School of Physical Sciences
PH 605 Thermal and Statistical Physics Part II: Semi-Classical Physics Quantum Statistics course-webpage: http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH605/PH605.html
Dr. Peter Blümler (
[email protected]) Room 123, Phone: 3228
Syllabus 3.
Semi-Classical Physics .............................................................................................2 3.1
The Boltzmann Distribution (derived!)........................................................................2 3.1.1 A simple example ..............................................................................................2 3.1.2 Generalisation....................................................................................................3
3.2
The Semi-Classical Perfect Gas ..................................................................................8 3.2.1 3.2.2 3.2.3 3.2.4
Definition of the semi-classical, mon-atomic, perfect gas ......................................8 Distinguishable / indistinguishable particles? .........................................................9 Contributions of different types of motion to Z1 .................................................. 13 The density of states ........................................................................................ 14
3.2.5 Partition function for translational motion, Z1tr .................................................... 17 3.2.6 3.2.7 3.2.8 3.2.9
3.3
Partition function of internal motion, Zint............................................................. 20 Partition function of (molecular) rotation, Zrot ..................................................... 21 Partition function of (molecular) vibration, Zvib ................................................... 24 Partition functions and comparison to experimental data ..................................... 28
Entropy and Energy of the Semi-Classical Gas .......................................................... 29 3.3.1 Entropy of a mon-atomic gas, the Sackur-Tetrode equation ............................... 29 3.3.2 The entropy of mixing-the Gibbs paradox.......................................................... 31 3.3.3 The principle of the equipartition of energy ........................................................ 32
3.4
Validity and Limit of the Semi-Classical Description .................................................. 35 3.4.1 The classical limit............................................................................................. 35 3.4.2 Maxwell velocity distribution in a classical gas................................................... 38 3.4.3 Rotational specific heat of diatomic molecules- ortho/para 1H2 ............................ 41
4.
Quantum Statistics................................................................................................... 45 4.1
Ideal Solids .............................................................................................................. 45 4.1.1 Einstein's theory of an ideal crystal................................................................... 45 4.1.2 Debye's theory of an ideal crystal..................................................................... 48
4.2
Quantum Statistics .................................................................................................... 59 4.2.1 4.2.2 4.2.3 4.2.4 4.2.5 4.2.6
4.3
Bose-Einstein statistics.................................................................................... 62 Fermi-Dirac statistics...................................................................................... 64 Comparison of Boltzmann, BE and FD statistics................................................ 66 Determination of α .......................................................................................... 68 Systems with variable particle number............................................................... 70 The Grand partition function, Z.......................................................................... 71
Application to Fermion/Boson-Systems .................................................................... 76 4.3.1 4.3.2 4.3.3 4.3.4 4.3.5 4.3.6
Free electrons in metals.................................................................................... 76 Pauli-paramagnetism....................................................................................... 83 The perfect photon gas - black-body radiation.................................................... 85 Bose-Einstein condensation ............................................................................. 88 Superconductivity and superfluidity, BEC........................................................... 95 Thermodynamics of stars ............................................................................... 103
PH 605: Thermal & Statistical Physics
page:2
3. Semi-Classical Physics
Recommended Books / Background Reading for this second part: • F. Mandl, "Statistical Physics" , Wiley, 1988 [QC175, 17 copies] • R. Baierlein: "Thermal Physics" , Cambridge University Press, 1999 [ISBN:0-521-65838-1]
3
Semi-Classical Physics
3.1
The Boltzmann distribution (derived!)
The Boltzmann distribution was introduced in the last section of part I (see Dr. Mallett’s script). recap: micro-state: macro-state:
certain assignment of particles to certain (energy) state realised by many micro-states („sum of micro-states“)
Ludwig Boltzmann concluded from the 2nd law of Thermodynamics that the macro-state with the most micro-states is the most stable in equilibrium (remember S = k B ln Ω ).
3.1.1
A simple example There are 3 (distinguishable, independent and identical) particles A, B and C. They are allowed to occupy 4 different energy states: ε 0 = 0, ε 1, ε 2 = 2ε 1 and ε 3 = 3ε 1 (e.g. harmonic oscillator). The total energy of the system amounts to 3ε 1. The occupation numbers are N0, N1, N2 and N3.
Now we are going to try to find the number of macro-states by which the system can be realised. 3
∑ Ni
E=
3
∑ N i εi
macro-state
N0
N1
N2
N3
I
2
0
0
1
3
3ε 1
II
1
1
1
0
3
3ε 1
III
0
3
0
0
3
3ε 1
i =0
i =0
We see that there are only 3 possible macro-states for the system. The next question is then: How many micro-states are possible to realise each macro-state? Note: We recall/realise that exchange of particles in the same micro-state doesn’t generate a new micro-state! energy state
macro-state I
ε3
A
B
C
-
-
-
-
-
-
-
ε2
-
-
-
A
A
B
B
C
C
-
ε1
-
-
-
B
C
A
C
A
B
ABC
ε0
BC
AC
AB
C
B
C
A
B
A
-
no. of micro-states
© Dr. Peter Blümler
3
macro-state II
6
School of Physical Sciences
macro-state III
1
University of Canterbury
PH 605: Thermal & Statistical Physics
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3. Semi-Classical Physics
Hence macro-state II has the highest statistical weight (Ω or thermodynamic probability or number of arrangements; often written as W for German „Wahrscheinlichkeit“ = probability) , Ω II = 6, in this example.
3.1.2
Generalisation
Inspired from this example we want to generalise this for N particles. For single occupation Ni = 1 it can be directly concluded that the highest statistical weight is given by all permutations, or Ω = N! However, if we consider cases in which the occupation number can eventually become larger than 1 (Ni > 1) we are overestimating by this method. This is because the permutation of particles in each individual micro-state doesn’t generate a new micro-state. Hence, we need the following correction: Ω=
N! N0 ! N1! N 2 !....
[3.1.1]
Note: The meaning of this equation can easily be checked on the previous example. 3! 6 macro-state I: ΩI = = =3 3 2! 0! 0!1! 2 3! 6 macro-state II: Ω II = = =6 3 1!1!1! 0! 1 3! 6 macro-state III: Ω III = = =1 3 0! 0! 3! 0! 6 We also know that in equilibrium the Boltzmann (entropy) equation tells us that the most probable is realised (for maximum entropy), or S = k B ln Ω max
[3.1.2]
Starting from these facts we now want to derive the equation for the Boltzmann-distribution: Given: N particles (distinguishable, independent and identical) in r energy states: ε 0 , ε1 , ε 2 , ..., ε r −1 with occupation numbers N 0 , N1 , N 2 , ..., Nr −1 additionally we can establish the following boundary conditions: a) total number of particles b) total energy
r −1
∑ Ni = N
= constant
[3.1.3]
∑ N i εi = E
= constant
[3.1.4]
i =0 r −1 i =0
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:4
3. Semi-Classical Physics
our goal is summarised in eq. [3.1.2]: We have to find the maximum statistical weight, or Ω=
N! r −1
→ maximum
[3.1.5]
∏ Ni ! i= 0
To simplify this task, we realise that when Ω has a maximum, lnΩ also must have a maximum (because the logarithm is a monotonic function). This enables us to use Stirling’s approximation: ln N ! = N ln N − N Hence, eq. [3.1.5] becomes:
for large N:
r −1
[3.1.6]
r −1
N! ln Ω = ln r −1 = ln N ! − ln N i ! = ln N! − ln N i! i =0 i =0 Ni!
∏
∏
i =0 [ 3 .1 .6 ]
=
N ln N − N −
∑
r −1
∑( Ni ln N i − N i )
i =0
Note: we will later see that for realistic conditions the last step (applying Stirling’s formula to Ni, hence Ni = large) is satisfied. ln Ω = N ln N − N −
r −1
∑ (N i ln Ni − Ni ) =
N ln N − N −
i =0
r −1
∑ N i ln Ni
i= 0
+
r −1
∑ Ni
i1 =2 03
= N [3 .1 .3 ]
ln Ω = N ln N −
r −1
∑ Ni ln N i
[3.1.7]
i =0
The maximum statistical weight (eq. [3.1.5]) is then given for: ∂ ln Ω = 0 ∂N i N , E
[3.1.8]
Rather than differentiating with respect to the occupation numbers Ni (ä¥), it is instructive to consider small changes (symbol δ) of the occupation number. Hence eq. [3.1.8] becomes
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
− δ ln Ω = δ
∑
N i ln N i
page:5
3. Semi-Classical Physics
product rule
=
∑ N i δ1ln2N3i + ∑ ln N i δN i = 0 δN i Ni
− δ ln Ω =
∑ δNi + ∑ ln Ni δN i = 0
[3.1.9]
Equation [3.1.9] can be combined with the constant boundary conditions in eqs. [3.1.3] and [3.1.4].
∑ δNi + ∑ ln Ni δN i = 0 ∑ δN i = 0 ∑ ε i δN i = 0
i)
− δ ln Ω =
(maximum)
ii)
−δN =
(const. must not change)
iii)
−δE =
(const. must not change)
The easiest way to solve such an equation system or to combine the conditions is the method of „undetermined (Lagrange) multipliers“ (recap: M. Boas: „Mathematical Methods in the Physical Sciences“, 2nd ed., Wiley 1983, page 174ff.). This gives:
∑ δN i + ∑ ln N i δNi + λ∑ δN i + β∑ ε i δNi = 0 ∑ δN i [1 + ln N i + λ + βεi ] = 0
[3.1.10]
where λ and β are the (yet) undetermined multipliers. The first term ( δNi ) in [3.1.10] can be chosen arbitrarily to be any number as long as the last two are chosen to fulfil ii) and iii) in the conditions above. But generally the following condition must hold: 1 + ln Ni + λ + βεi = 0 ln N i = − ( λ + 1) − βεi N i = exp [− (λ + 1)] exp [− βεi ] ≡ α exp [− βεi ]
[3.1.11]
with α ≡ exp [− ( λ + 1) ] . What is left to do? We have to find expressions for α and β. Determination of α: From eq. [3.1.3]
r −1
r −1
r −1
i =0
i =0
i= 0
∑ N i = ∑ α exp [− βεi ] = α ∑exp [− βεi ] = N α =
N r −1
[3.1.12]
∑e − βε i
i =0 © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
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3. Semi-Classical Physics
Determination of β: r −1
E =
From eq. [3.1.4]
r −1
∑ Ni ε i
r −1
= α
i =0
∑ ε i e − βε i
N
[ 3. 1. 12 ]
∑ ε i e − βε i
i =0 r −1
=
[3.1.13]
∑ e − βε i
i =0
i =0 r −1
N now a trick:
∑ ε i e − βε i
i =0 r −1
E =
∑e
r −1
∂ e −βε i r −1 ∂β i = 0 ∂ = − N r −1 =−N ln e − βε i ∂β i = 0 − βε i e 14243
∑
∑
∑
− βε i
i =0
i =0
[3 .1 .12 ] =
N α
N doesn’t depend on β, but α does, hence, E = −N
( )
∂ ∂ ln α ln α −1 = N ∂β ∂β
[3.1.14]
From eq. [3.1.2] we also know: S = k B ln Ω
[ 3 .1 .7 ]
=
[
r −1 k B N ln N − Ni ln Ni i =0
= k B N ln N − α
∑
∑e − βε i (ln α − βεi )] =
(
r −1 [3 .1 .11 ] = k B N ln N − α e − βε i ln α e − βε i i= 0
∑
[
k B N ln N − α ln α
)
e − βε i + αβ∑ ε i e −βε i ∑ 1424 3 14243
[ 3 .1 .12 ] =
N α
[ 3 .1 .13 ] =
S = k B [ N ln N − N ln α + β E]
]
E α
[3.1.15]
from the previous part (Maxwell relations: we recall, we use E here for the total internal energy -rather than U- because it typically used in QM notation) ∂U ∂E T = = ∂S V ∂S V ∂E ∂β V , N ∂E T = = ∂S V , N ∂S ∂β V , N [3 .1 .15 ] ∂S = kB ∂β V , N
[ − N ∂∂β ln α + E + β ∂∂Eβ 14243
V,N
[3.1.16]
]= k
B
∂E β ∂β V , N
[ 3 .1 .4 ] = − E
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
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3. Semi-Classical Physics
and with eq. [3.1.16] we get: ∂E ∂β V , N T= ∂S ∂β V , N
∂E ∂β V , N = ∂E k Bβ ∂β V , N β=
=
1 k Bβ
1 k BT
[3.1.17]
if we now insert this equation and eq. [3.1.12] into eq. [3.1.11]: ε N exp − i kBT N i = α exp [− βεi ] = r −1 ε exp − i k BT i =0
∑
we get the
Boltzmann-distribution: ε ε exp − i exp − i Ni k BT ≡ k BT = r −1 N ZN ε exp − i k BT i =0
[3.1.17]
∑
where ZN is called ‘partition function’ (Z from German: „Zustandssumme“ = sum of states of existence) We see, how a statistical distribution can be derived from very simple assumptions. The only assumption was eq. [3.1.2]: S = k B ln Ω max which is the statistical interpretation of the second law of Thermodynamics. We will apply the same formalism to obtain other (quantum) statistical distributions later.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
3.2
page:8
3. Semi-Classical Physics
The Semi-Classical Perfect Gas
Now we want to apply Boltzmann statistics to simple systems. The most simple system is the perfect (or ideal) gas.
