) -

)2it

2U'

2UU

' k)

=

2U

J

dk cos2 k

2U

2jU ) Io (-) exp (- 7EA

1

l

2U

J

.

7r(A -sink)

exp (-

)

2U

7rA

exp( 7r = exp-2U') 2U' 'T 7' 'T (I0(2U') -I2(2U)) = 211(2U')

7rA 2U').

Here we have used

f

cos

11

n

dk

In(x) -In+2(x) =

1nkexcosk = In(x),

2(nx

1)In+1(x)

Then we can calculate the magnetic susceptibility in the same way as for the XXX antiferromagnetic chain,

e - eo - 2h I a(A)dA + 8t1 i (27rU

a(A)-lB U'{R(A =

2Ulo(27rU)

exp(

A' u,

exp(

J A'

) +R(A

7rA )o (A)dA,

)}6(A')dA'

u'

(6.83)

2U' ).

Putting p(x) = 2U' exp(7rB/(2U'))o (xU' + B)/Io(ir/(2U')) we have e = eo - hlo

(Z

exp ( 2

U) Jo

p(x)dx

+8t10 (2U')Ii(2U') exp(-7rB/U')

f erx/2p(x)dx 0

p(x) = fCC{R(x - x') + R(x + x' +

2BLIB)

}p(x')dx' + e nx/2 (6.84)

Using the functions a(x) and b(x) defined in (3.60) we obtain the energy as function of B :

e=eo-4hexp( +8tlo( it it

2U'

)102U')a(U' it

2U' )h( 2U'

)e-lrB/u'b(B U'

(6.85)

6.3 1 1 U' expansion

101

Then the magnetization and magnetic field are 2U')(1+27KB)-1

o 7raot

h

m = 2aoIo(

'

(1

+ U/ ).

(6.86)

2U' 27rB 2U' ) exp(-UB) ao and bo were given in (3.72). Thus m/h at a small magnetic field is given by Z

m 1I0(U' 2

=h

+

7rtI1 (zU,)

(6.87)

27rB

91. In the non-half-filled These calculations were done by the case one can calculate the susceptibility, although one must solve integral author90,

equations83.

6.3 1 / U' expansion

We analyze the analytic properties of the gap (6.69) and the ground state energy (6.46) in the half-filled case. The following expansion is very useful: ((i!)2 ti

JO(w)J1(w) _ {E

J-o

=1

(-1)n(2n + 1)! w [n!(n + 1)!]2

n=o

.I !(I

(

i

+)1)! 2

1 J5

2n+1

(6.88)

2

(-1)'

J1 (0)) _

J

(2 ) 2ti}

w 2J+1

(6.89)

.

j=0J!(J+1)!

2

Then we can show (-1)n(2, + 1)!

00 Jo(w)J1(w)-2 xmdw o

x

- d 22n+1{n!(n + 1)!}2

CO

n=0

°° w2ne 2xmdw

=

/p

00

,

n L=O 1

1

(-1)n(2n + 1)!!(2n - 1)!! 1 x2n+1 {(2n+2)!!}2

1

2,1;-z2) = x{F(-2, °° J1(w)e 2xmdw = (-1)n(2n-1)!!

Jo

co

n=0

(6.90) 1

22n+1(n + 1)! x2n+1 (6.91)

The Hubbard model

102

Here F(a,

x) is the hypergeometric function. Using

+1)=e2Uw-e 4vw+-...,

1/(e2u(o

we find

AE = 2t[U' -1 + 2 n=1

1(-1)n-12nU' (F+ (2nU')2

- 01' (6.92)

E/Na = -A - 4tE(-1)n-1 n U'(F(-1 , 1 ; 1 ; 00 22 n=1

- (n U')2 )1

(6.93)

As functions of U', AE and E/Na have branch points at U' = +i/n, n = 1,2 . .... U' = 0 is the accumulation point of branch points. On the other hand expansion from U' = oo or 1/U' expansion gives AE

=2tU'-2t+4t[21n2 U-1-Z 3)C33)(1-22)U-3+...

+(-1) r-1 (

C(2r - 1)

(2r - 1)!!

(2r)!!

) 2r - 1

- 22r-2 )U'-2,+1+ ...]' 1

(1

(6.94)

E/Na=-A-4t[(2)21n2 U-1+(-1) r - ( 1

(2r-1)!! )

(

2r - 1

(2r)!!

(2.44)2_33)(1-22)U-3+...

1

22r-2

1

) '-2r+1 U

+ ...

.

(6.95)

These series converge at I U' l > 1. The expansion from U' = 0 is not convergent. So it is possible that we can get physically meaningful expansion from U' = co.

6.4 Perturbation expansion in the half-filled case Let us consider the perturbation expansion of the general Hubbard Hamiltonian, putting

o=U

niT nil - A(nit + nil) - h(niT - nil),

1=a

i<j

ti,j(cT 16 cj,

+

(6.96) (6.97)

All sites are occupied by an electron with up or down spin. Then 2Na states degenerate at ti,.j = h = 0, A = U/2. The energy eigenvalue is -UNa/2. Then

6.4 Perturbation expansion in the half-filled case

103

in this subspace of the Hamiltonian a state is represented by Jul, 62, ..., 6Na). In the first order perturbation the matrix element is always zero. In the second order perturbation one electron goes to the next site if the next site is occupied by an electron with opposite spin. This intermediate state

has the higher energy -UNa/2 + U. In the final state one electron returns to the vacant state and the unperturbed energy is again -UNa/2. Then the energy of the ground states is described by the effective Hamiltonian: tUijtji(6i

effective = -UNa/2 +

6j - 1).

(6.98)

i<j

Anderson showed that the XXX antiferromagnetic Hamiltonian can be derived from an itinerant electron model3. This is just the Heisenberg antiferromagnet with spin 1/2. Generally speaking the perturbation expansion of the L-fold degenerate ground state is given by L x L matrices. One can show that the odd-order elements are always zero. The fourth order term of the effective Hamiltonian is

U-3t4(1-6i'6j)+ i<j

i<j
i
tijtjktkltli(5(6j 6k)(6i . 6l) + 5(6i . 6j)(6k 61)

-5(6j'6k)(6i6l)-6i'6j-6j'6k-6k.ai -61'61-6i'6k-6j"61+1)].

(6.99)

The third term gives the four spin interactions. This term cannot appear for the one-dimensional nearest neighbour Hubbard model because the square path is impossible on this lattice. This term was derived by the authorlol On the two or three dimensional lattice this term plays an important role. For the one-dimensional model the effective Hamiltonian becomes *'effective = -ANa + rr > (4Si . Si+l - 1) 4

6

+U3E{4(1-4Si.Si+l)+(4Si-Si+2-1)}+O(Us). (6.100)

Then the energy per site is given by E Na

= -A+

+ 0'3 t (

t ((Si . Si+1)o - 4) U'

(Si Si+1)o

4

+

(Si . Si+2)o 16

+ 3) + p ( 64/

t U'S

(6.101)

The Hubbard model

104

where (...)o is the correlation function of linear Heisenberg model. Comparing with (6.95) we have 1

( Si

Si+1)o = 4

- In 2 = -0.4431471806,

(6.102)

(Si Si+2)o = 4 - 41n 2 + 4 (3) = 0.1820393099.

(6.103)

Thus we derive the second neighbour correlation function of XXX antiferromagnet from the ground state energy of the Hubbard model.

6.5 Asymptotic expansion from U = 0 As shown in the previous section the U = 0 point is a singular point and the usual expansion is not possible. But we need asymptotic behaviour of the energy gap and ground state energy in the half-filled case. For this purpose the following formula is very useful, 00

1:-1)n-1(

b

b2 + (na)2 - na) = -- +

+

[cos./(ay)2b2

( )dy, y

bla

n=1

a, b > 0.

(6.104)

The 1.h.s. is given by the following contour integral 1

c 2i sin zrz

b2 + a2z2 - az}dz.

{

(6.105)

The contour is taken so that it surrounds the positive real z axis. We deform

this path to the imaginary axis. See Fig. (6.4). We should note that the integrand has a pole at z = 0 and branch cuts at z = iy, jyj > b/a. Thus we get (6.104). Substituting this into (6.92) we have

DE = 4t

/-

-

(2U')--l dy. sinh?ry

y2

1/2U'

(6.106)

Then at U' << 1 this behaves as AE -- 8t exp(-7r/(2U'))

J

e-"`

4t

t(t + U'-1)dt :t

e-"/(2U').

(6.107)

7E

Then the energy gap is exponentially small for small U'. To calculate the ground state energy we should note the following identity, 1

1

F(- 2, 2 ;1; -x2) =

0 2 fir/2 71

1 + x2 sin2 Od¢.

(6.108)

6.5 Asymptotic expansion from U = 0

105

(b/a) i

- (b/a) ii

C

Fig. 6.4. Contour integral on the complex plane. The path is deformed to the imaginary axis.

Then equation (6.93) is transformed as it/2

2

(E/Na+A)/(4t) = --

00

(-1)i( (nU')2 + sine-nU')d. (6.109)

f

n=1

Substituting (6.104) we have

_

2 R

j.12 d

U'

(4

- sin

- Ln/U'

2

y2 - (sin

U')2

sinh(my)

dy)

(6.110)

If we change the order of integration of this double integral we have

_

U' 4

1

it

00 G(U'y) - JO sinh(iry) dy'

where 2

x

.

O 7E

2

il

x2 - t2 T-- -t2 x2

t2

at 0<x<1,

1-t2dt, at x>1.

(6.112)

The Hubbard model

106

E/Na U'

-1

-2

1

Fig. 6.5. Ground state energy of the half-filled Hubbard model as a function of U' - U/(4t). Line a is the expansion from U' = oo up to U'-21 of (6.95). Line b is the asymptotic expansion from U' = 0 up to U'6 of expansion (6.116). Dots are the result of numerical integration of (6.46).

This is written in complete elliptic integrals {E (x)

- - x2)K(x)}, (1

at 0 < x < 1,

G(x) =

(6.113)

2E(z 1)/x, at x > 1. it At IxI < 1, G(x) is expanded as G(x) =

2n

')0((2n - 1)!!)2

2nn!

n =1

(2n

-

x2n.

(6.114)

1)2

We can show that 00

JO

=

x2n

00

sinh(7rx)

x2n(e-7cx + e 37rx

dx = 2 J o

+e

57cx

+ ...)dx

2(2n) n2 n+1! (1 + 3-(2n+1) + 5-(2n+1) + ...)

(1 = 2(2n)! n2 n+1

- 2-(2n+1))Mn + 1).

(6.115)

Then we obtain an asymptotic expansion from U' = 0, (E/Na + A)/(4t) _ 7T

- -4ir3)

+

4

U2

931; (5)

_ 32it5

U /4

6.5 Asymptotic expansion from U = 0

(((2n - 3) ! !)3(2n - 1)(22n+1 - 1)) t(2n + 1) U12n (2n - 2)!!22n

107

- ....

7E2n+1

(6.116)

This is an asymptotic expansion and the convergence radius is zero. This was obtained by Economou and Poulopoulos22. Metzner and Vollhardt68 calculated analytically the second-order term of the perturbation expansion of the ground state energy and certified that it coincides with the U12 term of equation (6.116).

Part two Finite temperature integral equations for un-nested systems 7 Repulsive delta-function bosons 7.1 Uniqueness of the solution

In chapter 2 we consider the problem of bosons interacting via a repulsive delta-function potential with periodic boundary conditions. The ground state energy is obtained from a Fredholm-type integral equation. For this model Yang and Yang123 found a non-linear integral equation which gives the free energy at a given temperature. This theory is important because it consists of the basic theory for thermodynamics of other solvable models. The quasi-momenta kjs are determined by equation (2.30) for a set of quantum numbers I j. The equation is N

kjL=27rIj-

i

k1 -kl2tan-,

j = 1,...,N,

(7.1)

C

1=1

where the Is are integers (half-odd integers) for odd (even) N. This is equivalent to the following equations : aB(k1, k2, ..., kN) ak j

= 0,

B(kl, k2, ..., kN) = 2 1: 0 - 21rI jkj + > 61(kj - ki), j<1

61(x) = xtan-1(x/c) - 2 ln(1 + (x/c)2).

(7.2)

The extremal point of the function B is the solution of (7.1) and gives an eigenstate of the Hamiltonian. Let us consider the following N x N matrix: Bit =

02B

ak jaki =

a j1 (L +

c2

2c

+ (kj - km)2) 109

2c

c2 + (kj -

ki)2.

Repulsive delta function bosons

110

This matrix is always positive definite because 2c

2 uiBijuj = L E ui +E

(uj l<j c2 + (kj - ki)2

If

UI) 2

>_ 0,

(7.4)

for an arbitrary real vector {uj}. The function B is a concave function in N-dimensional space. Thus solutions of equation (7.1) for given set of {I j} are unique.

7.2 Holes of quasi-momenta and their distribution function In the limit of c -+ oo this set of eigenstates is complete. Let us consider the following function:

tan- i k

h(k) = k + L

kJ .

(7.5)

The position of a hole is defined by h(kh) =

27t

L

x unoccupied (half-odd) integer.

We denote the distribution function of holes as ph(k) and that of particles as p(k). In the thermodynamic limit we have h(k) = 27r fk p(t) + ph(t)dt.

(7.6)

Differentiating this yields 27r(p(k) + ph(k)) = 1

+ 2 fo0c2 cp(k)dk' +(k-k)/2 '

From equation (2.25) we have

e = f k2p(k)dk,

n=

f

p(k)dk.

(7.7)

(7.8)

The entropy of the distribution between k and k + dk is the logarithm of the number of orderings of Lp(k)dk particles and Lph(k)dk holes In [L(p(k) + ph(k))dk] ! [Lp(k)l! [Lph(k)]!

= Ldk [(P(k) + Ph(k)) ln(p(k) + ph (k)) -p(k) In p(k) _ ph(k) In ph(k)].

Here we use the Stirling formula ln(n!) _- n(lnn - 1). Thus the entropy per unit length is s =

f:(P(k) 00

+ ph (k))

ln(p(k) + ph(k)) - p(k)ln p(k) - ph(k)In ph(k)dk. (7.9)

7.3 Thermodynamic equilibrium

111

7.3 Thermodynamic equilibrium

The free energy per unit length f = e - Ts must be minimized under the condition that n is constant. f and n are functionals of p(k) and p'(k). Thus we should determine p(k) and ph(k) such as to minimize f - An. Next

we look for a parameter A such that n takes on its required value, in the variational calculus A is called a Lagrange multiplier. At the minimum point the variation of f - An must be zero for any infinitesimal variation of functions

0=6

(k2

- A)p(k) - T{(p(k) + ph(k)) ln(p(k) + ph(k))

-p(k)In p(k) - ph(k)In ph(k)}dk

_

{k2-A-TIn(P(k)+Ph(k))}bp(k) l p(k)

-T 1n(P(k)p h(k)Ph(k) 6ph(k)dk.

(7.10)

6p(k) and 6 ph(k) are not independent. From (7.7) we have

-8 p(k) +

b ph(k)

c2 + (k

it

-

Substituting this into (7.10) one obtains 6P(k){k2 -

0=J

A - T ln(

00

p(k)

/°°

T

it j

k')2

dq c2 + (k - q)2 In 1 +

P(q))} dk. Ph(q) (7.11)

This equation must hold for any arbitrary infinitesimal change of the function p(k). Thus

T 1n(Ph(k)) = k2 P(k)

-A-

Tf it

00

c

c2 + (k - q)2

ln(1 + P(q) )dq,

(7.12)

Ph(q)

must be satisfied at thermodynamic equilibrium. If we put e(k) = T ln(ph(k) /p(k)), this becomes

e(k) = k2 -A-

T

f

c2 + (k -

q)2

In (I+ exp ( e7 ))dq.

(7.13)

e(k) can be determined by iteration. This function has physical meaning as the excitation energy for an elementary excitation, as will be shown in the

Repulsive delta function bosons

112

next section. From equation (7.7) we have

27rp(k)(1 +exp(e(T))) = 1 +2c

0

c2+(gkdgq)z

(7.14)

From this equation we can determine p(k). If equation (7.13) is differentiated with respect to the chemical potential A, one obtains a linear integral equation ae(k, A) OA

_ -1

dq

c

1

ae(q, A)

c2 + (k - q)2 1 + exp(E(k))

(7.15)

OA

Comparing this with (7.14) we find ae(k, A)

-

OA

-27rp(k) (1

((k))) _ -2n(P(k) + Ph(k))

+ exp (e

(7.16)

The entropy density (7.9) is

s = f[(p + ph) ln(1 + exp(-e(k)/T)) + p(k)e(k)/T]dk. The free energy density is

f = f(k2 - e(k))p(k) - T(p(k) + ph(k)) In (1 +

(7.17)

Substituting (7.13) into (7.17) one obtains

f = f[A +

T

q)2ln(1+exp\

I c2+(kc

e(T)))]p(k)

-T(P(k)+ph(k))In(1+exp( e(T)))dk

f

+exp(-e(k)))

_ Ap(k)dk - T f ln(1 x [p(k) + ph(k)

-I

T 1c2 + (k

- q)2 P(q)dq]dk.

(7.18)

Substituting (7.7) we have a very simple expression for the free energy density:

f =An- T

J

2dk

In (1

+exp(-e(T))).

(7.19)

The thermodynamic potential density g = f - An is:

g(T,A) = -T fln(1 +exp(-e(T))) d.

(7.20)

For a given temperature T and chemical potential A we can calculate e(k)

7.4 Elementary excitations

113

through the non-linear integral equation (7.13) and obtain the thermodynamic potential g from (7.20). All thermodynamic quantities are derived from g through the thermodynamic relations n

S

OA,

e=g+An+Ts.....

O1,,

(7.21)

The pressure p is -g. 7.4 Elementary excitations

We consider the change of energy and momentum when one particle is moved from thermodynamic equilibrium. Assume that the l-th particle is removed and one particle is added between the m-th and m + 1-th particles,

III >12>13>...>IN}-* {I1 >I2>...>Im>III >Im+l >...>Il-1 >I1+1 >...>IN} The Bethe-ansatz equations for the original state are Lkj = 27rIj

N

-

2 tan -I

kj - kn,

j= 1, ..., N.

(7.22)

c

n=1

For the excited state they are kj

Lk = 2ir11 - 2 tan-

2

c

tan-1

-

kn

jrl

C

(7.23)

Lki

=

27rIi

-

2 tan-1

- k'n

k't

(7.24)

c

n#m

The change of momentum and energy are

AK =k-kl+T(kn-kn), nil

AE = kit - ki + E(k'ri - kn).

(7.25)

n#l

Subtracting (7.22) from (7.23) we have

AkjL + E

2c

c2 + (kj

- kl)2 (Akj - Akl)

k1- kl = 2 tan k1- kl - 2 tans 1

C

c

(7.26)

Repulsive delta function bosons

114

Then equation for the back-flow J(k) = p(k)OkL is h

(1 + p(k) ) J(k;ki,kl) (tari-1 k c

=:

kl

- I 7r(c2 + (k - k')2) J(k,; kl, kl)dk'

- tank c kl)

(7.27)

.

Energy and momentum changes are OK (ki, kl) = ki - kl + JJ(k;k,ki)dk, L\E(k', kl)

= kl2

- ki + f 2kJ(k; kl', kl)dk.

(7.28)

It is clear that AK(kl, kl) = AE(ki, kl) = 0. The differential of these with respect to k' is OAK

aki OAE I

=1+

f u(k; kl)dk,

- 2k' + f 2ku(k;kl)dk,

(7.29)

where

u(k;kl') =

a ak,J(k;kl',kl),

l

(1 + il(k))u(k; k')

- f a(k - k')u(k'; k')dk' = a(k - k').

(7.30)

Using this integral equation we find that u(k;kl) is given by the infinite series: u(k;ki) = 1 + q(k) [a(k

+

- k') + f dki 1(+ (ki)a(ki - ki)

a(k2 - k') +...]. f f dkldk21 + q(ki) 1 + rl(k2) a(k

ki) a(ki - k2)

(7.31)

If we substitute this into (7.29), we get an expression for OAK and O'J in Ok' terms of an infinite series: OAK l

+

= 1 + f dkia(ki - ki) 1

+ s(ki)

f f dkidk2a(ki - ki)1 + s(ki) a(ki - k2) I + r1(k2) 1

1

1

°kE = 2ki + f dkia(k' - k1)1 +

...,

r1(ki)2ki

l

+

f f dkidk2a(k'-ki)1+1(ki)a(ki-k2)1+q(k2)2k2+.... 1

1

7.5 Some special limits S u bs titu ti ng

115

thi s i n to (7 . 29) we ge t an express i on for

a0K(i,kl) =

l+f a (ki-k

i)

1+rl(k1)

1

aAE(ki, k1) ak1

= 2ki + f a(ki - k1)

OAK

an d OAE Ok;

Ka (1,k1)dk

a

1

3AE(ki, kl)

1 + i1(ki)

Oki

:

1,

dki

(7.32)

On the other hand we have equations for 27r(p(k) + ph(k)) and ae(k)/ak from (7.7) and (7.13): 27r(P(k) + ph(k)) = 1 +I a(k - k') 27(P(k') + Ph(k/)) dk,,

1+q(k')

ae(k)

= 2k + f a(k - k')1 +

ae(k')

dk'.

(7.33)

The solution of non-singular linear integral equation is unique. Thus a0K(k;,k1) =2ir(P(ki)+Ph(k;)), ak;

a/E(k;,k1) ak;

_

ae(k;) ak;

(7.34)

Integrating these one obtains AK(ki, k1) = 27r I k' P(k) + ph(k)dk,

AE(ki, k1) = e(k;) - e(k1).

(7.35)

Thus e(k) = T ln(ph(k)/p(k)) has a physical meaning as the energy of an elementary excitation. In the limit of T - * 0 this excitation coincides with the excitation calculated in §2.1.7.

7.5 Some special limits 7.5.1 The c = co limit In the thermodynamic limit the system is equivalent to ideal spinless fermions. In this limit c/(c2 + (k - q)2) in the integrand of (7.13) is zero. Thus e(k) = k2 - A. From equation (7.14) we have p(k)

= 27r exp[(k2 -A)/T] + 1

(7.36)

From (7.20),

g = -2 f dkln(I+ exp(_k

2T_

A)).

(7.37)

Repulsive delta function bosons

116

This is equivalent to (2.8).

7.5.2 The c = 0+ limit In this limit the integration kernel c/(c2 + (k q)2) can be replaced by 7r6(k - q). Equation (7.13) becomes e(k) = k2 - A - T ln(1 + exp(-e(k)/T)). We obtain (7.38) e(k) = T ln(exp((k2 -A)/T) 1).

-

-

The Gibbs free energy per site g, pressure p, p(k) and ph(k) are k2

g= -p=T J In 1

- A11 dk

_h,i_.

1

(7.39)

P(k) = 2n exp((k2 - A)/ T) - 1

These results coincide with those for ideal bosons in §2.1.1.

7.5.3 The T = 0+ limit Generally speaking e(k) is a monotonically increasing function of k2. At T = 0+, assume that e(qo) = 0. Thus p(k) = 0

for

k2 >

qo,

ph(k)

= 0

for

k2 < qo.

Equations (7.13) and (7.7) at k < qo are C

e(k) = k2 - A + 7T

f

Ro

Lqo

2icp(k) = 1 + 2c 9o

90

C2+

-

p(q)dq q)2. c2 + (k

(7.40)

The last equation is equivalent to the Lieb-Liniger equation (2.32). The above theory was introduced by Yang and Yang 123. They solved the Very surprisingly it seems that this simple nonequation linear equation gives the exact free energy in the thermodynamic limit of repulsive bosons. The next problem is to find the thermodynamic Betheansatz equation for other soluble models. As the next simplest case we treat the S = 1/2 XXX chain. numerically124.

8

Thermodynamics of the XXX chain

8.1 String solution of an infinite XXX chain

The author obtained the thermodynamic Bethe-ansatz equations for the XXX mode193. We consider the wave function in equation (3.18) assuming that N is infinity. Particle coordinates nj move from -co to co. Assume that Skn < 3kn_1 < ... < 3k1. The wave function is written as follows

M M a(n1, n2,..., nM) = (ZlZ2..ZM)nl E A(P) fJ(fJ zPl)n,+t-n3, P

1=2 I=I

A(P) = e(P) fl(xpj - xpj - 2i), zj = e°kj = <1

x+i

(8.1)

xj - i

In the infinite system, a satisfies the following boundary conditions: I

n1_

lnm _ r

a(nl, n2, ..., nM) I < oo,

lim a(nl, n2,..., nM)I < (. M

nM-r+

(8.2)

From this condition we find that Izlz2...zMI = 1 and that A(P) = 0 if one of I Iliy--j zpII is greater than one. If .Zkn < Zkn_1 < ... < sk1, we have

A(I) * 0, A(P * I) = 0,

IZ1Z2...Znl = 1.

(8.3)

These conditions are satisfied only if

xj = a + (n + 1 - 2j)i,

j = 1, 2,..., n.

(8.4)

From Izlz2...zMI = I a±n` a-ni I = 1, we find that a must be real. The condition M

(H zI) = I 1=j+1

a+(n-2j)i oC - ni 117

< 1, j = 1, 2,..., n - 1,

Thermodynamics of the XXX chain

118

is automatically satisfied for general n. Thus strings with arbitrary length are possible for the XXX Heisenberg chain. Moreover the following type of string is impossible:

x1=a+(2+/3)i, x2=a+/3i, X3=a-/3i, X4=a-(2+/3)i,

(8.5)

where a,/3 are real and /3 * 1.

8.2 String hypothesis for a long XXX chain

Now we return to the problem of a length N chain with periodic boundary condition. We assume that all rapidities x j belong to bound states with n = 1, 2. .... For bound states of n xs the real parts of all the xs

are the same and the imaginary parts are (n - 1)i, (n - 3)i,..., -(n - 1)i within the accuracy of O(exp(- 5N)). This assumption seems to be too strong and there are some counter examples in some extreme cases. This is a very controversial point of the thermodynamic Bethe-ansatz equations for soluble models, except for the repulsive boson case, which has no string solutions. But equations obtained using the string hypothesis seem to give the correct free energy and other thermodynamic quantities.

Consider the case where Mn bound states of n xs exist. We designate the xs as nxa>j

a = 1, 2, ..., Mn,

xa'1 = xa + i(n + 1 - 2j) + deviation.

(8.6)

From (3.17), N n,j e (xa)

xa,j - xm

11 (m,/3)#(n,a)

e(

x2 aj - x), e(

l

Xn,j - xm

l

m+isl

a,j'

X

j = 1,2,...,n.

(8.7)

j'*j Here e(x) - (x+i)/(x-i). The last product is delicate because the numerator or denominator may become very small. If we take the product of these n equations, these delicate terms are cancelled and we have n

eN(xa/n) = II j=1

eN(xa,j)

_

[J

(m,/3)/(n,a)

Enm(xa - xv ),

(8.8)

8.2 String hypothesis for a long XXX chain

119

where Enm(x)

/

x

\

x

1/

for n / \ e2 (2) e2 (x) 4...e2 (2n -

\ /

x

x

\

m,

for n = m.

2) e (2n)

The logarithm of these equations gives

N0(xa/n) = 2irIa +

Onm(xa (m,f)(n,a)

- X,),

(8.10)

where

0(x) - 2tan 1(x),

(8.11)

and

®nm(x)

x 0( Inx -mj\ )+201 In-mI+2

for

x )+0I\n+m x \n+m-2

)+...+20(

n * m,

20 2 +20(4X) + ... + 20 (2nx 2)

+ 0 (2n)

for

n = m. (8.12)

In is an integer (half-odd integer) if N - Mn is odd (even) and should satisfy: 06

III <

1

2

00

(N

- 1 - E tnmMm),

tnm ° 2min(n, m) - 8nm.

(8.13)

m=1

We can prove the number of sets {In} is CM - CM_1 under the condition

M=

1 nMn. For details see Appendix B. The energy of this state is given by:

E({Ia}) = N(-h - 4) +

E((xa)2

+ n2

+ 2hn).

(8.14)

n,a

We can construct wave functions through (3.3), (3.5) and (3.12) at S = SZ = N/2 - M. Wave functions at SZ = S - 1, S - 2,..., -S are obtained by applying the operator Stotai The energies for these cases are E - 2h, E 4h,..., E - 2(N - 2M)h. Then the total number of states which are generated by the string assumption and descending operator at SZ = N/2 - M is CM. Therefore the total number of states is 2N. This coincides with the true total number of states. It is expected that all eigenfunctions constructed in the

Thermodynamics of the XXX chain

120

above mentioned way should be a complete set. The partition function of this system is written as follows:

_ - [N/2] 1 -ex p(-2(N + 1 - 2M)h/T ) M=O

1 - exp(-2h/T)

l

tI

exp[-T-'E ({Ia})].

