Surface Physics Exercises in Quantum Mechanics

ARDB Technical Note 132 - Draft - 10/27/97

Surface Physics Exercises in Quantum Mechanics David H. Whittum In this note we review some amusing quantum mechanics problems relating to surface phenomena. The selection is a bit eclectic, but this is just a tech note after all.

Outline Introduction: Copper and Quantum Mechanics 1. Bound-States of a Point Charge Above a Conducting Surface 2. Field Emission from a Plain Vanilla Surface 3. Field-Emission Scaling from the Uncertainty Principle 4. RF Field Emission from a Plain Vanilla Surface 5. More Exercises Introduction It is helpful to review a couple of items at the outset. Before discussing a surface, it is good to consider an example of a solid; we pick copper. In addition, a short review of quantum mechanics may be helpful. Copper For the atomic physicist, Copper is a noble metal, with an atomic number Z=29, atomic mass A=63.5 amu, 1 the orbital configuration [ Ar ]3d 10 4 s , and ground level 2 S1 / 2 , with ionization energy 7.478 eV . As far as the solid state physicist is concerned, pure copper forms a facecentered cubic (fcc) lattice. This means that a perfect crystal of copper may be thought of as depicted in Fig. 1, when viewed up close, and in the bulk, beyond the surface,

FIGURE 1. Pure, single-crystal copper forms a face-centered cubic lattice.2

Algebraically, an fcc lattice is an array of points consisting of all integer combinations of the 1Recall that 1 amu is one atomic mass unit, defined as 1/12 the mass of the 12C carbon atom, or 1.6605x10-27 kg. 2 N. W. Ashcroft and N. D. Mermin, Solid State Physics, (Holt, Rinehart & Winston, Philadelphia, 1976) 1

following basis vectors, a a a ( yˆ + zˆ ), a2 = ( xˆ + zˆ ) , a3 = ( xˆ + yˆ ) , 2 2 2 and for copper, the dimension a=3.61Å. In a pure crystal of copper the density of conduction-band electrons is ne = 8.47 × 10 22 cm −3 , just one "free" electron per copper atom. The remaining 28 electrons per atom are bound, or "valence" electrons. If we convert the density of free-electrons to a volume per electon, then the radius of a sphere having the same volume is rs=1.41Å, at room temperature, and atmospheric pressure. The Fermi energy is ε F = 7.00 eV, the Fermi temperature is TF = 8.16 × 10 4 °K , the Fermi wavenumber is kF = 1.36 × 108 cm −1 and the Fermi velocity is VF = 1.57 × 108 cm/s . The work function is, depending on the crystal plane, 4.59 eV (100), 4.48 3 eV (110), 4.98 eV (111), 4.53 eV (112). Let us recall the particulars concerning the concept of Fermi energy. One can find these considerations in a solid-state text (my favorite is Ashcroft and Mermin). Recall that electrons are fermions, governed by the Pauli exclusion principle. If an ensemble of electrons is confined to a system at zero temperature, they will fill available states, starting with the lowest energy state, 2 electrons per state (corresponding to the two possible values for spin), until all N electrons are accomodated. The energy of the highest energy filled state is then called the Fermi energy ε F , and it may be related to the Fermi wavenumber kF according to a1 =

h 2 kF2 . 2m

εF =

In depicting electron states in a solid, we look into the bulk of the solid, and consider a cubic volume of length L, and volume V=L3. Momentum eigenstates are separated by ∆k = 2π / L and thus the number of eigenstates in a k-space volume Ω is Nk =

Ω 1 = ΩV . 3 ∆k 8π 3

Let us fill these states, 2 electrons per state, until the highest wavenumber, kF , is reached. Then we are filling a k-space volume Ω=