3.2.1
Definition of the semi-classical, mon-atomic, perfect gas semi −2 classical 144 443, QM to determine energy levels
mon atomic 14 4−2 44 3,
no internal energy or structure
perfect gas 1424 3
number of particles is much smaller th an available energy levels (no degeneracy)
This is some hybrid between quantum-mechanical and classical behaviour, however we assume to be able to explain more effects more precisely than in the pure classical picture. Definition: • identical gas particles (molecules or atoms) • only very weak (no) interaction between the particles (particles are separated, low density), (in energetic terms this means that the potential energy of interaction is negligible compared to their kinetic energy V<
N=
∞
∑ Nr
[3.2.1]
r =1
• the individual energies -each particle can exist in- are: • forming a complete set of discrete quantum states, with no.
ε1 ≤ ε 2 ≤ ε 3 ≤ K ≤ ε r ≤ K 1, 2,
• hence the state of the system is: N1 particles in (quantum) state 1 with energy ε 1 N2 particles in (quantum) state 2 with energy ε 2 N3 particles in (quantum) state 3 with energy ε 3 M M M M M Nr particles in (quantum) state r with energy ε r M M M M M • it has a (total) internal energy of E = E( N1 , N 2 ,K , N r , K) =
© Dr. Peter Blümler
School of Physical Sciences
∞
3, ..... , r,....
∑ N r εr
[3.2.2]
r =1
University of Canterbury
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3.2.2
page:9
3. Semi-Classical Physics
Distinguishable / Indistinguishable Particles?
Because the particles are chosen to be non-interacting, we can select a single particle (independent of the others) to represent a typical particle. Hence, Z1 (T , V ) =
micro-state:
∞
ε
[3.2.3]
r =0
Z N (T ,V ) =
macro-state:*
∑ exp − kBrT
E
∑exp − k BrT
[3.2.4]
r *however, the latter we do not know how to calculate!
For the calculation of the partition function, Z, we have to distinguish two very different cases: a) the system consists of distinguishable, identical, non-interacting particles (i.e. the exchange of particles results in a new state, e.g. ideal crystal)
≠
state a
state b
1
2
3
4
5
6
1
2
3
4
5
6
7
8
9
10
11
12
7
8
9
10
11
12
13
14
15
16
17
18
13
14
16
15
17
18
19
20
21
22
23
24
19
20
21
22
23
24
25
26
27
28
29
30
25
26
27
28
29
30
b) the system consists of indistinguishable, identical, non-interacting particles (i.e. the exchange of particles does NOT result in a new state, e.g. ideal gas) state a
© Dr. Peter Blümler
=
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state b
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PH 605: Thermal & Statistical Physics
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3. Semi-Classical Physics
case a) distinguishable particles (crystal of N particles): E1 = E2 = E3 = E4 =
ε1 (1) + ε1 ( 2) + ε1 ( 3) + K + ε1 ( N )
ε 2 (1) + ε1 ( 2) + ε1 ( 3) + K + ε1 ( N ) ε1 (1) + ε 2 ( 2) + ε1 ( 3) + K + ε1 ( N ) ε1 (1) + ε1 (2) + ε 2 ( 3) + K + ε1 ( N ) M = M M M L M energy level
particle indicator
Hence, from eq. [3.2.4]: Z N (T ,V ) =
E E exp − r = exp − 1 k BT k BT over all
∑
E + exp − 2 k BT
E + exp − 3 k BT
+ K
states
ε (1) + ε1 ( 2) + ε1 (3) + K + ε1 ( N ) = exp − 1 k BT ε (1) + ε1 (2) + ε1 (3) + K + ε1 ( N ) + exp − 2 k BT ε (1) + ε 2 ( 2) + ε1 (3) + K + ε1 ( N ) + exp − 1 + K k BT ε1 (1) ε (2) ε (3) ε ( N ) exp − 1 exp − 1 L exp − 1 exp − k T k T k T k T B B B B ε (1) ε1 ( 2) ε (3) ε ( N ) exp − exp − 1 Lexp − 1 + exp − 2 k T k T k T k T B B B B
Z N (T ,V ) =
ε (1) + exp − 1 k BT
ε ( 2) ε (3) exp − 2 exp − 1 k BT k BT
ε ( N ) Lexp − 1 + K k BT
This sum of products runs to infinity, so we can be sure to find all possible permutations of energy levels and particle indicators. Hence, we can replace it by a product of sums: Z N (T ,V ) =
ε1 (1) exp − k BT
+ L ⋅ ε1 ( 2) ε ( 2) ε ( 2) + exp − 2 + exp − 3 + L ⋅ exp − kBT kBT k BT LLL ε (1) ε (1) + exp − 2 + exp − 3 k BT k BT
ε (N) ⋅ exp − 1 k BT
© Dr. Peter Blümler
ε (N) ε (N ) + exp − 2 + exp − 3 k BT k BT
School of Physical Sciences
+ L
University of Canterbury
PH 605: Thermal & Statistical Physics
or
page:11
3. Semi-Classical Physics
∞ ε ( j) exp − r k BT j =1 r =1 N
∏∑
Z N (T ,V ) =
[ 3. 2. 3] N = Zj j =1
∏
[3.2.5]
if all the particles are identical ⇒ Z j = Z1 for distinguishable particles ZN
N
∏ Z1
=
= Z1N
[3.2.6]
j =1
case b) indistinguishable particles (gas of N particles): What is going to happen, if we would use the same formula to calculate the partition function of the N particles in a gas?
?
?
Z N (T ,V ) =
Z1N
?
To illustrate that this is WRONG, we consider two indistinguishable particles (N=2): ?
Z 2 = Z12 = =
∞
ε exp − r1 k BT r1 =1
∑
εr ∞ ⋅ exp − 2 r2 =1 k BT
∑
∞
ε exp − r k BT r = r1 = r2 144424443
∑
+
particles both in the same state
ε r + ε r2 exp − 1 k T B r1 ≠ r2 1444424444 3
∑∑
particles in different states
However, the particles in the second term are counted twice, but since the particles are indistinguishable this doesn't generate a new state, because: ε r + ε r2 exp − 1 k BT
ε + ε r1 = exp − r2 k BT
(there is no experimental method to distinguish these two states in a gas, because the particles are not fixed in space) Hence: Z 2 = Z12 counts the particles in different states falsely twice! We can correct for that by: Z2 =
© Dr. Peter Blümler
∞
ε exp − r k BT r = r1 = r2
∑
+
1 2
∑∑ r1 ≠ r2
School of Physical Sciences
ε r + ε r2 exp − 1 k BT
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3. Semi-Classical Physics
So how about a generalisation for N particles? The formula in [3.2.6] would count all possible permutation of energy levels wrongly. Hence we have to correct by N! ZN =
ε r + ε r2 + L + ε rN ε 1 [3.2.7] exp − N r + LLL + K exp − 1 k T N ! k T B B r 44 r ≠ r ≠ L ≠ r 1 2 N 1 42444 3 1 424 3 14444444244444443 ∞
∑
∑ ∑
all particles in the same state
some particles in same states
all particles in different states
This would represent the correct formula, where in the central term we would have sums over situations, where e.g. just two particles are in the same state (hence to be corrected by 1/2) and such were 3 are in the same state (hence correction by 1/6) etc...This is very messy! Now we can use the semi-classical arguments from the definition in section 3.2.1. The probability is very small for such a gas, that any single particle state is occupied by more than one particle! (We are somehow discussing a caricature of the classical continuum of energies, which however has to be understood as being split up into discrete (countable) -QM-like- levels of infinite number) • since there are infinite levels but a finite number of particles ⇒ most states are empty • or very few states are occupied by a single particle • and an insignificantly small number of states contains more than one particle (degeneracy ≈1)
semi-classical gas: number of particles << number of energy-states
This argument greatly simplifies eq. [3.2.7], because only the last term counts!
for indistinguishable particles:
ZN
=
ε 1 exp − r N! r k BT
∑
N
=
Z1N N!
[3.2.8]
Now we are ready to calculate the different contributions to the single particle partition function, Z1
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
3.2.3
page:13
3. Semi-Classical Physics
Contributions of different types of motion to Z1?
Thermo-dynamics: determined by the thermally activated dynamics of particles! Hence, we have to deal with motion! Different types of particle motion:
1) translational motion 2) internal motions 2a) rotation 2b) vibration [ 2c) electronic excitation]
translation
rotation
The energy of a state r is then:
εr =
ε atr {
state of translatio nal motion
and
ε Z1 = exp − r k BT r
∑
vibration + ε int b {
[3.2.9]
state of internal motion
int ε tr a + εb = exp − k BT a b
∑∑
[3.2.10a]
but the types of motion are independent! hence,
Z1 =
ε int ε tr exp − a exp − b k BT k BT 443 b a 442 443 1 1442 Z int Z1tr
∑
∑
[3.2.10b]
where Z1tr is the sum over all energy states of translational motion of the particle (independent of structure) and Zint is the sum over all energy states of internal motion of a molecule (since they are all identical, but depends on its structure... atomic vibration = translation) © Dr. Peter Blümler
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PH 605: Thermal & Statistical Physics
page:14
3. Semi-Classical Physics
Summarise: the molecular partition function: Z1 = Z1tr ⋅ Z int
Z1tr :
[3.2.10]
will apply to any perfect gas, independent of structure (e.g. He, Ne, O2, N2, CO, CO2, H2O, C2H5OH, etc.) p2 ε tr = 12 mv 2 = with p = p 2m • classically any value of p is possible • semi-classically: there might be restrictions due to QM
Zint :
depends on the internal structure, which is the same for all identical particles. Because the particles do not interact Zint doesn’t depend on the macroscopic state (P, N, V) of the system.