(8.15)

1 aJ

The free energy is given by G = - T In Y. We define functions hn(x) by Onm(x - xa ).

hn(x) = 0.(x) - N-1

(8.16)

(m,a)

We can define holes in an n-string sea by the solution of

2tJf/N = hn(xq),

(8.17)

where JJ are omitted integers or half-odd integers. 8.3 Thermodynamic Bethe-ansatz equations for the XXX chain

In the thermodynamic limit, we define distribution functions of n-strings and holes of n-strings as pn(x) and pn(x). The numbers of strings and holes between x and x + dx are pn(x)Ndx and pn(x)Ndx. Thus we have x

27r

co

°O

Pn(t) + Pn(t)dt = On(x) -

1 Dnm(x - Y)Pm(Y)dy.

(8.18)

M=1 _00

Differentiating with respect to x, we obtain the integral relation 00

an(x) = Pn(x) + Pn(x) +

(8.19)

Tnm * Pm(x).

Here Tnm(x) is a function defined by I aln-mi (x) + 2ain-,,1+2(X) + 2ajn-mi+4(x) + ... Tnm(x) = {

+2an+m-2(x) + an+m(x)

I. 2a2(x) + 2a4(x) +... + 2a2n_2(x) + a2n(x)

for

for

n

m,

(8.20)

n = m.

an(x) is a function defined by __ an(x)

1

n

x2 + n2, ao(x) = 8(x).

(8.21)

We denote the convolution f cc a(x - y) b(y) dy as a * b(x) for two arbitrary functions a(x) and b(x).

8.3 TBA equations for the XXX chain

121

The energy per site is

- (4 + h) +

e

gn(x)Pn(x)dx, n=1

g,(x) - 2nJan(x) + 2nh.

(8.22)

The total entropy per site is: 00

S= n=1

I

h(X)

00

pn(x)1n(1

+pn(x)ln(1

+ Pn(x))

+

Pn() x

pn(x))dx.

(8.23)

e - Ts should be minimized at thermodynamic equilibrium. Consider the functional variation of the free energy with respect to p,(x) and pn(x) 00

0=6e-TBs=

f dx [Kn(x) - T ln(1 + pn(x))] bPn(x)

-T ln(1 + pn(x))bpn(x)

(8.24)

From equation (8.19) we have cc

apn(x) _ -Spn(x) -

Tn

(8.25)

5P7(x).

m=1

Substituting these into (8.24) yields

0=T

00

n=1 f{

gn(x)

00

- In tln(x) + E Tnm * ln(1 + gml(x))}6pn(x)dx, (8.26) m=1

where qn(x) =- pn(x)/pn(x). Thus we have integral equations for an infinite number of unknown fl (x), In in(x)

=

gnTx)

+

Tnm *ln(1 +

(8.27)

qMI (x)).

M=1

The free energy per site becomes as follows :

f

nn=1

=e-Ts=-(4+h)+fgnp,,-T[pnln(l+q,)+Phln(l+qn-1)]dx.

We eliminate pn using (8.19) :

f

-(j

00

f ln(1+in1)an(x) n=1 00

+pn [In 11n

- nT - V Tnm * In(1 + qm1)] dx. M=1

(8.28)

Thermodynamics of the XXX chain

122

Using (8.27) one sees that the inside of the brackets on the r.h.s. is zero. So we have

f = - (4 + h) - T

f an(x) ln(1 + qn l(x))dx.

(8.29)

n=1

From the n = 1 case of (8.27) we have ln(1 + 111)

=

27rJa1(x) + 2h 7

+ E(al-1 + at+l) * ln(1 + qi 1).

(8.30)

1=1

Operating f dxs(x)f on this equation yields

J

dxs(x) ln(1 + rj1) =

2T

f

J

T

00

+

h

s(x)al(x)dx +

ai(x) ln(1+ q1 i(x))dx. l=1

Then equation (8.29) is transformed as follows:

f = J (In 2 - 4) - T I s(x) ln(1 + ii1(x))dx.

(8.31)

Solutions i7n of (8.27) are functions of x, J, T and h. Differentiating (8.27) with respect to J yields 2irJan(x) - 1 a,n * 1 atim 8.32) T. T qn aJ + (1 + ?m)lm aJ

Comparing this with (8.19) we have T 1 at1 n Pn = 27r (1 + tin)t1 aJ

h

P. + Pn

n

T a In tin 2,r

aJ

.

(8.33)

By the definition (8.20) we have: al * (Tn-l,m + Tn+l,m) - (ao + a2) * Tn,m = (an-l,m + an+1,m)al.

(8.34)

Using equations (8.27) and (8.34) yields (ao + a2) * In iii (x) =

27rJa1(x)

T

+ al * ln(1 + p2(x)),

(ao + a2) * In 1in(x) = al * ln(1 + iin-1(x))(1 + lln+1(x)), n = 2,3 . ....

(8.35)

(8.36)

Equations (8.35) and(8.36) are not sufficient to determine all the qn(x), as they do not contain h. Taking the n = 1 case of (8.27), In ill =

27rJal(x) + 2h 00 +a2*ln(l+t1l 1)+I:(aj-1+aj+1)*ln(1+r1i --1). (8.37) T j=2

8.4 Some special cases and expansions

123

Substituting (8.35) and (8.36) we can eliminate 13, j < n for a given integer n,

2h

q_

7, = an * In 11n+i - an+i * ln(1 + in) - an+2 * ln(1 + -11) 00

-

(al-i + al+i) * ln(1 + ii 1)

(8.38)

1=n+2

Thus we have 2h

In rln+l =

1

+ ai * In I1n + a2 * ln(1 + q n+i )

T

(al-n-1 + ai-n+1) * ln(1

+ 111 1)

(8.39)

1=n+2

For large n, ln(1 + in1) = o(n-2) and therefore: limoo

- al * In 11n =

In 1n+l

r

lim

In rln

T

,

- 2h

n

n-*oo

2h

(8.40)

(8 . 1)

T

Thus the following equations determine 11n,

=

27LJ

s(x) + s * l n( 1 + 112(x)), T In 17,(x) = s * ln(1 + qn-1(x))(1 + 1n+l(x)), In

qt(x)

lim In n->

n

=

2h

( 8 . 42 )

(8.43) ( 8 . 44 )

.

n

where

s(x) = 4sech

2 ),

s * f (x) _

fs(x_y)f(y)dy.

(8.45)

8.4 Some special cases and expansions

8.4.1 The J/T -40 case In the limit J/T 0 and h/T > 0 we can expect that 11n(x) is independent of x, because there are no x dependent terms in equations (8.42), (8.43), (8.44) and (8.45). As f dxs(x) = 1/2, equations (8.42), (8.43), (8.44) and (8.45) become 11n = (1 + 11n-1)(1 + 11n+1),

(8.46)

111 = 1 + 112, lim In 11n/n = 2h/T.

(8.47)

Thermodynamics of the XXX chain

124

Equation (8.46) is a difference equation of second order. It is similar to a differential equation of second order and contains two arbitrary parameters. The general solution of this equation is: a-lz-n

azn -

2

(8.48)

Z - z-1 Parameters a and z are determined by (8.47) and we have a = z, z = exp(h/ T) and qn

_

-

(sinh[(n + 1)h/ T] 2 - 1 sinh[h/T]

t1n=(n+1)2-1

for

for

It > 0

h=0.

(8.49)

Substituting this into (8.31) we obtain the free energy, magnetization and entropy:

m = 2s, = -Of /ah = tanh hl T, s = -3f/0T = In [2 cosh(h/ T)] - (h/T)tanh(h/T). (8.50) f = -T In [2 cosh hl T],

At h = 0 the entropy per site is In 2. This corresponds to the fact that the number of states per site is two. 8.4.2 High-temperature expansion or small J expansion For the XXX chain system we can perform the high-temperature expansion of the free energy density from the definition, f/T = -N-1InTrexp(---Y/T). (8.51) This is expanded as a power series of 1 / T. Assume that VP = o +1 and 1 commute with each other. Then the exponential of ye' can be o and expanded as follows,

exp(-. / T) = exp(-,*o/ T) (1

-

T-'-'Y' + T-2

2 1

-+

(8.52)

Thus,

f/T =-N-1 InTrexp(-,o/T)+ (NT1) +

i)

- 3(

where (X)

-

(

2)(°1) +201)3 -+ 3!NT3

_ Tr exp(-.YPo/ T)X Trexp(-moo/T)

2)

jye,

2!NT2 (8.53)

(8.54)

8.4 Some special cases and expansions

125

If we set

Si ,1 = -J

o = -2h l=1

N

S1 S1+1 + Si S +1 + Si Si+1,

(8.55)

l=1

for the Hamiltonian (3.1), we obtain the J/T expansion of the free energy at fixed h/T :

f/T = -ln(2cosh h/T)- 4T tanh2(h/T) 2

32T2

(3 + 2 tanh2(h/ T)

- 3 tanh4(h/ T)) + 0((J/T )3).

(8.56)

The calculation to higher orders can be done by the use of a linked cluster expansion. Higher order terms are polynomials of tanh(h/T). Apparently the expression for the free energy in (8.50) coincides with the first term of the above expansion. Writing ln(rln + 1) as the expansion l n ( rln ( x

)+1)=1n(a 1-1) +Y00f (')n ( JT ) n l=1

sinh2(h(n + 1)/T) sinh(hn/T) sinh(h(n + 2)/T)'

an =

(8.57)

we obtain an expansion of inrln(x): In rfn ( x ) = In

an

In

1

+

Tan fn

(1))2 2

(anfn2) +(anOC n)(f

2 )

+o ((T)

3

).

(8 . 58)

Substituting these expansions into (8.42), (8.43), (8.44) and (8.45) and taking first order terms in J/T, linear integral equations for fn1>(x) are obtained, al f 11 (x) - s * f 2 (x) = 2ts(x), anfnl)(x) - s * (fnl)1(x) +fn+1(x)) anfn(x) = 0 lim

n-*oo

n

(8.59)

= 0,

.

(8.60)

(8 . 61)

The r.h.s. of these equations are the inhomogeneous terms of the integral equations. The Fourier transform of these equations is (ebwl

+ e +0I)anfnl)(w) = n-1(CO) +fn+1(w).

(8.62)

Thermodynamics of the XXX chain

126

The general solution of this difference equation is

f"'(0)) = sinh((n + 2)h/ T) nlwl A(te) Isinh((n + 1)h/T) sinh((n + 2)h/T)enlwl +B(w) I sinh((n + 1)h/T)

- sinh(nh/T) e (n+2)1.1 sinh((n + 1)h/T) - sinh(nh/T)

I

e(n+2)Iw1

sinh((n + 1)h/T)

I

(8.63)

From the boundary conditions we have 7r

A(``') = cosh h/T'

B(co) = 0.

(8.64)

Thus,

f(1) ( 1

)

f1 (x)

[sinh 3h/T elwl cosh(h/ T) sinh 2h/ T R sinh 3h/ T f cosh(h/T) Lsinh2h/Ta1(x) it

- sinh h/T

e_31w11

sinh 2h/ T sinh h/ T

'

sinh2h/Ta3(xW].

(8.65)

Substituting this into (8.31) we obtain the second term of the J/T expansion (8.56). The higher order terms can be calculated by solving the linear integral equations f o r f (n2), f ( 3 ) , .... The equations are similar to (8.59), (8.60) and (8.61) except for the inhomogeneous terms, which are given by lower order fn(l). 8.4.3 The low-temperature limit At low temperature In nn diverges as I/ T. So we should define the following functions :

en(x) = T In qn(x)

(8.66)

The integral equations become

el(x) = 2icJs(x)+s * TIn(1

+exp(E2(x))1

T

en(x) = s * T ln(1 +exp\enTx))) (1 + exp(fn lim

n-*oo

T1

En(x) = 2h. n

(8.67)

The free energy expression becomes

f = - (4 + h) - T

f an(x) ln(1 + exp(-sn(x)/ T))dx n=1

8.4 Some special cases and expansions

J (ln 2 - 4)

-TJ

127

s(x) ln(1 + exp(ei(x)/T))dx.

(8.68)

The T = 0 limit of these equations is el (x) = 27rJs(x) + s * e2 (x), en(x) = s * (en--1(x) + n+1(x)),

lim en(x) = 2h,

(8.69)

n

n->oo

j + h) +

f = (4

00

f

an(x)en(x)dx

n=1

= J (In 2 - 4)

- Js(x)e(x)dx. for en(x) > 0 for en(x) < 0.

for en(x) > 0

(8.70)

for en(x) < 0.

In the ferromagnetic case J > 0 we have en(x) =

21rJan(x) + 2hn,

n=1,2, ...,

(8.71)

and therefore f = -(4 + h). This is the ground state energy of the ferromagnetic case.

In the antiferromagnetic case J < 0 we have en(x) = en (x) = an-1 * e1 (x) + 2(n - 1)h,

n = 2,3 . ....

(8.72)

el(±B) = 0.

(8.73)

The equation which determines e1 is

e1(x) = -2irIJIs(x) + h +

f

R(x - y)el(y)dy,

YI>B

In the limit of h -* 0, B becomes infinite. We have e1(x) = -27rlJIs(x) and f = J(ln 2 - These results coincide with those of chapter 3. 4).

8.4.4 The fugacity expansion

In the case of very large h the free energy can be expanded as a series in z = exp(-h/T). z is called the fugacity. From equations (8.22) and (8.27) we have expansions of nn1 as follows : 1

nn

=

z2n

exp

(- 21rJ an(x)) exp [-Tnm * (nm 1 - 211-2 + -...) ]

.

(8.74)

Thermodynamics of the XXX chain

128

The expansion of qi 1 and X121 up to z4 is -1

III

= z2exp (-

T 27rJ

a1(x)) 7rJ

x (1- z2 f a2(x - y) exp(-2T ai(Y))dy) + 0(z6), 27rJ

1121 = z4 exp(_ T a2(x)) + 0(z6).

(8.75)

As nn 1 becomes small, (8.29) is more convenient than (8.31),

f

1 dxan(x) (qn1

h-T

211n +

n=1

4 Substituting (8.75) we obtain

f = -4

- h - z2T fai(x)exp(_ai(x))dx -z4T { f a2(x) exp a2(x)) - I ai(x) exp (-4T al(x) dx - dx dyal(x) exp V-2T (al(x) + al (Y))) a2(x - Y)} + O(z6). (_27rJ

1TJ

(8.76)

Putting x = tan u, y = tan v, we have

(f + 4 + h) /T = zee KIo(K) +

+ e-K/27o(K/2)

" e2K(1-c0Swi c0Sw2)(1 - COS C01 COS w2)

1 7r2

z4{-Ze-2KIo(2K)

0

+O(z6),

0

1 - 2 cos wl COS Cot + COS2 0)1

dcoldw2} (8.77)

where K = J/T, wl = 7 + u + v, w2 = u - v and Io(x) is a modified Bessel function. This result is the same as that of Katsura48. The strings are stable in the chain of infinite length with the condition that there are only a few down-spins. Some counter examples of the string assumption are found in some special limits. Nevertheless the thermodynamic Bethe-ansatz equation seems to give the exact free energy even in the case

when the density of down-spins is comparable to the density of up-spins. This non-linear integral equation contains an infinite number of unknown functions. To solve this equation one needs to do numerical calculations by computer. For the conventional analysis we can use the high-temperature expansion method or the exact diagonalization method. For the investigation

8.4 Some special cases and expansions

129

of the low-temperature thermodynamics of solvable models this method is the only way. For the spin 1/2 ferromagnetic XXX chain, the susceptibility T-' and the specific heat behaves as T. By numerical calculation diverges as using thermodynamic Bethe-ansatz equations, it was established that y = 2

and a =

1/2109,81

9

Thermodynamics of the XXZ model

9.1 Thermodynamic equations for the XXZ model for A > 1 Gaudin derived a set of thermodynamic Bethe-ansatz equations for A > 130 The wave function for M down-spins in an infinite lattice becomes

M M .f (ni, n2,..., nM) = (ziz2...zM)ni E A(P) fJ(11 P j=2 l=j e' kj = sin (x j + l)

zj =

zP1)n'+1-nj,

sin2(xj-i)

(9.1)

This corresponds to (8.1). From the normalizability condition of the wave function we have

A(I) * 0,A(P

I) = 0,

IZ1Z2...ZMI = 1

M I

ft zl I < 1, j = 1, 2,3,..., M - 1.

(9.2)

l=j+i

These conditions are satisfied only if

xj=a+(M+1-2j)i, j=1,2,...,M, -Q
(9.3)

We can show that 7M

zt

sin 02 (a - iM)

(cosh /(M - 2j) - cos Oa < 1, 1 < j < M - 1,

v

cosh On - cos 4

for arbitrary M. Thus a string with arbitrary length is possible for the XXZ chain for IAI > 1. In the case JAI < 1 the string condition is more 130

9.1 Thermodynamic equations for the XXZ model for A > 1

131

complicated than for the case IAI > 1. From (4.11) we have the following equation corresponding to (8.8): n

en (xa) = II

eN(xa,j)

=

j=1

Enm(xa - x'). fl (m,13)(n,a)

(9.4)

Here sin ±(x + in)

en(x)

sin 22 (x - in)

e1_1

,

(x)ejn-m1+2(x)ejn-mi+4(x)...en+m-2(x)en+m(x)

for

Enm(x) =

n

m, n = m.

for

e2

2

xn is the real part of the a-th string for strings of length n. The logarithm of (9.4) is NOn(xa) = 27LIa +

O" nm(xa (m,J3)*(n,0)

where

(tan) 2 +271 9n(x) = 2 tan-1

- xv ),

x¢

no

tanh

+ 7rl, L

27r

[OX

J

and

Onm(x)

=

Bin-mi(x) + 20jn-1+2(x) +... + 20n+m-2(x) + On+m(x) for n in,

262(x) + 264(x) +

...

+ 202 _2(x) + 82n(x)

for

n = m.

The function 6n(x) is a quasi-periodic function which satisfies 0n(x + 2jQ) = 6n(x) + 27rj, j = integer.

We consider the energy of the general eigenstates which is given by the set of quantum numbers {Ia }, E({Ia }) = N

It -

JA)

+ E (27rJ inh c5 an(xn) + 2hn),

(9.9)

n,a

where an(x)

=

1

0 sinh no

27r cosh no

- cos Ox

(9.10)

Thermodynamics of the XXZ model

132

The partition function of the XXZ model is as follows: [N/2]

_

I+ (1 - 6N,2M) exp-

(N - 2M)h1 exp[-T- E({Ia})]. n

T

M=O

{ia }

(9.11)

Corresponding to (8.16) we define the following functions: hn(x) = Bn(x) -

(9.12)

XT). (m,a)

Using this function we can determine the position of holes for n-strings. We set distribution functions of particles and holes of n-strings as pn(x) and pn(x). By the equation (9.7) we have conditions for these two kinds of functions, (9.13)

+ E T nm * Pm(x)

an(x) = Pn(x) + Pn(x

m

Here an was defined in (9.10) and a,n-m, (x) + 2aln-mi+2(x) + 2a,n-mi+4(x) + ...

+2an+m-

I

Tnm(x)

-

for

(x) + a n+m (x)

2a2(x) + 2a4(x) +

... +

2a2n-2(x) + a2n(x)

n

for

m

I

(9.14)

n = M.

Here the meaning of a convolution of two periodic functions a and b with periodicity 2Q is redefined : Q

a * b(x) =

a(x - y)b(y)dy.

(9.15)

Q

The energy per site is 00

e = - (J4 + h) + n=1

I

Q

gn(x)Pn(x)dx, Q

2irJ sinh 4

(9.16)

an(x) + 2nh.

gn(x) =

The entropy per site s is 00

S=E n=1

Q

Pn(x)In(I +

Pn(x)) +pn(x)ln( I Pn(x)

+ Ph(x))dx.

(9.17)

Pn(x)

The condition of minimizing the free energy e - Ts yields equations for Ox) = Pn(x)lp.(x), lnqn(x) -

gn(x) 1

+

Tnm * ln(1 + rimi(x)).

(9.18)

9.2 Theory for the I A I < 1 XXZ model

133

This set of equations is equivalent to the following: 2ttJ sinh In rl1(x) _

s(x) + s * ln(1 + 112(x)),

(9.19)

Inrln(x) = s * ln(1 + q,-1(x))(1 + qn+1(x)), lim

(9.20)

lnrln-2h n

n-*co

(9.21)

T'

where

s(x) = 4

E

2nQ)

sech(7r(x

n=-oo

(9.22)

2-

The free energy per site is

f = -( JA + h) - T

a

0°

Q

f an(x) ln(1 +

gn1(x))dx.

(9.23)

Q

n=1

Corresponding to (8.31) we have another expression for the free energy:

f= J

27r s

h0

- T f QQ s(x) ln(1 + 111(x))dx. (9.24)

Q a1(x)s(x)dx --

4 Equations (8.42)-(8.44) and (9.19)-(9.21) have almost the same structure. 93 These equations are called the Gaudin-Takahashi equations3o,

9.2 Theory f o r the J A I < 1 XXZ model

9.2.1 String solution of the infinite XXZ model with I A I < 1 The shapes of strings for JAS < 1 is quite different from those for Al J> 1. Takahashi and Suzuki108 proposed a condition on the strings and constructed thermodynamic integral equations. Later Hida37 and Fowler and Zotos27

derived these conditions from the normalizability condition of the string wave function for an infinite chain. For JAI < 1 there are two kinds of strings, one of which has its centre on the real axis and the other is centred on the poi axis, xj = a + (n + 1 - 2j)i,

(9.25)

xj = a + (n + 1 - 2j)i + poi.

(9.26)

We designate that the string of the former type has parity v = 1 and that the latter has parity v = -1. Applying the normalizability condition to the form (9.25) yields sinh 2Y (a + i(n -

n

11

1=j+1

1

2j))

sinh Z(a - in) 2

I

V

cosh ya - cos y(n - 2j) cosh ya - cos yn

Thermodynamics of the XXZ model

134

Thus cos yn < cos y(n 1>1

n

11 zt =

2j) for j = 1, 2, 3,..., n - 1. For (9.26) we have

cosh (a + i(n -

2j)) = cosh(a - in) 2 2

I=j+1

cosh ya + cos y(n - 2j) cosh ya + cos yn

and therefore cos yn > cos y(n - 2j) for j = 1,2, 3, ..., n - 1. Then from the normalizability condition we get

0
(9.27)

This is equivalent to exp(iri([(n

- j)y/ir] + [jy/ir])) = v, JY

JY

7r

7E

for j = 1, 2,..., n - 1,

(9.28)

for j=1 , 2 ,

(9.29)

n - 1

where [x] denotes the maximum integer less than or equal to x (Gauss' symbol). For rational po = 7r/y, (9.29) is a strong condition. If po = nl/n2 and n1 and n2 are co-prime, a string greater than nl cannot satisfy at least one of (9.29). Thus n >- n1 + 1 strings are forbidden. Moreover for a n = n1 string, the momentum is always 0 or it. So this string also has no meaning in thermodynamics. Next we seek the number n and parity v which satisfies (9.28) within n < n1. Equation (9.28) is equivalent to

[(n j)Y] +

]

[

= [(n- j- 1)Y] + [(j +1)Y]

7r

7r

(Mod 2),

7C

j = 1,2,...,n-2,

[(n-1)yl = 1-v 7r

As [(n

J

2

(Mod 2).

- j)y/7r] - [(n - j - 1)y/7r] is 0 or 1 and [jy/ir] - [(j + 1)y/n] is 0 or

-1, we obtain [(n _j)Y]

+ [jy ] = [(n- i- 1)Y] + [(j +1)y], j = 1,2,...,n-2. 7E

7E

7E

These are strong restrictions on the parity v and the length n. The above conditions are equivalent to the following conditions which were given in reference108. The length n of a string should satisfy n-1

2 E[jy/ir] = (n - 1) [(n - 1)Y/it],

(9.30)

j=1

v sin{(n - 1)y} > 0.

(9.31)

9.2 Theory for the I A I < 1 XXZ model

135

This condition was first introduced by the author108 under the assumption that these strings form a complete half-filled state. Later Hida37 and Fowler and Zotos27 showed that conditions (9.30) and (9.31) can be re-derived from the normalizability condition of the wave function for N -> oo and finite M. For a given value of A (or y) we can determine the series of n which satisfies the condition. This series becomes finite if y/7r is a rational number, and in this case the number of unknown functions becomes finite. We consider the y = 7r/v case. For n = 1, both v = 1 and -1 are possible. For n = 2, 3,..., v -1 only v = +1 states are possible. These excitations have the following energy and momentum: sin y sin(ny) E = -2J v cosh(ay) 2nh, - cos ny +

K = -i In 1 _

sinh 2 ((xy + i(1- v)ir/2 + iny)

sinh z(ay + i(1 - v)ir/2 - iny)J

(9.32) (9.33)

The energy and momentum have the following relation:

E = -J sin y

cos ny - cos K + 2nh. sin ny

(9.34)

The momentum is restricted to the region

IKI<7r-ny-7[ny,) for v=1, 7r > JK J > 7r - ny - 7c ny ]) for v I

7E

Then for n = v the energy and momentum are always zero. Only one state is obtained from this string solution. So we exclude this n = v state from the thermodynamics of this case. So v string states (1, +), (2, +),..., (v -1, +), (1, -)

play important roles. Especially at A = 0, y = 7r/2, v = 2 we have only the string states (1, +), (1, -). These are single states at momentum IK I < 7r/2 and IK I > 7r/2.

Next we consider the y = it/(v1 + 1/v2) case. The following set of strings satisfy the condition (9.30) and (9.31), (1, +), (2, +),..., (vi - 1, +), (1, -), (1 + vi, +), (1 + 2vi, -), ..., (1

+ (V2 - 1)V1,

(-1)v2-1),

(U1, (-1)v2).

Then v1 + v2 strings are necessary to describe the thermodynamics of this case.

For a general rational number between 0 and 1, we can express it by a

Thermodynamics of the XXZ model

136

Table 9.1. Length nj, parity vj and qj of strings for some rational values of y/71

j

5

1

1

2

2

3

3

4

4

5

1

-

3 16

1

+ + + +

-

4

1

3

2

2

3

1

4

-1

1

+ + + +

6

6

7

11

+ -

8

5

+

1

13

5+

69

13/3 10/3 7/3 4/3

-3/3 -2/3 -1/3 1/3

1

2 3

4 1

6 11 5

9

21

10 11

37 53

12

16

-

5+ 1

3+

+ + + +

56/13 43/13 30/13 17/13

+ -

-13/13 -9/13 -5/13

+ + -

4/13 3/13 2/13 1/13

-

-1/13

continued fraction with length 1:

=

11

IVi

11

11

+ IV2 + ... + IV, VI, V2, ..., vi-1 > 1, vi > 2.

(9.35)

We define the following series of numbers y-1, yo, Y1, , Yi and mo, ml,..., mi as

Y-1 = 0, yo = 1, Y1 = v1 and Yi = Yi-2 + viyi-1, i

mo=0, mi=

V

(9.36)

k=1

The general rule to determine the parity v and length n is as follows: nj = yi-1 + (j - mi)Yi, vj = (-1)[(ni-1)/Po] for mi < j < mi+1, (9.37) nm, = Yi-1, vm, = (-1)i.

The number of strings is mi. We give examples for some rational numbers in Tables (9.1) and (9.2). We put x as the real part of strings with parity vj and length nj. a takes values from 1 to Mj. We find the following relations for these series of numbers:

nj = [(1 - 26m,,.i)n.i-1 + nj+11, for mi < j< mi+1 - 2, 2 < 1, nj = (1 - 25m,-,,j)nj-i + nj+1, for j = mi no = 0, nm, + nm1-1 = Yi.

(9.38)

9.2 Theory for the JAI < 1 XXZ model

(a) y = it/5 PO =

(b) y = 47r/5 PO

5

(c) y=37r/16

4

_ 16 PO

3

(e) y = 137r/69 PO

5

(d) y=137r/16

_ 16 PO

137

_ 69

(f) y = 56ir/69 Po

13

13

_ 69 56

_ Fig. 9.1. Strings on the complex plane for given values of po = it/y. If we change y to it - y, the strings are almost the same. But imaginary parts of yxj shift by it.