4π 3 kF , 3

and we have N electrons, so that N = 2 Nk F

1 4π 3 kF3 =2× 3V kF = 2 V . 8π 3 3π

Evidently Fermi wavenumber may be related to the density of free electrons, 3

Physics Vade Mecum, A Physicist’s Desk Reference, Herbert L. Anderson, ed., (American Institute of Physics, New York, 1989) pp. 306-317 and pp. 318-335. 2

ne =

N k3 = F2 . V 3π

Let us also notice that the Fermi energy is referred to the ground state. Thus if the ground state is separated from the potential at infinity (i.e., well beyond the surface of the solid) by an amount eV0 < 0 , then the highest energy level referred to that at infinity is,

ε F + eV0 = −eφ , where eφ is the work function, the amount of energy required to liberate an electron. For copper, taking a value eφ ≈ 4.65 eV as representative, and using ε F ≈ 7.00 eV, one has V0 ≈ −11.65 V. Quantum Mechanics Recall that in the quantum mechanical description of a one-electron system, and neglecting r spin, we describe the state of the system with a complex wavefunction Ψ(r , t ) , a function of space and time coordinates. The evolution of the system is governed by the Hamiltonian operator H , HΨ = ih

∂Ψ , ∂t

and we may express H as H=−

r h2 r 2 ∇ + V (r ), 2m

with V the potential. The quantity

ρ P = ΨΨ∗ , may be interpreted as a probability density. Associated with this is a law of conservation of probability, taking the form, r r ∂ρ ∇ • J P + P = 0, ∂t with r r r h Ψ ∗∇Ψ − Ψ∇Ψ ∗ . JP = 2m i

{

}

r It is helpful in analyzing the behavior of such systems, to determine the energy eigenstates ψ E (r ) such that r r Ψ(r , t ) = ψ E (r )e − iEt / h , is a solution, i.e., 3

r  h2 r 2 Hψ E = − ∇ + V (r )ψ E = Eψ E .   2m A general solution may then be represented as a superposition of energy eigenstates, r r Ψ(r , t ) = ∑ cEψ E (r )e − iE / ht . E

Let us turn next to consider some amusing problems at a conducting surface. 1. Bound-States of a Point Charge Above a Conducting Surface A recent article4 notes that the bound states of a point charge above a conducting plane form a Rydberg series. A hand-waving explanation of this observation would consult the picture of an image charge and discuss the two-body problem, with constrained center of mass coordinate. In this note we take a more pedantic approach to check the result quoted in the article. The image potential of an electron above a conducting plane takes the form5 e2 V (z) = − , 4z

(z>0)

with z the distance above the plane. We enumerate bound states of energy E. Schrodinger’s equation is  h2 2  ∇ + V ( z ) Ψ = EΨ , −  2m  and the problem is separable due to invariance under translations parallel to the surface, r r r Ψ(r⊥ , z ) = ψ ( z )eik ⊥ • r⊥ .

r r The coordinates in the plane are r⊥ = xxˆ + yyˆ , and the transverse momentum in this state is hk⊥ . We have r   h2 d 2  h 2 k⊥2  + V ( z ) ψ =  E − ψ, −  2 m dz 2  2 m   or, in normalized coordinates,  d2 1 2  2 + − q ψ = 0 . ξ  dξ  The notation is 4

U. Hofer, et al., “Time-Resolved Coherent Photoelectron Spectroscopy of Quantized Electronic States on Metal Surfaces”, Science 277 (1997) pp. 1480-1482. 5 J. D. Jackson, Classical Electrodynamics (John Wiley & Sons, New York, 1975) Ch. 2. 4

ξ=

z , z0

with z0 twice the Bohr radius, z0 = 2

° 2 h2 = r ≈ 1 . 1 A. e me 2 α 2

Here

α=

e2 1 ≈ , hc 137

is the fine structure constant, and re =

e2 ≈ 2.82 × 10 −13 cm , 2 mc

is the classical electron radius. We have adopted the notation r h 2 k⊥2 1 2 2 2 E= − α mc q , 2m 8 defining the variable q. Let us solve this equation for positive ξ . For ξ >> 1/ q our system takes the approximate form  d2 2  2 − q ψ ≈ 0 ,  dξ  and solutions finite at large ξ take the form ψ ∝ e − qξ . (Notice that solutions with imaginary q are possible; these correspond to unbounded motion). This suggests looking for a solution in the form