Both however depend on the temperature, which is responsible for which energy-levels are occupied/activated On the other hand, temperature is defined by these motions, so we might look at this from just the other side of the equation!
z 3.2.4
V=∞
The density of states
For this we recap the QM-treatment of a particle in a 3D-box:
V=0
y
x Therefore we have to solve the time-independent 3D Schrödinger equation −
h2 ˆ 2 ∇ ψ( x, y , z ) + V ( x, y, z ) ψ ( x, y, z ) = E ψ ( x, y, z ) 2m
for a particle in a 3D box with the following boundary conditions (see figure): 0 for 0 ≤ x ≤ a1 and V ( x, y , z ) = ∞ for all other
0 ≤ y ≤ a2 and
0 ≤ z ≤ a3
requiring ψ ( x , y, z ) = 0 on all faces of the box (V = ∞). [you might refer to http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH502/PH502.html] © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:15
3. Semi-Classical Physics
This has the solution: ψ n1 , n 2 , n3 ( x, y , z) = En1 , n2 , n 3 =
h2 n12 8m a12
n πx n πy n πz 8 sin 1 sin 2 sin 3 a1a2 a3 a1 a2 a3 n2 n 2 + 22 + 32 with n1 , n2 , n3 = 1, 2, 3,K a 2 a3
which simplifies by substitution with the wave-vector, k: n2 n2 n2 πn πn2 πn3 k 2 = π 2 12 + 22 + 32 and k = 1 , , a a a a a a 1 2 3 1 2 3
[3.2.11]
we can further simplify by assuming a1 = a 2 = a3 = a (cubic box) Vbox = a3
volume:
and from de Broglie’s equation we get: p = hence
k =
[3.2.12]
h 2π with k ≡ λ λ
2π p h
or
k =
[3.2.13] p h
[3.2.14]
We want to describe and calculate the problem of how many states are in a certain volume in kspace! This is because, in k-space the different energies/states defined by the quantum numbers (n1, n2, n3) are equally spaced (see figure), which makes the calculation much easier.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:16
3. Semi-Classical Physics
Note that the spheres in this figure do represent states (not particles!). The volume of each quantum state in k-space is then: π Vk = a
3
[3.2.15]
The density of lattice points (or allowed k-vectors) is: 3
1 a = Vk π
ρk =
[3.2.16]
To calculate the density of the states we want to know the number of such normal modes of standing waves with a k-vector k, whose magnitude lies in the interval k to k+dk. This number can be obtained (approximated) by calculating the number of k-lattice points which lie between two spherical shells (radius1 = k and radius2= k+dk, = hollow sphere of thickness dk). This will give approximately the correct answer for a dense lattice. Because a (macroscopic box) is big, the assumption is valid. Careful: n1, n2, n3 >0 only one (positive) octant of the shells must be counted (k x, k y, k z >0) This octant has a volume: V octant =
+
1 8 {
4 πk 2 123
dk {
positive surface thickness octant
Therefore, the number of modes of standing waves with k between k and k +d k is N ( k ) dk =
(
)
[3 .2 .16 ] 1 4 πk 2 dk a 3 k 2 dk 3 [ 3. 2. 12 ] k 2 dk box 4πk 2 dk ρ k = = a = V 8 8 π3 2π 2 2π 2
density of states in k-space: N ( k ) dk
=
k 2 dk 2π 2
[3.2.17]
V
using eqs. [3.2.13] and [3.2.14] k can be substituted for ε, v, λ
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:17
3. Semi-Classical Physics
Now we are ready to determine the single-particle partition function Z1 for the different type of motions....we start with the translational.
3.2.5
Partition function for translational motion, Z 1tr
For translational motion we have three degrees of freedom (cf. eq. [3.2.11]). ε tr = ε nx + ε ny + ε nz 1 2 3
[3.2.18]
hence the energy can be separated by components. The (energy-) eigenvalues of the Schrödinger-equation in the previous part are given by: ε ni =
h 2 ni2
8mai2
with i = 1, 2, 3
[3.2.19]
or for a simple cubic box (ai = a) ε ni =
h 2 ni2 [ 3 .2 .11] h ki2 = with i = 1, 2, 3 2m 8ma2
[3.2.20]
and with eq. [3.2.3]: Z1tr
ε = exp − r kBT r
∑
εx + ε y + ε z ∞ n n2 n3 = exp − 1 k BT n i =1
∑
εx n tr Z1 = exp − 1 k BT n1 =1 ∞
∑
with i = 1, 2, 3 and ni = 1, 2,3, 4...
εy ∞ n exp − 2 n 2 =1 k BT
∑
∞ εz n exp − 3 n1 =1 k BT
∑
because there is no preferred direction of motion, the three sums are equivalent, and we have to solve: Z1tr
∞ ε = exp − n n =1 kBT
∑
3
with ε n =
h2 k 2 h2 n2 = 2m 8ma 2
[3.2.21]
Now we have to consider the density, N(k), of the k-values (degeneracy of the state). For large (macroscopic) dimensions of the box (a >> 1) the ε n are very close (semi-classical continuum), so that (in good approximation) the sum in eq. [3.2.21] can be replaced by an integral. But careful: Equation [3.2.17] was already obtained for counting the density of k-values (k x, k y and k z) in a three-dimensional fashion. Hence, we have to drop the power of 3 when using the density of states in 3D-k-space (otherwise we would count the same twice!)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
Z1tr
∞ ε = exp − n n =1 k BT
∑
[ 3. 2. 17 ]
=
V
=
∞
3
∞ ε ≅ N ( k ) exp − n k BT 0
∫
ε k exp − n 2π 2 0 k BT V
∫
∞
∫
page:18
3. Semi-Classical Physics
2
(
dk
[3.2.22]
∞ h 2k 2 [ 3 .2 .20 ] V dk = k 2 exp − 2mk BT 2 π2 0
)
k 2 exp − ak 2 dk
2π 2 0 14442444 3
∫
dk
h2 2mkBT
with a ≡
1 π 4a a
Partition function for translational motion (single particle): Z1tr =
π
3
(2mk BT ) 2 3
V
=
2π 2 4 h
3
(2πmk BT ) 2 3
V h
[3.2.23]
Alternative: A more direct approach uses the solution of the Schrödinger-equation. This already includes the possible degeneracy of states (so we can drop inclusion by the density of states in k-space) ∞ ε Z1tr = exp − n n =1 k BT
∑
3
3 ∞ ε ≅ exp − n dn kBT 0
∫
∞
ε solve the integral: exp − n k BT 0
∫
we substitute: x 2 ≡ ∞
ε exp − n k BT 0
∫
n2 h2 2
8ma kBT
[ 3. 2. 21] ∞ h 2n 2 dn = exp − 8ma2 k T B 0
∫
⇒ dn = ∞
dn
a 8mk BT dx h
( )
a a dn = 8mkBT exp − x 2 dx = 2πmkBT h h 0 4 1 4244 3
∫
π 2
∞ ε tr Z1 = exp − n k BT 0
∫
3
3 3 [3 .2 .12 ] V 3 a3 a dn 2 2 qed! = 2 π mk T = ( 2 π mk T ) = ( 2 π mk T ) B B B h h3 h3
This direct approach avoids the „historical“ argument about density of states, which however is quite useful for a variety of problems!
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:19
3. Semi-Classical Physics
Now we are ready to apply our knowledge!!! Before we do this we recap how thermodynamic properties can be derived from the partition function. (cf. part I of lecture) internal (average) energy:
∂ ln Z ∂ ln Z U = E = E = k BT = k BT 2 ∂ ln T V ∂T V
[3.2.24]
entropy:
∂ ln Z S = k B ln Z + ∂ ln T V
[3.2.25]
pressure:
∂ ln Z P = k BT ∂V T
[3.2.26]
Helmholtz free energy:
F = E − TS = − k BT ln Z
[3.2.27]
enthalpy:
∂ ln Z H = E + PV = k BT ∂ ln T
Gibbs free enthalpy:
∂ ln Z G = F + PV = − k BT ln Z − ∂ ln V
+ ∂ ln Z V ∂ ln V
T
T
[3.2.28]
[3.2.29]
A perfect gas of N indistinguishable particles (or states) has a partition function (eq. [3.2.8]) Z1N . This in combination with eq. [3.2.23] allows us the determination of the partition N! function due to the translational motion of the gas particles. ZN
=
Hence, the internal energy of the gas can be calculated using eq. [3.2.24]: ∂ ZN ∂ ln Z N E tr = k BT 2 ln 1 = k BT 2 N! ∂T V ∂T V [ 3. 2. 23 ]
=
k BT 2
N
3N 3 ∂ 1V ln (2πmkB ) 2 T 2 ∂T N! h3 144424443
doesn't depend on T ≡ C
∂ 3N 3 N ∂ ln T = k BT 2 ln C + ln T = k BT 2 ∂T 2 2 1∂2 T3 1/T
Translational energy for N particles of a perfect gas: E tr = 32 k B N T
© Dr. Peter Blümler
School of Physical Sciences
[3.2.30]
University of Canterbury
PH 605: Thermal & Statistical Physics
page:20
3. Semi-Classical Physics
~ Or for 1 mole (N = NA) of perfect gas: E tr = 32 k B N A T = 32 RT (R = (ideal) gas constant = 8.3145 J/mol K) This is -for instance- in total agreement with kinetic gas theory (see QC1).
Specific heat (due to translational motion) for N particles of a perfect gas: ~ ~ tr ∂E tr CV = ∂T
3.2.6
= 3 R = const. 2 V
[3.2.30]
Partition function of internal motion, Zint
What do we know? From section 3.2.3 we recap: eq. [3.2.9] ε = ε tr + ε int =
ε tr +
rot ε{
+
rotational contribution
and we had shown (eq. [3.2.10]) that we know: Z1 =
ε
vib ε{
vibrational contribution
+
elc ε{
contribution of electronic excitation
Z1 = Z1tr Z int
∑ exp − k BrT r
In the chosen semi-classical limit, we can again assume or consider a large number of energy eigenvalues (states). The ε r are combinations of these eigenvalues (we sort them), which are independent (see also separation of variables in QM-problems) ε1 = ε1tr + ε1int =
ε1tr + ε1rot + ε1vib + ε1elc
int tr rot vib elc ε 2 = ε tr 2 + ε2 = ε2 + ε2 + ε2 + ε2 LLetc.LL
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:21
3. Semi-Classical Physics
and we can use the same arguments as in getting from eq. [3.2.9] to [3.2.10], or Z1 =
=
ε tr ε int exp − r exp − r k T k T r 4424B43 1 r 4424B43 1 Z1tr Z int
∑
∑
ε tr
ε rot
ε vib
ε elc
∑ exp − kBrT ∑ exp − kBr T ∑ exp − k rBT ∑ exp − k Br T
r 442443 1 r 442443 1 r 442443 1 r 442443 1 Z1tr Z rot Z vib Z elc
Partition function due to internal motions: Zint = Z rot ⋅ Z vib ⋅ Z elc
[3.2.31]
where the partition function of rotational energy, Z rot , vibrational energy, Z vib , and due to electronic excitation*, Z elc , are all independent of the macroscopic state of the system! *not calculated in this lecture
3.2.7
Partition function of (molecular) rotation, Zrot
We recap the QM-description and solutions (for details see http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH502/PH502.html) 3D-Schrödinger-equation:
Λˆ 2 Yml (θ, φ ) = − l ( l + 1) Yml ( θ, φ)
Hamiltonian in spherical polar co-ordinates (Legendrian) Legendrian: Λˆ 2 =
wavefunction: spherical harmonics
1 ∂ ∂ 1 ∂2 sin θ + sin θ ∂θ ∂θ sin 2 θ ∂φ 2
eigenvalues: E l = l ( l + 1)
h2 2I
with moment of inertia: I = µr 2 , where µ is the reduced mass. Quantization conditions:
© Dr. Peter Blümler
l = 0, 1, 2, K and
−l ≤ m ≤ l
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:22
3. Semi-Classical Physics
ε lrot = l ( l + 1)
Hence,
h2 2I
[3.2.32]
Because m is not present in eq. [3.2.32], but spans − l ≤ m ≤ l -we remember- that each energy level is (2l+1)-fold degenerate. ∞
∞ ε lrot l (l + 1) h 2 Z rot = (2l + 1) exp − = ( 2l + 1) exp − 2 I k BT k BT l =0 l =0
∑
Hence,
∑
If the moment of inertia (or mass) is big* or the temperature high, the energy levels are very close again (semi-classical argument), and we are allowed to replace the sum by an integral. ∞ l (l + 1)h 2 dl Z rot = ( 2l +1) exp − 2 I k BT 0
∫
we substitute: x ≡
(
)
l (l + 1)h 2 h2 h2 = l 2 + l ⇒ dx = (2l + 1) dl 2 I k BT 2I k BT 2 I k BT
∞
∞
[
]
2I k T 2I k BT 2 I k BT 2 I k BT −x −x ∞ Z rot = ( 2l + 1) exp (− x ) 2 B dx = e d x = − e = 0 424 3 h (2l + 1) h2 0 h2 1 h2 0 1
∫
∫
Partition function for (molecular) rotation: Z rot =
2 I k BT h2
=
8π2 I k BT h2
[3.2.33]
Comment on symmetry considerations: for diatoms (e.g. N2, O2) there are additional degenerate states (only two degrees of rotational freedom) In order to correct eq. [3.2.33] a symmetry factor, σ, is included, so that 2I k BT Z rot = (we will discuss this later) σ h2
*) not true for instance for 1H2 and 2H2 (we will see later!) © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:23
3. Semi-Classical Physics
At the bottom of page 13, we had stated that Zint is the sum over all energy states of internal motion of a molecule. We also see that eq. [3.2.33] doesn’t depend on macroscopic parameters (like V, cf. eq. [3.2.23] ), which is good, because otherwise our arguments in section 3.2.3 would have been wrong! We can consider these molecules as individuals. Each is rotating independently of the others! It doesn’t matter if there is 1, 2 or N of these molecules (the rotation of molecule A does not interfere with that of B and vice versa). So if we would (somehow magically) exchange their energies or positions, we would generate a new micro-state (e.g. state 1: Molecule A at position 1 has rotational energy ε A and B at position 2 has energy ε B ; state 2: Molecule B at position 1 has rotational energy ε B and A at position 2 has energy ε A). [This is different for the translational motion! Why?]