Thermodynamics of the XXZ model

138

Table 9.2. Length nj, parity vj and qj of strings for conjugate values of y/ir in previous table 4

1

1

2

2

3

3

4

4

5

1

13 16

1

5

1

+ + +

-4/4 -3/4 -2/4 -1/4

1

2 3

+ -

6

6

+ + +

7

11

+

8

5

4

1/4

1

4+

+ -

3

-13/13 -10/13 -7/13 -4/13 3/13 2/13 1/13

-

_

56 69

1

1+

-1/13

9 10

1

2 3

+ + + +

4 1

6 11

+ -

5

21

37 53

11 12

+ +

16

1

1+--T 4+-Zs+a

-56/56 -43/56 -30/56 -17/56 13/56 9/56 5/56

-4/56 -3/56 -2/56 -1/56 1/56

9.2.2 Scattering phase shift among strings Corresponding to (8.8) and (9.4) we have a Bethe-ansatz equation for strings as follows : MI Mk

=-H

{ej(xa)}N

II Ep (xa

k=1/3=1

- xk).

(9.39)

where ej(x) = g(x; nj, vj),

(9.40)

nj-1

for nj = nk,

g(x;2nj,vjvk) JJ g2(x;21,vjvk) l=1

(9.41)

EJ,k( ) = J g(x; (nj + nk), vjvk)g(x; I nj - nk 1, vjvk) min(nj,nk)-1

x

fl

g2(x;Inj-nkI +21,vjvk) for nj

nk,

1=1

g(x; n, +)

sinh (x + in) in)' g(x' n' -)

= sinh 2v x

-

_

cosh 2 (x + in)

cosh v2(x - in))

(9.42)

The logarithm of (9.39) yields MI Mk

NOj(xa) = 2irI3 +

O jk(xa k=1 #=1

- 4),

a = 1, 2, ..., Mj,

(9.43)

9.2 Theory for the JAI < 1 XXZ model

139

where

O(x) = f (x; nj, vj), Ojk(x) = f(x; l nj - nkl,vjvk) + min(nj,nk)-1 f(x;nj + nk,Vjvk) + 2

f(x;Inj - nkI + 2i,VjVk),

(9.44)

i=1

and

f (x; n, v) = {

for ny/1r = integer, 0 2v tan-' {(cot(ny/2))v tanh(yx/2)} otherwise.

('9 45

)

The quantity Ii is an integer (or half-odd integer) for Mj odd (or even), which is located in the region MI

III <

N9 j(oo) - > Mk0 jk(oo) 27z

.

(9.46)

k=1

The function f (x; n j, v j) is a monotonically increasing function for m2i < j < m2i+1 and a monotonically decreasing function for m2ti_1 < j < m2i.

9.2.3 Bethe-ansatz equation for the XXZ model with I A I < 1 Following Yang and Yang we define particles and holes of strings. We obtain an integral relation between the distribution functions p j and p,4 of particles and holes of strings in the thermodynamic limit, mi

Tj,k * Pk(x).

(9.47)

(29r)-1 a -OJ,k(x).

(9.48)

aj(x) = sign(qj)(pj(x) + Po(x)) + k=1

Here

aj(x) =

(21r)-l d Oj(x),

Tj,k(x) =

The symbol a * b denotes the convolution of a(x) and b(x) as follows: a * b(x) =

00

a(x - y)b(y)dy.

(9.49)

The functions aj(x) and their Fourier transforms a#)) are written as

a j(x) =

1

ysinygj

21r cosh yx + cos yq j' aj(w)

_

sinhgjco sinh pouw'

qj = (-1)`(pi - (j - mi)pi+1), for mi < j < mi+1,

(9.50) (9.51)

where the pi, i = 0,1, ..., l are defined by: PO = 7z/Y,

P1 = 1, Vi = (Pi-1/Pi], Pi = Pi-2 - Pi-M-1-

(9.52)

Thermodynamics of the XXZ model

140

For the series q j we find the following relations :

qj =

1

2

[(

< - J-< m i+1 - 2,

1-28

qj = (1 - 26m,_,,j)gj-i + qj+1, for j = mi - 1, i < 1, qo = po, nmt +

(9.53)

nmt-1 = 0.

The Fourier transform of Tj,k(x) is given by T1,k(w) = Tk,j((O) 2sign(gj)coth(pi+1(9)sinh((po - lgjl)o))sinh(gkc))

=

sinh(pow)

+6j,m,-16k,m, - 6j,k,

for j< k, mi < j< mi+1.

(9.54)

At j = 1 we have Tl,k((O) = sign(g1)2 cosh coak(c)) - 81,k. + 82,02,m,.

(9.55)

The last term appear only at A = 0, y = 7r/2, po = 2. At j = ml we have Tm,,k(w) = -Tm,-l,k(w),

sinh((po - 2pi)w)

l

Tm,,mr(w) = -Tmi_1,mi(c)) =

sinh(po(o)

(9.56)

The energy and entropy per site are given by e=

-(J4 + h) + E j=1

2irJ sin y gj(x) - -

'

f

gj(x)Pj(x)dx, (9.57)

aj(x) + 2njh, y

and MI

h

s=E foo j=1

(9.58)

Pj

Pi

To minimize the free energy density e - Ts with respect to p j, we have

6(e - Ts) _

f

gj(x)Spj(x) - T {Spj ln(1 +

Pj

6p ln(1 + Pdx. PM

The variation of (9.47) gives Spy

= -6pj - sign(qj) E Tjk * bpk.

9.2 Theory for the Al I< 1 XXZ model

141

Thus we get

6(e - Ts) = T

f dxbpj(x) J

-In(PL) + p.i

sign(gk)TJ,k * In(1+ h)}. Pk

k

At the thermodynamic equilibrium one obtains the following non-linear equations determining qj(x) = po(x)/po(x): m,

In gj(x) = gj(x)/T + E sign(gk) Tk,j* ln(1 + qk 1(x)), j = 1, ..., mi. (9.59) k=1

The free energy is given as follows: m,

f

=e-Ts=-(4 +h)+Ef J=1

00

pj(x)[gj(x) - T lnrh(x)] - T(pj+

ii 1)dx.

(9.60)

If we substitute (9.59) into the first bracket and (9.47) into the second, the Tj,k terms are cancelled and we get JO

m

,

f = -(4 + h) - T

aj(x) ln(1 + g-1(x))dx.

(9.61)

j=1

If one uses the j = 1 case of equation (9.59) and (9.55) one obtains

f =_ JA - sign(qt) 27rJ sin y Y

/ a1(x)st f

(x)dx

-T f sI(x)[ln(1 + q1(x)) + 62,m, ln(1 + g21(x))]dx. (9.62)

The second term in the bracket is non-zero only at A = 0. From equations (9.50)-(9.52) we get the following relations: aj - si * ((1 - 28m,-,,.i)a.i-1 + aj+1) = 0

for mi_t < j < mi - 2, am _1 - (1 - 26mj-i,m,-1)Si * amt-2 - di * amt-1 - si+1 * amt = 0 for i < 1, am,-1(x) = -am,(x) = Si * am,-2,

(9.63)

Thermodynamics of the XXZ model

142

where

-

ao(x) _ 6(x),

Si(x) = di(x)

dco

e'wx

7zx _ -sech-, 4pi 2pi -co 41L cosh(pico)

L00 do) eicox cosh((pi

- pi+1)w)

47r cosh(piw) cosh(pi+i(o)

(9.64)

Using (9.54) one can show the following relations: TJ,k - si * ((1 - 26mi-i,j)TJ-1,k + Tj+1,k)

_ (-1)i+1(6j-1,k + 6j+1,k)Si for mi-1 < j < mi - 2, Tm,-l,k - (1 - 2Vmi-i,mi-1)Si * Tmi-2,k

- di *

Tmi-1,k

-Si+1 * Tmi,k = (-1)i+l(6mi-2,kSi + bmi-1,kdi - 6mi,kSi+1)

for i = 1, 2,..., l - 1, Tmi-1,k = -Tm,,k = SI * Tmi-2,k + sign(gk)Sm,-2,ks1,

(9.65)

with TO,k = 0. Using (9.63) and (9.65) one can rewrite (9.47) as follows: * h h Pj + PJh = Si (PJ-1 + Pj+1) for mi_1 <_ j < mi - 2,

PM;-1

+Pmi-1 = Si * Pmi-2 + di * pmi-1 - Si+1 * Pmi,

h Pmi-1 + Pmi-1 = Pm, + Pm, = sl * Pm ,-1

(9.66)

with po = 6(x). Equations (9.59) are rewritten as 21rJ sin y

In qj

= (1-

26mi-,,j)Si

* ln(1 + 11j-1) + si * ln(1 + 11j+1),

for mi-1 < j < mi - 2, j

ml - 2

In 11mi-1 = (1 - 26m,_i,m,_1)si * In(1 + 11m,-2)

+di * ln(1 + nm,-1) + si+i * ln(1 +'im), for i < l lnhlm,-2 = (1 - 2bm,_i,m,-2)si * ln(1 + 11m,-3) +Sl * ln((1 + 11m,-1)(1 + 11m,1)), 1n 11m,-1

- ylh/T = ylh/T - ln11m, = sl * ln(1 + 11m,-2)

(9.67)

Then if we write In K(x) = In rlm,-l -y1h/T we obtain integral equations with

9.3 Some special limits

143

MI-1 unknown functions -27ryTny6(x),

ln(1 +rio)

_

In qj = (1

- 28mj_,j)si * ln(1 + mj_i) + si * ln(1 + rid+1)

for mi-1 < j < mi - 2, j * mi - 2, In fm,-1 = (1 -

1)Si * ln(1 + '1m;-2)

+di * ln(1 + nm,-1) + si+1 * ln(1 + qm,) for i < 1, In elm,-2 = (1 - 26,,,,m,-2)Si * In(1 + hm,-3) +si * ln(1 + 2 cosh(yih/T)ic + K2), In K(x) = si * ln(1 + fm,-2).

(9.68)

9.3 Some special limits

9.3.1 The T -* oo or J

O limit

In equations (9.68), In(1+io) becomes zero and the rlj(x) are all independent of x. Thus we get the following difference equation: qj-i)i-2am_,,;(I rid _ (1 + + i1j+i) for m1-1 < j < mi - 2, j mi - 2,

11m,- = (1 + for i < 1, '1m,-2 K2

= (1 +

l1m,-2)1-2Sm;_,,mj_i

gym,-3)1-2sm,

(1 + F1m,-i) ln(1 + qm,)

+ 2 cosh(yih/T )K + K2),

= (1 + i1m,-2)

(9.69)

The solution of this set of equations is

(sinh(ni + yi-1)h/T')2 sinh(yi-1h/T) for mti-1 < j < mi, j < mi - 2,

17i

K = sinh(nm,_2 + yi-1)h/T sinh(y1_1h/T)

(9.70)

For j = 1 we have ij = (2 cosh h/T )2 - 1. Substituting this into (9.62) we find the free energy,

f/T = -ln(2coshh/T). At h = 0 this gives that the entropy per site is In 2, as it should be.

(9.71)

Thermodynamics of the XXZ model

144

9.3.2 The case J>0, We define ej(x) = T In fi(x) and e (x) = T ln(1 + i (x)). One can show that e j, j * 1 is always positive. The equation (9.59) gives ei(x)

27rJ sin y

a(x, 1) + 2h - Ja(x - y, 2)ei (y)dy.

(9.72)

Y

If e1(x) < 0 at lxi < B and el(x) > 0 at lxi > B, then one obtains a linear integral equation for p1(x), B

P1(x) + fB a(x - y, 2)Pi(y)dy = a(x, 1).

(9.73)

This is equivalent to the Fredholm type equation obtained in (4.40).

9.3.3 The caseJ>0, 0
a(x, 1) + 2h - f a(x - y, 2)ev2+1(y)dy.

(9.74)

Y

If e(x)v2+1 < 0 at lxi < B and ev2+1(x) > 0 at jxl > B, then one obtains a linear integral equation for pv2+1(x), 1B

Pv2+1(x) +

B

a(x - y, 2)Pv2+1(y)dy = a(x, 1).

(9.75)

This is equivalent to the Fredholm type equation obtained in (4.40). 9.3.4 The A = 0 case

Here we have y = ir/2, v1 = 2, 1 = 1 and m = 2. Then from the first and last equations of (9.67) we have

In q1(x) = 2h + 4Js1(x)/T, In 112(x) = 2h - 4Jsi(x)/T.

(9.76)

Substituting these into (9.61) or (9.62) we get the free energy, which coincides with (1.19).

10

Thermodynamics of the XYZ model

10.1 Bethe-ansatz equation for the XYZ model

The low temperature behaviour of the XYZ model using thermodynamic Bethe-ansatz equations was investigated by the author ...... In the Betheansatz equation (4.15), the function sinh(yx/2) is merely replaced by the elliptic theta functions. We should consider equation (5.54),

i)) N HI(iC(xi - i)))

N12

-e

-27riv'/po

H1(iC(xl

11 Hi(il;(xi

- xj + 2i))

- xj - 2i))'

N/2

po = K(l)/C.

xi = Qv' + ipov, Q =

(10.1)

In the limit l -* 0 it becomes

(sin(i(xI + i)) N = sin(iC(xi i)))

-

-e_27riv'/po

1

j_1

sin(il;(xi

- xj + 2i)) xj - 2i))

(10.2)

This is an equivalent equation to (4.15), if we assume that the xis are finite

and v' = 0. C becomes y/2 and Kl becomes it/2. So it is natural that we assume the same type of strings which can be determined by po =

xj = a+(n+l-2j)i, xj = a+(n+1-2j)i+poi, -Q
(10.3)

The solution becomes doubly periodic. So we consider the distribution of solutions for -Q < 91x < Q and -po < sx < po. It is expected that the same kind of strings appear in the case of the XXZ model at 7r/y = po. We 145

Thermodynamics of the XYZ model

146

can determine the njs and qjs as (9.38) and (9.51), MI Mk

= -exp(-2iriv'/po) JJ [J Ej,k(xa

{ej(x,

k=1#=1

-

mt Mj

1

(10.4)

njxa, j=1 a=1

where ej(x) = g(x; nj, vj), H1(i?(x + in)) g(x; n, +) = in))' in)) in)) nj-1

-H1(Kj +

g(x;n,-) =

(10.5)

111(Ki +

(10.6)

g(x;2nj,vjvk) JJ g2(x;21,vjvk) for nj = nk, i=1

Ej,k(x) = j

g(x; (nj + nk), vjvk)g(x; Inj min(nj,nk)-1

- nkl, vjvk)

g2(x;Inj-nkl+2l,vjvk) for nj

x

nk.

1=1

(10.7)

Taking the logarithm of (10.4) we have NOj(xa) = 2icIa

- 2irv'/po +

m/ Mk

Oj,k(xa - xk), o' = 1, 2, ..., M1,

(10.8)

k=1 /3=1

where

8j(x) = f(x;nj,vj), Ojk(x) = f(x; lnj

- nkl,vjvk) +

min(n j,nk)-1

f(x;nj+nk,Vjvk)+2

f(x;Inj-nkl +2i,vjvk),

(10.9)

i=1

and f (x, n, v) is defined by 00

f(x,n,v) = f (x, n, v) + E f (x

- 21Q, n, v) + f (x + 21Q, n, v).

l=1

f (x, n, v) was defined in (9.45). An eigenstate should be identified by the set

10.1 Bethe-ansatz equation for the XYZ model

of quantum numbers

147

From (5.74) the energy must be

E _ -NJZR -

J

Z

N12

2Isn2t;

) ; a(xi,1) 1=1

=-NJR- JZ1LSri2 r 1

R - 41 1 -

j

x

a),

j=1 a=1

(a(0,1) + a(Q,1))],

[qj

1

aj(x) -

nsn2

MI Mi

sinh(gj21l/Q)

cos(7rjx/Q)], +2 2Q p0sinh(po7rl/Q)

(10.10)

where qj was defined in (9.51). The number of zeros must be N/2, mj

N/2 = > njMj.

(10.11)

j=1

Thus the energy per site is given by

e= -JzR -

m'

JZicsn2

Q

j=1

f Qaj(x)pj(x)dx.

(10.12)

The entropy per site is Q

h

Q

Pi

I=1

Pi

(10.13)

From (10.9) we have the relation between p3(x) and p (x) : mj

aj(x) = sign(qj)(pj(x) + Po(x)) + ETJ,k * Pk(x).

(10.14)

k=1

Moreover from (10.11), 1

m'

2 = m - > nj j=1

rQ

J

pj(x)dx.

(10.15)

Q

Next we need a Lagrange multiplier to guarantee the condition (10.15). One should minimize e - Ts + 2hm under conditions (10.14), and after that the multiplier h should be chosen so that (10.15) is satisfied. Just in the same way as (9.59) was derived, we get the following integral equations for qj(x) =

lnrlj(x) = gj(x)/T +>sign(gk)Tk,j * ln(1 +11k 1(x)), j = 1,...,m1. (10.16) k=1

Thermodynamics of the XYZ model

148

Here *, gj(x) and Tjk are Q

f * g(x) = I f (x - y)g(y)dy, Q

gj(x) - -

JZirsn2C

aj(x) + 2njh,

00

TJ,,k (x )

2Q

>

e ianx/Q 7

T j,k (x - 2nQ).

j,k ZQ) =

n=-,

n

The quantity g - e - Ts + 2hm is given as follows: g(JZ, T, h) = -JZR Q

MI

+1: fpj[gj() - T Ingj(x)] - T[pj + j=1

ln(1 +q 1)dx

Q

m

,

_ -JzR - T > sign(gj) j=1

aj(x)ln(1 + q 1(x))dx.

(10.17)

Q

Corresponding to (9.62) this is sign(g1)irJssn2

g = -JR + h -

Q

a1(x)s1(x)dx

Q

-T f Q s1(x)[ln(1 + 111(x)) + 62,,n, ln(1 + g21(x))]dx.

(10.18)

Q

Then m should be determined by M

2 Oh

2

2

s1()(1

+111(x))-1

Ohx)dx.

(10.19)

Q

The equation (10.16) is also equivalent to the following block tridiagonal equations : ln(1 + 11o) _

-7Jzsn2C6(x)

CT In q j = (1- 28m,_,, j)si * ln(1 + qj_1) + sti * ln(1 + q j+i ),

for mi_1 < j<mi-2,j=ml-2 In qm,-1 = (1 - 26m,_1,m,-1)si * ln(1 + qm,-2)

+di * ln(1 + qm,-1) + sj+1 * ln(1 + qm,), for i < I

In q,,,-2 = (1- 26,,j_,,,,-2)S1 * ln(1 + qm,-3) +sl * ln(1 + 2 cosh(ylh/T )1c + k2), In K = SI * ln(1 + qm,-2).

(10.20)

In this equation the parameter h appears only in the cosh(ylh/T) term.

10.2 Some special limits

149

So rll(x,h) is an even function of h and alh=o = 0. Using (10.19) we find m = 1/2 and the condition (10.15) is satisfied at It = 0. Thus equation (10.20) becomes irJZ sn2

ln(1 In rl1 = (1

-

b(x)5

CT 2bm,-1,1)Si * ln(1 + ill-i) + si * ln(1 + gl+i )

for mi_1 < j<mi-2,jrml-2, In nm,-1

= (1 - 2bm,-i,m,-1)Si * ln(1 + 11m,-2)

+di * ln(1 +'1m;-i) + si+1 * ln(1 + rlm;) for i < 1, In 11m,-2

= (1 - 2bm,_j,m,-2)Sl * ln(1 + 11m,-3)

+2s1 * ln(1 + 1c),

In i = s1 * ln(1 + 11m,-2).

(10.21)

Corresponding to (9.62) the free energy is

f = -J,R - sign(g1)

irJZ s sn2

Q

a1(x)s1(x)dx Q

Q s1(x)[ln(1 + 111(x)) + b2,m, ln(1 + n21(x))]dx. -T J-Q

(10.22)

We can calculate the free energy of the XYZ model in a zero external field.

10.2 Some special limits

10.2.1 The T

oo or J -* 0 limit

In equations (10.21), ln(1 +rlo) becomes zero and the r1j(x) are all independent of x. Thus we get the following difference equation: >7 = (1

+11j-1)1-2ami_1,J(1

+11j+1)

for mi-1 rlm;-1 = (1 for i < 1,

rim,-2 = (1 + KK2 = (1 + 1lm,-2).

+11m;-2)1-26 m._1,mi_1(1

+11m,-1)ln(1 + fm,)

1lm,-3)1-2Sm!_1,m, z(1

+ 2K + K2),

The solution of this set of equations is

(10.23)

Thermodynamics of the XYZ model

150

_

-`

(nj+yi-1 l2

Yi-1

1

for mi-i <j < mi, j< ml-2, K =

nm,-2 + Yi

(10.24)

Y1

We have rl1 = 3 for po 2 and K = Ill = n2 = 1 for po = 2. Substituting this into (10.22) we find the free energy,

f/T = -1n2.

(10.25)

This gives that the entropy per site is In 2, as it should be.

10.2.2 The JZ > 0, Jx < 0 and T -f 0 limit We define e3(x) = T In rl j(x) and e (x) = T ln(1 + 11 j (x)). One can show that e j, j * 1 is always positive. The equation (10.20) gives 61(x)

_

-7rJzsn2C

a(x,1) -

Q

a(x - Y, 2)e1(Y)dy.

(10.26)

ej(x) = 0, j > 1.

(10.27)

Solving this we have e1(x) _

7rJsn2Cs1(x),

Substituting this into (10.22) we have

f = -JzR - icJZsn2C

j/'Q a1(x)s1(x)dx.

(10.28)

Q

This is just the ground state energy given in (5.80).

10.2.3 The JZ > 0, JX > 0 and T

0 limit

Here we have 2 > po > 1 and v1 = 1. One can show that ej, j

V2 + 1 is

always positive. The equation (10.20) gives -,Jzsn2C eve+1(x) =

a(x,1) -

Q

a(x - Y,

2)evz+1(Y)dy.

(10.29)

Solving this we have eve+1(x)

7rJ'sn2

s2(x), e3(x) = 0,

j > V2 + 1.

(10.30)

10.2 Some special limits

151

10.2.4 The Jx = 0 case Here we have y = it/2, vl = 2, 1 = 1 and m = 2. Then from the first and last equations of (9.67) we have

lntli(x) = 4Jsi(x)/T, lnriz(x) = -4Jsi(x)/T.

(10.31)

Substituting these into (9.61) or (9.62) we get the free energy, which coincides with (1.19) at h = 0.

11

Low-temperature thermodynamics

11.1 The XXZ model

11.1.1 The XXZ model at 2h > J(1 - A), A < 1 Johnson and McCoy investigated the low-temperature thermodynamics at A > 1 using the thermodynamic Bethe-ansatz equations45. This paper did not treat the thermodynamics of the XXZ model at JAI < 1 or the XYZ model. In this case the ground state is a completely ordered state. The elementary excitations are bound states of down-spins. One down-spin state has an energy momentum relation as follows :

el (K) = 2h + J(0 - cos K).

The lowest energy state is K = 0. Then in this case the system has the energy gap Eg = 2h - J(1 - A) and the low-temperature specific heat behaves as T-3/2e Eg/T

The magnetization m = -Of /8h behaves as

1-

O(T-1/2e-Eg/T)

The system is almost completely magnetized.

11.1.2 The case J(1 - A)/2 > h > 0

At A < -1 and h < he the system has an energy gap and magnetization begins at h = hc. Here h, is obtained by (4.35). Thus we can give the following phase diagram for the low-temperature behaviour of the XXZ model in magnetic field. The shaded region is the gapless phase, and 152

11.2 Roger's dilogarithm and specific heat at h = 0

153

generally speaking the low-temperature specific heat is proportional to the temperature. The solution of the thermodynamic Bethe-ansatz equation is as follows: e1(x) = -As(x) + s * e2 (x),

en(x) = an-1 * el (x) + 2(n - 1)h.

Only the n = 1 solution is relevant to the low-temperature thermodynamics. The other excitation has a finite energy gap and its contribution to the thermodynamics is exponentially small. Inside the shaded region, the specific heat is proportional to the temperature T. The coefficient is inversely proportional to the velocity of excitations. On the boundary of the gapless and gapfull regions the velocity becomes zero and the specific heat is proportional to T 1/2. This is because the dispersion of excitations is

11.2 Roger's dilogarithm and specific heat at h = 0 11.2.1 Specific heat of the XXX antiferromagnet

Consider the entropy density at low temperature. As pn(x) = p,(-x), we can write the entropy density as follows : 00

s = 2 E or pn(x) ln(1 + in(x)) + pn(x) ln(1 + in 1(x))dx.

(11.1)

n=1

From equations (8.42), (8.43) and (8.44) we find that the integral equation for rln(x) at low temperature is In 11i(x) =

- 1' -nxl2 e + s * ln(1 + 112(0,

lnrln(x) = s * ln(1 + nn-1(x))(1 + 1n+1(x)), lim

n-aco

In 11n

n

-=0.

(11.2)

For x -+ oo, 11n = (n + 1)2 - 1, and for x _- 0, Y1n _- n2 - 1. Pn(X) is given by

the differential of fn(x) with respect to J, Pn(x) = 27r(1 + 1n(x))11n(x) 3J

Differentiating equation (11.2) with respect to J, we have the linear integral

Low-temperature thermodynamics

154

equations,

-

1 Oiii

R -7cx/2

qi aJ

Te

1 011;

= S*

ni OJ

li

a,i2

1

+s 1+r12 0J' x11;_1

1

+

aq;+1 OJ

1

1 +'i;+i

1 + t,_1 OJ

} (11.3)

n1 hn aJ =0.

On the other hand, differentiation with respect to x yields similar integral equations, 1 1 aI1 = n21 _7rx/2 * 012 e s 2T + 1 + 112 Ox 171 ax

=

1 aq;

ni

Ox

lim

1

*

1

s

1 + hli_1

aqn

n->oo n11n ax

1 ag;+1 ai,_1 + ax 1 + nj+1 ax

= 0.

(11.4)

Thus we find in

2 Onn

aJ

ItJ ax

Then equation (11.1) becomes:

s=

2T

2

Jo

1

12-1

7L2J

21

n+1)2 -1

00

Y(1 + Y)

1 Y( 1

+

I n + y) +

1

1+Y

In(1+

ln(1+Y)+1+ ln(1+dy. Y)

Y

l)dy Y

(11.5)

Y

Putting u = 1/(1 + y) we have

s= 2T /i 1 In ic2JJo u

1-u + 1-u In 1u du 1

1

1 1

nJJ0 U In 1ludu 4T 1 1 2J1+22+32+42+... 1

2T

3J

(11.6)

Thus the low-temperature specific heat per site is given by Os CTOT

j

2T

We find that the low-temperature specific heat is proportional to the temperature.

11.3 The ferromagnetic chain and modified spin-wave theory

155

11.3 The ferromagnetic chain and modified spin-wave theory

11.3.1 Numerical analysis of the thermodynamic Bethe-ansatz equation The low-temperature behaviour of the XYZ model was analyzed by means

of the thermodynamic Bethe-ansatz equations. The XXZ case is gapless for JAI < 1 and the low-temperature specific heat is proportional to the temperature96 h

oT

y = cos-1(-0),

3J siny,

0
(11.7)

For A < -1 and A > 1 the system has a gap and the specific heat behaves as exp(-a/T). The coefficient diverges as A -* 1, namely at the ferromagnetic XXX point. Yang and Yang 119-122 calculated the longitudinal magnetic susceptibility using the Bethe-ansatz method,

_ T1 ox

4y

(11.8)

J(7r-y)7rsiny

This also diverges in the limit A 1. The next problems are the low-temperature behaviours of the magnetic

susceptibility and specific heat of the ferromagnetic XXX model. In the analytical calculation of the Bethe-ansatz equations, the simplest XXX case remains unknown. This problem was investigated by many authors. Let us

assume that C ' T-" and X - T. Baker et al.4 estimated y - 1.6 using the series expansions from high temperature. Bonner and Fisher16 estimated

that a _ -0.4 and y _- 1.8. Quantum Monte Carlo calculation yielded y _- 1.3 - 1.75

But the numerical calculation of the thermodynamic

Bethe-ansatz equation gave the final answer that a = -0.5 and y = 2.0 81, 82,109,116 From the data at T > 0.002J, the following low-temperature expansions were obtained109

f = J{-1.042(T)3/2 + 1.00()2 X=

-

J-01{.1667(j)2

-0.9(T)5/2

-

+O(T)II

-

+ 0.581 - (T)3/2 + 0.68 (T)1 + O (j)1/2} (11.9)

11.3.2 Spin-wave calculation of the ID ferromagnetic chain For a long time it was believed that the spin-wave theory is not applicable for one- or two-dimensional Heisenberg models where there is no long range order at finite temperatures. But the first term of the free energy in equation (11.9) is the same as that of the conventional spin-wave theory, although the

Low-temperature thermodynamics

156

second term is completely different. This means that the spin-wave theory is valid in some sense. In the conventional spin-wave theory the chemical potential of the spin-wave is proportional to the external magnetic field. In zero magnetic field there appear infrared divergences in the calculation of physical quantities. If we assume that the chemical potential should be determined by the condition of zero magnetization, we can suppress this divergence and the calculation of physical quantities becomes possible. Using the mean field type approximation, the author found the following expansion of the free : energy102,103

f

T[L

(2) ( T ) 1/2 (27x)1/2 \2SJ

T + 4S2J T 3/2 2 1 '(2) + o(T )], 8 (2,r)) (2SJ)

C(2)

1

+(2S2 (27x)1/2

x=

8

3S

4

JT- 2 2

1

+S2 _27r)

i(2) ( T ) 1/2

- S (27x)1/2 2SJ 3

[1

(2SJT

+ 0(T3/2)].