ψ = f (ξ )e − qξ . Then

ψ ′ = f ′e − qξ − qfe − qξ ,

ψ ′′ = f ′′e − qξ − 2 qfe − qξ + q 2 fe − qξ ,

and Schrodinger’s equation reduces to d2 f df f − 2q + = 0. 2 dξ dξ ξ We look for a series solution, 5

∞

f (ξ ) = ∑ ck ξ k , 0

and find c0 = 0 , and ck +1 =

2 qk − 1 ck . k (k + 1)

Notice that the series truncates if 2qk=1 for some k. On the other hand, if the series does not truncate, then, for large k we have ck +1 ≈

2q c, (k + 1) k

and in this case the solution for f differs by at most a finite polynomial from the function defined by Taylor coefficients (for some constant A), ck

k 2q) ( ≈A .

k!

This latter series is summable and gives f ≈ Ae 2 qz , corresponding to a divergent solution. We conclude that a finite solution requires truncation of the series, and therefore a discrete spectrum. We denote qn =

1 , 2n

and the corresponding eigenfunction,

ψn = e

− qnξ

n

∑c ξ , n k

k

1

with ckn+1 =

k / n −1 n ck . k (k + 1)

The corresponding spectrum is r r h 2 k⊥2 1 2 2 1 En k⊥ = − α mc 2 . 2 m 32 n

( )

The constant 1 2 2 α mc ≈ 0.85 eV , 32 6

is 1/16th of the Rydberg energy, 1 R = α 2 mc 2 ≈ 13.6 eV. 2 It will be convenient for later work to abbreviate

ε0 =

1 1 R = α 2 mc 2 ≈ 3.40 eV. 4 8

For perturbations respecting translational invariance parallel to the surface, and where symmetry permits, one expects to uncover the equivalent of the Lyman series ( n → 1), 0.64 eV (1.9 µm -"α ") , 0.76 eV (1.6 µm -" β ") , etc., converging to 0.85 eV (1.5 µm ) . The Balmer series ( n → 2) corresponds to 0.12 eV (10.5 µm -"α ") , 0.16 eV (7.8 µm -" β ") , etc., converging to 0.21 eV (5.8 µm ) . The Paschen series ( n → 3), corresponds to 41 meV (30 µm ) , 61 meV (21 µm ), converging to 95 meV (13 µm ) . In converting to eV we made use of hω = mc 2

re 2π 1.24 eV = , α λ λ ( µm )

to relate the transition energy to wavelength of the emitted photon, as for single-photon emission. Note that room temperature corresponds to kBT = kB × 300°K ≈

1 eV ≈ 26 meV . 38.7

It would be amusing to revisit this problem starting from the Dirac equation. Also, the treatment of z<0 is a bit fuzzy. To smoothly connect the potential outside to that inside, we should pay attention to the constellation of charges in the vicinity of the surface. 2. Field-Emission Scaling from the Uncertainty Principle A fundamental problem for high-gradient linacs is field-emission, the extraction of electrons from a conducting surface in the presence of a large field. Let us review here the problem first considered by Fowler and Nordheim. 6 We consider a conducting plane, to which is applied a DC electric field E0. This field lowers the potential beyond the surface, making it possible for electrons to tunnel through and be emitted. Let us determine the conditions favorable to emission with a scaling argument. In the course of the tunneling process, the uncertainty, ∆p, in the electron momentum is related to the potential according to 2 ∆p ) ( eφ ≈

2m

⇒

∆p ≈ (2 meφ ) . 1/ 2

From Heisenberg’s principle, the uncertainty in position is

6

R. H. Fowler and L. W. Nordheim, Proc. Roy. Soc (London) A 119 (1928) 173. 7

∆x ≈

h h ≈ 1/ 2 . 2 ∆p 2(2 meφ )

This corresponds to an uncertainty in energy on the order of ∆ε ≈ eE∆x . When this uncertainty is on the order of the work function, ∆ε ≈ eφ , one expects to find electrons being emitted. In this case 2(2 m) (eφ ) eEh 1/ 2

h eφ ≈ eE∆x ≈ eE 1/ 2 2(2 meφ )

⇒

3/ 2

≈ 1.