Z rot, N = (Z rot )N
Hence,
[3.2.34]
How about the thermodynamic properties of rotational motion? internal energy (due to rotational motion): ∂ ln Z rot, N ∂ ∂ N = kBT 2 E rot = k BT 2 ln Z rot ln Z rot = k BT 2 N ∂T ∂T V ∂T V V [3 .2 .33 ]
=
∂ 2I kB ∂ 1 kBT 2 N ln 2 T = k BT 2 N ln [CT ] = k BT 2 N ∂T T V ∂T h Rotational energy for N particles of a perfect gas:
or for 1 mole (molar):
and
© Dr. Peter Blümler
E rot = k B NT
[3.2.35]
~ E rot = RT
[3.2.36]
~ ~ rot ∂E rot CV = ∂T
= R = const. V
School of Physical Sciences
[3.2.37]
University of Canterbury
PH 605: Thermal & Statistical Physics
3.2.8
page:24
3. Semi-Classical Physics
Partition function of (molecular) vibration, Zvib
Again, we recap the QM-description and solutions (for details see http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH502/PH502.html) 1D-Schrödinger equation for harmonic oscillator: −
h2 ˆ 2 1 ∇ ψ( r ) + mω2 r 2 ψ (r ) = E ψ( r ) 2m 2
2 has the solutions: ψ s ( ξ) = N s H s (ξ) e − ξ 2
(
)
(
with ξ = 2π mν h x = mω h x
)
and eigenvalues E = s + 12 hν = s + 12 hω
with s = 0, 1, 2, K
We now apply the same procedures as before:
(
)
(
)
ε svib = s + 12 hν = s + 12 hω and
εs Z vib = exp − vib k BT s =0 ∞
∑
with s = 0, 1, 2,K
(
)
∞ s + 1 hν = 2 exp − k BT s =0
∑
[3.2.38]
[3.2.39]
However, in this case we cannot simply replace the sum by an integral! Why not? We will see, when we calculate the separation of the energy levels, that even for a semi-classical gas, they are too far apart to justify such a step. (In the next chapter, we will calculate these separations and check our procedures, the integration would require temperatures, where -in ordinary molecules- the chemical bonds would disintegrate). In order to get a physically meaningful result, we -unfortunately- have to solve the sum! First we simplify the argument by substituting x ≡
Z vib =
hence eq. [3.2.39] simplifies to
hν k BT
∞
∑ exp (− (s + 12 ) x )
[3.2.40]
s= 0
then we multiply eq. [3.2.40] with e − x : Z vib e − x =
∞
∑ (( s= 0
))
exp − s + 12 x exp( − x ) =
∑ exp (− (s + 32 ) x ) ∞
[3.2.41]
s =0
Now we subtract eq. [3.2.41] from [3.2.40] and get:
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
(
)
page:25
3. Semi-Classical Physics
∑ exp (− (s + 12 ) x ) − exp (− (s + 32 ) x ) ∞
Z vib 1 − e − x =
s1 = 044444424444443
−x − 3x e 2 −e 2
− 3x − 5x +e 2 −e 2
− 5x −7x +e 2 −e 2
+K
we see that when we calculate the sum, all terms cancel except for s = 0 (exp(-x/2)). Hence,
(
Z vib 1 − e
−x
)= e
−
x 2
x exp − 2 = Z vib = 1 − exp (− x )
1 x x exp + − exp − 2 2
re-substitute: Partition function for (molecular) vibration: hν exp − 2 k BT Z vib = hν 1 − exp − k BT
1
=
hν hν − exp − exp + 2k BT 2k BT
Z vib , N = (Z vib )N
and*
[3.2.42]
[3.2.43]
*due to the same arguments which lead to eq. [3.2.34]! Alternatively: Z vib =
∞
∞ 1 x exp − x s + = exp (− sx ) exp − 2 2 s =0 s = 0
∑
∑
∞
∞
x x (exp( − x) )s = exp − exp (− sx ) = exp − 2 s =0 2 n =0
∑
∑
with exp( − x ) < 1 for x > 0
The sum can be expressed as a geometric series (α ≡ exp(-x), hence α < 1), for which ∞
∑α s
s =0
Hence, Z vib =
© Dr. Peter Blümler
= 1 + α + α 2 + α 3 + .... =
1 1− α
for α < 1
exp( −x / 2) 1 = x 2 1 − exp( − x ) e − e− x 2
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PH 605: Thermal & Statistical Physics
page:26
3. Semi-Classical Physics
How about the thermodynamic properties? internal energy:
∂ ln Z vib , N E vib = k BT 2 ∂T
∂ ∂ N = k BT 2 ln Z vib = k BT 2 N ln Z vib ∂T V ∂T V V
hν exp − 2 k BT hν ∂ ∂ hν 2 = k BT N ln = k BT 2 N − − ln 1 − exp − ∂T hν ∂T 2k BT k BT 1 − exp − k BT hν exp − kh νT Nhν exp − h ν 2 hν k BT B 1 2 k BT = k BT N + = Nhν + 2 2 2 kBT 1 − exp − kh νT 1 − exp − khνT B B
Vibrational energy for N particles of a perfect gas: E vib =
1 Nhν Nhν + 2 exp + hν −1 k BT
[3.2.44]
~ for the molar energy, E vib , we simply have to replace N in eq. [3.2.44] by NA. So, what would have happened, if we would have chosen an integration rather than a summation? Let’s check on a deriving a hypothetical partition function from eq. [3.2.40]: W R O N G !!
hyp Z vib
∞
∫ ((
= exp − s + 0
1 2
∞
) x) ds = e ∫ exp (− sx) ds = e − x2
− 2x
0
∂ ∂ ~ hyp hyp 2 Evib = k BT 2 N A ln Z vib = k BT N A ∂T ∂T V
e − xs − x
∞
= 0
e
− x2
x
− x − ln x = k T 2 N ∂ A 2 B ∂T
− h ν − ln h ν k T 2 k BT B
1 1 = k BT 2 N A h ν 2 + = N A hν + RT T 2 2 k BT ~ hyp ~ vib,hyp ∂Evib and CV = = R = const. ∂T V (which also corresponds to the equipartition theorem, which will be introduced later! This is just for revision!)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:27
3. Semi-Classical Physics
Experimental data for validation of the chosen approach: ~ Specific heat of oxygen gas at room-temperature: CV300K (O 2 ) = 0.03R From IR-spectroscopy: ν (O 2 ) = 4.67 ⋅ 1013 Hz ~ Hence, CVvib,hyp is certainly wrong! ~ ~ vib ∂Evib how about eq. [3.2.44] CV = ∂T
−1 = N hν ∂ exp + h ν − 1 A ∂T k BT V
h ν exp hν k T ~ vib k BT 2 B CV = N A hν 2 exp + h ν − 1 k T B
hν ~ plug the experimental values in: CVvib = R k BT
[3.2.45]
2 exp khνT B 2 + h ν − 1 exp k BT
hν 6.62 ⋅10 −34 Js 4.67 ⋅ 1013 Hz = = 7.47 k BT 1.38 ⋅10 − 23 J/K 300K ~ 1754 CVvib = R 55.7 = 0.031 R (excellent agreement!) (1753)2 1 Nhν , 2 of the harmonic oscillator, only a fraction of the higher vibrational states is excited (at normal temperatures). So, the semi-classical argument is not valid in this case! Closer inspection of equation [3.2.44] reveals, that besides the zero-point energy term,
However, we can „hypothetically“ check, if eq. [3.2.44] and the deduced eq. [3.2.45] give the correct „semi-classical“ argument by rising the temperature (T →∞) Therefore, we do something typical, we introduce a characteristic temperature, Θ ≡
hν kB
(for the oxygen-example above: Θ = 2240 K)
() ( (T ) )
()
Θ 2 2 exp Θ ~ vib Θ T T CV = R =R 2 2 T exp Θ − 1 1 − exp − Θ exp Θ
© Dr. Peter Blümler
(
School of Physical Sciences
( T ))
(T ) University of Canterbury
PH 605: Thermal & Statistical Physics
page:28
3. Semi-Classical Physics
Θ 2 critical (ΘT )2 → 0 ( ) T = R lim T → ∞ (1 − exp (− Θ ) 2 exp (Θ ) terms only T → ∞ (1 − exp (− Θ ) 2 T T 14424T4 3 123
~ lim CVvib = R lim
T →∞
→0
→1
Θ) − Θ De L' Hospital ( 1 T T = R lim = R lim = R lim =R T → ∞ 1 − exp (− Θ ) T → ∞ − Θ exp (− Θ ) T → ∞ exp (− Θ ) T T T T 1 424 3 2
2
→1
3.2.9
Partition functions of the perfect gas and comparison to experimental data
3 RT 2
~ CV 3 R 2
h2
RT
R
1
1 N A hν N A hν + 2 exp ( x ) − 1
mode translational rotational vibrational hν with x ≡ k BT
~ E
Z1 3
(2πmk BT ) 2 3
V h
8π 2 I k BT
e x 2 − e− x 2
Rx 2
ex
(e − 1) x
ε Θ≡ i kB
we can introduce characteristic temperatures, generally by
2
[3.2.46]
hν h2 and Θ rot = 2Ik B kB Note: The characteristic temperature can be associated with a fraction of excited states at a temperature T (for T = Θ ⇒ e-1 ≈ 37% or about a third) hence: Θ vib =
experimental values at 300K mode
wavelength
spectroscopy
Θ [K]
rotational
0.05 - 1 cm
micro-wave
1.5-30
vibrational
1-2 µm
IR
800-1500
100-700 nm
UV/VIS
104-105
electronic excitation.