(11.10)

At S = 1/2 these become

f = T[-1.0421869 (7 )1/2 + (T /J) - 1.2320919 (7) x = T2 [6 +0.5825974(7)1/2+0.6788396(7)

3/2

+ O(T2)] , +O(T3/2)].

In the third term of the free energy the spin-wave result is apparently different from thermodynamic Bethe-ansatz calculations, equation (11.9).

This means that the spin-wave theory works to give correct results for the first few terms of the low-temperature expansion. Yamada115 gave the low-temperature expansion of the correlation length by the Bethe-ansatz method for the quantum transfer-matrix. The low-temperature expansion also coincides with the spin-wave result. The thermodynamic Bethe-ansatz method was used for the analysis of the quasi one-dimensional Heisenberg ferromagnetl10

11.4 The antiferromagnetic XXX model

At A = 1-, y = 0+, and we have x = 4/(Jir2). The low-temperature behaviour of the susceptibility shows a strong logarithmic singularity. The

11.5 The XYZ model at IJx I < Jy < JZ

157

Jx

0.7 r

0.6

0.5

0.4

0.3

0.2

0.1

0.5

0

1

1.5

T/J 2

Fig. 11.1. Magnetic susceptibility of the antiferromagnetic Heisenberg chain as a function of temperature. There is a logarithmic anomaly at low temperature.

susceptibility has an infinite slope. This point has been found by the Betheansatz calculation of the quantum transfer matrix23. Recently this singularity 69 Such a logarithmic anomaly is found has been experimentally also in the antiferromagnetic ladder lattice with three legs using the quantum Monte Carlo verified89,

calculations28.

11.5 The XYZ model at JJYJ < Jy < JZ

In the case Jy < JZ the ground state of the Hamiltonian (5.41) has a gap. Then the free energy at low temperature is written as follows:

f (T) - f (0) = T-112 exp(-Of/T ).

(11.12)

It was pointed out that there are two gaps for the free energy at Jx > Ospinon

=

096,

(11.13)

Low-temperature thermodynamics

158

Jy/Jz

Jx/Jz

Fig. 11.2. The plane of J,,/J, and Jy/JZ and the energy gap. In region X the spinon gap given by equation (11.13) is dominant in free energy. In region Y the free energy has a spin-wave gap given by equation (11.14).

Aspin-wave =

Jsn(2 C, l)Kuu'

Sri

(Ku(K1 - C), u,)

(11.14)

where 1, C and u are determined by 1 = 11 J2 z

- J2 ,

0 < < K1,

Jx/JZ =

x

Ku

= Kl

.

(11.15)

U1

We call these the spinon gap and the spin-wave gap. These are the minimal energies of elementary excitations. For Jx < 0 we have a spinon gap (5.92) only. As shown in Fig. 11.2, the spinon gap is lower than the spin-wave gap in a wide region X. But near the point F(1,1) there is a region Y where the spin-wave gap which was defined by (5.102) becomes lower. So the free energy at low temperature at Jx > 0 is represented by these two gaps, as was shown in Ref. (xcvi). On the line Jx = Jy (line FI in Fig. 11.2) we have Ospinon =

JZ

- Jy /2,

Ospin-wave = Jz - Jy

The crossover of the gap in free energy occurs at Jy/JZ = 0.6. On the other

hand, at Jz = Jy (line AF-F in Fig. 11.2) we have 1 = 0,1' = 1, K1 = 00 and g' = 0. So the gaps become zero. The free energy behaves as f(T) = f(0) cT2 except at the point F. At the point F (Jx = Jy = Jz) we have

-

f(T)=f(0)-cTis

Part three Finite temperature integral equations for nested systems

12 S

= 1/2 fermions with repulsive potential in the continuum

12.1 Derivation of the thermodynamic equations 56 and Takahashi94 derived the thermodynamic Bethe-ansatz equation for delta-function fermions. Recapping the equations (2.97) and (2.98), Lai55,

M kj a=1

N Aa

- Aa+ic' a

1

-ic'/'

- kj+ic'll

7=1,...,N,

(12.1)

M7rAa-AQ+ic

11'\Aa-kj-ic')

ic

a = 1, ..., M.

(12.2)

If we have one solution of this equation we can construct an eigenfunction 'P(xisl, x2S2, ..., XNSN)

'E

of the Hamiltonian (2.1)

n

i=1

02

ax?

i<j

with S = SZ = 2 - M. We can generate N - 2M + 1 states by the application

of the spin descending operator S-. The energy eigenvalue is N 1 kJ2. Conjecture 1. If a set of solutions (k1, k2, ..., kN ; A1, A2, ..., AM) of (12.1) and (12.2) contains a complex k (or A), its complex conjugate k (or A) is in the set of ks or (As). This means that the distributions of ks and As are symmetric with respect to the real axis. We can prove that for a symmetric state, the ks must be real for the repulsive case (c > 0). If -3kj > 0 the absolute value of the l.h.s. of (12.1) is less than unity, but the absolute value of the r.h.s. is greater than

unity. Then 3k j > 0 must be excluded. If sk j < 0 the absolute value of 159

160

S = 1/2 fermions with repulsive potential in the continuum

the l.h.s. is greater than unity, but the absolute value of the r.h.s. becomes less than unity. Then we have Zk j = 0 for c > 0. For the c < 0 case we have strings with pairs of ks. In the thermodynamic Bethe-ansatz equations the treatment of the ks at c > 0 becomes very simple. Next we consider the distribution of As on the complex plane. The equation (12.2) resembles the equation for the XXX model if we read 2Aj/c as xj in (3.17). In the limit c -+ co, the kjs remain finite and the two equations become equivalent. It is very natural to think that we have string states as in §8.2. We write the As as AaJ Aa'3 = Aa + (n + 1

- 2j)c'i + deviation,

j = 1, 2, ..., n.

(12.3)

Assume that there are Mn strings. The equation (12.1) becomes 00 Mn

exp(ik jL) = fJ fi e (

-

kj

An

nc/

n=1 a=1

), j = 1, 2,..., N.

(12.4)

We consider the product fN1

1

e(A" - kjlJ

(12.5)

net

j=1

By (12.3) this is transformed as N

n

Annj

11rje j=1 1=1

- kj

(12.6)

C'

Substituting (12.2) we have 11e(Aal -A

(12.6) = 11{- II 1=1

e 1=1

C

(m,li) h=1

(An"-Am'h

(m,f)r(n,a) h=1

(12.7)

C

Using (12.3) we have N

H e(

j=1

An_k nC

t

An-Am

ao M

) =-H

Enm(

m=1fl=1

a

C

), a = l, 2,...,M,, n > 1,

(12.8)

where Enm(X) is defined in (8.9). The logarithm of (12.4) and (12.8) is Mn °° kjL=27rIj-EE0(k j - Aa ), j=1,2,...,N, net

n=1 a=1

(12.9)

12.1 Derivation of the thermodynamic equations

EB( AancI K J 1 = 27rJa +

Mm

161

/Ana-Am/

l

j=1

C/

a = 1,2,...,Mn, n > 1,

(12.10)

Here 0(x) = 2 tan-1 x, -7L < 9 < it and the function Onm(x) was defined in (8.12). Ij are distinct integers (half-odd integers) for even (odd) M1 +M2+...,

Ij - M1 + M2 +... (mod 1).

(12.11)

The Ja are distinct integers (half-odd integers) for odd (even) N - Mn, which satisfy

Ja -N-Mn+ 1 (mod 1), 1 Ja < 1(N - 1 - > tnmMm), tnm = 2 min(n, m) - Snm.

(12.12)

Given a set of integers {Ij,Ja} which satisfies (12.11) and (12.12), we can determine a set of kj and An through equations (12.9) and (12.10). For a set of integers {Ij,Ja}, there is a set of omitted integers which satisfy (12.11) and (12.12) but is not contained in {Ij, Ja }. We define holes of k and holes of An as solutions of

Lh(k) = 2t x (omitted I), Ljn(An) = 27z x (omitted Jn), where 00 M,

1

h(k)-k+L EEB(kncAa)

(12.13)

n=1 a=1

Jn(A)

1N

(

A-kj1 °°

Mm

= j, E e \ nc) - L M=I fl=1 E),,,

A-Am c,

(12.14)

We consider a very large system. We put the distribution functions of ks and Ans as p(k) and CA), and those of holes as ph(k) and 6n(k). By the definition of holes we have: h(k) = 2n(p(k) + ph(k)),

(12.15)

dkln(k) = 2ir(6n(k) + an(k)).

(12.16)

dk

S = 1/2 fermions with repulsive potential in the continuum

162

Equations (12.13) and (12.14) are written as follows : 0(kk')6

h(k)=k+

nc

n

i

0(k

n (k)

_ k') P

Hereafter we denote f

n (k')dk' ,

fO

(k')dk' -

(12.17)

(k

nm

c

k') am (k')dk

(12 . 18)

dk as f dk. Substituting (12.17) and (12.18) into

equations (12.15) and (12.16) we have 1

_

2 = P(k) + P (k) -

an * 6n(k),

h

(12.19)

n=1 00

an * p(k) = 6n(k) + 6n(k) +

Tnm * am(k),

(12.20)

m=1

where an is a function defined by an(k) =

njc'j I7G (nc')2 + k2

,

(12.21)

ao(k) = 6(k),

and

aln_m1(k) + 2aln-.1+2(k) + 2aln-mi+4(k) +... +2an+m-2(k) + an+m(k) for n * m, Tnm(k) 2a2(k) + 2a4(k) +... + 2a2n_2(k) + a2n(k)

for

n = m.

The grand partition function of this system at magnetic field h and chemical potential A is written as follows, [N121

=

sink (N-2M+1)h

N=O M=O

sink 1,

exp (- T (EN,M,{r,Jn} - AN)), {j

T

Jn}

(12.22)

where EN,M,{r,Jn} are eigenvalues of the Hamiltonian

_n

i=1

a2

axi2

i<j

with total spin N/2 - M. In other words the state belongs to the representation By the steepest decent method the Gibbs free energy at a given temperature T, chemical potential A and magnetic field h is given by [2M1N-2M]

G(T, A, h) = -T In E = E(p) - AN(p) -h(N(p) - 2M(un)) - TS(p, Ph, 6n, 6n).

(12.23)

12.1 Derivation of the thermodynamic equations

163

We have L =

L =

f k2p(k)dk,

f p(k)dk,

00

M

L =

(12.24)

n J 6n(k)dk, n=1

S

L

= 00

+ n=1

f {(p + ph) In(p + ph) - p In p _ ph In ph}dk f {(an + an)1n(an + an) - an In an - an In an }dk. (12.25)

Thus the free energy is a functional of functions {p, ph, an, It must be minimized under the conditions (12.19) and (12.20). At the thermodynamic equilibrium the variation of G must be zero, an}.

00

0 = bG/L = f (k2 - A - h)b p(k)dk +

2nh f ban(k)dk n=1

ph)

-T f {bpln(p Pph) +bphln(p

}dk

p

{banin(an+an)+banin(an

-TE f

an

n= 1

hIn)}dk. a

(12.26)

From equations (12.19) and (12.20) we have 00

00

bph=-by+Ean*ban, bah=an*bp-ba-ETnm*bam. n=1

m=1

Substituting these into equation (12.26) we have

_ Jlk2-A-h TL T bG

+

00

n=1

f

{ 2nh

T

Co

h

6n) } b pdk

In(

p

n=1

_ In ( 6 n ) - an * In (1 + an

00

P) p

h

+ E Tnrn * ln(1 + 6n') }6andk.

(12.27)

am

m=1

Thus we have a set of coupled non-linear integral equations for C(k) ph(k)/p(k) and 114) = an(k)/an(k) as follows: In C(k) =

k2

-A I

an*

h n=1

(12.28)

S = 1/2 fermions with repulsive potential in the continuum

164

In qn(k) =

2T - an * ln(1 + -1(k))

00

+

Tnm * ln(1 + qm1(k)), n = 1,2,....

(12.29)

m=1

Equations (12.19) and (12.20) are rewritten as

(1 + (k))p(k) = 27 + E an * un(k),

(12.30)

n=1 00

an * p(k) = (1 + 11n(k))6n(k) + E Tnm * 6m(k).

(12.31)

m=1

From the thermodynamics the pressure is given by

p = -G/L.

(12.32)

Using equations (12.23), (12.25), (12.27) and (12.29) one obtains

p = T f ln(1 + C-'(k)) dk.

(12.33)

This is the same expression as the one for free energy of repulsive bosons in (7.20). The equations (12.27) can be written as follows: k2

In

-A

- R * ln(1 + -1)

- s * ln(1 + qi

T lnrii(k) = s * ln((1 + 112)/(1 + C-1)),

),

In fin(k) = s * ln(1 + qn-1)(1 + t1n+1), lim In

n- oo

where

4

n = T,

(12.35) (12.36) (12.37)

n

s(x) = c, sech - , R(x) = a1 * s(x) _ 2c'

(12. 34)

f

a1(x - y)s(y)dy.

(12.38)

oo

To derive these we should note the following relations: s*(an-1+an+i) = an, s*(Tn-l,m+Tn+l,m) = Tnm-(Sn-1,m+5n+i,m)S. (12.39)

12.2 Some special limits

12.2.1 c -* 0+ At first we consider the limit c -- 0+. Here the width of the functions s(x) and R(x) becomes zero and we can replace them by 1125(x). Then equations

12.2 Some special limits

165

(12.34)-(12.37) are written as follows :

In (k) =

k2

_ TA

- - ln (1 + -1(k))(1 + rh (k)) ,

(12 . 40)

In qi(k) = 2 ln((1 + rl2(k))/(1 + -1(k))),

(12.41)

In rln(k) = 1 ln(1 + in-1(k))(1 + qn+1(k)),

(12.42)

lim

In 11n(k)

n

- 7,2h

( 12 . 43 )

Equation (12.42) is a difference equation. The general solution is

(anb -anb-1 2

71n =

a-a 1

-1.

(12.44)

a and b are arbitrary parameters which should be determined by other conditions. From (12.43) we have a = exp(Ihl/T). Then the solution of (12.42) and (12.43) is written as qn (k)

sinh f (k) + nh/T 12 ) sinh h/T

_

- 1.

(12 . 45)

Substituting this into (12.41) we have 1

1 + -1

_ (sinh f (k)) 2

(12.46)

h/T

Thus equation (12.40) becomes In

sinh f (k)

sinh(h/ T - f (k))

-

k2 -A

(12 .47)

T

Solving this equation we have 1 1 + exp[(k2 - A + h)/T] f (k) = 2 1n ( 1 + exp[(k2 - A - h)/ T] )

1+

(12 . 48)

u1 = (1 + e (k2-A+h)VT)(1 + e (k2-A-h)/T),

p = T f dk ln((1 +

e-(k2-A+h)/T)(1

+e

(12.49)

(k2-A-h)/T))

(12.50)

The last equation coincides with (2.56) at h = 0. Thus we can reproduce the known results for the c = 0 case.

166

S = 1/2 fermions with repulsive potential in the continuum 12.2.2 c -+ oo

In this case the rlj(k) are independent of k, because s(x) and R(k) are infinitely wide functions. Then equations (12.35)-(12.37) become In q, = 2 ln(1 + r72), In rh = 2 ln(1 + r1j-i)(1 + qj+1), lim Inn n

n->c'

=T

(12.51)

The solution of this set of equations is hn

_ rsinh(n+1)h/T\2 1. l sinh h/ T

(12.52)

From equation (12.34) we have k2-A

(k) = e T

1

ehIT

+ ehlT

p = T f 21r ln(1 +2cosh(h/T)e

(12.53)

13 S

= 1/2 fermions with an attractive potential

13.1 Derivation of the thermodynamic equations

In the ground state, each A is accompanied by two complex ks. This is regarded as a pair of up-spin and down-spin electrons,

ka=Aa+ilc'I,

a=1,2,...,M'.

k2=Aa-iIc'I,

At finite temperature there appear As which are not accompanied by complex ks. These As form strings similar to those in the repulsive case, Aa>j

= An + (n + 1 - 2j) I

j=

I,

1, 2, ..., n,

a = 1, 2, ..., Mn.

Thus an eigenstate is characterized by N - 2M' real ks, M' real A', and Mn An. We have M = M' + E' 1 Mn. From equation (12.1) we have (A' a -An

(A' -A'

expi(ka+ka)L= He(

21c'1')In

/f#a

xe ka

m,/3,j

e(

)

21c'I

(13.1)

leAa)e(ka leA,,

From equation (12.2) we have 1

r

2

r

e(ka l-A,)e(ka IcA"

N-2M'

_n

Ar - An, j e(a21c'l

r

e( A"

j=1

(13.2)

m,(i,j

Substituting this into equation (13.1) we have N-2M'

_

j=1

-1c' I

exp(2iAaL)

k

167

)

M''

r

11 a=1 e`

A' - Ar

a 2 c'

).

(13.3)

S = 1/2 fermions with an attractive potential

168

Here we have represented an unpaired k as k j. From equation (12.1) we have M/

00 Mn

1

exp(ikjL) = 11 e(kj - A«) a=1 -ICI And from equation (12.2), N-2M'

11 e( j=1

-

e(kj - Aa)

fj 11 n=1 a=1

oo Mm

Aa - k 1) '

Aa - Am

_-H

HEnm( m=1/3=1

nIC I

(13.4)

nl c I

).

IC

(13.5)

The logarithms of equations (13.3)-(13.5) are N-2M'

2A«L=27rJa+j=1E 0(

1. _ kj)+E0A' M' a-A (

-IC,

(kj

kjL=27rIj+E0(kj-Aa)+00

-ICI

a=1

)

N-2M' / An - k , eI a ' \ nI C I j=1

cO Mm

(13.6)

--na(13.7) nIC

n=1 a=1

An -Am

enm( a

= 21CJa +

),

-21c'I

/3=1

I

(13.8)

)

' CI

m=1 /3=1

Here Ja is an integer (half-odd integer) for N - M' odd (even), I j is an integer (half-odd integer) for M+ M1 + M2 +... even (odd). Ja is an integer (half-odd integer) for N - Mn odd (even) and satisfies: 00

k1

IS m=1

We define functions j'(A'), h(k), jtt(An), i

i

'

1

j(A)=2A -L{

N-20

A'O(-A'i

j=1

oo Mn

_

_A/

(13.9)

-2Ic'I An

na)+Ee(k n lc' ) n=1 a=1

l

(13.10)

c

a=1

in(Att) = L{

)+Efl=1of

j=1

M'

N-2M'

kjMA'

An

_ k l)

e( nlcil

-

co

Mm

> onm(

m=1/3=1

0I.

An - Am Ic'I

(13.11)

Holes of A', k and An are defined as solutions of

j'(A') = 2ir x (omitted J'), h(k) = 27r x (omitted I), jn(An) = 27r x (omitted Jn).

(13.12)

13.1 Derivation of the thermodynamic equations

169

In the limit of a very large system we define the distribution functions of A', k, A' as a'(k), p(k), an(k) and those of the holes as a'h(k), ph(k) and an (k). Using the relations:

dj'(k) dk

d_) d_n(k)

dk

= 2ir(a'(k) + a'h(k)), = 2ir(P'(k) + P h(k)), = 27r(an(k) + an(k)),

we have equations for a', a'h, p, ph, an, an : 1

it II

=a'+a'h+a2*a +al*P, = p+ph+al * a'+Ean * an, n

=an+an+ETnm*am.

an

m

Here njc'j

an(k) _

,K (k2 + (nc')2)1

ao(k) = b(k),

aln_ml(k) + 2aln-ml+2(k) + 2a,n-ml+4(k) +

+2an+m-2(k) + an+m(k))

for

n

...

m,

Tnm(k) 2a2(k) + 2a4(k) +... + 2a2n-2(k) + a2n(k)

for

n = m. (13.16)

The grand partition function is [N/2]

_

sinh (N-2M+1)h T h

N=O M=O

sinh hT

exp

{rE

AN - EN,M,{y,rjn}

(13.17)

T

The Gibbs free energy is given by

G(T,A,h)=-TIn.=E(p,a')-AN(p,a') -h(N(p, a')

- 2M(a', an)) - TS(a', a'h, P, Ph, an, an),

(13.18)

S = 1/2 fermions with an attractive potential

170

where

E(p, u')/L = f k2p(k) + 2(k2 - ci2)u'(k)dk,

(13.19)

N(p, a')/L = f p(k) + 2o'(k)dk,

(13.20)

M(u', an)/L = f u'(k) +

(13.21)

nun(k)dk, n=1

S(6', uth

APh, an, un)/L = f{(a' + ofh)ln(aI +

- o'lna' - oPhlnaPh}dk

f

+ {(P + ph)ln(p + ph) - pln p - phln ph}dk {(un + an) ln(un + un) - un In un -

+

Ch

n

In

un}dk.

n

(13.22)

By the condition of minimizing the free energy, we have

0 = bG/L = f 2(k2 - ci2)ba'(k) + (k2 - A - h)bp(k)dk 00

+ j 2nh f bun(k)dk n=1

r

-T J {ba ln(

+u'h ut

ut

+ath

) +bu'hln( 0h ) h

+bp In (p

h

p ) }dk

) +bphIn(p Pp

-T

P

00

}b6nln(6n+6n) +b6nln(an h6n)}dk. n=1

(13.23)

an

an

From equations (13.13), (13.14) and (13.15) we have bath

=-but -a2*but- a1*bp, 00

bph=-bp - al *

bu'-Ean*bu" n=1

h

an = an

p-bun-1: Tnm m=1

Substituting these into equation (13.23) we have a set of coupled non-linear

13.2 Some special limits

171

integral equations for ,' = a,h/6', C = ph/p, and rin = 6n/a, as follows: 2(k2

Inti'= In

-

ci2

- A)

T k2

=

-A

+a2*ln(1+'1 ')+al*ln(1+ co

-h+ al * ln (1+ n '-1 )-

T

an

*ln( 1 +hn

1

),

n=1

T + an * ln(1 + C-1) + E Tnm * ln(1 +'m1). 00

In qn = 2

(13.24)

m=1

The pressure p and Gibbs free energy G are given by

p = -G = T J ln(1 +rli-1)dk + T f ln(1

(13.25)

L 7E The equations (13.24) can be written as follows: In rl'

=

2(k2

-Tci2 -A) + a2 * ln(1 + n'-1) + a1 * 1n(1 + C-1),

lns*In(1+,i')-s*ln(l+11), In rl1 = s * (ln(1 + -1) + ln(1 + 0),

n=2,3,...,

In 11n = s * (ln(1 + rln_1) + ln(1 + nn+l )),

lim

Inrln=2h n

(13.26)

.

T From equations (13.13), (13.14) and (13.15) we have a set of linear equations n-*co

for a', p, 6n, 1

I -6'(1+rl')+a2*a'+al*p, 1

= p(1 + C) + al * a' + E an * an,

271

(13.27) (13.28)

n

an *p=(1+'1n)an+ETnm*am.

(13.29)

m

13.2 Some special limits 13.2.1

In this limit we can put al = a2 = 8(x), s =

28(x). Then equation (13.26)

becomes '1' = exp(2(k2T A))(1 + 11'

+ rl1

'

+

+

S = 1/2 fermions with an attractive potential

172

111 =

(1 + C-1)(1 + 112),

hn =

(1 + 11n-1)(1 + rln+l), 2h

In nn lim n-*oo n

=

(13.30)

T

The general solution of the second, third, fourth and fifth equations is nn =

f2 (n)

- 1, -l = f2(0) - 1,

1 1

1

1' = f 2(-1),

bZn - b-1Z-n f(n)

(13.31)

z _Z_1

Here b is a function of k and z =_ e-h/T. Substituting these into the first equation we can determine b(k), 1 + Z-1e(k2-A)/T b(k) = z2

1+

ze(k2-A)/T

Substituting these into (13.25) we have

p=T

dk

ln(1 + e(k2-A+h)/T)(1 + e

(k2-A-h)/T)

(13.32)

At h = 0 this result coincides with the pressure of non-interacting fermions (2.56).

13.2.2 T-+0+ In the low-temperature limit we introduce the following quantities: e'(k) = T In r1'(k), K(k) = T In '(k), en(k) = T In rin(k).

We can show that en(k) > 0. Then in the limit of T determined by:

0, e'(k) and K(k) are

e'(k) = 2(k2 - A - ci2) - a2 * e'-(k) - al * K (k), K(k) = k2 - A - h - al * e'-(k).

(13.33)

Here we used (13.24). From these equations we can show that e' and K are monotonically increasing functions of k2. We define parameters B and Q by e'(B) = 0 and K(Q) = 0. r1' and C are zero in the region [B, -B] and [Q, -Q], respectively, and infinity outside these regions. From equations (13.13)-(13.15) we have a set of linear integral equations in the

13.2 Some special limits

limit T -+ 0,

1 = 6'(k) +

B

J B a2(k - k')a'(k')dk'

+ fQ ai(k - k')p(k')dk', 1

27r

B

= p(k) + LB al(k - k')u'(k')dk'.

These are equivalent to equations (2.120) and (2.121).

173

14

Thermodynamics of the Hubbard model

14.1 Strings of the Hubbard model The author obtained the thermodynamic Bethe-ansatz equation for the Hubbard mode195. It is expected that there are strings of As of arbitrary length and real independent ks by analogy with the repulsive delta fermion problem. But for the thermodynamics we must consider pairs of fermions and bound states of pairs of fermions. A pair of fermions is given by

= 7t - sin-'(A' + U'i), ka = 7t - sin-' (A' - U'i).

k«

(14.1)

Here we take the branch of sin -1 as -7t/2 < 91 sin-1 x < 7t/2. The energy of this state is given by E = -2t(cos k1 + cos k2) = 4t91

1 - (A'06

- U'i)2.

(14.2)

This value of this excitation is always positive, and is of the order of U when U is large. Thus for the low-temperature thermodynamics this excitation is

not relevant. But in the high-temperature limit one cannot get the correct entropy per site (In 4) without these excitations. This excitation resembles the pairs of fermions which appear in attractive delta-fermions. In the Hubbard model there appear bound states of n pairs. This excitation has 2n ks and n As,

Aai=Aa +(n+1-2j)U'i, j=1,2,...,n, ka = it - sin-'(A ' + niU'), ka = sin-'(A + (n - 2)i U'), a ka = it - k2 ka = sin-' (Aa + (n - 4)iU'), ka = it - k4 174

14.1 Strings of the Hubbard model 2n-2 k2n-1

= sin-'(An - (n - 2)iU'), = 7r - k2n-2

a

a

k22n

175

,

r

= 7r - sin-1(Aa - niU').

(14.3)

The energy of this excitation is 2n

E = -2t E cos ka = 491 1- (Aa - nU'i)2.

(14.4)

j=1

The other excitation is strings of n As,

Aai = An + (n + 1 - 2j)U'i.