We may express this in normalized form using E=

mc 2 1 (α / 2)4 µ , e re

ε0 =

eφ eφ =8 2 2. ε0 α mc

1 2 2 α mc , 16

In terms of normalized variables, our condition is, after some algebra, 2  eφ    µ  ε0 

3/ 2

≈ 1.

In practical units, using µ ≈ 3.11 × 10 −2 E0 (GV/m ) , and eφ / ε 0 ≈ 4.65 / 3.40 ≈ 1.37 (for copper), we may write this as E0 ≈ 100 GV/m. At this field level one would expect the orderly bound states characteristic of the solid to lose their integrity. In a similar manner we can estimate the tunneling time, h h (2 meφ ) ∆t ≈ ≈ ≈ eE 2 ∆ε 2eE∆x

1/ 2

α m c  eφ  ≈   2eE  ε 0 

1/ 2

.

For copper, with eφ / ε 0 ≈ 1.37 , this gives ∆t ≈ 10 −14 s / E(GeV/m ) , quite a short time scale, when the field is large. 3. Field Emission from a Plain Vanilla Surface Let us consider next a more quantitative approach to field emission, being interested in the scaling for the emitted current density as a function of applied field. Continuing with the notation of the Sec. 1, Schrodinger’s equation may be cast in the form, for ξ > 0,  d2 1  2  2 + − q + µξ  ψ = 0 , ξ  dξ 

( ξ > 0)

We will consider the simplest of surfaces in this section, and discard the image potential. In this equation, 8

µ=

eE0 re −2 4 ≈ 3.11 × 10 E0 ( GeV/m ) . 2 mc (α / 2)

In the first approximation, we are interested in the limit µ << 1. The quantity q 2 is related to the energy of this eigenstate according to r h 2 k⊥2 1 2 2 2 E= − α mc q . 2m 8 For ξ < 0, we consider a propagating solution, our picture of a conduction band electron,  d2 2  2 + Q ψ = 0 .  dξ  The quantity Q2 is related to the potential within the surface, V0 < 0 , according to r h 2 k⊥2 1 2 2 2 E − V0 = + α mc Q . 2m 8 This is to say that Q2 + q 2 =

−V0 = ρ2 . R/ 4

Solutions for ξ < 0 are just plane waves,

ψ = A+ eiQξ + A− e − iQξ . Let us solve Schrodinger’s equation in the region ξ > 0,  d2  q 2  µ ξ  dξ 2 +  − µ   ψ = 0 .    We make the change of variables,   q2 η = µ1 / 3  − ξ  ,  µ so that  d2   2 − η ψ = 0 .  dη 

9

This we recognize as Airy’s equation, with solutions Ai and Bi. Asymptotic forms are7 1 Ai(η > 0) → π −1 / 2η −1 / 4 e −ς , 2 π Ai( −η < 0) → π −1 / 2η −1 / 4 sin ς +  ,  4 1 Ai ′(η > 0) → − π −1 / 2η1 / 4 e −ς , 2 π Ai′( −η < 0) → −π −1 / 2η1 / 4 cos ς +  ,  4

Bi(η > 0) → π −1 / 2η −1 / 4 eς ,

π Bi( −η < 0) → π −1 / 2η −1 / 4 cos ς +  ,  4 Bi ′(η > 0) → π −1 / 2η1 / 4eς ,

π Bi′( −η < 0) → π −1 / 2η1 / 4 sin ς +  ,  4

and we abbreviate 2 ς = η3 / 2 . 3 Our interest is to match to an outgoing solution in the region ξ > 0, and this corresponds to

ψ (ξ ) = Bi[η(ξ )] + iAi[η(ξ )] .