• for normal temperatures we can certainly neglect the electronic part. • rotational levels are ‘completely’ excited (semi-classical argument valid) • vibrational levels are partially excited (semi-classical argument not valid)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:29
3. Semi-Classical Physics
direct comparison for different molecules
O2 *) HCl H2 *)
Θvib [K] 2240 4100 6000
Θrot [K] 2 15 85
mass
*) for homonuclear, diatomic molecules we will learn (at the end of this first part) that eq. [3.2.33] is not entirely correct (especially at low temperatures).
3.3 3.3.1
Entropy and Energy of the semi-classical Gas Entropy of a Mon-atomic Gas, the Sackur-Tetrode-equation
For a mon-atomic, perfect gas (e.g. noble gases: He, Ne, Ar...) without electronic excitation, all we have to consider is the energy due to translational motions. (Vibrations or rotations of a single atom do not generate new micro-states). In section 3.2.5 we have seen how from eq. [3.2.23]: Z1tr = and eq. [3.2.8], Z N
=
(Z1tr )N
V h3
3
(2πmk BT ) 2
N!
the internal energy of translational motion was calculated to be E tr = 32 k B N T ([3.2.30]) Now we want to calculate the entropy. For this purpose, we start with eq. [3.2.27] for the Helmholtz free energy: F = E − TS = − k BT ln Z N N tr N 1 V 3 Z1 2 ( ) = −k BT ln = − k BT ln 2πmkBT N! h 3 N! 3 V = −k BT − ln N ! + N ln 3 (2πmkBT ) 2 h
( )
using Stirling’s approximation (eq. [3.1.6]) 2πmkBT F = −k BT N − N ln N + N ln V + N ln h2
© Dr. Peter Blümler
School of Physical Sciences
3 2
University of Canterbury
PH 605: Thermal & Statistical Physics
page:30
3. Semi-Classical Physics
3 2 E − F 2 π mk T B S tr = = 32 k B N + k B N 1 − ln N + ln V + ln T h2 This is the
tr
Sackur-Tetrode* equation 5 V 3 3 2πmk B S tr = kB N + ln + ln T + ln N 2 2 h 2 2
[3.3.1]
*Otto Sackur and Hugo M. Tetrode ca. 1912
it is important, because it has no arbitrary (substance specific) constants. Hence it can be applied to any atomic gas! example: entropy for vaporisation of Ne: first we calculate the change of entropy by going through all phases of Ne: (experimental data) a) solid: (assuming S(0K) =0) T = 0K →Tmelt = 24.55K: Tm
∆S =
∫ C P d ln T = 14.92 J mol 0
b) melting at Tm: ∆S m =
-1 -1
K
(using Debye‘s theory for CP - see later)
∆H m 334 J/mol = = 13.64 J mol -1 K -1 Tm 24.55 K Tb
c) liquid: Tm → Tboil = 27.2K ∆S =
∫ CP d ln T = 3.85 J mol
-1 -1
K
Tm
d) boiling (vaporise) at Tb: ∆S b =
∆H b 1758 J/mol = = 64.62 J mol -1K -1 Tb 27.2 K
The total change of entropy from a-d is then: S(27.2K) = 96.40 J mol -1 K -1 The Sackur-Tetrode-equation gives: S(27.2K) = 96.45 J mol -1 K -1 (again assuming that at 0K Ne is a perfect ‘gas’ with S=0) This is an amazing verification of our statistical treatment!
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
3.3.2
page:31
3. Semi-Classical Physics
The entropy of mixing-The Gibbs Paradox
From the Sackur-Tetrode equation the entropy is correctly described as an extensive property of a gas (S ∝ N). In the early days of Statistical Thermodynamics long discussions resulted from the fact, that the indistinguishability of the gas molecules was not understood and not taken into account (see discussion on page 12). These discussions and problems are summarised in the Gibbs-paradox (of historical importance only!) 1) Consider a box divided into two equal volumes, V. One part is empty (vacuum) and the other contains gas particles. What is the change of entropy when the separator is removed as illustrated?
a)
b)
separator removed Z1tr =
V h
3
3
(2πmk BT ) 2
(
V → 2V hence Z1tr → 2 Z1tr
)
∆F = Fb − Fa = − k BT ln 2 Z1tr − ln Z1tr = − k BT ln 2 since E tr doesn’t depend on V and the temperature is constant: ∆S = k BT ln 2 (as expected) 2) Now consider the same arrangement, but the two volumes initially being filled with two different gases.
a)
b)
separator removed Now each gas experiences an increase in entropy of ∆S = k BT ln 2 . Hence to total change in entropy is ∆S = 2 kBT ln 2 3) Finally consider the two volumes initially being filled with the same gas.
a)
b) separator removed
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:32
3. Semi-Classical Physics
The original argumentation considered the gas particles to be distinguishable. Then case 3) is identical with 2) and one would get a change of entropy of ∆S = 2 kBT ln 2 , although nothing has changed! This is the paradox! The correct argumentation realises, that the particles are indistinguishable! Before removing the separator:
ZN ZN Za = 1 ⋅ 1 N! N!
After removing the separator:
Zb =
(2 Z1 )2 N (2 N ) !
or using eq. [3.2.27]: Z N ZN Before removing the separator: Fa = − k BT ln 1 ⋅ 1 = − k BT 2[N ln Z1 − N ln N + N ] N! N! ( 2 Z1 ) 2 N = − k BT [2 N ln 2Z1 − 2 N ln (2 N ) + 2 N ] After removing the separator: Fb = − k BT ln (2 N ) ! and ∆F = − kBT 2 N [ln 2 + ln Z1 − ln ( N ) − ln 2 + 1 − (ln Z1 − ln N + 1)] = 0 hence ∆S = 0 , which is correct!
3.3.3 Theorem:
The Principle of the Equipartition of Energy A classical thermodynamic system in thermal equilibrium at a temperature T (high), in which the energy of the whole system depends on the square of one phase-space coordinate* (e.g. momentum, position) has an energy 12 kBT associated with it. * alternatively: this co-ordinate appears squared in the associated Hamiltonian ⇒ each degree of freedom contributes 12 kBT to the total energy.
Question:
If this is true, why then eq. [3.2.36] in comparison to the three degrees of rotation in the figure on page 13?
Derivation of the theorem: Co-ordinates in phase-space for N particles: x1 , x 2 , K, x N ; y1 , y2 , K , y N ; z1 , z 2 , K, z N ; p1x , p2x , K, pNx ; p1y , p2y , K, pNy ; p1z , p2z , K , pNz all together: 6N co-ordinates. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:33
3. Semi-Classical Physics
We want to simplify this by introducing a general co-ordinate ηi ηi ≡ x1 , K, x N ; y1 , K, y N ; z1 , K , z N ; p1x ,K, pNx ; p1y , K, pNy ; p1z , K, p zN with i = 1K6 N and − ∞ < ηi < ∞ From the theorem we know the condition for the energy of the ith state: ε i = Aηi2 + Eother
[3.3.2]
where Eother represents other energy contributions, which are independent on ηi . The number of molecules, Ni, (population) with a certain energy, ε i, is then given by the Boltzmanndistribution: ε exp − i Ni k BT = N ZN
[3 .3 .2 ] =
Aηi2 exp − Eother exp − k T k BT B 6N Aη2i exp − Eother exp − k T k BT B i =1
[3.3.3]
∑
We are not interested in the contributions of Eother , because we only want to calculate the probability (or contributions) associated with ηi . We can get rid of Eother by summing over all Ni just for these energies Eother or realising (assuming) that it doesn’t depend on the state. This will give us the desired probabilities, P( ηi ) , associated with ηi : Aηi2 exp − kBT P( ηi ) dηi = ∞ Aη2i dη exp − k BT i −∞
[3.3.4]
∫
In this equation we have also replaced the sum over states by an integral over co-ordinates. We are allowed to use the integral because the theorem is stating a classical system. Since the energy depends on the co-ordinate, ηi , integration over all co-ordinate-space also counts all energies. Then the average energy, E , of the system is given by: ∞
∞
E=
© Dr. Peter Blümler
∫
Aη2i −∞
P(ηi ) dηi =
∫
Aη2i dη exp − k BT i 2 Aηi dη exp − k BT i −∞
Aη2i −∞ ∞
∫
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
∞
with the given determined integrals:
∫
page:34
3. Semi-Classical Physics
2 − ax 2
x e
−∞
1 dx = 2a
π and a
∞
∫
2
e − ax dx =
−∞
π for a > 0 a
k T πkBT A B 1 2A A = k BT qed! we get E = 2 πk BT A alternatively we can also calculate the average energy by calculating an ensemble average of the individual energies: ∞
E = εi =
Aη2i
=
ε Aηi2 exp − i k BT −∞
∫
∞
ε exp − i k BT −∞
∫
∞
=
∫
dηidother
dηi dother
∞ ∞ Aηi2 Aη2i Eother dηi exp − dηi exp − dother Aη2i exp − k BT k T k T B B −∞ = −∞ see above ∞ ∞ 2 Aη2i Aηi dηi exp − Eother dother exp − exp − dη k BT k BT i k BT −∞ −∞ −∞
∫
Aηi2 −∞ ∞
∫
∫
∫
∫
Hence the molar average energy for molecules can be estimated using the equipartition theorem:
(
)
(
)
~ 1 E = RT 3 + f rot + f vib 2 and the molar specific heat:
~ 1 CV = R 3 + f rot + f vib 2 translation
f rot = 2
for linear molecules
f rot = 3
for non-linear molecules
f vib
depends on number of chemical bonds and angles in the molecule (but is seldom active at normal temperatures)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:35
3. Semi-Classical Physics
~ The following graph shows CV versus T for a diatomic molecule
~ CV [R] 2 32
7 2
trans.+rot.+vibration
5 2
translation+rotation
3 2
translation
translation of 2 atoms
diatomic molecule Θ rot
Θvib
Tdissoc
T
The thermodynamic properties of polyatomic molecules is determined by the structure (predictions are possible, but more difficult). However, the same basic ideas can be applied, but special care has to be taken when a degree of freedom is accessible (excited). There even exist situations where a vibrational degree of freedom can „mutate“ into a rotational degree of freedom. E.g. Hindered rotation due to voluminous parts of the molecule at low temperature generates vibration around the bond. At sufficiently higher temperature there is then enough energy to do perform the rotation (CV then decreases with increasing T)
T
3.4 3.4.1
Validity and Limit of the Semi-classical Description The Classical Limit
We want to derive the limits for a semi-classical treatment. Have a look at section 3.2.1 on page 8 to recap the definition. For this purpose we will analyse translational contributions to the energy of a system only.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:36
3. Semi-Classical Physics
Equation [3.2.23] gives the partition function for translational motion: 3 V Z1tr = ( 2πmkBT ) 2 h3 The probability, Ps, of a particle being in a particular state s (with energy ε tr s ) of translational motion is given by the Boltzmann distribution ε tr exp − s k BT Ps = tr Z1
[3.4.1]
When there are N particles, the mean occupation number, n s , of energy levels (average number of particles in state s) can be expressed as n s = NPs
[3.4.2]
Now we use the definition of the semi-classical system. In particular the fact, that the number of particles is much smaller than the number of available energy levels (see page 8 and 12), hence n s << 1
for all s
[3.4.3]
and ε tr N exp − s k BT [ 3. 2. 23 ] n s = NPs = = tr Z1
ε tr Nh3 − 32 (2πmkBT ) exp − s V k BT
<< 1
[3.4.4]
If that is the condition for all s, then it must be somewhat independent of the energy of each state over a broad temperature range. Hence we request the following to be true for all states s:
Condition for the (semi)-classical regime: 3 Nh 3 (2πmk BT )− 2 V
<< 1
[3.4.5]
Historically this was true, because in the original work of Boltzmann and Maxwell (before Planck’s „discovery“ of h) h → 0. We also recognise satisfactorily that high temperatures (T) and low densities (N/V) are necessary to fulfil eq. [3.4.5] (in complete agreement with the definitions and discussions in part 3.2).