(14.5)

Essler, Korepin and Schoutens showed that this classification of the Betheansatz eigenstates gives 4N states24. We should construct equations for the M'n Aa s, N - 2 i nMn kjs and Mn Ans. The equations for these are derived from (6.29) and (6.30),

p( )(

ex ik jNa

e

sin k j - An,

(n,a)

nU'

)e

(sin k j - An,

(14.6)

nU'

(n,a)

2n

exp(iNa

ka'1) l=1

= exp(-NQ(sin-1(Aa + in U') + sin-' (A 'n - in U')

--Nl re(A« - sinkj) rl Enm (Aa U nU'

Am

1111

j=1

(m,$)

e(Aa - sinkj)

N

11 j=1

nU

(14.7)

),

= - fi Enm(Aa - Am

r

(14.8)

Ur

(m,fl)

Taking the logarithm of these equations we have 00 M n=1 a=1

kjNQ=27rIj-EEO(sinkj 'Aa)-EE0(sinkj A«)

nU

n=1 a=1

nU

'

(14.9)

NQ(sin 1(A« + in U) +siri 1(Aa - in U')) = 21rJ« N-2M' A_ sinkj) + Onm( Aa A' a U' nU' j=1 (m,/i)

+ E ,(

-

(14.10)

Thermodynamics of the Hubbard model

176

N-20 e(Aa-sinkjl nU'

j=1

)

=21TIn+ E Onm(AaA_l a

U'

(MP

/

(14.11)

Here I j, Ja, Ja should satisfy the following conditions: 00

(E(Mi + M;) even)

integer,

Ij=

i=1

- odd integer,

half Jfn - (

Jn -

0

Mil) odd) i=1

integer, (N - Mn odd) t half odd integer, (N M'n even) integer, (N - Mn odd) { half odd integer, (N - Mn even)

-

-

00

IJa I < 2(Na-N+2M'-EtnmMm), m=1

IJ« I < 2 (N

- 2M' -

00

tnmM'),

(14.12)

m=1

where tnm = 2 min (n, m) - 6nm. A set of integers {I j, Ja, Ja } satisfying this

condition gives an eigenstate IT) with particle number N and down-spin number M. We can generate (Na - N + 1)(N - 2M + 1) states using the rland S- operators defined in (6.10) and (6.14). The grand partition function is given by

E(T, A, h) = E exp(-

E-AN

I j J' J"n

T

e(2A-U)/T + e2(2A-U)/T +... x (I + + x (1 + e 2h/T + e 4h/T +... + e2(N-2M)h/T )

e(N.N)(2A-U)I T

)

- U/2)/T sinh(N - 2M + 1)h/T sinh h/T sinh(A - U/2)/T -N M x E exp(-E - ANa + (Na - N)U/2 + h(N - 2M)) (14.13) _

Na

sinh(Na

[N121

- N + 1)(A

T

I j,j,n,.Ian

The energy E and momentum K of this state is given by N-2M'

E= E (-2t cos k j - h) + E 4t9l 1 - (Aa - in U')2 j=1

(n,a)

14.2 TBA equation for the Hubbard model

177

00

+2h

nMn,

(14.14)

n=1

(14.15) (n,a)

(n,a)

14.2 Thermodynamic Bethe-ansatz equation for the Hubbard model

In the case A < U/2, h > 0 we have

G=-TIn 7=E-AN -TS,

(14.16)

E/Na = L_2tcosk - h)p(k)dk 00

00

+

49R VI

- (A - inU')26n(A)dA + 2h E n

n=1

n=1

J

rn(A)dA,

00

rn

N/Na = ,f p(k)dk + E 2n f an(A)dA, n=1

f{(P+Ph)ln(p+ph)-plnp-phlnph}dk

S/Na

+> .f{(an + an) ln(an + 6n) - an In an - 6n In a }dA n

+

f{(cr, + cr ) ln(6n + 6n) - 6n In 6n - 6n In 6n}dA. n

(14.17)

From equations (14.9)-(14.11) we get 1

2- = p(k) + ph(k) 00

- cos k (E '

f

n=1 J

an(sink n

00

an(sin k - A)(u'n(A) + 6n(A))dA

(14.18)

00

- A)p(k)dk = 6n(A) + an (A)

00

+ m=1 1

i

1

91

(14.19)

Tnm * am(A)5

1 - (A- nUi)2

-

it

an(sin k

- A)p(k)dk

00

= n (A) + 6n (A) +

Tnm * Qm(A) m=1

(14.20)

Thermodynamics of the Hubbard model

178

The condition 6G = 0 gives the equations for l; = ph/p, qn = un/6n and i

1h

On - an law In C(k) =

-2tcosk-h-A T

00

co

1 + r1(A) an(sink-A)1n(1+, 1(A))dA,

n=1

(14.21)

rln

In rln(A) = 7, - f dk cos kan(sin k - A) ln(1 + -1(k)) it 00

+ E Tnm * ln(1 + t1m1(A)),

(14.22)

m=1

In tln(A) =

f

-J

4031 - (A - nU'i)2 - 2nA T dk cos kan(sin k - A) ln(1 + l; -1(k))

77

00

+ Y, Tnm * ln(1 + im 1(A)).

(14.23)

m=1

Substituting (14.18), (14.19), (14.20), (14.21), (14.22) and (14.23) we have an expression for the Gibbs free energy per site,

G/Na = -T

f7r

ln(1 + C-1(k))

dk 27r

00

-T E

91

1

1 - (A - nU'i)2

n=1

ln(1 + gn1(A))-, it

(14.24)

Using the n = 1 equation of (14.23) this is transformed as E0 - A - T {J po(k) ln(1 + ((k))dk g = Na 7E

+f

'00

ao(A)ln(1+n1(A))dA}.

(14.25)

Here E0, po(k) and ao(A) are the energy, p(k) and a1(A) at T = h = U/2-A = 0, (half-filled, zero-field ground state),

Eo/Na = -4t t

s(A - sink) 2-,

6o(A) _ p o(k)

Jo(w)J1(w)dw Icc 1 + exp(2U'(o)' dk

1

I + cos k

27t

f

oo

al (A - sin k)ro(A)dA,

(14.26)

14.3 Some special limits

179

and

s(x) -

I sech

7rx

4U' 2U'* Equations (14.21), (14.22) and (14.23) are transformed as follows: °°

kA(k)

in (k) = T + J

dAs(A - sink) Inc

(14.27)

1 + n1 (A)

+

(14.28)

s(A - sink) ln(1 +(k)) cos kdk,

(14.29)

Inrli(A) = s * ln(1 + 112(A))

-f

7

7r

In ii (A) = s * ln(1 + qz(A))

-f

7

s(A

- sink) ln(1 + C(k)) cos kdk,

(14.30)

7r

In rln(A) = s * In{(1 + q,-i(A))(1 + qn+1(A))}, n>>-2,

(14.31)

In in (n) = s In{(1 + n-1(A))(1 + 7n +1(A))}, n > 2, In in(A) 2h hm n T

(14.32)

U - 2A T

In qn' (A) 11m

n

(14.33) (14.34)

Here xo(k) is defined by

- 4t f dAs(A - sin k)9I 00

KO(k) =- -2t cos k

1- (A

00

7r/2

-J

/2

s(A - sin k) In

( I+

1+C-1(7r-k)

n/2

-J

1(k)

s(A - sin k) In (

1 + (k)

- U'i)2.

(14.35)

coskdk, coskdk.

(14.36)

1 + C(i- k) ) From equation (14.35) we have ic0(k) - k0(t - k) = -4t cos k. Then at ski < 7r/2 we have lnC(k) < lnC(ir k). Thus the last term of (14.29) is negative and that of (14.30) is positive. The equation for n/ resembles that for the antiferromagnetic XXX model, and that for tl1 resembles the 7r/2

-

ferromagnetic XXX model. 14.3 Some special limits oo 14.3.1 The limit U In this limit we get from (14.23) that nn = 00.

Thermodynamics of the Hubbard model

180

From (14.22) we have

2T + O(U-1) +

In i7 (A) =

Tnm * ln(1 + gm1(A)).

(14.37)

m=1

This equation is equivalent to the J = 0 case of the XXX model, and we have rin(A)

=f2 (n)- 1, f(n) = sinh((n + 1)h/T)/ sinh(h/T).

(14.38)

Substituting this into (14.21),

In (k)

-2t cos k - h - A T

-2tcosk -A

T

-

ln(1 + qn 1 ) n=1

- ln(2 cosh h/T ).

(14.39)

Using (14.24),

GIN,,=-T

f

7rln(1

+2cosh(h/T)exp[(2tcosk+A)/T])dk.

(14.40)

The particle number and magnetization are

cosh(h/T)X, N = -OG(TA ,h) = h) 2Sz = -OG( = sinh(h/T)X A'

ah

X = Na f

2e(2tcosk+A)/T

dk

1 +2cosh(h/T)e(2tcosk+A)/T 27r

(14.41)

Thus S, IN = 2 tanh(h/T). This means that the magnetization is the same as for free spins with S = 1/2.

14.3.2 The limit U - 0 p and

are functions of k. But we define p_(A), p+(A), C_(A) and +(A) as

follows :

p_(A) =

1

P+(A) =

1 1

1

A2

p(sin-1 A), _(A) = (sin-1 A), P(ic - sin-' A),

C+(A) = cv(lr - sin- ' A),

A2

(14.42)

14.3 Some special limits

181

where I siri-1 Al < ir/2, JAI < 1. Then the integral equations (14.28)-(14.35) become In C+ = t(2 ±

2)T-1

In'li = 1 In 1 +

1 - A2 +

±

In q, =

1+

+

(14.43)

1+

(14.44)

i2 ln(1+112),

(14.45)

(1

),

2

1

2

In (1 )

1 1

2

lnrl1 =

2

2ln(1 +in-1)(1 +11n+1),

(14.46)

lnqn = 21n(1 +71'n-i)(1 +t1n+1),

(14.47)

- -2h n T In rln - -2A lim In 11n

lim

(14.48)

n-->c

n-3co

(14.49)

T

n

The solution of equations (14.46)-(14.49) is

f(n)

Iln = f2 (n) 11n =

f2(n)

-

1,

azn - a-1z-n z - z-1

f'(n) = bwn - b-1w-n

w-w-

=e-hlT

z 1,1,

= eA'T.

(14.50)

a and b are functions of A at JAI < 1. Substituting this into (14.43)-(14.45) we have

+ = f'(1)lf(1), C- = x-2f'(1)lf(1), 1+ x = exp 1+

1

-,

T (14.51)

From these equations, one can show

f(-1) + f'(-1) = 0, xf(0) = f'(0). We obtain the relation b2

-

a2(w + X -1Z-1) a2(w-1 + x-1z-1)

- (w + x-1z)

- (w-1 + x-1z)

We can eliminate b from this equation and obtain an equation for a2. The

Thermodynamics of the Hubbard model

182

final result is

(1 + x-1zw)(1 + x-1zw-1)

a=

(1 + x-lz-lw)(1 + x-iz-1w-1)' (1 + xzw)(1 + xz-lw) (1 + xzw-1)(1 +

b=

(14.52)

xz-lw-1)

Using equation (14.26) we have

G/Na =

-T

j

--4B - A 1

A2

,7s

ln(1 + -(A)) + 2 ln(1 + g(A))dA 7E

From (14.50), (14.51) and (14.52),

(1 + -(A)) 1 + p1(A) = f(1) + x-2f'(1) = {(1 + x-lwz) (1 + x-1wz-1)(1 + x-lw-lz)(1 + x-1w-1 z-1)}1/2. Thus the thermodynamic potential per site is

G/Na = -T fr

dk

ln(1 + e(2tcosk+A-h)/T)(1 + e(2tcosk+A+h)/T)

(14.53)

This result coincides with the result for free band electrons.

14.3.3 The limit T -+ 0 We write

k(k) = T lnC(k), el(k) = T lnrlj(A), ej'(A) = T lnq'(A).

From the equations we find that el, e2, ... and e2, e3, ... are always positive. Then in the limit T -* 0 these equations become

x(k) = -2t cos k - h - A + fai(sink El(A) = 2h +

- A)ei (A)dA

f dk cos kal (sink - A)x-(k)

- f dA'a2(A - A')ei (A').

(14.54)

Functions a2(A), o3(A)... and 61(A), 62(A), ... are zero. We denote points where

K(k) changes sign by +Q, and points for el(A) by A = +B. The equation for p(k) and a1(A) becomes

p(k) =

1

2n

+ cos k

f

B B

a1(sink - A)a1(A)dA,

14.3 Some special limits

f

B

Q

a, (sink - A)dk = o1(A) + fB a2(A Q

183

- A')dA'.

(14.55)

This is equivalent to the Fredholm-type integral equation in (6.37)-(6.39)

14.3.4 The limit t -+ 0

In this limit the first term on the r.h.s. of (14.28) is a constant, U/(2T). Then C(k) has no k dependence and is a constant. The last terms of equations (14.29) and (14.30) are zero. Thus ni and rl1 are also constants. The integral equations become difference equations which appeared in the thermodynamics of the XXX model. The solutions are

n

- /sinh(n + 1)h/T 12

sinh h/T J - 1' _ sinh[(n + 1)(U/2 - A)/T] 2 - 1 " - ' sinh[(U/2 A)/T] _ U cosh [(U/2 - A)/ T] cosh h/ T 2T + Substituting these into (14.25) we have

l

-

G/(NaT) =

_ -ln[1 +

(U/2T

A)

e(2A-v)VT

(14.56)

-ln[2cosh h/T+2cosh (U/2-A)/T]

+ 2e(A-v12)IT cosh(h/T)].

(14.57)

The low-temperature thermodynamics of the Hubbard model was investigated by the author98. Actual numerical calculation of this set of integral equations was done by Kawakami, Usuki and Okiji49

Part four The quantum transfer matrix and recent developments

15

The transfer matrix and correlation length 15.1 The transfer matrix for the Ising chain We consider the simplest case of a one-dimensional system,

_-JzESfSf 1-2hES1.

(15.1)

l

l

The partition function of this model is

Z = Trexp(-`1) _ T

exp( -alas+1 +Hal) = TrTN, Uj=+1

4

K = J/T, H = h/T.

(15.2)

Here T is the transfer matrix, T

-

exp(K/4 + H) exp(-K/4 + H) exp(-K /4 H) exp(K /4 - H)

-

(15.3)

The partition function is given by

Z = Trexp(- /T) = 21 212 = eK/4(cosh H ±

sinh2 H + eK ).

(15.4)

X1 and 22 are eigenvalues of this transfer matrix. The column corresponds to the state of al and the row corresponds to the state of a1+1. This is an example of the well-known transfer matrix method for classical systems. We can calculate the free energy per site of an infinite system by finding the largest eigenvalue,

f = -T ln(eK14(cosh H + 185

sinh2 H + eK )).

The transfer matrix and correlation length

186

The two-point correlation function (S

Trexp(-/T)Sj'Sj'+n

-

Tr exp(-/ T) N(SZ)21

At

is given by TrTjSZTnSZTN-n-j

TrTN

+ 1 (SZ)22 + (21-n12 + A2 -nA7)I(Sz)I212

(15.5)

AN

where (SZ)ab = (a1 SZ 1 b). In the limit of N -+ oc this becomes (1ISzll)2

(15.6)

+ 1(11S'12)12(22/21)n.

Then the two-point correlation function decays exponentially as A + B The correlation length is given by (ln(/l//2))-1. Generally speaking the largest eigenvalue of the transfer matrix gives the free energy per site and the correlation length is given by the ratio of the second largest eigenvalue and the largest eigenvalue. This method is applicable to the two-dimensional Ising model. 15.2 The transfer matrix for the classical Heisenberg model The classical Heisenberg model is given by the following Hamiltonian: N

= -J Dni ni+i + nn +1 + ni ni+i), 1=1

(n')2 + (nj )2 + (nl )2 = 1.

(15.7)

Nakamura and Fisher investigated this problem 71,25 . The partition function is given by an angular integral over 01 and (pi,

_ Z

47r

exp [T

(15.8)

sin 01 sin 01+1 cos(91 - p1+i) + cos 01 cos 01+1].

Then the partition function is given by: Z = A1 [1 + (221A1)+ (23/)1)N + ...1, where IA11 >- 1221 >- ... are

= Tf

eigenvalues of an integral operator T,

027r

4

(15.9)

f7T

dq,'

d cos 0'K (0, p, 0', (p')f (0', T') = Al.f (0, T),

K(0, cp, 0','p') = exp T [sin 0 sin 0' cos(q - q') + cos 0 cos 0']. (15.10)

15.2 The transfer matrix for the classical Heisenberg model

187

This operator commutes with the angular momentum operators, lx = -i [sin cp ae - cot 0 cos (p

0 ],

(P

ly = i[coscp

a (15.11)

lZ

and the total angular momentum operator 12 = lz + ly + 1

Thus this . operator should be diagonalized by spherical harmonics. The kernel K can be expanded in terms of spherical harmonics, U

00

ji (7,) Ylm(O, (P)YIm(6', P').

K(6, ( P , 0', (P')

(15.12)

m=-l

1=0

Here the spherical Bessel functions jl(z) and spherical harmonics Yim(O,'p) are defined by

1 d lsinz jl(z) = (_1)1z1(__) z Ylm(8, (P) =

eim(p(-1)

( z ) = ( 1 (1- z 2)

m+Im2

(cos0),

ml dlmi

(15.13)

P1 (Z).

2

The Legendre polynomials Pi(z) are defined by dl (z2

1

P1(z) =

nil! dzl

-

1)1.

The largest eigenvalue Al of T is sinh(J/T)/(J/T) and its eigenvector is Yoo. The free energy per site is

f = - In 2 = -T ln(sinh(J/T)/(J/T)). TN

(15.14)

The energy per site is given by

e=

T

-J[coth T

- Ti.

(15.15)

The specific heat per site is I.

C

OT =

-

[sinh(J/T)JIT

(15.16)

188

The transfer matrix and correlation length

The two-point function is given as follows: s(k) = (n'n' nl+k) =

TrT N-k nZTk nZ

(15.17)

TrTN

For k < N this is given by (l,mInz 1O0)12

s(k) =

Al.

k

(15.18)

)

,m

From the recursion relation 1

ZPIm(Z) =

21+1

1P1+1(Z)

+ 21

1

-

(l+1)2-m2

12

4(l + 1)2 - 1 Yl+1,m(0, (P)+V

412

(l, ml nz lO0) _

PI-1(Z),

(15.19)

M2

l Yl-l,m(O, (P) (15.20)

J 3 bm,ool,l,

2io = cosh(J/T)/(J/T) - sinh(J/T)/(J/T)2.

(15.21)

and

s(k) = 3 (coth(J/ T) - (T /J))k.

(15.22)

Thus the magnetic susceptibility per site is given by 00

1

1 + coth(J/T) - (T/J)

x = 1 [s(O) + 2 E s(k)] = T k=1 3T 1- coth(J/T) + (T/J)

(15.23)

16

The spin 1/2 XXZ model in a magnetic field

16.1 The diagonal-to-diagonal transfer matrix

Here we give a set of equations which determines the free energy and correlation length of the S = 1/2 XXZ chain in a magnetic field in the z-direction. The Hamiltonian is given as follows:

t = -J

N

N

{SSA+1 + S

2h

ASS

j=1

S, I=1

J>O,

-oo
(16.1)

The quantum transfer matrix for this model is equivalent to that of the diagonal-to-diagonal transfer matrix of the six-vertex model. This eigenvalue problem is a special case of the inhomogeneous six-vertex model. The XYZ chain in a zero magnetic field is also solvable, N

'Y19ix Sj+1 + JYSY S+1 + JZS151+1},

Jx < Jy < J,

(16.2)

j=1

The thermodynamics of quantum 1D systems is transformed to some kind of classical 2D strip by adding an imaginary time axis. The width of the strip is proportional to the inverse of the temperature. The most popular method is the checker board decomposition. The Monte Carlo method is applicable and one can calculate low-temperature properties of such systems 6,38 Betsuyaku defined the quantum transfer matrix in the real space The numerical treatment of the matrix is useful for the calculation of thermodynamic quantities 14. The transfer matrix can be used for unsolvable models like spin 1 chains". The dimension of the matrix increases exponentially as the Trotter number M increases. The Trotter number is the number of slices along the imaginary time axis. So in this numerical method, it is impossible to treat the very low-temperature .

direction14,15

189

The spin 1 12 XXZ model in a magnetic field

190

showed that the eigenvalue problem of the quantum transfer matrix for the XXZ chain in a magnetic field is solvable by the Bethe-ansatz method. It is equivalent to that of the diagonal-to-diagonal transfer matrix case.

Koma51,52

of the six-vertex model. Usually the six-vertex model is treated by the row-to-row transfer matrix. Bariev5, Truong and Schotte113 had showed that the six-vertex model can be treated also by the diagonal-to-diagonal transfer matrix. Koma calculated the Bethe-ansatz equation numerically and got the free energy and the susceptibility of the ferromagnetic XXX chain. The results coincide completely with those of the thermodynamic integral equation method109 Suzuki, Akutsu and Wadati showed that the XYZ problem is reduced to the diagonal-to-diagonal transfer matrix of the eightvertex model and this can be treated by the inhomogeneous eight-vertex model86.

The partition function Z of the system described by (16.1) is approximated as follows :

Z = Tr{exp(-1/(MT)) exp(-2/(MT))}M, Jy1 =- E J{SjxSjx+1+SYS+1+OS,fSj+1}+h(S+Sj+1), j=odd

Jt'2

E J{SjxSjx+1+SYS+1+ASS+1}+h(S+S+1). j=even

(16.3)

Then the partition function is given by the six-vertex model with 2M x N bonds,

N M

Z=

JJ JJ B(a2i+j,j62i+j+1,j; 62i+j,j+la2i+j+1,j+1) {c} j=1 i=1

Here 61,,,, ; l = 1, ..., 2M, m = 1, ..., N gives the direction of bonds and takes

values +1 and -1. B is given by the 4 x 4 matrix,

B(61o2; 611 62) =

a b

b

0

0

c

0

0

a

0

0 c

a 0 0 0

0 b'

(16.4)

'

(JA

4MTh)'

=exp(-4MT)sink(2MT)' b = exp JO+4h), _ J =exp ( 4MT c = exp (-4MT) cosh (2T)

(16.5)

16.1 The diagonal-to-diagonal transfer matrix

191

If we put 6l,,,, -* (-1)i+m+16i,m, Z is given by

N M Z = E JJ fIA(621+j,j62i+j+1,j;621+j,j+162i+j+1,j+1), {Q} j=1 i=1

``1(6162; 6162) = -

a

0

0

0-

0 0

c

b0

b

c

0

000a

-

hen in the case N = 2M x integer, we have

Z = TrTN,

T (al, 62, ..., 62M; 6i, 62, ..., 62M )

A(6162; 02M01)A(6364; 6263)...A(62M-162M; 62M-262M-1)

This 22M x 22M matrix is the diagonal-to-diagonal transfer matrix of the six-vertex model. The eigenvalue problem of this transfer matrix is a special case of the inhomogeneous six-vertex model on the square lattice which was treated by Baxter'. The relation with the diagonal-to-diagonal six-vertex model and inhomogeneous six-vertex model is shown in Fig. (16.1). Consider an inhomogeneous six-vertex model with the following columndependent Boltzman weights : al = pih(v + vi + q) bi = piw-1 h(v + vi - q) bi = picoh(v + vi - rl)

ci = pih(2i),

l = 1,...,L.

(16.6)

Here L is the number of columns. h(u) is u, sin(u) or sinh(u). In this case the transfer matrix T(v) is defined in 2L-dimensional space, T = TrR1(o a )R2(62, a2)...RL(6L, at), f,

Ri(++) = ( al

Ri(-+)

=

0

0

I

0) , Rl(--) _

10 01

0),

a)

.

(16.7)

The space is divided into subspaces by the number of down-spins k. Without

loss of generality we can put k < L/2. In this subspace we can construct a

The spin 1 12 XXZ model in a magnetic field

192

Bethe-ansatz wave function with k parameters u1,

I`) _

..., uk,

f(Y1,Y2,...,Yk)6Y,6Y2...6YkI0), k

f(Y1,Y2,...,Yk) = >A(P)11 F(Yj;upj), P

j=1

L

Y-1

F(Y;u) = wY f[ h(u+v1 +rl) 11 h(u+vi -,1), 1=y+1

1=1

A(P) = e(P)

h(up j - Up - 2ij).

(16.8)

j<1

The periodic boundary condition is satisfied by

O(uj + 0

k CO L-11

O(uj - rl)

h(uj - u,n + 2rl)

h(uj - u,n - 2rl)'

L

O(v) _ rl pih(v + vi).

(16.9)

1=1

The corresponding eigenvalue is given by T(v) = w-L+ko(v

_ )Q(v + 2q) + wk4(v + Q(v)

rl)Q(v - 2q)

Q(v)

k

Q(v) = 11 h(v - uj).

(16.10)

j=1

To solve the diagonal-to-diagonal transfer matrix we should consider an inhomogeneous six-vertex model, the Boltzman weights of which are given by

al=c1=1, b1=b'1=0 for even al

1,

=exp(-4MT) sinh(2MT)' b1 = exp (

b'1 =exp(

JA 4MT- 4h),

JA+4hl JA l J exp(-4MT cosh(2MT/ll I'cl = 4MT /

for odd 1.

(16.11)

The conditions (16.6) are satisfied if we put h

L = 2M, w = ±exp(MT ), v = 0,

hh'(0))

_ +sinh(MT)' MT

and

p1 = 1/h(2rt), v1 = rl

for even

1,

(16.12)

16.1 The diagonal-to-diagonal transfer matrix

P1 =+

h(vl + ij)

bb'

± h(vi -,1)

h(vI

bb'

=

193

exp

for odd 1.

( - 2MT ) sink (2MT

(16.13)

Then we have the Bethe-ansatz equation for uj, j = 1, ..., k,

(h(u1+2r1)h(ul+;+v1M=-w_2Mj h(ul-u,n+2q) h(ul)h(ul

- i + v1)

(16.14)

h(ul - um - 20

m-1

The eigenvalue is given by A = T (O) = wko(,1)

Q

_Q(2))

k

h(m

fl

= aMwk k

2q)

h(um

M=1

(16.15)

)

We put rl + v1 = 2a and xl = ul + a. By these transformations equations (16.14) and (16.15) become

1h(xl-a)h(xi+a+21)1M=-ex 2h)kh(x1-xl+2rl) h(xl+a)h(xl-a-2q) p ( T 11_11h(xi-xl-2q)' I = 1,2,...,k, A = exp(

(16.16)

h(2a) M

JA)

hk

MT + 4T

h(2a

2q)

h(xi - a - 2i) 11 h(xl a) 1=1

-

(16.17)

For JAI > 1 we put h(x) = sin x,

rl = id/2,

a = 2 tanh-1(tanh 0 tanh

0 = sign(A) cosh-1 1

sinh(JA/2MT) sinh(J/2MT)

'

(16.18)

2MT

for A = +1 we put h(x) = x,

a=

q = isign(A)/2,

i tanh 2

J 2MT

,

(16.19)

and for JAI < 1 we put

h(x)=sinhx,

rl=id/2, 8 = cos

a = 2 tan-' (tan 0 tank 2MT

/.

1

sinh(JA/2MT) sinh(J/2MT) )' (16.20)

Without loss of generality we can put k < M, because the case of k > M

is reduced to the k < M case by reversing spin directions and putting

The spin 1 12 XXZ model in a magnetic field

194

h - -h. As all of the elements of the matrix T are non-negative, the largest eigenvalue (with respect to the absolute value) is real and positive. As the matrix T is asymmetric, the eigenvalues are not necessarily real. For the largest eigenvalue, k is equal to M. We write this eigenvalue as A. The exact free energy per site is given by the limiting value of AM as M oo,

f = -T In( lim AM).

(16.21)

M-co

Next we consider two-point functions S11(n) = (Si'Sj'+n) and SZZ(n) (Si'Sj'+n). SYY(n) - (S S +n) is the same as S"x(n). In the case of an infinite chain, these are written as (OIUXTnR-nUxRn10)

Sxx(n) = (OITn l0)

SZZ(n) = (OIUZTnR-nUZRnI0) (OIT"IO)

Here R is the shift operator, 2MSS

Ux

= 2lax)61'61 V61'6j j=2

2M

1

UZ =

2(6Z)6],6i 11 6-j,6,., j=2

R

-

661,62662 63...662M,6;

ax and aY are Pauli matrices. (0I and 10) are the left and right eigenstates of the largest eigenvalue. R2 and the transfer matrix T commute. So we put the corresponding eigenvalue of R2 as exp(iK j) of the eigenstate 1j). K j is given

by 2icl/M and is the total momentum of the state j. Sxx(n) at n = even is given by

Sxx(n) = 7,(OlUxlj)(jlUxl0)(

-Kj)nl2)

Then if Ij) is the largest eigenvalue which satisfies (01Uxlj)(jlUxl0)

0, the

correlation length xx is given by 1/ ln(Ao/IAjl) xx is determined by the largest eigenvalue in the sector k = M ± 1, AM' -1. While ZZ is determined

by the second eigenvalue in the subspace k = M, A. A' M-1 is always real. But AM is not necessarily real, xx = M m 1/ ln(AM/IAM-11),

ZZ = l m0 1/ ln(AM/IAM ).