( ξ > 0)

We match the solutions at

η0 = η(ξ = 0) =

q2 . µ2 / 3

Matching conditions are

ψ (0 − ) = A+ + A− = ψ (0 + ) = Bi(η0 ) + iAi(η0 ) , dψ − dψ − dη 0 ) = iQ( A+ − A− ) = 0 )= Bi ′(η0 ) + iAi ′(η0 )) = − µ 1 / 3 ( Bi ′(η0 ) + iAi ′(η0 )) . ( ( ( dξ dξ dξ The solution is  i  1 iµ 1 / 3 iµ 1 / 3 + A+ =  Bi(η0 ) + Bi ′(η0 ) Ai ′(η0 ),  Ai(η0 ) + 2 Q Q  2  1/ 3 1/ 3  i  1 iµ iµ A− =  Bi(η0 ) − Bi ′(η0 ) +  Ai(η0 ) − Ai ′(η0 ) . 2 Q Q  2  In the limit of not too large a gradient, we have η0 >> 1, and may evaluate these coefficients using the asymptotic forms A+ ≈

1  −1 / 2 −1 / 4 ς iµ 1 / 3 −1 / 2 1 / 4 ς  η −1 / 4 eς π η e ≈ π η e + 1/ 2 2 Q  2π

7

 iµ 1 / 3η1 / 2  , 1 + Q  

M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions (Dover, New York, 1972), p.446. 10

η −1 / 4 eς  iµ1 / 3η1 / 2  A− ≈ . 1 − Q  2π 1 / 2  For very large gradients, the coefficients must be evaluated numerically. Routines for this purpose may be found in Numerical Recipes.8 With the solution in hand, we may derive a figure for the net flux far to the right of our surface, and compare that to the incident flux, to derive a transmission probability, T. Probability flux, evaluated to the left of the surface, is JP z =

∗ ∂ h A+ eiQξ + A− e − iQξ A+ eiQξ + A− e − iQξ + c.c. 2m i ∂z

(

) (

)

=

h A+ eiQξ + A− e − iQξ 2 m iz0

=

hQ hQ A+ A+∗ + A− A+∗ − A−∗ A+ + A− A−∗ ) + c.c. ( A+ + A− )( A+∗ − A−∗ ) + c.c. = ( 2 m z0 2 m z0

=

hQ ( A+ A+∗ − A−∗ A− ) m z0

(

) (iQ)( A e ∗

+

iQξ

)

− A− e − iQξ + c.c.

Probability flux, evaluated far to the right of the surface, is JP z =

( [

]) ∂∂z ( Bi[η(ξ )] + iAi[η(ξ )]) + c.c.

h Bi η(ξ ) + iAi η(ξ ) 2m i

]

[

∗

1/ 3  ∗ µ h = Bi(η) + iAi(η))  (  ( Bi ′(η) + iAi ′(η)) + c.c. 2m i  z0 

≈

∗ hµ 1 / 3 Bi(η) + iAi(η)) ( Bi ′(η) + iAi ′(η)) + c.c. ( 2 m iz0 ∗

hµ 1 / 3  −1 / 2 −1 / 4 π   π  exp − i ς +    iπ −1 / 2η1 / 4 exp − i ς +   + c.c. ≈ π η   2 m iz0  4   4  ≈

hµ 1 / 3 m z0 π

The transmission probability is then hµ 1 / 3 J (ξ > 0) µ1 / 3 m z0π T = Pz = . = J P z + (ξ < 0)  hQ  Qπ A+ A+∗ ∗   A+ A+  m z0  In the limit of not too large a gradient ( η0 >> 1), this is W.H. Press, S. A. Teukolsky, W. T. Vetterling, and B. P. Flannery, Numerical Recipes, The Art of Scientific Computing, (Cambridge University Press, Cambridge, 1992). 8

11

T≈

4e −2ς .  Q µ 1 / 3η01 / 2   1/ 3 1/ 2 +  Q   µ η0

This may be made more explicit using η01 / 2 = q / µ 1 / 3 ,  4 q3  T≈ exp − . 3 µ Q q   +   q Q 4