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:37
3. Semi-Classical Physics
Now we want to compare these results with some QM-considerations. Therefore, we start with de Broglie’s law: λ =
h = p
3 and from eq. [3.2.30] for N = 1 ε tr = k BT 2 2m ε tr h
h 3m k BT
λ =
hence,
[3.4.6]
This we substitute into eq. [3.4.5]: N h3 V (2πmk T ) 32 B
3
3
N 3 2 h3 N 3 2 3 = = λ << 1 3 V 2π (3mk T ) 2 V 2π B
[3.4.7]
1
V 3 It is useful to introduce an average distance of the particles d ≡ , then N λ3
3
3 2 << 1 22 π 3 d 3 1
[3.4.8]
≈ 0. 33
Hence, we can approximate that in the (semi)-classical limit: λ3 << d 3
In words:
[3.4.9]
The de Broglie wavelength has to be small compared to average distance of the particles. Then the wave-properties of the particles can be neglected (a quantummechanical interference of the wavefunctions can be neglected). We are allowed to treat the constituents as particles obeying classical (Newtonian) mechanics.
example: He gas at room temperature. Density ca. 1020 atoms/cm3 ⇒ m = 6.7 10-24 g ⇒ (from eq. [3.4.6], T = 300K)
d ≈ 2 · 10-7 cm λ ≈ 8 · 10-9 cm
We see that for ordinary conditions the classical limit from eq. [3.4.9] is fulfilled! The limit is violated e.g. in condensed He (liquid), electrons in metals etc (see part 4). Generally, we have to consider to abolish the classical limit for • low mass • low temperature • high densities
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
3.4.2
page:38
3. Semi-Classical Physics
Maxwell velocity distribution in a classical gas
As a last example of classical treatment, we want to derive Maxwell’s velocity distribution for a perfect (classical) gas with our new statistical instruments. (recap 1st year material) From eq. [3.4.4] we directly get the mean number of particles with a momentum of magnitude p: 2 ε tr 3 Nh 3 ( 2πmk BT )− 2 exp − s with ε trs = p V 2m kBT 2 3 Nh 3 n ( p) = ( 2πmk BT )− 2 exp − p V 2mkBT
ns =
Hence the probability, P(p), of finding a single particle having a momentum between p and p+dp is: P( p) dp =
1 N ( p) n ( p ) N
[3.4.10]
where N(p) is the density of states.
hence:
k 2 dk
=
from eq. [3.2.17]: N ( k ) dk
N ( p )dp
2π =
2
and from eq. [3.2.14]: k =
V
2π p h
2
V 2π 2π V p dp = 4π p 2 dp 2 h 3 h 2π h
[3.4.11]
and eq. [3.4.10] becomes: P( p) dp =
1 N
Nh 3 p 2 2 − 32 4 π p d p ( 2 π mk T ) exp − B 2 mkBT V h3 V
3 p 2 P( p) dp = 4π p 2 (2 πmkBT )− 2 exp − dp 2mkBT
[3.4.12]
This is Maxwell’s momentum distribution! Before we continue, we check if eq. [3.4.12] needs an additional normalisation constant! (we only considered the magnitude of the momentum, hence integration from 0 to ∞) ∞ ∞ p 2 − 32 2 P ( p ) dp = 4 π (2πmkBT ) p exp − dp 2mk BT 0 0
∫
∫
3
= 4π (2πmk BT )− 2
© Dr. Peter Blümler
3 π (2mkBT ) 2 = 1 already normalised (single particle) 4
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:39
3. Semi-Classical Physics
In order to obtain a velocity distribution, we substitute p = mv: into eq. [3.4.12]: m2v 2 3 P( v) dv = 4π m 2 v 2 (2πmkBT )− 2 exp − 2 mkBT
mdv
Maxwell’s velocity distribution (3D) m P( v) dv = 4π 2πkBT
3
mv 2 2 2 dv v exp − 2 k T B
[3.4.13]
(see also graphs next page) Check your abilities in deriving the following from eq. [3.4.13]: 2 k BT m
v m.p. =
most probable speed (max. of P(v))
∞
∫
v = vP(v ) dv =
mean speed:
0
second moment:
v
2
[3.4.14] 8k BT πm
[3.4.15]
∞
k T = v 2 P (v ) dv = 3 B m
∫
[3.4.16]
0
note: this is equivalent to E =
and the rms (root-mean-square) speed: v rms =
v2 =
1 3 m v 2 = kBT 2 2
3k BT m
[3.4.17]
As a final test, we also want to check, if a Maxwell energy distribution will give the correct result. E=
p2 2m
⇒ p = 2mE and dp =
m dE 2E
3 2 mE in eq. [3.4.12]: P( E ) dE = 4π 2 mE (2πmkBT )− 2 exp − 2mk BT
m dE 2E
3 2 (k BT )− 2 E exp − E dE π k BT Maxwell’s energy distribution!
P( E ) dE =
∞
3 2 Hence, the mean energy is E = E = E P( E ) dE = (kBT )− 2 π
∫
0
© Dr. Peter Blümler
School of Physical Sciences
∞
3 E E 2 exp − dE k BT 0
∫
University of Canterbury
PH 605: Thermal & Statistical Physics
E substitute: x ≡ and k BT
∞ 3
∫ x2e
−x
3 π 4
dx =
0
3 3 2 E= (kBT )− 2 (k BT ) 2 k BT π
page:40
3. Semi-Classical Physics
∞ 3
∫ x 2 exp (− x ) dx = 0
2k BT 3 3 π = k BT qed! 2 π 4
The following figure shows Maxwell’s velocity distribution versus a dimensionless abscissa (normalised with respect to v ). v m.p. and v rms are also indicated.
Maxwell velocity distribution 0.08
vm.p.
0.06
P(v) [%]
vrms v
v
0.04
v 0.02 0.00 0.0
0.5
1.0
1.5
2.0
2.5
v v The next graph shows Maxwell’s velocity distribution for gases (various masses) at 300K
CO 2 Maxwell distribution of different gases at 300 K
Air
0.20
P(v) [%]
0.15
Ne
0.10
He
0.05
0.00
0
© Dr. Peter Blümler
1000
v [m/s]
School of Physical Sciences
2000
University of Canterbury
PH 605: Thermal & Statistical Physics
page:41
3. Semi-Classical Physics
The next figure shows the temperature dependence of the Maxwell distribution.
P(v) [%]
100 K 200 K
Maxwell distribution for Air at different temperatures
300 K
0.2
400 K 500 K
0.0 0
3.4.3
500
v [m/s]
1000
1500
Rotational Specific Heat of Diatomic Molecules Ortho/Para-Hydrogen
In this last section of part 3 we want to explore the influence of quantum-mechanical effects on the specific heat of homonuclear, diatomic molecules. The most drastic case is 1H2. We take this as a preparation for the next part, where we want to study quantum effects in more detail. 1
Inucl =½
PROTON
Fermion
combined wavefunction must be antisymmetric
2
Inucl =1 Inucl =½
DEUTERON TRITION
Boson Fermion
combined wavefunction must be symmetric combined wavefunction must be antisymmetric
H H H
3
We will learn more about Fermions and Bosons in the next part!! for 1H2 (Hydrogen-MOLECULE!) The wavefunctions for nuclear spin (σ) and the combined rotational wavefunctions for the electron (ψ) must satisfy: nucl rot σ12 ψ12
rot = − σnucl 21 ψ 21
1
2
Hence, the molecular wavefunction is antisymmetrical (Fermion) and obeys Pauli’s exclusion principle. In other words: The total wavefunction (incl. spins) must be antisymmetrical with respect to the interchange of position. No two electrons can have identical set of quantum numbers. (W. Pauli)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:42
3. Semi-Classical Physics
Summary of properties for 1H2: name
nuclear spin
σ nucl
ψ rot
J *)
total wavefunction
ortho-1H2 para-1H2
↑↑ (triplet) ↑↓ (singlet)
symmetrical antisymm.
antisymm. symmetrical
odd even
antisymm. antisymm.
*) combined rotational quantum number (orbital + spin) for electron
EXCURSUS: Linear combination and symmetry of wavefunctions: The symmetry of wavefunctions can be defined and tested by introducing an exchange operator, Xˆ , (or ‘swap-labeling’ operator = swaps the labels of a combined wavefunction) Xˆ ψ12 ≡ ψ 21
and Xˆ 2 ψ12 = Xˆ ψ 21 = ψ12 ⇒ Xˆ 2 = ˆI
with Xˆ we can now test linear combinations of wavefunctions for their symmetry: symmetric:
ψ s = ψ12 + ψ 21
antisymmetric:
ψ a = ψ12 − ψ 21
e.g. spin I = ½
ˆψ = ψ +ψ = +ψ test : X s 21 12 s test : Xˆ ψ s = ψ 21 − ψ12 = − ψ a
symmetric: αα , ββ , αβ + βα antisymmetric: αβ − βα
Generally, diatomic molecules possess (2I+1)2 of such product functions. 2I + 1 of these are symmetric (mI1=mI2, e.g. αα , ββ ). Remaining 2I(2I+1), of which half are symmetric (e.g. αβ + βα ) linear combinations; and the other half are antisymmetric (e.g. αβ − βα ) linear combinations. Hence, symmetric antisymmetric
with the ratio:
Ns I +1 = Na I
N s = (2 I + 1) + N a = I ( 2 I + 1)
2I ( 2 I + 1) = ( I + 1) ( 2 I + 1) 2
3
for hydrogen 1H2 (with I = ½)
Ns 2 3 = = Na 1 1 2
N 2 for deuterium 2H2 (with I = 1) s = Na 1 Conclusion: 1 mole of 1H2 at high temperatures consists of 34 mole of ortho (triplet) hydrogen (o- 1H2) and 14 mole of para (singlet) hydrogen (p-1H2). When the 1H2-gas is cooled down from room temperature, these fractions of o- 1H2 and p- 1H2 are conserved, because the different types of hydrogen do not easily convert into each other. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:43
3. Semi-Classical Physics
Of course at very low temperatures (< 20K) the molecules want to enter the state of lower enegry (J = 0 or para). However, this means that the wavefunctions of ortho somehow have to mutate to para. This is not directly allowed (or must lead over a dissociated molecule) and thus this process of ‘ortho/para conversion’ is very slow (≈ days). [The rate can be increased by introducing catalysts or (paramagnetic, e.g. O2-traces) impurities, which e.g. introduce local magnetic fields to alter the wavefunction via dipolar coupling]. This also explains the extremely high characteristic rotational temperature for 1H2 of Θrot =88K (see page 29). Heat of ‘ortho/para conversion’ (J = 1 →J = 0) for 1 mole of 1H2 ≈ -1100 J Heat for liquidification of 1 mole of 1H2 ≈ +600 J The ‘ortho/para conversion’ boils off reasonable amounts of hydrogen!! For the calculation of thermodynamic properties of 1H2 we have to deal with two different substances (o- 1H2 and p- 1H2): N ( Z1tr Z1rot Z1vib ) for a heteronuclear, diatomic molecule: Z N =
N!
for a homonuclear, diatomic molecule: Here 1H2 1
Z NH 2
N N vib tr rot vib ( Z1tr, ortho Z1rot Z ) ( Z Z Z ) 1, para 1, para 1, para , ortho 1, ortho = (34 N ) ! (14 N ) ! 3 4
1 4
Translational and vibrational parts are the same for ortho- and para- 1H2, hence Z1tr, ortho = Z1tr, para = Z1tr and 1
Z NH 2
and
( =
)
N Z1tr Z1vib
N! 14243 perfect gas
vib vib Z1vib , ortho = Z1, para = Z1
(Z1rot,ortho ) N (Z1rot,para ) N (14 N ) ! (34 N ) ! 3 4
N!