(16.22)

16.1 The diagonal-to-diagonal transfer matrix

195

It seems that AM becomes complex in some regions for A < 1 and h * 0. The second eigenstate belongs always to the k = M - 1 subspace for the JAI <- 1 case. But for JAI > 1, it belongs to the k = M subspace if Ihj is sufficiently small. At k = M, equation (16.16) can also be written as

- 2ii) h(x1 + a)h(x1 - a - 2r1)h(x1 + 2q) -2h + h(xi - xj + 2q)h(xl - 2q) ln{ T h(x1 - xj - 2rl)h(x1 + 2r1) h(xi - a)h(xi + a + 2r1)h(x1

M In

_

27r

l

- 12) i (16.23)

j=1

We write the largest eigenvalue as AM and

% x1>% x2>...>91 XM, xl = -xM+1-l,

l = 1, 2, ..., M.

(16.24)

We put the solution of (16.23) as y1 to discriminate from solutions of the largest eigenvalue. For A'M-1 we remove yM. The corresponding equation is R y1 >

...

h(y1

M ln{ h(Y1 + ln1

Yl = -YM-1, 1 = 1, 2, ..., M - 1, - a)h(yi + a + 2i)h(yi - 211) _ 2h T + a)h(Y1 - a - 2ij)h(Y1 + 2rl) 1

> 91 yM-1,

h(2,j + Y1) }

j

-

M-1

h(21-yi)

+

(16.25)

ln{

j=1

h(Y1-yj+2t1)h(Y1-211)

h(Y1 - yj - 2q)h(Y1 + 211)

.

(16.26)

For the second largest eigenvalue in the k = M subspace A,2Y1, we have Mln {

h(z, - a)h(z1 + a + 2r1)h(z1 - 211) h(z1 + a)h(z,

_ -21r(l - 1)i +

h(zi

- a -

h(z1

}

- z1 + 2r1)h(2r1

- z,)

+ ln{ h(z1 - z1 - 2r1)h(2r1 + z1) T - zj + 2r1)h(zi - 2') 2h

1 { h(z1- zj M In

211)h(zl + 2ij)

2r1)h(z, + 211)

(16.27)

In the case J > 0, A > 1, and It sufficiently small, z1 lies on the R1z1 =7r/2 axis,

it/2=91z1>9IZ2>...>RzM>-7r/2, Z1 = -ZM+2-1,

l = 2, 3,..., M,

(16.28)

If the magnetic field h is sufficiently large and A < 1, AM is not necessarily real. In this case the correlation function S22(n) is oscillating depending on the phase of AM and decays exponentially with the correlation length zz.

The spin 1 12 XXZ model in a magnetic field

196

Thus we can calculate AM, AM' -1 and AM using these equations and equation (16.17). The correct eigenvalue is obtained in the limit of infinite M. In Koma51 and Yamada's115 calculation this limit is taken via numerical extrapolation. Here we take the limit of M oc in equations (16.17) and (16.16) analytically. oo, equations a is of the order of J/MT. Then in the limit of M (16.23)-(16.28) become 2Jsg(x1)

= 21r (1

T

- 21) + 2hiT +

1

00

i

j=1

>ln[f(xl,xj)f(xi,-xj)1,

l = 1,2,...

(16.29)

for the largest eigenvalue AM and 2Js

T g(yi) =

2hi

T+

h(2rl + yj)

1

i

[ln( h(2q -

yi)

00

+Eln(f(y1,yj)f(yi,-Yj))1,

l = 1,2,3,...

(16.30)

j=1

for A' M-1 and 2Jsg(zl) 2hi + 1 [ln(-f(zl,zo)) = 27E1 + T T i 00

+ln(f(zl,zj)f(zz,z-j))], 1 =...,-1,0,1,2,...

(16.31)

for A. Here

JS =M->oc lim -iaMT =

for

JAI > 1,

for

JAI = 1,

for

JAI < 1,

and

g(x)

h'(x)

- h(x)

_ 1 h'(x + 2j) _ 2 h(x + 2rl)

- 2;) 2 h(x - 2rl) 1 h'(x

(16.32)

16.1 The diagonal-to-diagonal transfer matrix

sin 2x

cothx

f(x,y)

2x - cosh sinh 2x - cos 20 '

197

for

JAI > 1,

for

JAI = 1,

for

CAI < 1,

h(x - y - 2nih(x + 2i)

(16.33)

(16.34)

h(x-y+2rl)h(x-2t)*

When M becomes large, a becomes small as O(J/(4MT)). Then the form of equation (16.17) is not appropriate to take the limit M - oo. For this purpose the following equations are useful: MT(h,(-2t1)+

h(-2!)) = (h(2rl)+2ah'(2rl))M+

8l -a

1111

h(-2n)

h'(-2i) 11 h'(-2q) +

a

6

1=2

_ h(2;1) + 2ah'(2;) M

-

)

2a

8i = -ia cot

"(l_M 2)

(h(21))M 2a '

2a

h(2r) M

-

21

(16.35)

)

), 8 - -ia cot M) .

(16.36)

Using these we rewrite equation (16.17) as follows:

AM l = CD + 11 M h(x 1=1

h(x1 - a - 2q)

1

- a){h'(-2r ) + h(-2r )/(bi - a)}' l

1

(16.37)

h CDAM-1 = exp MT) h'(-2q) 1

M-1

X 1-1 M A2M

h(y1 - a - 2q)

h(y1

- a){h'(-2q) + h(-2rl)/(b - a)}'

(16.38)

1

h(zM - a - 2rl)

CD- h(zM M-1

- a)h'(-2q) h(z1 - a

- 2q)

X 11 h(zj - a){h'(-2t7) + h(-2t)/(8i - a)}'

(16.39)

The spin 1 12 XXZ model in a magnetic field

198

Table 16.1. Boltzmann weights of vertices for inhomogeneous six- or (eight-) vertex models which correspond to Hamiltonians (16.1) and (16.2) even 1

XYZ odd l

e

XXZ odd l

a

1

bl, br

0

eanrT cosh aNrT

eaMT , eaMT

1

e aM cosh aMT

eaMT cosh ZMT

Cl

++

aNiT sinh

Jx-J

Jz

di

4MT

0

eaMT

e aMT sinh 2MT

Je}ah

JA-4h

z-J sinh J4MT

0

where C = eJo/4T+h/T (h(2a)(h(2rl) + 2ah'(2rl))\M 2ah(2n + 2a) M h(2q) D± _ I1 (

-

h(2)+2h'(2))

(16.40)

16.2 The limit of an infinite Trotter number From equations (16.29)-(16.31) we have

xl - 6

and yi - 6

(16.41)

.

Then terms in the product of (16.37) and (16.39) approach unity if l >> 1,

M - l >> 1 and M >> 1. In the limit of infinite M equation (16.37) becomes AM

= 2eh/T cosh(JA) flh 4T

h(21 -xi)h(2tl +x1)JS h(xi)h(xl)h2(2m)((T7r(l

- z))2 + ('-4°)2)

(16.42)

16.2 The limit of an infinite Trotter number (a)

199

(b)

T

x

x

T

(d)

(c)

x x

T

T

(e)

x

T

Fig. 16.1. Equivalence of two-dimensional lattices. (a) The thermodynamics of a chain is reduced to that of a checkerboard. (b) The x axis and i axis are exchanged. (c) This problem is equivalent to the six-vertex (or eight-vertex) problem with a diagonal-to-diagonal transfer matrix. (d) This is equivalent to the slanted lattice. (e) The slanted lattice is equivalent to an inhomogeneous six-vertex (or eight-vertex) model where even columns have special Boltzmann weights (a = c = 1, b = d = 0).

The spin 1/2 XXZ model in a magnetic field

200

Using

cosh x = M, + (0741 - 2)))2 1=1

we have

Al

h(2rl - x1)h(2 + x1) (

2ehlT

M

1=1

JS

1

T7r(l -

h(xi)h(x1)h

)2

(16.43))

2

(16.44)

In the same way we obtain J 00 h(2rl - yt)h(2,1 + y1) JS AlM-1 = 2ehIT 4 T h(y1)h(y1)h2(2rl)

J

A2 M

2ehIT

JD h(zo - 2rl)

h(2rl - z1)h(2rl - z-1) 11

4

.

h(z1)h(z-1)h2(2rl)

JS

2

(T l . (16.45)

Equations (16.30)-(16.31) and (16.43)-(16.45) do not contain the Trotter number M and are simpler than (16.23)-(16.28) and (16.17). We should note that z1 = -z1 does not necessarily hold. If we set p1 = tanh 0/ tan x1, for JAI > 1, p1 = 1/x1,

q1 = tanh 6/ tan Y1,

q1 = 1/y1,

p1 = tan 0/ tanh x1, for Al J< 1,

r1 = 1/z1

for

q1 = tan 0/ tanh Y1,

ri = tanh 9/ tan z1 JAI = 1,

r1 = tan 0/ tanh z1 (16.46)

equations (16.29)-(16.31) become

J

2hi

2TG(pt) = 7-, +2m(l

- 2) + 1

1

00

In[L(p1,pj)L(p1,-Pj)], j=1

(16.47)

J

2hi

1

1 + iq1

i

[ln(l - iq1

2TG(g1)_ T

00

+

ln(L(g1, q j)L(g1,

(16.48)

j=1

J 2T

2hi

1

G(r1) = 7 + 27rl + 1 [In(-L(ri, ro)) 00

+

ln(L(r1, r j)L(r1, r_ j))] , =1

(16.49)

16.3 Analytical solutions for special cases

where

G(x) = A

x(x2 + 1

- A-2)

201

L(x, y) = iy + (1 - A-22 (1- ix))

-iy+(1-A- /(1+ix))'

X2 + 1

(16.50)

Equations (16.43)-(16.45) become 00

AM

=

JA 2eh/THTT(41rTtl

- 1/21

i hT() = 2e A'-i

J

AM

2(1 + ipi)(1 - ip1),

fJ(JA)2 00

00

JA

(16.52)

+ igl)(1 -

JA

(16.51)

2

= 2ehIT (- ) (1 + igo)1(4; A 1) (1 + iq)(1 + iq-i) l=i

(16.53)

These equations are convenient for numerical calculation because they do not contain complex functions. 16.3 Analytical solutions for special cases 16.3.1 The Ising limit In this limit we have

J = J, /A,

A --* oo,

Then equation (16.47) becomes pi

=

In

D

-

4irT(l 1/2) lJzl

tanh 0 = 1.

(16.54)

(4h + 2TD)i

(pl - i)(PI + i) ]

[f j=l (p, + i)(p3 - i)

Jz

,

( 16 . 55)

(16.56)

.

Substituting (16.55) into (16.5 6), we find that eD

- cosh((Jz + 4h + 2TD)/(4T)) cosh((Jz

16.57)

- 4h - 2TD)/(4T))

Solving this equation with respect to eD, we have eD

exp(Jz2T2h

=

)(/sinh2

T +e-Jz/T + sinh T

(16.58)

Using this and equations (1 6.51) and (16.54), we have AM

= exp(4T) (cosh T + jsinh2

+e-J=/T).

T

(16.59)

The spin 1 12 XXZ model in a magnetic field

202

So the free energy per site is

f = -T In[exp(4T) (cosh h +

sinh2T +e-J=/T)].

(16.60)

Equation (16.48) gives

ri =

(4h+2TD')i

47rTl

- i)(r_j - i)

l + iro 1 - iro ) Then we have e D1

+i)(r_j +i)

l

(16.61)

- sinh((Jz + 4h + 2TD')/4T) sinh((JZ - 4h - 2TD')/4T)'

(16.62)

and therefore eD'

= exp(Jz2T2h) (/sinh2 T + e-J=/T - sinh h ).

(16.63)

Equation (16.52) gives AM

h

- isinh2 h +e =/T).

T

T

= exp(4T,) (cosh

(16.64)

Using (16.22), (16.59) and (16.64) we find that the correlation length is ZZI

rcosh h/T + sinh2 h/T + e-J=/T 1 = In I cosh h/T

-

(16.65)

.

sinh2 h/T + e-Jz/T J

Equations (16.60) and (16.65) match the results for the Ising chain, (15.4) and (15.6).

16.3.2 The XY limit

From (16.16) we have 0 = 7r/2,

h(x) = sinhx. Then all logarithms in

equations (16.29)-(16.31) vanish. Equation (16.29) gives tank xi

-

tanh xi =

47rT

J

and 1

tanhxi

4hi (J - 1) +

(27rT(- 1)+2hi

(27rT(l- Jz)+2hi )+

2

) + 1.

16.3 Analytical solutions for special cases

203

Substituting this into equation (16.43) we have 00

AM = 2 exp

1

fen

(E 27r J

-

dx In [1 +

o

( °T (

l

+ 2h)211

(

z))2 J J

2n

= exp [ _-

f dx In (2 cosh (J co 2 ,+ 2h \ 11

(16.66)

Here we used the identity 1

ln(a + a2 --b2) =

2m

27r

f dx ln(2a + 2b cos x).

(16.67)

o

Thus

f2

f =-T 2 J

2h))

dx In (2 cosh (J co 2Tx

(16.68)

This is equivalent to (1.19). In the same way, 1

tanh yj

_( 27r TJl+ 2hi ) + 1

2n

ex

Al`W_1 W-1

(

dx In

- 2T p( 27r Jo

27r T l+ 2hi 2

) +l,

J

sinh[(J cos x + 2h)/2 T] ( (J cos x + 2h)/2 T

The inverse of the correlation length is 1 -i XX = 2it

2n o

dx ln(coth I(J cos x + 2h)/2T 1),

(16.69)

for 21h1 < J and 2h

J

+

(12h) 2 - 1)

1

2n

2it

J cos x + 2h 2T ), (16.70)

dxIn(coth I

for 21 h1 > J. Results (16.68)-(16.70) are consistent with known results.

16.3.3 The T = h = 0 case For very small T, the distribution of xls becomes dense. The mean distance is of the order of T/JS. At h = 0 all the xis are on the real axis. Denote the distribution function of the x1s as p(x). p(x) obeys the equation 7r T

JS

K x

T p(t)dt + J, fK 12i In [f (x, y)f (x, -y)] p(y)dy = g(x). T Jo

(16.71)

204

The spin 1 12 XXZ model in a magnetic field

We define the function F(x),

7rT fx

I

is

at x > 0,

p(t)dt,

(16.72)

-F(-x), at x < 0. By partial differentiation of equation (16.71), one has

-

F(x)

q(x) -

q(x - y)F(y)dy = g(x).

K

K

7h'(x

1

27ri

- 2q) for

it cosh 20 - cos 2x 1 sign(A)

77 x2+1

for

JAI > 1,

(16.74)

JAI = 1,

sin 20

1

h'(x + 2rl) h(x + 2q)

h(x - 2q)

sinh 20

1

for

it cosh 2x - cos 20

(16.73)

JAI < 1.

The Fourier transform of g(x) and q(x) is g(n) _ -7ri sign(n)(1 - e-H2n0l),

NO)) _ -7ti sign(w"() )(1 - eI°cosh('),

g(w) _ -mi coth

mr

2-

- 20)w/2

sin h 7tw/2 (16.75)

sign(A)e-12"01

sign(0)e 1wI, sinh[(mr

for for

- 20)w/2]

sinh(mr(o/2)

IA > 1, IAI = 1,

for

Al C< 1. (16.76)

The function g(x) has a pole at x = 0, but we can define the Fourier integral

by the principal integral. One should note that the Fourier transform becomes discrete at IA > 1 because the interval of integration is finite. Thus

16.4 Numerical calculations of the XXZ model

205

we have

F(x) = cot x,

F(n) = -7risign(n),

for A> 1, F(x) = 1,

F((o) = -7risign(co),

x

for A= 1, = -iri tanh

F(co)

0)(0

((7r

l

2

for IAA < 1, F(w) = -7ri tanh(w/2),

sinh ( 7r(x /( it 0) 0))'

F(x)

F(x) =

for A=-1,

sinh irx'

00

F(x) _ E

F(n) = -7ri tanh n9,

n=-oo sinh[ic6(x

A < -1.

for

RO

-

nnr)]

,

(16.77)

From equation (16.43) the ground state energy per site e is

e = - li m T In AM = jS

_s 71

K K

-

K In{ JO

h(2nx )h(2q) }F'(x)dx F2(x)h2(x)h2(2q)

F'(x) + g(x)F(x)dx.

(16.78)

Thus

JA A

-J(4 +

A > 1,

for

4 1

-

for

0°

sinh(Bc)) dw) J ssinh(ir(o) cosh[(7r - 0)co] JAI < 1,

A2

(16.79)

e =

A2

- 1 1: 00

e-2nj0'tanh Inol)

for A< -1.

n=1

These results coincide with the known exact ground state energy per site of the XXZ model in a zero field, as expected. 16.4 Numerical calculations of the XXZ model In equations (16.47)-(16.49) it is possible to calculate each term of the

sum, but in each term one must calculate a logarithmic function.

So

The spin 1 12 XXZ model in a magnetic field

206

we would like to take the product of each term and after that take the logarithm of it. The logarithm is, however, a multi-valued function. In actual numerical calculation the imaginary part of In x is fixed at [it, -it], but there is no guarantee that this choice of branch is appropriate. So we transform equations (16.47)-(16.49) as follows:

J-{2

4T

4hi

pi- V(JO+ J* + 1 In [exp (7 (A V-1(pi) - G(pi))) fl L(pi, pj)L(pi, -Pd)] }), 00

j=1

(

4hi

qi

x(1 + ri

4 4T

= V ( JO + T { ir l +

=V

4ni

iql

)

Ji

1

Zi i n [exp ( 27, (AV-1( qi )

fiL( qi,qj )L( qi, _ j )]

16 80 ) .

- G( qi )))

1> 1 ,

(16 . 81)

j

4T 1 Ji { it l + I n [e xp ( 27, (AV-i( ri ) - G( ri )) ) + JO 2i 00

x(-L(ri,ro)) fi L(ri,rj)L(ri,r_j)] }).

(16.82)

j=1

The function V(x) is chosen as follows:

at A> 1

( V(x) = V-1(x) = x V (x) = tan 0/ tanh

V-i(x) _

9

(7E

-

sinh-1

it/(ji - 0) sinh

(71

B)x

V(x) = it/sinh-1(m/x),

V-1(x) _ it/ sinh(it/x)

V(x) = tank 0/ tanh (2 V-1(x)

_

at

tank-1 tai e ) ,

IAI < 1

BB

em (u) to-1(2K (u)x 7r/2K

tn((it/2K) tanh- (tanh 0/x), u) K(u')/K(u) = 0/ii,

at A=_1

u)/ ,

at

A < -1. (16.83)

Here tn(x,u) - sn(x, u)/cn(x, u) is an elliptic function. K (u) is a complete elliptic integral of the first kind with modulus u. u' is 1 - u2. In these equations the imaginary part of the logarithms is fixed at [it, -it]. These functions are obtained from the distribution of roots (16.77) at zero temper-

16.4 Numerical calculations of the XXZ model

207

ature. In actual numerical calculation we determine the first several hundred pis. The higher pls are approximated by

pi = 4Tir(l - 1/2)/JA + const. x i and the sum is estimated accurately.

Iterative calculations of equations (16.80) and (16.81) converges very

rapidly for arbitrary h, A and T. This yields AM and AM_1. In contrast, equation (16.82) for AM is sometimes unstable for h

0.

It stands

hcritical is dependent on A and temperature. It seems that equation (16.82) gives a stable solution at least in this region. I AM I > I AM_1 at J A I > 1 and hcritical > I h I

We give the susceptibility and specific heat as functions of temperature for A = 1 and A = -1 in a zero field in Tables (16.2) and (16.3).

208

The spin 1 12 XXZ model in a magnetic field

Table 16.2. Susceptibility and specific heat of the S = 1/2 ferromagnetic XXX chain T /J specific heat Jx 0.025 0.050 0.075 0.100 0.1250 0.1500 0.1750 0.2000 0.2250 0.2500 0.2750 0.3000 0.3250 0.3500 0.3750 0.4000 0.4250 0.4500 0.4750 0.5000 0.5250 0.5500 0.5750 0.6000 0.6250 0.6500 0.6750 0.7000 0.7250 0.7500 0.7750 0.8000 0.8250 0.8500 0.8750 0.9000 0.9250 0.9500

441.5 132.56 67.218 42.028 29.41 22.08 17.39 14.180 11.867 10.137 8.802 7.746 6.894 6.1935 5.6094 5.1160 4.69470 4.33125 4.01505 3.7378 3.4930 3.2756 3.0813 2.90686 2.74945 2.6068 2.4770 2.35856 2.25000 2.15023 2.05825 1.97323 1.89445 1.82126 1.75312 1.68953 1.63009 1.57440

0.0847726 0.103884 0.1144228 0.1210856 0.1255503 0.1285785 0.1305907 0.1319037 0.1327730 0.1333757 0.1338067 0.1340965 0.1342357 0.1341973 0.1339523 0.1334778 0.1327611 0.1317999 0.1306011 0.1291786 0.1275514 0.1257419 0.1237738 0.1216711 0.1194573 0.1171547 0.1147838 0.1123635 0.1099107 0.1074401 0.1049650 0.1024966 0.1000445 0.09761703 0.09522095 0.09286195 0.09054464 0.08827272

16.4 Numerical calculations of the XXZ model

209

Table 16.3. Susceptibility and specific heat of the S = 1/2 anti ferromagnetic XXX chain T /J Jx specific heat 0.0250 0.0500 0.0750 0.1000 0.1250 0.1500 0.1750 0.2000 0.2250 0.2500 0.2750 0.3000 0.3250 0.3500 0.3750 0.4000 0.4250 0.4500 0.4750 0.5000 0.5250 0.5500 0.5750 0.6000 0.6250 0.6500 0.6750 0.7000 0.7250 0.7500 0.7750 0.8000 0.8250 0.8500 0.8750 0.9000 0.9250 0.9500

0.4419 0.4484 0.45393 0.45925 0.4647 0.4705 0.4770 0.48412 0.4920 0.5004 0.5093 0.51839 0.52746 0.53630 0.54472 0.55256 0.55970 0.566066 0.571606 0.57629 0.580142 0.583156 0.585372 0.586830 0.587577 0.587663 0.587142 0.586066 0.584487 0.582454 0.580016 0.5772157 0.5740957 0.5706945 0.5670477 0.5631878 0.5591444 0.5549446

0.0167 0.03386 0.05148 0.06998 0.0898 0.1117 0.1359 0.16234 0.18999 0.21776 0.244473 0.269086 0.290837 0.309245 0.324084 0.335345 0.343176 0.347838 0.349658 0.3489939 0.3462115 0.3416629 0.3356767 0.3285502 0.3205466 0.3118945 0.3027889 0.2933943 0.2838467 0.2742577 0.2647172 0.2552966 0.2460515 0.2370242 0.2282464 0.2197403 0.2115209 0.2035972

17

The XYZ model with no magnetic field

17.1 The transfer matrix for the XYZ model The partition function of (16.2) at temperature T is approximated by that of the eight-vertex model on a 2M x N checker board lattice, as shown in Fig. (16.1). The transfer matrix in the real direction is a 22M x 22M matrix. It is defined by: T(a1,62,...,62M;61,62,...,62M) 6263)...A(a2M-1°2M; a2M-2a2M-1), = A(aia2; 4M61)A(a3a4;

A(+ + ++) = A(- - --) = a, A(+ + --) = A(- - ++) = d, A(- + +-) = b, A(+ +-) = A(- + -+) = c, A(+ a = eJZ14MT sinh (Jx + Jy \ b _ ejz/4MT cosh (Jy - Jx

-

-

4MT J'

c = eJ,/4MT

4MT

'

cosh('), d = _eJ,14MT sinh (Jy - Jx 4MT

4MT

(17.1)

The other elements of A are zero. The free energy per site and the correlation length are given by:

f = -T lnA0,

= 1/ln(jAo/AiI),

(17.2)

where A0 and Al are the largest and the second largest eigenvalues of T. This transfer matrix is that of an inhomogeneous eight-vertex model on a square lattice with 2M columns, T = TrR1(61, o )R2(a2, a )...R2M(62M, a2M), Cal

\0

0I

Ri(-+) _ (di 0)

,

210

Ri(--) _

(bi

di

/

a)

,

(17.3)

17.1 The transfer matrix for the XYZ model a21 = C21 = 1, a21-1 = a,

b21 = d21 = 0

b21-1 = b,

211

and

C21-1 = c,

d21-1 = d.

(17.4)

The weights of this square lattice are given in Table (16.1).

17.1.1 Baxter's theory for the inhomogeneous eight-vertex model According to Baxter8 the inhomogeneous eight-vertex model is soluble if the Boltzmann weights on the l-th column are given by al = pl0(211)0(vl - 11)H(vl + tl), b1 = pi0(2rl)H(vi -11)0(vi + il), c1 = piH(211)0(vi - 11)0(vl + 11),

di = piH(21)H(vl - rl)H(vl + rl).

(17.5)

Here H and O are Jacobian elliptic theta functions. pi and vi are dependent

on l but 1 and the modulus k should be common for all columns. In his theory, the matrix T(v) with the parameter v is defined by changing v1 v1+v in equation (17.5). Matrices T(v) and T(v) commute with each other for any

v and v'. Define the function 2M

4(v) = fJ p10(0)H(v - vl)O(v - vi).

(17.6)

1=1

One can construct a matrix Q(v) with parameter v which satisfies [T (v), Q(v)] = 0,

[Q(v), Q(v')] = 0,

T(v)Q(v) = 4(v + 11)Q(v - 211) + 4(v -11)Q(v + 2rl).

(17.7)

T(v) and Q(v) are simultaneously diagonalized. An eigenvalue of Q(v) is factorized as follows:

Q(v) = exp(_ iv7rv/K) fi h(v - wj),

h(x) - H(x)O(x).

(17.8)

j=1

The sum of the wjs should satisfy

M

2M

w j - E v1 } = 2K (v" + M + even integer) + ivK'.

{2

j=1

(17.9)

1=1

v and v" are integers. v" is 0 or 1 in the symmetric or antisymmetric subspace, respectively, with respect to reversing all arrows. K and K' are

212

The X YZ model with no magnetic field

the complete elliptic integrals of the first kind with modulus k and respectively. The corresponding eigenvalue of T(v) is given by M

h(v - wl - 2q)

T (v ) = O( v + r ) e '.',glK H

h(v - wi)

1=1

)e tv7rq/K

v

1 - k2,

T7 h(v - wl + l1-1 h(v - wi)

(17. 10)

17.1.2 Transcendental equations We apply the above theory to the eigenvalue problem of the quantum transfer matrix given in equation (17.4). For even columns we have V21 _',

(17.11)

P21 =

For the odd columns we have

P21-i = b(0(2i)H(vi - rl)O(vi + i1))-', a/c = sn(vi + ri)/sn(2q), b/c = sn(vi - rl)/sn(2r1), V21-1 = v1,

d/b = ksn(2rl)sn(vi + rl).

(17.12)

The modulus k, 1 and vl are determined by k

=

AZ-Az

AZ -Ay

VAZ -A x + AZ -A2

sn2(21l) = cd/(kab),

sn(tl + v1) = sn(2rl)a/c.

(17.13)

Here we put AZ = a2 + b2 - c2 - d2 = 2 sinh (2MJ Ay

2(ab - cd) = 2 sinh(2MT

Ax

2(ab + cd) = 2 sinh (2MiT

T

(17.14)

From equation (17.2) we find that d is negative and therefore sn(2ri) and sn(rl + vi) are pure imaginary,

A2 -A 2 + A2 _ A2 Z

i

Ax + A x

y

Z

Y,

0 < 2t1/i < K'.

17.1 The transfer matrix for the X YZ model

213

By this parametrization equation (17.5) is satisfied. The function 0 in equation (17.6) becomes O(V) O(v)

M bO(0)h(v - rl)h(v - vi) 0(2q)H(v1 - !)O(vl + 17)h(2q) }

(17.15)

Putting v = 0 in equation (17.10), we have the eigenvalue of T, h(-wl + 2q) =

A = T (O) = e-ivmi/K 4)(-t1) M

h(-wl)

1=1

X

iv7cry/K

M M h(-wl + 2i) ] 11 h(-wl)

bO(0)0(2g - (i + v1))h(v1 +

O(i + v1)0(2i)h(v1 - i)

(17.16)

The wls are solutions of the following equation: h(w1)h(w1 + rl - v1)

h(wl - 2q)h(wl -

q - v1)

M = -e-2iv7rry/K M h(w1 X1=11 h(wi

- w1 + 2n).