The kinematic dependence may be made more explicit,  4 ( ρ 2 − Q2 )3 / 2  4q ρ 2 − q 2  4 q 3  4Q ρ 2 − Q 2 T≈ exp − exp − . = ρ2 ρ2 µ  3 µ  3  Next we would like to convolve this probability, with the incident flux corresponding to the spectrum of momenta characteristic of conduction band electrons in copper. For copper at zero temperature, this distribution is simple; the distribution in momenta, r r Q k = k⊥ + zˆ z0 r is uniform over a sphere in k -space, with radius, kF . Thus for example, the maximum Q is given by, Qm = kF z0 ≈ 1.36 × 108 cm −1 × 1.06 × 10 −8 cm ≈ 1.44 , where we evaluate this for copper. The quantity,

ρ=

−V0 11.65 eV ≈ ≈ 1.85 , ε0 3.4 eV

(with slightly different values depending on the choice of crystal plane), and we employed ε 0 = R / 4 ≈ 3.4 eV. r To simplify our notation, let us define a normalized coordinate in k -space, r r r ˆ . The probability flux in z associated with a momentum state is Q = kz0 = k⊥ z0 + zQ J Pz =

hQ , m z0

and the quantity with the classical analog of current density is then

12

jz = −ene

r r hQ 3 d Q T Q ∫ m z0 Q >0

( )

1 4πQm3 / 3

(

 4 ρ 2 − Q2 r 4Q 2 ρ 2 − Q 2  dQ exp − 2 ∫ ρ µ Q >0  3

h 3 = −ene 3 4π Qm m z0

3

)

3/ 2

   

,

where the integral is over the half-sphere of radius Q m , in Q-space. Employing spherical coordinates, this may be expressed as 3 h jz = −ene 2 Qm3 m z0

 4 ( ρ 2 − Qr2 χ 2 )3 / 2  4Qr2 χ 2 ρ 2 − Qr2 χ 2 exp − . ∫0 Q dQr ∫0 d χ ρ2 µ  3 

Qm

1

2 r

Let us introduce

ρˆ = ρ / Qm ,

µˆ = µ / Qm3 ,

and observe the relation to work-function. Fermi energy is just the energy of the most energetic occupied state at zero-temperature, referred to the ground-state, 1 ε F = α 2 mc 2Qm2 = ε 0Qm2 = EF − V0 = −eφ − V0 , 8 where E F is the energy of the most energetic occupied state, referred to a potential of zero at infinity, beyond the surface. This relation may be expressed as −eφ εF = Qm2 = + ρ2. ε0 ε0 Thus 2

 ρ eφ 1 eφ ρˆ − 1 =   − 1 = . = 2 ε 0 Qm ε F  Qm  2

Our integral may be expressed as 1

1

jz = j0 ∫ dr ∫ d χ r χ 4

0

0

2

 4 ( ρˆ 2 − r 2 χ 2 )3 / 2  ρˆ − r χ exp − , µˆ  3  2

2

2

with j0 = −ene

6hQm . m z0 ρˆ 2

Let us make this expression a bit more explicit. The number density is 13

k3 1 2m ne = F2 = 2  2 ε F   3π 3π  h

3/ 2

1 α 6  εF  =   24π 2 re3  ε 0 

3/ 2

.