1 4
[3.4.18]
We have to determine (see section 3.2.7 and eq. [3.2.46]) Z1rot , para =
∞
Θ −6 ( 2 J + 1) exp − J ( J + 1) rot = 1 + 5e T J = even
∑
Θrot T
+ 9e
Θ − 20 Trot
+K
[3.4.19]
( 0 , 2, 4 ,...)
Z1rot , ortho =
∞
∑
J = odd (1, 3 , 5 ,...)
© Dr. Peter Blümler
Θ −2 ( 2 J + 1) exp − J ( J + 1) rot = 3e T
Θrot T
School of Physical Sciences
+ 7e
Θ −12 Trot
+K
[3.4.20]
University of Canterbury
PH 605: Thermal & Statistical Physics
page:44
3. Semi-Classical Physics
Unfortunately, these sums (infinite limit) cannot be expressed by an simple analytical expression (at least I don’t know one!). Hence, we have to compute them (see below). From eqs [3.4.19] and [3.4.20] then ZN (eq. [3.4.18]) can be computed and then ∂ ln Z N ∂E E = kBT 2 and from that CV = ∂ T V . ∂ T V This was performed here by a short MAPLE-program (see below). The next figure shows the result.
( )
~ CV [ R − 1]
T Θ rot Figure: Simulation Specific heat of ortho- (blue) and para- (red) hydrogen. The ordinate is CV in units of [R-1], the abscissa is dimensionless in units of T/Θrot. The 3:1 mixture of ortho/para is shown in black. Note that all three curves converge in the classical limit CV = R (see eq. [3.2.37]). This happens at about T/Θrot≈5 or 420 K.
You can download the MAPLE-routine from the course-webpage: http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH605/PH605.html
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
4
Quantum Statistics
4.1
Ideal Solids
page:45
Before we are going to derive the statistical description of quantum systems, we want to see how ideal solids can be described. This description is a good compromise between semi-classical treatment and special (QM) assumptions. The ideal solids (crystals) are idealised by assuming: • having no defects • not rotation or translation of the atoms • have only vibrational energy Hence we have to derive the vibrational energy of the crystal, which we want to compare with experimental results.
4.1.1
Einstein’s Theory of an Ideal Crystal
Assumptions: 1) Crystal lattice of perfect symmetry (isotropic potential field) 2) Only the motion (vibration) of a single particle is considered. It’s neighbours are fixed (no vibrational coupling) 3) The vibrational displacements are very small (Hooke’s law is valid)
Hence, each atom in the lattice has 3 degrees of vibrational freedom. A crystal of N atoms can therefore be described as 3N harmonic oscillators, which should have a single characteristic (Einstein-) frequency ν E.
From part 3.2.2 we recognise the atoms in the lattice as distinguishable, hence we can readily apply the solutions from part 3.2.8:
From eq. [3.2.42]:
and from eq. [3.2.44]
© Dr. Peter Blümler
hν E exp − 2 k T B Z Nvib = hν E 1 − exp − k T B E vib =
3N
3 3 Nhν E Nhν E + hν 2 exp E −1 k BT
School of Physical Sciences
[4.1.1]
[4.1.2]
University of Canterbury
page:46
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
When we also define a characteristic temperature, Θ E , the Einstein-temperature, via (cf. eq. [3.2.46]) hν E ΘE ≡ [4.1.3] kB We get the following expressions for the molar energy of the crystal and it’s specific heat. ~ 3 3RΘE E E = RΘ E + Θ 2 exp E − 1 T ~ ~ E ∂E E CV = ∂T
and
[4.1.4]
2
Θ E exp Θ E T = 3R T 2 exp Θ E − 1 V T
[4.1.5]
(see also eq. [3.2.45], factor 3) To test the results of Einstein’s theory, we want to calculate the limits at T = 0K and T → ∞. ~ ~ We expect for T → 0, that CVE → 0 and for T → ∞, that CVE → 3R (Dulong-Petit’s rule). Furthermore, experimental data (see later this section) reveal a T 3 dependence when approaching 0K. a) Einstein’s theory in the low limit (T → 0) 6→ 7∞8 64→ 7∞4 8 2 Θ Θ E exp E we can argument that the denominator approaches infinity T ~ T faster than the nominator, hence the limit is 0 (as lim CVE = lim 3R required). Alternatively, we can use L’Hôspital’s rule 2 T →0 T →0 exp Θ E − 1 (three times): T 144244 3 →∞
once:
Θ2 − 2 E3
→ 64 7∞4 8
Θ3 Θ exp E − E4 T
ΘE ΘE Θ 2E exp + T T T T 2T 2 3R lim = 3R lim Θ Θ Θ T →0 T → 0 exp Θ E − 1 − 2 exp TE − 1 E2 exp TE T 43 4 T 142 4 →∞
twice:
−
ΘE T2
−
→ 6 7∞8
Θ 2E
T3 3R lim Θ T →0 ΘE − 2 exp TE T
ΘE T = 3R lim Θ T → 0 exp E T
1+
1424 3 →∞
three times: 3R lim
T →0
−
ΘE T2
Θ Θ − E2 exp TE T
1 =0 T →0 exp Θ E T 4 142 3
= 3 R lim
→∞
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:47
4. Quantum Statistics
~ This time we are also interested in the functionality with which CVE approaches 0K. Therefore, we can neglect the "1" in the denominator of eq. [4.1.5] because in the low limit Θ exp TE >> 1 and we find: 2 ~ Θ Θ CVE (T → 0) = 3R TE exp − TE
[4.1.6]
THIS DOES NOT RESEMBLE THE EXPERIMENTAL DATA (T 3 dependence)! b) Einstein’s theory in the high limit (T → ∞) This we have already calculated for the perfect gas in the high limit (see page 28). The only difference is the factor 3 and we get the correct answer (Dulong-Petit’s law).
(ΘT )2 → 0 = De L'Hospital 1 = 3R lim = 3R 2 T →∞ T → ∞ exp (− Θ ) (11 −4exp ( − Θ )) exp (Θ ) T 1 424 3 T 424T4 3 123
~ lim CVvib = 3 R lim
T →∞
→0
→1
→1
~ The graph on the right shows CVE as predicted from eq. [4.1.5].
Hence the problem with Einstein’s theory occurs at low temperatures. The following table gives some values the Einstein-temperature, Θ E , for some solids:
Einstein model ~ CVE [R]
The insert is a magnification for low temperatures.
3
2 0.01
0.005
1
0
0.04
0.06
0.08
0.1
T/ΘE 0
Silver:
Θ Ag E = 200 K
Lead:
Θ Pb E ≈ 60 K
Diamond:
ΘC E = 1300 K
1
2
3
4
The Einstein temperature for diamond is extremely high and would correspond to vibrations with a frequency of ν E ≈ 1012 Hz, which is in the infra-red! The assumption (no. 2), that all atoms vibrate independently at ν E must be wrong! Because this energy (or frequency or temperature) corresponds to relatively high temperatures (see table) there must be a distribution of energies (or frequencies).
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:48
4.1.2
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
Debye’s Theory of an Ideal Crystal
Improvement of Einstein’s theory by introduction of a range of frequencies (instead of a single ν E) with an upper limit ν D ≡ Debye -frequency. Physical interpretation: At low temperatures there are always some allowed frequencies, so that k BT > hν. These can be excited and thus CV is larger than in Einstein’s theory. There are no longer 3N independent oscillators, but the frequency distribution describes coupled oscillators. EXCURSUS: Coupled oscillators (see also coupled pendula)
x x1
x2
extension of the equation for harmonic motion by a coupling term = 0 for independen t oscillator s λ = ≠ 0 for coupled oscillator s &x&1 + ω 2 x1 + λx 2 = 0 and x&&2 + ω2 x 2 + λx1 = 0 contribution of oscillator 2
contribution of oscillator 1
This differential equation system can be solved by introducing normal co-ordinates q1 ≡ x1 + x2 and q 2 ≡ x1 − x2
(
)
(
)
⇒ q&&1 + ω2 + λ q1 = 0 and q&&2 + ω2 − λ q2 = 0 From the solution we then get two independent normal (oscillation) modes with frequencies ω2 ± λ . Analogous treatment of N coupled oscillators would provide a general solution, but can no longer be solved analytically (Computer simulations). However, Pieter Debye found a way to avoid these complications.
© Dr. Peter Blümler
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University of Canterbury
PH 605: Thermal & Statistical Physics
page:49
4. Quantum Statistics
Therefore, he used the following approach: • Start with a distribution of oscillation frequencies (distribution of phonon wavelengths of sound waves). • The lattice is able to oscillate with frequencies much smaller than ν E. • The physical idea behind this, is that the specific heat of the crystal is stored in many standing phonon waves (normal modes). • Take Einstein’s approach of quantization of energies and degrees of freedom, but not all frequencies are equally possible neither is it just one (distribution of frequencies/wavelengths). • The wavelength λ is large compared to the atomic distances (λ >> d). Hence the atomic structure can be ignored. The crystal behaves like an elastic continuous body. The following graph compares the distribution of frequencies for the Einstein and Debye model. We do not yet know the shape of the Debye-distribution, but we know that its area must be normalised to the number of states (=3N) and must decay towards zero and has an upper limit defining the Debye-frequency.
N(ν) Einstein:
N(ν)
3N
Debye:
density of states = area = 3N
? ν
νE
νD
ν
We need to calculate the internal energy of the crystal: E=
∞
∑ N (ν)dν
ν =0
Eνvib
∞
≅
∫ N (ν) Eν
vib
dν
[4.1.7]
0
where N(ν)dν is the distribution (or density) of frequencies (i.e. the number of allowed modes with frequencies between ν and ν+dν), of which we know that ∞
∫ N (ν) dν = 3 N
[4.1.8]
0 © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:50
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
The density of frequencies, N(ν)dν, can be deduced from the density of states (cf. eq. [3.2.17]) N (k )dk = Hence, with k =
k2 2π
2
Vdk
with k =
2π λ
and νλ = c
2π 2π ν and dk = dν c c N (ν )dν =
2π 4πV ν 2 V dν = dν c 2π 2 c 2 c3
4 π2 ν 2
[4.1.9]
In these equations, c is the velocity of the phonons (speed of sound). But we remember, that in a solid the velocity of sound propagation depends on the direction of propagation, and that we have to distinguish between: • longitudinal waves ⇒ compression of lattice • transverse waves ⇒ shear of lattice (see figure below)
This is comparable to different polarisation directions of light waves (see later in ‘black-body radiation). Obviously, there are different forces acting on the lattice atoms for transverse and longitudinal movements. Hence, the different energies generally also cause different propagation speeds. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:51
4. Quantum Statistics
One can define a mean sound velocity, c , weighted by the degrees of freedom of the longitudinal (1 degree of freedom) and transverse (2 degrees of freedom): 1 3
c
=
3 c
3
=
1 3 clongitudin al
+
2
[4.1.10]
c 3transverse
which applies to all lattice vibrations of considerably low frequencies (Hooke’s law). Debye’s simplification: • no distinction between clongitudinal and ctransverse, hence the speed of sound, cS, is clongitudin al = c transverse = cS
[4.1.11]
• this applies to all frequencies (somewhat continuous spectrum) Equation [3.2.17] (density of states) was derived (cf. section 3.2.4, semi-classical argument) under the assumption that the k-values are closely spaced. Small k-values correspond to long λ, longer than the distance, d, of the lattice points. From an atomistic point of view, there must be a lower limit for λ → d (λmin or ν max). This can also be explained by the minimum wavelength of the sound-waves the lattice can propagate: 1) A crystal with an interatomic spacing of d cannot propagate waves with a wavelengths less than 2d, because the highest frequency is reached when directly neighbouring atoms move in antiphase (see figure).