- wi - 2rlY

(17.17)

If we put a = -1(11 + v1) and x1 = wl + a, equations (17.16) and (17.17) become

h(xi - a)h(xi + a + 2i))M = -e 2iv7rrl/K T7 h(xi - x1 + 2i) h(xl + a)h(xl - a - 271) f1 h(x1 - x1- 20' j=1

(17.18)

A

_ bO(0)O(2i + 2a)h(2a) M

- O(2a)O(2r1)h(2a + 2n)i

-ivitr1/K e

M h(x1 - a - 2r1) 11 h(xl a)

-

(17.19)

In the case Jx = 0, Ax is zero and we have

k=

1 -

1 - (Ay/AZ)2 sn(21l)

1+

1 - (A/AZ)2

i

211 = W12.

(17.20)

k

Using the quasi-periodicity of h(x),

h(x + iK') =

- exp(-ir(K' + 2ix)/2K)h(x),

and equation (17.9), we write equation (16.36) as follows:

(h(xl - a)h(xi + a + 2,7)l M `h(xl + a)h(xi - a - 2q))

"+M

(17.21)

The largest eigenvalue belongs to v" + M = even, and the second largest

The X YZ model with no magnetic field

214

eigenvalue belongs to v" + M = odd. So we have

h(xl - a)h(xl + a +

exp [2 7r (l

- 1/2)i/M] ,

(17. 22)

e xp [2 7r (l

- 1)i/M]

(17. 23)

h(xi + a)h(xi - a - 2j)

h(Yi - a)h(Yi + a + 2q)

=

h(yi + a)h(Yi - a- 2q)

.

Hereafter we write the yis as solutions of (17.18) for the second eigenvalue. The xis and yis are all real and satisfy the following conditions: l = 1,2,...,M,

xi = -xM+1-1,

yl = K,

l = 2,3,..., M.

Yi = -YM+2-1,

(17.24)

We can set v = 0 for these eigenvalues. These properties hold in the region IJxI < Jy. So one can determine the xis and yis for fixed M and T. In the case of even M, equation (17.18) is written as follows, M/2 T,

gM(xi) = 2ir(l - 1/2) - i E ln[f (xi, x3)f (xi, -x>)], j=1

J,gM(Yi) = 2ic(l - 1) +

[ln(-f(yi,K))

M/2-1

+ E ln(f (Yi, y.i)f (yi, -y.i))],

(17.25)

j=2

where 1

gM(x)

2aln

h(xi+a)h(xi-a-2ri)h(xi+2ri) hx 2 ( l- ahx ) ( +a+211)hx( l- q) l

'

h(x - y - 2,)h(x + 2i)

- h(x JS - -2iaM. f( x'Y )

- y + 2rl)h(x - 2i)'

(17.26)

Using the Liouville theorem one can show f (x , y)f (x , -y) f (x , K)

-

1 - tdn2(y)/tdn2(x - 2q)

= 1 - tdn2(y)/tdn2(x + 2n)'

(17.27)

tdn(21 + x)

tdn 2

hr2(0)h(x - y)h(x +

h2(x)h2(y)

(17. 28)

x)' y) _

1

1

tdn2(y)

tdn2(x)

tdn(x) - sn(x)/(cn(x)dn(x)).

(17.29) '

(17.30)

17.1 The transfer matrix for the X YZ model

215

17.1.3 The limit of M oo The exact free energy per site is given by the limiting value of A as M co. The correlation length is given by the ratio of the largest and the second eigenvalues. We take the limit M -* oo in equations (17.25)-(17.26) and (17.19) analytically. In this limit we have k-4

XJ2z -

J2x - J2z - J2y Jy - Jy

Jy - JX +

Jz - Jy

Jz +

JZ

sn(2n) --* i

JX + Jy

Jz - Jx +

Jy

J --+

- Jy

s s

8

-

Jztdn(2rl) 4i

(17.31)

As a is 0(M-1), the function gm(x) in equations (17.25)-(17.26) becomes

-

h'(x) 1 h'(x + 2i) h(x) 2 h(x + 2rl) 2 tdn2(2q) sn(2x1) tdn2(2n) tdn2(x1)

lim gm(x) = g(x) -

M-co

- 1 h'(x - 2q) 2 h(x - 2rl) (17.32)

-

Then equations (17.25)-(17.26) are written as follows : 2Js

7, 9(x1) = 27r (l 2Jsg(Yi)

T

-2+i 1

1

+

In j=2

In 1

1 - tdn2(xj)/tdn2(xi - 2q) 1 - tdn2(x1)/tdn2(xi + 21l) 1'

[In(tdn(2rl + Y1))

= 2m(l - 1) +

00

a0

1

i

tdn(2rl - Y1)

1 - tdn2(yj)/tdn2(Y1 - 2ri) (1 tdn2(Y.i)/tdn2(Yi + 2q))]

(17.33)

-

From these equations we can calculate the limiting value of xl and yl of equations (17.25)-(17.26) as M - oo. The form of equation (17.19) is not appropriate to take the limit M -* oo. For this purpose the following identities are useful: h(-211)) l1=11

II(h'(-211)+

6-a

- (h(211)+2ah'(2ij))M+ (h(2rl))M 2a

M

h'(-2h) rj (h'(-2n) + 1=2

h(-211) 6 1 - a)

2a

'

The X YZ model with no magnetic field

216

h(2rl) + 2ah'(2q) M C ) 2a

8i =

h(2rl) M

-

(17.34)

2a

l=

2)),

1)).

(17.35)

Using these we rewrite equation (17.19) as follows: lMl

h(21l)

AO = C[1 +

h(2t1)+2ah'(2q))

h'(0)h(x,-a-2rl)

T`-T

x 11 h' 2 h 1=1 (xl - a){ (- n) Al

8+11)l(I 2

(17.36)

a)}

M h'(0)h(yl - a - 2n) h(2q) + 2ah'(2q)) I h(y1- a)h'(-2q)

=C[1-

h(27)

M

x TT l=2

+ (h

h (yl

i h (0)h(yj - a - 217)

-

a ){ h' (

- 'l ) + (- rill 8- - a) } 2

h

2

( ,

(17.37)

'

where

C=

1bO(0)0(217 + 2a)h(2a)(h(27) + 2ah'(217)) M

0(2a)0(27)h'(0)2ah(21 + 2a)

From equations (17.25)-(17.26) we have

xl = 6 + 0(1-3) + 0((M -

l)-3),

and

y,

6+

0(l-3)

+ 0((M -

l)-3).

Then terms in the product of (17.36) and (17.37) approach unity if l >> 1, M - l >> 1 and M >> 1. In the limit of infinite M, the equations (17.36) become AO = 2exp(

00

Jsi

4T +

(2 O'(2t1) T 0(2r7)

_

h'(21) h(217)

)) cosh

Jsi h'(2q)

T h(27)

h'2(0)h(21- x,)h(217 + x,)JS 2)2 + (Jsih'(2t1)1h(2tl))2)).

x Wh2(x,)h2(2il)(T27r2(l

-

Using H'(217)/H(2rl)

- 0'(217)/0(217) = 1/tdn(21), 00

cosh x = fl(1 + x2/(ir(l 1=1

- 2))2),

J

17.1 The transfer matrix for the X YZ model

217

and equation (17.29) we have 00

2

AO =

W

- xi)h(2rl + x1)J2

hr2(0)h(2rl

-

h2(xi)h2(2q)T27r2(i

1=1

_

Jz

tdn22ri

l2(

1/2)(

21J(4ttT(l

1)2

tdn2

l

17.38) (17.38)

/

Using 00

sinhx = x [J(1 +

x2/(irl)2),

1=1

we find in a similar way that Jz

Oo

rl

n1=

- yi)h(2g + yi)JS

hr2(0)h(2q h2(Y1)h2(211)T27r2(l -1)2 y2

Jz = 2() 4T fl( 47r T(l - 1))2(1 - do

(17.39)

"0

1=2

i

The free energy and the correlation length are given by

f = - T In Ao

-T ln2 - T

Jz

00

1/2)

1=1

l2( \1

tdn22tl

l

tdn2xl (17.40)

-1

= In

In[(

Al

1=1

JZ

z

1/2)) ('

47rT(l

00

JZ l 2( r -In (4T/ -Eln[`47rT(l-1)/ \1 ll

If we put p1 = -itdn2tl/tdnxl,

tdn2xl)] tdn22t1

2T

27r(l - 1/2) +

G(q1) = 2ir(l - 1) +

i

E In [L(p1, pj)L(p1, -pj) =1

(ln(L(g1, 0))

00

+ j In[L(gi, gj)L(gi, -qj)1), j=2

(x2 + Dx)(x2 + Dy)

G(x) = x

x2 + 1

(17.41)

qi = -itdn211/tdnyl, equations (17.33)

become as follows :

2T G(pi) =

l

tdn2Y1)J

The X YZ model with no magnetic field

218

(x2 + Dx)(x2 + Dy) - ixAxAy - iy(1 + x2)

L(x, y)

(x2 + Dx)(x2 + Dy) + ixAxAy + iy(1 + x2)

Ax = Jx/J, Dx =

Ay = Jy/JZ, 1 -A2 , Dy = 1 - Ay.

(17.42)

We find that the solution of (17.42) by iteration converges very rapidly. For j >> JZ/(47rT), pj is about 4irT(j - 1/2)/JZ. So one can replace the sum for large j by an integral. AO and Al are

Jz

00

Ao

=2][1 (

Al

J 24T

2

=i 47rT(j - 1)

(l+ p ),

00

(17.43)

Equations (17.42)-(17.43) are very convenient for numerical calculations because they do not contain complicated functions such as elliptic functions. 17.2 Special cases and numerical methods

17.2.1 The T -* 0 limit At very small T, the distribution of x1s becomes dense. The mean distance is of the order of T/JS. Put the distribution function of xis as p(x). Equation (17.33) becomes:

T

J fK 2i In [f (x, y)f (x, -y)] P(y)dy = g(x).

JT I K P(t)dt +

(17.44)

The function g(x) has a pole at x = 0, so the inhomogeneous term of the integral equation for p(x) has a second order pole. It is difficult to apply the Fourier transform technique for p(x), so we define the function F(x),

7rT /x JS

F(x)

p(t)dt,

at x > 0,

x

(17.45)

-F(-x), at x < 0. By partial differentiation, equation (17.44) becomes F(x) q(x)

- fx q(x - y)F(y)dy = g(x), K

1

h'(x - 2i1)

h'(x + 2ij)

27ri

h(x-2rl)

h(x+21l)) +KK'

2ig

17.2 Special cases and numerical methods

219

Using the imaginary transformation and the infinite product representation of the elliptic theta functions, we have 7rx2 h(x) = j02(0)exp (-2KK')

00 sinh(7r(2jK - x)/K') sinh(7r(2jK + x)/K') ' sinh 2 27r K K'

7rx

x sinh

K

Thus the functions q(x) and g(x) in equation (17.32) become 00

sin(20)

1

q(x) = E q(x - 2mK), q(x) = K' cosh(27rx/K') + cos 20' m=-oo 00

g(x) =

g(x - 2mK), m=-

g(x) =

K'

{coth(7rx/K)

sinh(27rx/K') - cosh(27rx/K') + cos 20 }

Here we put 0 - 7r(1 + 2irl/K'). The Fourier transform of these equations is q(x)einnx/Kdx =

qn = I KK _ sinh(n(r - 22)) sinh(nr)

gn = L

K

g(x)ein7rx/Kdx

f Cj(x)einnx/Kdx co

= -7ri (coth(n r)

K

i - 7rK' , 2K

A

=-

ni- coshsinh(nr)

K'0

(17.46)

2K

The function g(x) has a pole at x = 0. We define its Fourier transform by a principal integral. Then we have the Fourier transform of F(x), Fn = Sn/(1 - qn) = -7ri tanh(n1).

(17.47)

Thus,

F(x) _

F(x - 2mK),

F(x) =

m=-oo

7r2

2

0cosech7' 7' x

0.

K

Introducing a new modulus u,

0K' 27rKk

- Ku Ku'

(17.48)

220

The X YZ model with no magnetic field

we write F(x) and its derivative in terms of elliptic functions,

F(x) = Kk ctn ( Kk , u) ,

ctn(x) =- sn(x) ,

JS

P(x) = 7r T(Ku2dnKux,u) Csn\Kux,U// Kk Kk Kk /

(17.49)

2.

From equation (17.38) we get the energy per site:

e = - li m T 1nAo =

JS jK

ln{

hr2(0)h(21 - x)h(2rl + x)

F2(x)h2(x)h2(2rl)

,

}F (x)dx.

By partial integration we have r

2J,

K

e =

F'(x) + g(x)F(x)dx.

Jo

7r

Here we used F(x)h(x) = h'(0)+O(x2) and h(2q-x)h(2q+x) = h2(2rl)+O(x2) near x = 0. As F and g have a pole at x = 0 we put

F(x) - w(x),

u(x)

v(x) = g(x) - w(x),

w(x) - 2K cot 2K .

We find w _ -irisign(n) and that u(x) and v(x) have no singularity. Thus K

e= 71S

f

u(x)v(x) + u'(x) + w(x)(u(x) + w(x)) + w2(x) + w'(x)dx.

K

As w2(x) + w'(x) = -(7r/2K)2 we have e

_

JS7r

2K

00

JS

rr_

+ 00

-J "

2K

_

Luuv-u

I 1 + 2

E

+ wu(u-, + v-n)]

sinh(n(i - 21))

sinh(ni)

+2

sinh(n(i - A))

]

sinh(ni) cosh(n2) (17.50)

This coincides with the exact ground state energy per site of the XYZ chain obtained in (5.80).

17.2.2 The JX = 0 case (anisotropic XY chain) In equation (17.42) all interaction terms vanish and it becomes: Pl

2 Pi +1-0 2

Y

2 P1

+1

= jX1'

al =- 4irT(l

17.3 Numerical calculations

221

So we have (Ay + a2 - 1)2 + 4al )/2

P12 = (Ay + ai - 1 +

and equations (17.43) yield:

0 (1 + (1 + A2)ai2 +

(1 + (1 + A2)ai 2)2 - 4ai 4A2 2

Ao = 2H{

(17.51)

11

Using the formula 2n j 07r

1

ln(a + Ja2 - b2) =

ln(2a + 2b cos x), 27E

we have 1

2n

In Ao =1n 2 + 2L

°°

dx E ln[1 + (1 + Ay + 2Ay cos x)ai 2] t=1

f27r

1

In{2cosh(4T

Jo

27r

JZ +Jy +2JZJycosx)}dx.

In the same way we obtain A1, 1 j21t lnA1 = In{2sinh(41T 27t

JZ +Jy +2JZJycosx)}dx.

The free energy per site and correlation length are

f=

2 f21t In{ 2cosh(47

JZ +Jy +2JZJycosx)}dx,

(17.52)

In{coth(4T JZ + Jy + 2JZJ cos x) }dx.

(17.53)

1

=

j21t

Equation (17.52) coincides with equation (1.47) at h = 0. 17.3 Numerical calculations

Using equations (17.42) and (17.43) we calculate the free energy and cor-

relation length for arbitrary (Jx/JJy/JZ). In Table (17.1) we give these quantities for (J/J,, Jy/JZ) values of (0.8, 0.8), (0.5, 0.5), (0.5, 1.0) in tables A, B, C, respectively. We take pi and q, up to l = 1024. The sums for larger l in equations are approximated by integrals. It seems that the correlation length becomes longer as the temperature goes down,

-

(17.54)

The X YZ model with no magnetic field

222

Table 17.1. Free energy and correlation length according to equations (17.42)-(17.43) for different anisotropy parameters (Jx/JZ,Jy/JZ) (A) (0.8,0.8) T /JZ 0.03 0.04 0.05 0.07

(B) (0.5,0.5)

(e - f )lJZ

T /JZ

(e - f )lJZ

E-6 11019.5 4.688 E-5 906.813 2.1972 E-4 203.643 1.4456 E-3 37.7358 0.10 6.37579E-3 11.1653 0.20 4.07554E-2 2.87522

0.03

2

E-8 9.28

0.04 0.05 0.07 0.10

8.2 8.96

E-7 2.514 E4

1.32996

0.40

0.115028

0.40

4.35

0.139038

E5

E-6 2884.67 1.522 E-4 243.033 1.4141 E-3 38.1380 0.20 2.28744E-2 4.52924 1.537295

(D) (-0.5,0.6)

(C) (0.5,1.0)

T/JZ -f/JZ 0.03 0.2752456 20.61248 0.04 0.2758122 15.42187 0.05 0.2765430 12.29748 0.07 0.2785067 8.700840 0.10 0.2827654 5.954982 0.20 0.3085682 2.738899 0.40 0.3964288 1.352457

T/JZ -f/JZ 0.03 0.3201312 139.6721 0.04 0.3203167 63.6121 0.05 0.3206097 37.96712 0.07 0.3215180 19.46736 0.10 0.3236850 10.59361 0.20 0.3383583 3.831694 0.40 0.4041916 1.609889 (F) (-1.0,1.0)

(E) (1.0,1.0)

T/JZ -f/JZ 0.03 0.2546415 10.04573 0.04 0.2569907 7.741370 0.05 0.2595858 6.341317 0.07 0.2653749 4.717322 0.10 0.2752379 3.473658 0.20 0.3149567 1.971316 0.40 0.4126261 1.175348

T /JZ

-f lJZ

0.03 0.4434480 18.3789 0.04 0.4436825 13.8750 0.05 0.4439843 11.16418 0.07 0.4447916 8.054780 0.10 0.4465170 5.71082 0.20

0.456958

2.94813

0.40 0.5028667 1.535774

It is expected that spin-wave excitation will not affect the long range correlation. But spinon excitation drastically destroys the long range order. Then it is expected that A = Ospinon. On the contrary, the energy is determined by the short range order. So we can expect that A f = min(Aspinong Aspin-wave).

(17.55)

On the line AF-F of Fig. (11.2) the system is gapless and the correlation length behaves as

' 1/T.

18 Recent developments and related topics

18.1 Numerical analysis of the S = 1 chain From analogy with the S = 1/2 Heisenberg antiferromagnet, people expected the general S antiferromagnet to have gapless excitations. But Haldane predicted that for integer S the system has an energy gap, and is gapless 34 for half-odd integer S, based on field theoretical Several numerical calculations were performed on the Hamiltonian, considerations33,

N S

Si+l,

(18.1)

SN+1 = S1-

i=1

The diagonalization method up to N = 1417 for S = 1 could not give a decisive answer regarding the existence of the gap. Nightingale and Blote75

calculated the energy gap of the N = 32 chain using the diffusion Monte Carlo method. They estimated the gap as 0.41J in the limit N = oo. The author calculated the correlation functions using the world line Monte Carlo method104. He showed that it decays algebraically for the S = 1/2 chain and exponentially for the S = 1 chain. The estimation of the correlation length is = 4.5 + 2. Nomura76 did a more elaborate calculation and estimated as = 6.1. Monte Carlo calculation of the elementary excitation was done for the N = 32 chain105. The asymmetry of this spectrum with respect to Q = t/2 axis was found. This excitation spectrum is justified by the neutron scattering experiments64. Recent diagonalization calculations53 support the

existence of a gap. The actual S = 1 Heisenberg chain is modified by an anisotropy effect. Thus the general Hamiltonian is SIYS+1 +ASti S+1) + D(Siz)2, i=1

223

J > 0.

(18.2)

224

Recent developments and related topics

For A = 1 and D = 0, the existence of a gap was established by the quantum Monte Carlo method. The next problem to solve is how the gap is destroyed

by the anisotropy. On the line D = 0, the gap vanishes at A = 1.17. The inter-chain coupling also destroys the gap and induces long range order. The critical value is estimated by a simple mean field theory. It must be emphasized that the S = 1/2 quasi one-dimensional system always has three-dimensional order at low temperatures. In the analysis of the S = 1 case described by the Hamiltonian (18.2), the Monte Carlo method and the diagonalization method are both very useful. The estimation of physical quantities as N = oo can be done through Shank's transformation. The determination of Z2, where the system becomes gapless, is very difficult because the gap becomes small very slowly near the critical point. This transition is of the Kosterlitz and Thouless type. In the S = 1/2 case the two-point function (Sl S+r) decays algebraically.

But for the S = 1 case the two-point function decays exponentially. This state is far from the ordered state. Thus a weak perturbation of interchain coupling cannot make the ordered state. There should be a critical interchain coupling which separates the ordered and disordered ground state. Ni(C2H8N2)2NO2CIO4, abbreviated NENP, is expected in the disordered phase because of small interchain coupling. On the other hand CsNiC13 has a bigger interchain coupling and thus is ordered at low temperatures. This critical interchain coupling is estimated by Sakai and Takahashi80. At zero temperature the magnetization is zero in a weak magnetic field. Magnetization appears suddenly at the critical field. The magnetization curve and differential magnetic susceptibility were calculated by the exact diagonalization method107. These results coincide very well with those of experiments In the magnetized state the system should be described by the conformal field theory or Luttinger liquid theory because the system is gapless. In this region, spin correlation functions decay algebraically. It is shown that the

central charge is unity and the correlation exponents are functions of the magnetization. Numerical calculation of exponents was done at A = 1 and D = 0107. The S = 1 Heisenberg model is not soluble. But if a biquadratic term is added the system becomes mathematically tractable in some cases,

__

N

Si . Si+1 - fl(S1 . Si+1)2.

(18.3)

i=1

The point f3 = -1 was solved by Sutherland85 using the Bethe-ansatz

18.1 Numerical analysis of the S = 1 chain

225

In method. The point /3 = 1 was solved by Takhtajan and both cases the system is gapless. The point /3 = -1/3 was investigated by Babujian112.

Affleck, Kennedy, Lieb and Tasakil. The ground state is known at this point and the system has a gap. The excited states are not soluble. This point is different from the Bethe-ansatz soluble point. It is expected that the state

has a gap in the region -1 < /3 < 1. The elementary excitations of the S = 1 Heisenberg chain were calculated by the quantum Monte Carlo method and the diagonalization method. The lowest excitation energy is at momentum n. This spectrum is not doubly periodic nor symmetric with respect to the axis Q = 7r/2. These points are different from the S = 1/2 chain. The scattering intensity of these lowest energy states is calculated. In the region 0.3ir < Q < it, there is a strong delta function peak in the dynamical structure factor. On the other hand for 0 < Q < 0.3ir the lowest energy state is the lower edge of the continuum. It is expected that in this region the lowest energy state is a scattering state of two elementary excitations near Q = it. Then it is expected that the gap at Q = 0

is twice the gap at Q =7r. In the actual NENP system there is anisotropy and the situation becomes more complicated. Details are discussed in Ref. 106.

Appendix A The Young tableau and the theorem of Lieb and Mattis

In this section we consider a system of one dimensional particles which interact via a many-body potential. Consider

y

2

ax?

+ V(xl, x2, ..., xN).

(A.1)

The potential is a symmetric function with respect to the exchange of particles. There are N ! permutation operators Xp of variables. It is evident

that these operators commute with the Hamiltonian and form a group. Generally speaking, if a Hamiltonian commutes with each element of a group one can choose a complete set of eigenfunctions, each of which transforms like an irreducible representation of that group. A group formed

by all N-order permutations is called a symmetry group of order N and abbreviated as SN. The irreducible representation of SN is represented by a Young tableau. Consider a function of N variables, 4(xl, X2,..., xni Ixnl+1, ..., xni+n2lxni+n2+1, ..., XN);

(A.2)

nl > n2 >_ ...,

where we mean that 0 is separately antisymmetric in the variables xl, X2,...' xn, in the variables xn1+l, ... ) xn1+n2 and so on. If the bars cannot be moved to the right, i.e.

{1->

1'(nl+n2+...+nj_1+k,nl+n2+...+n1-1+1)}(Y

k=1

= 0,

(A.3)

for arbitrary l > j, we say that the function 0 transforms like an irreducible representation which corresponds to the Young tableau [

1 n l -n2 2n2-n3 ...]

226

.

The Young tableau and the theorem of Lieb and Mattis

227

For example if 4 (xl, x2l x3) satisfies (A.3), two functions 41(xl, x2, x3) = 0(x1, x2Ix3) and ¢2(x1, x2, x3) = 4i(xl, x31x2) form a base of an irreducible representation [1, 2] of S3. In fact we show

(02),P12 (02) = (0 1) (02) = C-1 O) C001

2) =

P13

P(3

3 1

(0 -1) (02)'P23 (02) 1

(0012(0

(01

(02)'

(02)'

0)

(02)'

- -1) (02) 1) - C 11 01) (02)

2)

1

3

PC2

3

This is an irreducible representation by 2 x 2 matrices. The many-body wave function of N particles with spin value s is a function of spatial coordinates x1, ..., xN and spin coordinates s1, s2, ..., SN. Spin coordinates take the discrete values s, s 1, ..., -s. We put the number of particles with the same spin value as nti with nl > n2 > ... > n2s+1. The eigenfunctions of (A.1) can be written as

-

N! nl !n2!...n2,+1 !

(A.4)

lp(xlS1,x262,...,xNSN) = E Oj(x1,x2,...,xN)Gj. ,,II,,

j=1

A typical spin function for particles with spin value s is

GI = (s,s,...,sls- 1, s - 1,...,s- II...I -S,-S,...,-S).

(A.5)

The spatial wave function 01 is separately symmetric (antisymmetric) for bosons (fermions) for variables x1, x2, ..., xnl and xnl+l, , xnl+n2 and so on. The other 4 js are obtained if we require that (A.4) is totally symmetric or antisymmetric. Theorem : Let us define the lowest eigenvalue of (A.1) as E (Y) when the

wave function transform like an irreducible representation defined by a Young tableau Y. If two Young tableaux Y = (1'l'2, 2n2-n3, ...) and Y = (114 n2,2n'2

3,

,) have relations:

nl > nl,ni +n2 > ni +n2,ni +n2+n3 > ni

+n'+n3,...,

then E(Y) > E(Y'). This is a famous theorem due to Lieb and Mattis. From this theorem one can show that the one-dimensional fermions with 1/2 spin described by (A. 1) cannot be ferromagnetic. This theorem also holds for Hamiltonian (2.1).

Appendix B The number of string solutions

Here we prove the number of sets {I"j which satisfy (8.13) is CM - CM_1. It is clear that the number of sets is

ft(N_itiii\

1: 061+2062+...+MaM=M i=1 from the condition (8.13), where N

N

CM_(M)-

N(N-1)(N-2)...(N-M+1) M!

The summation (B.1) is rewritten as follows: (B.1)

=

am=o °°

X

00.

)

00

CN-2M+C aM

7(

OCM-1=0

(N-2M+2JM+aM-1)... aM-1

N-2M+2(063+2064+...)+062 062

062= 0

N-M+063+2064+... M-2062-3063-...

(B.2)

The last sum on the r.h.s. is the coefficient of expansion of

xM-3063-4064-say-...

(1 - x2)-(N+1-2M+2(063+2064+3065+...))(1 +

of the Taylor

x)N-M+063+2064+3065+...

This is the M-th order coefficient of

[(I - x2)(1 +

x)]M-1(I

-

x)-N+1

fl ((l -

228

x x)2(n-2) (I + x)(n-2) )

a

The number of string solutions

229

Using the relation

0(B+a1Xa=(1-X)-B-1, a

``

a=0

JJ

we have

(B.2) = the M-th order coefficient of M

(1 +

N+2M-1

ui 1(x))}

(B.3)

j=2

where uj(x) are functions determined by: (B.4)

(uj+l - 1)2 = ujuj+2,

and

=

=

1

U2

,

(1

- x)X31

+ x) (B.5)

U3

z The general solution of the difference equation (B.4) is

_ 2 _ baj - b-1a j uj -f1' fJ a-a 1

(B.6)

Parameters a and b are determined from the initial conditions (B.5) and we have

x3x+1

a=b=

2(1

zx).