Other terms are ε  Qm =  F   ε0 

1/ 2

−1

1  eφ  , 2 = 1 +  , ρˆ εF  

h 1 = α. m c z0 2

Thus 1 α 6  εF  j0 = −ec ×   24π 2 re3  ε 0  2

3/ 2

α 7 ec  ε F   eφ  = − 2 3   1 +  εF  8π re  ε 0  

ε  ×6× F  ε0  −1

1/ 2

 eφ  1 +  εF  

−1

1 × α 2 −1

2

 ε   eφ  ≈ 3 × 10  F  1 +  A/cm 2 εF   ε0   12

For copper, eφ ≈ 4.65 eV , ε F ≈ 7.00 eV, and using ε 0 ≈ R / 4 ≈ 3.40 eV , we have,

ε F 7.00 ≈ ≈ 2.06 , ε 0 3.40

eφ 4.65 ≈ ≈ 1.37 , ε 0 3.40

eφ ≈ 0.665 . εF

In addition, making use of only the very latest values for the fundamental constants,9

α ≈ 1 / 137.0359895 (61) , e ≈ 1.60217733( 49) × 10 −19 C, re ≈ 2.817940 92(38) × 10 −15 m,

me c 2 ≈ 0.510 999 06(15) MeV , c ≈ 2.99 792 458 × 108 m / s, π ≈ 3.141592 65359K,

one finds

α 7 ec ≈ 7.49 × 1011 A/cm 2 , 2 3 32π re so that j0 ≈ 7.6 × 1012 A/cm 2 . It is straightforward to evalute this integral numerically. We are interested in parameter values, ρˆ ≈ 1.85 / 1.44 ≈ 1.28, and in gradients from 1 GV/m to perhaps as large as 10 GV/m or µ ≈ 3.11 × 10 −2 − 3.11 × 10 −1 . This corresponds to µˆ ≈ 1.0 × 10 −2 − 1.0 × 10 −1. With such values of µˆ , an analytic estimate of the integral is worthwhile. With a change of variables, r ≈ 1 − ε r , χ ≈ 1 − ε χ , and a small argument approximation,

9

“Review of Particle Physics”, Particle Data Group, Phys. Rev. D, 54 (1996) 65. 14

1

jz ≈ j0 ρˆ − 1 ∫ 2

0

≈ j0

 4 ( ρˆ 2 − r 2 χ 2 )3 / 2  dr ∫ d χ exp −  µˆ  3  0 1

1  4 ( ρˆ 2 − 1)3 / 2  1 1/ 2  4 2 ρˆ − 1 exp − d ε ρˆ − 1) ε r + ε χ (  r ∫ dε χ exp − ∫ 3 µˆ  µˆ 0   0

(

2

). 

We approximate the exponential integrals with 1

∫ 0

∞

1/ 2  1/ 2   4  4 µˆ dε exp − ( ρˆ 2 − 1) ε  ≈ ∫ dε exp − ( ρˆ 2 − 1) ε  ≈ 1/ 2 . 2  µˆ  0  µˆ  4( ρˆ − 1)

We then obtain

 4 ( ρˆ 2 − 1)3 / 2    4 ( ρˆ 2 − 1)3 / 2  −1 / 2 2 µˆ 1 2 ρˆ − 1 exp − (ρˆ − 1) µˆ exp− 3 µˆ .  1 / 2  ≈ j0 2 ˆ 3 µ 16 ˆ   4( ρ − 1)    2

jz ≈ j0

2

This may be expressed as 1 ε  jz ≈ j0  F  16  eφ 

1/ 2

5/ 2 1/ 2  4  eφ  3 / 2   4  eφ  3 / 2 Q3  µ2 1  ε0   ε0  2 m exp −    ≈ j0     µ exp −   , Qm6 µ  4  eφ   ε F   3  ε F   3µ  ε 0  

or −1

α 7 ec  eφ   ε 0  jz ≈ 1 +    128π 2 re3  ε F   eφ 

1/ 2

 ε0     εF 

1/ 2

 4  eφ  3 / 2  µ exp−    .  3µ  ε 0   2

In practical units this is −6

jz ≈ 6.15 × 10 A/cm

2

(ε F / eφ )1 / 2 E 2 exp−6.838 × 10 7 (eφ )3 / 2  ,  E0  (eφ + ε F ) 0 

(eV and V/cm units)

with work function and Fermi energy in units of electron-volts, and electric field in units of volts/cm. [The expression given by Gomer 10 has the coefficient 6.2 × 10 6 A/cm 2 , still tracking down the source of the discrepancy.] In this, we made use of µ ≈ 3.11 × 10 −9 E0 (V/cm ) . For copper we have