(B.7)

As uj 1(x) = O(x1), we have 00

(1 - ui 1(x)) = 1 + O(xM+1), j=M+1

and therefore (B.3) = the M-th order coefficient of x)N-M{TT(1

(1 +

-u

(B.8)

1(x))}-N+2M-1.

j=2

Substituting 1- uj-1= f j-1 f j+1 /f we have: 00

u-'(X))

fI( 1- j

-

= (1

j=2

00

u-1 2 )

( j=3

f j-1 f j+1 2

fj

_(1) =

X2) ff3 2 f3

_ (1 - x2)a u3(x)

1 -

1 - 4x/(1 + x) *

lim

f n+1

nioo fn

Appendix B

230

Thus we have (B.8) = N - M + 1-th

x)N-Mr1`1

(1 +

2

-

order

coefficient

1 - 4x ) l

of

N-2M+1

(B.9)

1+xJ

Using the binomial theorem, the function on the r.h.s. is 00

2-N+2M-1

I-

-(-1)s+r (N - 2M + 1) (s/2 ) (4x)r(1 +

x)N-M-r

"0

`

s=O r=O

r

s

xN-M+1 for (4x)r(1 +

As the coefficient of we have

x)N-M-r is

zero for r * N - M + 1,

(N - 2M D-1)N-M+1+s 00 S + 11

(B.9) = 2N+1

s/2

1

J N-M+1J

S=O

N!(N-2M+1)!

1

= 2N-2M (N - M + 1)!(N - M)!

XE(-1)r r 00 =O

(N M)(2Nr

-2r).

J

(B.10)

J

The last sum becomes 00

r

r =0 =O

-2r)

M) (2N

(N

J

N

J

- x)-N-1 = N - 2M-th order coefficient of (1 + x)N-M(1 - x)-M-1

= N - 2M-th order coefficient of (1 -

N-2M

E

(N - M) !

x2)N-M(1

(N - M) !

1)N-2M r!(N - 2M - r)!M! - (1 + M!(N - 2M)!

Then we have finally

(B.1)=(B.10)=

N!(N-2M+1) (N-M+1)!M!

-

Nl (MN l

(Ml -\M1).

(B.11)

Appendix C The commuting transfer matrix and spectral parameter

For the transfer matrices defined by (5.3) and (5.24), we consider the condition that two matrices T(a, b, c, d) and T(a', b', c', d') commute. The transfer matrix is defined by the trace of a 2 ® 2 matrix, RI = W161 ®tl + W261 ®t2 + W361 ®t3 + W464 ®T4,

c+d

W1 =

2

, W2 =

c-d 2

W3 =

a-b 2

W4 =

a+b 2

(C.1)

Here N + 1 spin spaces are defined. The matrix T(a, b, c, d)T(a', b', c', d') is given by Tr fl, Ul, where U1 is a 4 x 4 matrix, U1

=

1

61

{WiW4(D14 +

W4W1 41 + 31

-

iW3W'I(D +6l +6l {W3W434 + W4W343 + 1W1W212 - iW2Wi21}

+64{WiW1l(D11 + W2W2I22 + W3W3(D33 + W4W444}.

(C.2)

Here we put 11 =-T' ®i 1. The matrix T(a', b', c', d')T(a, b, c, d) is given by Tr fJl VI with V1 =

61

{W'W4'14 + W,4 W1(D41 + iW' 2W3(D23

-

+6l {W2W424 + W4'W242 + 1W3W131

iW'3W2(D 32}

-

iW1W3F13}

+61 {W3W4(D34 + W4W3(D43 + iW'IW2(D12 - iw2W1(D21}

+64{W1 W1(D11 +

W2W2(D22

+ W3W3(D33 + W4W4(D44}.

If we can find a non-singular 4 x 4 matrix S which satisfies S U1 = VIS, 231

(C.3)

Appendix C

232

we can show T(a, b, c, d)T(a', b', c', d') = T(a', b', c', d')T(a, b, c, d).

If we put S=

x1T1

S Ul

- Vi s =

®ilt

+ x222 ®'L21 + x313 ® T3 + x4T4 ®i4 ,

we have

- W4Wix2 + WlW4x3 + W2W3x4) -2 3 ® i 2 (-W2W3x1 +W4Wix2 - W1W4X3 + +2ial {23 - W4W2X3 + W3Wix4) 2iol {'C2 ®i3 (-W3WZx1

W3W2X4)}

®21 (W2W4x1

W1W3x2

-'L 1 ® i 3 (-W2W4x1 - W3Wlx2 + W4WZx3 + W1W3x4)} +2iaj {21 (D(-W4W3x1 + W3W4x2 - W2WlX3 + WIW2x4) -'C® ®il (w4w3xl - W3W4X2 - WlW2'x3 + W2Wix4)}.

(C.4)

Then the following six equations should be satisfied : -W3W2X1 - W4Wix2 + WIW4x3 + W2W3x4 = O, -W2W3x1 + W4Wix2

- WlW4x3 + W3W2X4 = 0,

W2W4X1 - W1W3X2 - W4W2X3 + W3Wix4 = 0,

-W2W4x1

-

W3W11x2 + W4W21x3 + W1W3X4 = 0,

-W4W3x1 + W3W4x2

-

W2W1X3 + W1W2X4 = 0,

W4W3X1 - W3W4X2 - W1W2X3 + W2Wix4 = O.

(C.5)

From the first three equations we can determine the ratio of xl : X2 : X3 : x4. Substituting this into the last three equations, following three determinants must be zero : -W3W2 -W4Wi

det -W2W3 W2 W4

det

W1 W4

-W1W3

W2W3

W1 W4

-W4Wi W4W2

W3WZ W3Wi

-W2W4 -W3Wi

W4W2

W1W3

-W3W2 -W2W3

-W4Wi

W]W4

W1W4 -W1W3

-W4W1

W2W3 W3W2

-W4W3

W3Wi

/

WjW2

W2 W4

-W4W3

+W3W4

-W2W1

The commuting transfer matrix and spectral parameter

det

-W3W2 -W2W3 W2W4

-W4Wi WiW4

W4Wi

-W1W3

W4W2

W4W3

-W3W4

-W1W2

W1W4

233

W2W3 W3W2 W3Wi W2Wi

The first determinant is factorized as follows : -W ( 3W 4W'W'2 + W1 W2 W'W 3 4)( 1

3

4

1

W2W2-w2w'2-W2W/2+W2W/2 1 3 2 4 2

-W1 3 + W2 W3 + Wiw4 - W2W4 ).

(C.6)

Assume that the second term is zero. w4 is given by w4 = w3 + (w4 - w3 )(w12 - w2 )/(W1

- W2)

(C.7)

The third determinant is factorized as follows :

_ w2w2w'2 + w2w2w'2 (W'2 W' - w2w2w'2 3 + W'1 W' 4)(W2W2w12 2 4 1 4 1 1 1 4 2 3 3 3 w2w2w'2 w2w2w'2 + w2w2w'2 +x2w2w'2

2

1

3

-

4

2

-

3

1

4

2

1

3

4)

(C.8)

If one substitutes (C.7) into the second term, we have W1 W3 -W2W4

'2

2

W1 (x2

2

2 W1 - W2

2

2

2

2

W'2 (X2

2

W3) + W2 (x3 - W1 + W3 (x1 - W2))

(C.9)

Then we assume that the last parenthesis is zero and therefore w3 is given by

W3 2

12

W12(w3

- W2) + W2 (W1 - w3)

(C.10)

2 W1 - W22

Substituting this into (C.7) we have '2

W4

-

12

W12(W4

2

- W2) +W 2 (WI - W4)

(C.11)

2 W1 - W22

The second determinant is W/2

2(W2 W3W4WiI W3 W3 [W 1

4- W2) 2 + (W 1- W2) 4 + W 42(W22- W2)] ' 2(W2 +W1W2W W [W 2(W2 4- W2) + W 2- W2) 4 + W 42(W23- W2)] 22 ' 2(W2 +WlW2WI W44LW (W4 - W2) + W 1- W2) 4 + W 42(W23- W2)1 1

2

3

2

2

1

3

3

1

3

3

ff

2

1

2(w2

+W3 W4W2 W4 W 1

3

-

W2) + W '2(W2 2

2

1

-

w2) + w32(w 2 3

2

-

1

W21)1-

(C.12)

Substituting (C.10) and (C.11) into the above formula we find four brackets vanish. Then three equations are satisfied by only two relations (C.10) and (C.11). This means that we can generate a transfer matrix with one arbitrary

Appendix C

234

parameter except for the trivial scale factor. Conditions (C.10) and (C.11) are equivalent to 2 '2 '2 2 WI - W3'2 W1 - W22 W1 W2 W1 - W32 2 W3 - W42

'2

W3

-

'2

'2'

W4

2

2 W2 - W4

Using (C.1) we obtain (5.25) and (5.26).

'2

W2

-

'2'

W4

(C.13)

Appendix D The matrix Q(v)

Assume that 2Lri = 2m1K +im2K'. Here L, ml and m2 are arbitrary integers. We define 2N x 2N matrices QR(v) and QL(v) as follows: N

QR(v)

Tr

al/1

{

ri

S(aj,

/3jIv)},

j=1

N

QL(v)

«Ill

= Tr{n S'(aj, fjIv)}.

(D.1)

j=1

Here S(a j, /3 j I v) and S'(a j, l3 j I v) are L x L matrices with 2L non zero elements

as follows: Zo

z_1

0

z1

0

Z_2

0 0

Z2

0

0

0 0

0 0

0

0

0 Z1-L

ZL-1

ZL

(D.2)

S(+,/3lv) : zm = H(v +K +2mn)rflm, S(-, /3Iv) : Zm = O(v + K + 2mri)ifm,

S'(c*,+Iv) : zm = O(v - K - 2mn)iam, S'(a, -Iv) : zm = H(v - K - 2mri)T' m.

(D.3)

Tj m and T'0CM are arbitrary parameters. One can show that

T(v)QR(v) = 4(v - n)QR(v + 2h) + 4(v + n)QR(v - 2ri), QL(v)T(v) = 4(v -1)QL(v + 211) + 4(v + 11)QL(v - 2ij), 0(v) = {pO(0)H(v)O(v)}N,

(D.4)

QL(u)QR(v) = QL(v)QR(U)

(D.7)

235

(D.5) (D.6)

Appendix D

236

Assuming QR(v) and QL(v) are non singular we put Q(v) = QR(v)QR1(vo) = QL'(vo)QL(v)

Using (D.7) we obtain (5.31). The transfer matrix is given by N

(D.8)

[T(v)] 1p =Tr{[JR(aj,ljIv)}. j=1

T(v)QR(v) is given by N 1 [TQR]aja = Tr{ H U(aj, Nj)},

j=1

where U are 2L by 2L matrices, aS(+, a), bS(-, /1),

dS(-, /3) bS(+, /l) cS(+ , /3)

dS(+, fl),

aS(-, fl))

U(+, ) = C cS(-, U(

fl)

--

' '

(D.9)

We should look for a 2L x 2L matrix M such that M-1U(a,

fl)M =

B(a, fl) ) '

C(a, fl),

where A, B, C are L x L matrices. If this is satisfied we can decompose TQR as follows :

T(v)QR(v) = H1 + H2,

(D.10)

where N

[H1]aj# = Tr{fl A(aj, fij)},

(D.11)

j=1

N

[H2]1 = Tr{fJ

B(aj,fj)}.

j=1

If we set M to be the block triangular matrix, M

M-1 should be M1

_

EP 0

E

E -P 0 E

'

'

(D.12)

The matrix Q(v)

237

where E is the L x L identity matrix and P is the diagonal matrix,

P=

/Pi

0

0

P2

0 0

0

0

p3

0

0

.

0\ ,

PL-1 0

0

pj = k1/2sn[K + (2j - 1)q].

(D.13)

0

PL

We find A(oc, /3) and B(a, /3) are matrices of the type (D.2), A(a, /3) : zm = pOO (0)H(v -11)O(v - i1) (

B(a, /) : zm = pO(0)H(v + r7)O(v + ii)

xm

) q(a, /3, ml v + 2r7),

xm+1

(xm+1 xm

q(a, /3, m v - 211),

(D.14)

where

xm=OO(K+(2m-1)t1), q(+, /3, mlv) = H(v + q + K + 2mr/)tfm, q(-, /3, mlv) = @(v + q + K + 2mq)T#m.

Then we can write A(a, /3) = p0(0)H(v - r7)O(v - q)X-1S(a, /3l v + 217)X,

B(a, /3) = pO(0)H(v - q)o(v - q)XS(a, /3l v - 2q)X-1,

(D.15)

where X is the L x L diagonal matrix with elements xmam,,. We should note that X and P are independent of v. So we have equation (D.4). In the same way we can show equation (D.5). Finally we should prove (D.7). The matrix elements of QL(u)QR(v) are given by N

[QL(u)QR(v)h f = Tr{ 11 W (aj, /j l u, v)},

(D.16)

j=1

where the W are L2 x L2 matrices, [W(a,flu,v)]mm'inn' = E[S'(0C,Ylu)]mn[S(Y,/3IV)]mm'.

(D.17)

Y=±

The number of non-zero matrix elements is 4L2 and has the form as follows:

(H(u - K - 2j'r7)O(v + K + 2jq) + O(u - K - 2j'r7)H(v + K + 2jr7))iaj,ipj.

Appendix D

238

The function H(A)O(B) + O(A)H(B) is decomposed as f (A + B)g(A - B). The function g(x) satisfies

g(x) = g(-x) = -g(x + 4K). Then the ratio of the non-zero elements of W(u, v) and W(v, u) is

g(u-v-2K-2q(J+J')) g(v-u-2K-2i1(J+J')) Thus we find that [W(a,l3Iu,V)]mm'Inn' =Ym,m'[W(a,/Iv,u)]mm'Inn'lYn,n',

(D. 18)

Ym,m' = tm+m'tm-m'+1,

(D.19)

where

and the tm are defined by the recurrence relation tml tm+2 = g(u - v + 2K + 2mn)/g(v - u + 2K +

(D.20)

Equation (D.18) is written in matrix form as W(a,/3lu,v) = YW(a,/3lv,u)Y-1. Thus we can show (D.7).

(D.21)

Appendix E Special functions

AE.1 Elliptic integrals The complete elliptic integrals K(k) and E(k) are defined by dx

1

K(k) _

J

E(k) =

d4

it/2

(E.1)

(1 - x2)(1 - k2x2) = JU

f

kxd= [/2

I1

1 - x2

1 - k2 sin2 1 - k2 sin2

(E.2)

These are represented by the hypergeometric function,

K(k) = 2F(-2, Z,1;k2), E(k) = 2F(2, 2,1;k2).

(E.3)

AE.2 Elliptic functions

AE.2.1 Elliptic theta functions To define elliptic functions it is convenient to use the following elliptic theta functions through nome q with I q I < 1,

91(x, q) = -i

(-1)nq(n+1/2)2e(2n+1)xi,

(E.4)

n=-oo

q) = k92(x,

00

q(n+1/2)2e(2n+1)xi

(E.5)

gn2e2nxi

(E.6)

n=-co 00

93(x, q) = n=-oo

94(x, q) = E (-1)ngn2e2nxi n=-co 239

(E.7)

240

Appendix E

These functions are doubly quasi-periodic with respect to the shifts x -* x+7r and x -* x + iln(1/q). Then nome q characterizes the periodicity of elliptic functions. These satisfy 91(x + 7r, q) = -91(x, q),

(E.8)

92(X + r, q) = 92(x, q),

(E.9)

93(x + 1r, q) = 93(x, q),

(E.10)

94(x + 1r, q) = 94(x, q), 91(x - i In q, q) = -q-1e 2ix'91(x,

(E.11)

q),

92(x - i In q, q) = q-1e-2ix92(x, q), 93(X - i In q, q) = q-1e-2ix93(x, q), 94(x - i In q, q) = -q-1e2ix94(x, q).

(E.12) (E.13) (E.14) (E.15)

These functions are represented by infinite products, 00

g2nz-1)(1- q2n),

91(x, q) = -ig1i4(z - z-1) jl(1- g2nz)(1n=1

(E.16) 00

g2nz-1)(1- q2n),

92(x, q) = g114(z + z-1) jl (1 + g2nz)(1 + n=1

(E.17) 00

93(x, q) = fl(1 + q2n-1z)(1 +

q2n-1z-1)(1- q2n),

(E.18)

n=1 00

94(x, q) = fl(1 -

q2n-lz)(1

-

q2n-1z-1)(1

-

q2n),

(E.19)

n=1

where z = eix. For nome q, the corresponding modulo k and k' are determined by k

- (92(0,q))2 92(0, q) 2 - 93(0,q)

94(0, q) 2 k' = (94(0,q))2

(E.20)

93(0,q)

Elliptic integrals K(k), K(k') are K (k) = 2 (93(0, q))2, K (k') = 1n(2 q) (93(0, q))2.

(E.21)

Jacobian elliptic theta functions H(x, k), O(x, k), H1 (x, k), 01(x, k) are defined by

lrx

7rx

H(x,k) = 91(2Kq), O(x,k) = 94(2Kq H ( x, k) = 9 2( 1

7rx

1Cx

,q), O ( x, k) _ 9 3 ( ZK,q 1

(E.22) ( E . 23 )

Special, functions

241

By the imaginary transformation we have H( x, k) _

-7UX /K')'/' exp( xp( - 4KK' ) H (ix, k' ),

O 1 (x , k) = (K/K')112ex p

7rX

2

( 4K K' )0 i (i

H i ( x, k) = ( K/K' ) 1/2 exp(

( E . 24 )

x, k') ,

(E . 25)

ix, k' ),

( E . 26 )

O(x, k) _ (K/K')i/2exp(_4KK')Hi(ix,k').

(E.27)

4KK') O (

AE.2.2 Lionville's theorems

We call a complex function with two periods a doubly periodic function,

f(x) _ .f(x + w) = f (x + w'),"sco/co'

0.

We call a doubly periodic function which is rational an elliptic function. The parallelogram a, a + w, a + co, 'a + co + co' is called the fundamental period-parallelogram. The sum of order of poles in the fundamental periodparallelogram is called the order of the elliptic function. (i) (Liouville's first theorem) An elliptic function which has no pole in the period-parallelogram is a constant. (ii) (The second theorem) In the parallelogram the sum of the residues of poles is zero. (iii) (The third theorem) An elliptic function with order n takes arbitrary value n times in the parallelogram. (iv) (The forth theorem) The difference of the sum of poles and the sum of zeros is equal to one period. AE.2.3 Jacobian elliptic functions The Jacobian elliptic functions sn(x, k), cn(x, k), dn(x, k) are sn(x, k) =

cn(x , k)

1

H(x, k)

O(x, k)' Hi (x, k) k O(x, k)

= rL

(E.28)

(E.29)

Appendix E

242

dn(x,k) =

k'O1(x,k) O(x, k) '

(E . 30)

We sometimes write these as snk(x), cnk(x), dnk(x) or omit modulus k. These are doubly periodic functions,

sn(x + 2K) = -sn(x), sn(x + 2iK') = sn(x), cn(x + 2K) = -cn(x), cn(x + 2iK') = -cn(x), dn(x + 2K) = dn(x), dn(x + 2iK') = -dn(x).

(E.31) (E.32) (E.33)

These satisfy sn2(x) + cn2(x) = 1,

k2sn2(x) + dn2(x) = 1,

(E.34)

and sn(x)' = cn(x)dn(x), cn(x)' _ -sn(x)dn(x), dn(x)'

_ -k2sn(x)cn(x).

(E.35)

Addition theorems for these functions are sn(x)cn(y)dn(y) + cn(x)dn(x)sn(y)

sn(x + y) =

(E.36)

1 - k2sn2(x)sn2(y)

cn(x)cn(y) - sn(x)dn(x)sn(y)dn(y)

cn(x + y) =

(E.37)

1 - k2sn2(x)sn2(y)

dn(x)dn(y)

dn(x + y) =

- k2sn(x)dn(x)sn(y)dn(y) (E.38)

1 - k2sn2(x)sn2(y)

The elliptic amplitude function is a multi-valued function, am(x, k) = sin-i sn(x, k).

(E.39)

This function satisfies am'(x, k) = dn(x, k), am(x + 2K, k) = am(x, k) + it,

am(x + 2iK', k) = am(x, k).

(E.40)

AE.3 The gamma function and Riemann's zeta function F(x) is defined by

F(x) = f e ttx-ldt,

at fix > 1.

(E.41)

The following relations stand for this function:

F(x + 1) = xF(x), I'(1) = 1, ['(1/2) = fir.

(E.42)

Special functions

243

The following formula is useful for Wiener-Hopf factorization:

r (2

- x)

+ x) F ( 2

cosy'

(E.43)

,

(E.44)

7r

cos(x) =

\2-x/ sin(x) =

x

(E.45)

n

1,\1 +)>,\1 ((

((

The asymptotic behaviour at big IxI and argxj <7r is

F(x) = xx-7'e-x

27r 11 +

12x

(E.46)

+ O(x 2)] .

Riemann's zeta function is defined by 00

fi(x)

fix > 1.

x,

(E.47)

n=1

This function has the integral representation 1

fi(x)

= F(x)

0

et - 1 dt, fix > 1,

I

21-x)F(x) l

(E.48)

tx-1

00

1

-

fi(x) = (1

tx-1

0o

et + 1

dt, fix > 0.

(E.49)

There is a functional relation 21-xi-xF(x)

l; (1 - x) =

(E.50)

Values at special points are 4

2

2,

(2) = 6

22n-1ic2n

(2n) =

,

C(4)

,

C(1

(B)

(E.51)

= 90

- 2n) _ (-1)nBn,

(E.52)

(-2n) = 0 for n= 1,2 . ....

(E.53)

Here Bn is the Bernoulli number defined by =1+E00

exx 1+2 B1 =

1

B2

n=1

1

=

6' 30 C(x) has a pole at x = 1 and lim [ (x)

- x- 1 1

((2n)!1Bnx2n

]

1

5

B3= 42 , B4 = 40 , B5 = 66

....

= y = 0.57721 (Euler constant).

(E.54)

(E.55)

Appendix E

244

The Bose-Einstein integral function F(x, v) is defined by e v e- 2v 1 J00 tx-1 F(x, v)

+ ....

(E.56)

n

(E.57)

er+v - 1 dt = Ix + 2x

= F(x)

If x is not a positive integer, we have 00

F(x, v) = F(1 - x)vx-1 +

"

(x

-n )(

v

n.

n=

When x is a positive integer, we have X-1

F(x v) =

(x

1

- 1). X_'

Y

r=1

(x - n (-v)n

- In v] +

n.

n*x-1

(E.58)

AE.4 The Bessel function and modified Bessel function The Bessel function is defined by

(x)v

Jv(x) =

2

(-1)j(x/2)2j j_0 j!T(v + j + 1)

(E.59)

For integer v this function has the integral representation,

Jn(x) =

I

ei('0 'sinB)d8

2

(E.60)

The spherical Bessel function jn(x) is defined by VJ'+2I(x)

Jn(x)

_ (-1)Y (xdx)ISlxx

(E.61)

The modified Bessel function I,,(x) is defined by Iy(x) = e-w7r/2Jv(ix)

_

(x)v

(x/2)2j

(E.62)

The hypergeometric function F(a, /3, y ; x) is defined by ]['(a + n)h (/3 + n) xn

F(y)

F(a , /3 , y; x) = 1

1 a )1( P

=1+

063

1!y

x+

n-o

n !T( 7

a(a + 1)/3(/3 + 1)x2

2!y(y + 1)

+

+ ....

n)

(E.63)

The Legendre polynomials Pn(x) are defined by Pn(x)

=

I

2nn!

do 2 dxn(x - I).n

(E.64)

Special functions

245

The associated Legendre polynomials are defined by dn+lml

P'n(x) = n

1

2nn!

(1 - x2)lml/2

(x2 dxn+Im1

- 1)n.

(E.65)

The spherical harmonic function YIm(B, 0) is defined by Ytm(O,4) _

(-1)

1) (l - 1m1)!P'n'1(cos6)em0. 2

(l + Iml)!

l

(E.66)

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1290(E).

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Index

N-body problem, 14 S U(2) symmetry, 87 n operator, 88

des Cloizeaux-Pearson mode, 39, 40 descending operator, 159 diagonal-to-diagonal transfer matrix, 189 diagonalization method, 225 difference equation, 124, 143, 149 differential magnetic susceptibility, 224 distribution function, 15

addition theorem, 82, 242 AKLT model, 225 angular momentum operator, 187 anisotropic XY model, 7 anisotropy, 224 associated Legendre polynomial, 245 asymptotic expansion, 107 attractive fermions, 29

effective Hamiltonian, 103 eight-vertex model, 67 elliptic theta function, 71, 211, 239 energy gap, 56 entropy, 4

back-flow effect, 19, 39 Baxter, 70 Bessel function, 244 Bethe, 34 Bethe-ansatz, 11 binomial coefficient, 3 bipartite lattice, 86 biquadratic term, 224 block tri-diagonal equation, 148 Bogoliubov excitation, 19 bound state, 17, 37

four-spin interaction, 103 Fourier series, 50 Fourier transform, 38 Fredholm-type integral equation, 16 free energy, 10 fugacity, 12 fugacity expansion, 127 fundamental period-parallelogram, 241

Gaudin, 25 Gaudin-Takahashi equation, 133 generalized Bethe-ansatz, 25 Gibbs free energy, 12 grand canonical ensemble, 10 Griffiths, 42

bound state of pairs, 174 charge excitation, 95 checker board decomposition, 189 checker board lattice, 210 chemical potential, 10 classical Heisenberg model, 186 complete elliptic integral of the first kind, 239 complete elliptic integral of the second kind, 239 complex quasi-momentum, 16 conformal field theory, 224 conserved quantity, 85 continued fraction, 136 convergence radius, 107 convolution, 139 correlation function, 4 correlation length, 203, 223

Haldane gap, 223 half-filled state, 92 Heisenberg model, 34 high-temperature expansion, 124 hole distribution function, 110 hole excitation, 18 holon, 95 Hubbard model, 85 Hulthen, 37 hypergeometric function, 244 inhomogeneous eight-vertex model, 211 inhomogeneous six-vertex model, 189

delta-function boson, 10 251

252

irreducible representation of finite group, 226 Ising model, 46, 185 Ising-like antiferromagnet, 59 Jacobian elliptic function, 51, 241 Jacobian elliptic theta function, 71, 240 Jordan-Wigner transformation, 2, 3

Kosterlitz and Thouless transition, 224 ladder lattice, 157 Lagrange multiplier, 111, 147 Legendre polynomial, 187, 245 Lieb, 12 Lieb-Liniger equation, 15, 116 Lieb-Wu solution, 88 linked cluster expansion, 125 Liouville theorem, 76, 241 logarithmic anomaly, 45 logarithmic derivative, 73 long range order, 224 longitudinal correlation function, 4 low-temperature thermodynamics, 129 Luttinger liquid theory, 224 magnetic susceptibility, 41, 54, 188 magnetization curve, 41, 51, 53, 224 McGuire, 25 metal-insulator transition, 85 modified Bessel function, 244 modified spin-wave theory, 155, 156 modulus, 60 momentum distribution, 12 Mott transition, 85

nearly degenerate ground state, 57, 80 normalizability condition, 130, 133 pair of fermions, 167 pair-exchange operator, 22 parity of string, 134 particle distribution function, 110 particle excitation, 18 partition function, 3 Pauli matrix, 1, 73 periodic boundary condition, 10, 26, 35 permutation, 25 pressure, 11, 21

quantum Monte Carlo method, 155, 223 quantum transfer matrix, 156, 190 quasi-momentum, 15, 35 quasi-periodicity, 213 rapidity, 36 Roger's dilogarithm, 153

second neighbor correlation, 38 Shanks transformation, 224 shift operator, 35, 69 six-vertex model, 67, 189 specific heat, 4, 153

Index

spherical Bessel function, 187, 244 spherical harmonics, 187 spin coordinate, 21 spin excitation, 92 spin-wave gap, 82, 84, 158 spin-wave theory, 41 spinless fermion, 2 spinon, 40, 94 spinon continuum, 41 spinon gap, 81, 158 spontaneous order, 58 Stirling formula, 110 string, 18 string solution, 37, 48 sublattice, 85 Sutherland, 225 symmetry group, 226

Takahashi-Suzuki equation, 143, 149 Takhtajan and Babujian, 225 thermodynamic Bethe-ansatz equation, 116 thermodynamic limit, 11, 18 thermodynamic potential, 112 thermodynamic relations, 113 three-body problem, 13, 22 total angular momentum operator, 187 total momentum, 35 transfer matrix, 67, 185 transverse correlation function, 4 Trotter number, 198 two-point function, 7, 57, 186 unpaired fermions, 167 variational calculation, 111 Wick's theorem, 5 Wiener-Hopf factorization, 43 Wiener-Hopf type integral equation, 44 XXZ model, 34, 46 XY model, 1 XYZ model, 67 Yang, 25 Yang-Baxter relation, 14 Yang-Yang theory, 110 Young tableau, 226