10

Robert Gomer, Field Emission and Field Ionization, (Harvard University Press, Cambridge, 1961) 15

1 1  jz ≈ j0   16 1.37 

1/ 2

 1   2.06 

5/2

 4 3/ 2  µ 2 exp − (1.37)   3µ 

 2.14   2.14  11 2 2 ≈ 0.035 j0 µ 2 exp −  ≈ 2.8 × 10 A/cm × µ exp −   µ   µ   68.8  ≈ 2.5 × 108 A/cm 2 × E 2 (GeV/m ) exp − .  E(GeV/m )  Thus at 4 GV/m jz ≈ 125A/cm 2 . For large enough current density, one may become concerned with the cold surface approximation. 4. RF Field Emission from a Plain Vanilla Surface The typical situation of interest for high fields (>100 MeV/m), corresponds to a field varying sinusoidally in time at microwave frequencies. At such frequencies and fields electrons are interacting with a large number of photons, and photon energies are small compared to the energy level spacings, and we may continue to treat the electric field classically. In this section, we derive the amended scaling for an rf field, with period long compared to the tunneling time. As we’ve seen, tunneling time is short ∆t ≈ 10 −14 s / E(GeV/m ) , while even at 100 GHz, the rf period (10 ps) is relatively long in comparison. Consider a function f ( x ) = x 2 e −1 / x , with x = A sin θ and let us average over one-half cycle in angle θ, assuming A << 1. We expand 1 1 1 2 ≈ + ε , x A 2A where

θ=

π +ε. 2

We have f ≈

π

+∞

 ε 2  1 2 −1 / A  ε2   2 1 2 −1 / A A sin θ e exp − A e exp ≈ ( )   ∫−∞  − 2 A  ≈  π   2 A π π ∫0

This implies that our DC emission law, a j ≈ b E 2 exp −  ,  E takes the form

16

1/ 2

A2.5e −1 / A .

 1  j≈   2π a 

1/ 2

a b E 2.5 exp −  ,  E

when expressed in terms of rf amplitude. An additional factor of 1/2 appears here, since no emission occurs during one-half cycle of the rf. The net correction is essentially a “duty-cycle” correction, j ≈ ηrf j peak , amounting to the fraction of time during which the field is sufficiently near the peak to extract appreciable current,  E  ηrf =    2π a 

1/ 2

.

The approximation made in expanding the term in the exponent amounts to the assumption ηrf << 1 . To compute the time-averaged current density for a pulsed rf signal, one would need an additional duty cycle correction. For example, for a 1 µs pulse length, and a 60 Hz pulse repetition frequency, the additional duty cycle correction is 6 × 10 −5 . More explicitly, the average current density, within the rf pulse then takes the form −1

31 / 2 α 7 ec  eφ   ε  j ≈ 17 / 2 5 / 2 3 1 +   0  re  2 π ε F   eφ 

5/ 4

 ε0     εF 

1/ 2

µ

5/2

 4  eφ  3 / 2  exp −   .  3µ  ε 0  

In practical units,  68.8  j ≈ 1 × 10 7 A/cm 2 × E 2.5 (GeV/m ) exp − .  E(GeV/m )  Thus at 4 GV/m, j ≈ 10A/cm 2 . Wang and Loew quote a coefficient that equates to 1.6 × 108 A/cm 2 .11 [track down discrepancy, their result is closer than Gomer’s...] 5. More Exercises There are many more exercises one could cram in here: a general WKB treatment of tunneling, incoporation of the image potential, a dielectric inclusion layer, space-charge and thermal effects. Meanwhile, time is up for this note, and we must turn to other things...

11

G. A Loew and J. W. Wang, “RF Breakdown Studies in Room Temperature Electron Linac Structures”, XIIIth Internat’l Symposium on Discharges and Electrical Insulation in Vacuum, (1988) SLAC-PUB-4647. 17