© Dr. Peter Blümler
School of Physical Sciences
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
2) Alternative explanation via the sampling or (Nyquist) theorem (see PH506*): The sampling interval of the atomic lattice is the atom spacing, d. Hence, a wave with λ2 < d will be sampled 1 n 1 as λ1 ≥ d (where = + and n ∈ ¥, see figure). The sampling theorem also states λ 2 d λ1 λmin = 2d.
λ min = 2d
Both arguments yield:
c 2d
or ν max ≡ ν D =
[4.1.12]
Now we have all the necessary arguments to calculate N(ν)dν: νD
∞
From eq. [4.1.8]:
∫ N (ν) dν
=
0 νD
∫ N (ν) dν
[ 4 .1 .9 ]
=
0
νD
∫ 0
=
∫ N (ν) dν
= 3N
0
4πV ν 2 c
3
dν
∫
=
0
3 νD
12 πV ν c 3S 3 0
[ 4 .1 .10 ] ν D
=
4πV cS3
12πV ν 2 c 3S
dν =
12πV c3S
νD
∫ ν dν 2
0
ν 3D = 3 N
Hence, the average speed of sound in a Debye solid is c 3S =
4 πV 3 νD 3N
[4.1.13]
Debye-frequency 1
cS 3 3 νD = 4π (V N ) 13 1 424 3
1
3 3 cS = 4π d
[4.1.14]
d
* http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH506/PH506.html © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:53
4. Quantum Statistics
c We have to check, if ν D fulfils the requirement of ν max = : 2d 1
c c 3 3 cS ν D = = 0.62 S = 1.24 S ≈ ν max d 2d 4π d When we substitute eq. [4.1.13] back into eq. [4.1.9], we receive an expression for the distribution of frequencies in the Debye-solid: N (ν )dν =
4 πV ν 2 c3
[ 4. 1. 10 ]
dν
=
12πV ν 2 c 3S
dν
[ 4 .1 .13 ]
=
12 πV ν 2 9N dν = 3 ν 2 dν 4πV 3 νD νD 3N
Debye’s spectral distribution function of a 3D-isotropic, ideal crystal: N (ν )dν
=
9N ν 3D
ν 2 dν
[4.1.15]
Similarly we would get N 2 D (ν) ∝ ν
for a "two-dimensional lattice"
N 1D (ν ) ∝ const .
for a "one-dimensional lattice"
The 3D-spectral distribution function is shown in the next figure:
N( ν) [N/νD3 ]
8 6 4 2
ν/νD
0
0
0.5
1.0
1.5
νD
What we are after is an expression for the internal energy (cf. eq. [4.1.7]). But we have to be careful now, because N(ν)dν already contains the 3N degrees of freedom (or the density of states/degeneracy of the 3N oscillators). Hence we need the vibrational energy of a 1D-oscillator. νD
E=
∫ N (ν) Eν
vib,1D
dν
0
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:54
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
1 hν Eνvib,1D = hν + 2 exp h ν − 1 k BT Hence, the molar energy of the Debye crystal is Eνvib,1D is given by eq. [3.2.44] for N=1:
νD
ν D νD 3 3 ~ 9 N Ah ν ν 9N A h ν ν 2 ED = + ν d ν = d ν + d ν h ν − 1 ν 3D 0 2 ν 3D 0 2 exp h ν − 1 exp 0 k BT k BT
∫
9 N Ah ν4 = ν 3D 8
∫
νD
+ 0
9N A h ν 3D
νD
∫
∫
ν3
dν hν −1 exp 0 k2 BT 1444 444 3 no simple expression
νD
~ 9 9N A h ν3 ED = N A h ν D + dν 8 ν 3D 0 exp hν −1 k BT
∫
[4.1.16]
The integral doesn’t have an easy solution. However, MAPLE allows us to probe the expression at several point, resulting in the quasi-continuous figure below. You can download the MAPLE-routine from the course-webpage: http://wwwnmr.ukc.ac.uk/nmr/staff/pb/teach/PH605/PH605.html
~ CVD [ R ]
T/Θ D
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:55
4. Quantum Statistics
Like in Einstein’s model we want to investigate the high and low limit of Debye’s theory. Therefore, the expression in eq. [4.1.16] is somewhat simplified by introducing a characteristic temperature, ΘD, the Debye temperature. ΘD ≡
hν D kB
[4.1.17]
νD
~ 9 9N A h ν3 hence: ED = RΘ D + dν 8 ν 3D 0 exp hν −1 k BT
∫
a) Debye’s theory in the high limit (T → ∞) Θ for T → ∞ it results that T >> ΘD or h ν << 1 and TD << 1 . k BT νD
~ 9 9N A h ν3 lim E D = RΘD + lim dν 8 T →∞ ν 3D T → ∞ 0 exp h ν − 1 1k2 BT 3
∫
<< 1
The critical exponent is much smaller than 1 in this limit, hence we can expand it (only using the first two terms of the expansion series): νD
9 9N A h ν3 ~ ED (T → ∞ ) = RΘ D + dν 2 8 ν 3D 0 1 + hν + 1 hν + K − 1 k BT 2 k BT
∫
νD
9 9N A h k BT 9 9N A k BT ν 3D = RΘ D + ν 2 dν = RΘ D + 8 h 8 3 ν 3D ν 3D 0 9 Θ2 = RΘ D + 3RT + − 9 RΘ D + 3 R D ± K 8 20 T 8 14444 4244444 3
∫
higher terms of the expansion
∂E~ D ~D = 3R and hence, CV (T → ∞ ) = ∂T V
Θ2 + − 3 R D2 m K 20 T 14442444 3
to be exact, but these → 0
We see that Dulong-Petit’s rule is fulfilled by Debye’s theory in the high limit (see also figure on previous page)
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:56
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
b) Debye’s theory in the low limit (T → 0) This is the more interesting case, because Einstein’s model didn’t agree with experimental values here. νD
~ 9 9N A h ν3 lim ED = RΘ D + lim dν 8 T →0 ν 3D T → 0 0 exp hν −1 k BT
∫
k T k T To simplify, we substitute x ≡ hν hence ν = Bh x or dν = Bh dx and the integration limit k BT Θ hν x max = k DT = D >> 1 . Hence, the integration limit x max → ∞ in the low limit. B T
9N A h ~ 9 ED (T → 0) = RΘ D + 8 ν 3D =
k B4
T
ΘD →∞ T
4
∫
h4
9 k3 R ΘD + 9 N A k B T 4 B 8 ν 3D h3
e x −1
0 ∞
x3
x3
∫ e x −1
dx =
0 424 1 3
dx
9 T 4 π4 RΘ D + 9R 8 Θ3D 15
π 4 15
9 3 T4 R ΘD + π 4 R 3 8 5 ΘD 3 ∂E~ D ~D 3 4 4T 3 12 4 T and hence, CV (T → 0) = = π R 3 = π R ∂T 5 5 ΘD ΘD V =
So Debye’s model gives the desired T 3 dependence found in experimental data. Of course it also ~ yields CVD ( 0K) = 0 .
Debye’s T 3 law (for the low temperature limit) ~ CVD (T → 0)
=
12 4 T π R 5 Θ D
~ The following graph shows a close-up of CVD for low temperatures (calculated with the previously mentioned MAPLE program). When comparing to the insert of the graph on page 47 for the Einstein model, we recognise that Debye’s model predicts a less steep approach to 0K.
3
T ≈ 233.8 R Θ D
3
[4.1.18]
~ CVD [ R]
T/ΘD © Dr. Peter Blümler
School of Physical Sciences
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PH 605: Thermal & Statistical Physics
page:57
4. Quantum Statistics
The following figure directly compares both theories on the same experimental data. We clearly recognise that both models similarly well predict CV at high temperature. Debye’s theory clearly is the better match when approaching 0K.
Molar specific heat CV [cal/K/mol]: Experimental values (points) are normalised with respect to ΘE (Einstein-temperature, shown as × ) and to ΘD (Debye-temperature, shown as š). The theoretical curves are also shown (E = Einstein-function, D = Debye-function). The insert on the right shows the superiority of the Debye-theory at low temperatures. The following table shows the Debye-temperatures, ΘD, for various substances: substance Ag C (Diamond) Fe Pb NaCl CaF2 FeS2
ΘD [K] 225 1800 465 94.5 281 474 645
The table shows very different values of ΘD for different elements. If Debye’s theory is correct all the measured CV(T) values for these substances must fall on the same principal curve when T is normalised by the substance specific ΘD. Such a curve is called "master-curve". This is shown in the next figure. We recognise excellent agreement for most substances.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:58
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
Master-curve for the Debye-theory: Temperature dependence of CV for various solids versus an abscissa normalised by the individual ΘD (see table). From the previous table we realise that lighter elements (e.g. Diamond) have extremely high Debyetemperatures. The next table compares this with the atomic mass and the molar specific heat at 20°C. ~ 293 K substance ΘD [K] mass [u] CV [Jmol-1K-1] C (Diamond) Ag Pb
1800 225 94.5
12 108 207
5.9 25.1 26.8
~ Dulong-Petit’s rule predict a CV = 3R ≈ 24.9 Jmol-1K-1 in the classical limit (high T). From the table we see good agreement for the heavy elements. Diamond however does not. This is because the energy of vibration (harmonic oscillator) is:
(
)
E = hω s + 12 with ω = thus E ∝ m
k m
where k is the spring constant (bond strength)
− 12
Therefore, the lighter an element the more thermal energy (k BT) is needed to excite (activate) oscillations with the same frequency as for the heavier elements at the same temperature. The oscillations of the diamond bonds are even at room temperature not really excited (or ‘molten’... hence they are rigid, hence diamond is harder than lead at room temp.*) However, in the frame of Debye’s theory this is corrected via a substance-specific, characteristic temperature ΘD (then all materials show the same temperature dependence of CV, master-curve). *to understand as a rule of thumb. Of course structural effects (metal/non-metal) play also a role. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
4.2
page:59
4. Quantum Statistics
Quantum Statistics
When does quantum statistics apply? Or when do we leave the semi-classical limit? In section 3.4.1 we have already derived a condition for the validity of the (semi-) classical approximation: 3
N 3 2 3 λ << 1 V 2π
eq. [3.4.7]
1
V 3 λ << N
or in eq. [3.4.9]
The following table summarises relevant facts for some interesting materials:
(N )
(VN )1 / 3
material
mass [kg]
@ T [K]
λ [m]
1/ 3 d= V [m]
Air
4.8 · 10-26
300
1.9 · 10-11
3.4 · 10-9
180
N2 (liquid)
4.7 · 10-26
77
3.8 · 10-11
3.9 · 10-10
10
4
6.6 · 10-27
4.2
4.4 · 10-10
3.7 · 10-10
0.86
e! *)
9.1 · 10-31
300
4.3 · 10-9
2.3 · 10-10
0.053
He (liquid)
λ
*) conductive electrons in Cu
We clearly see that for light materials or low temperatures (like electrons or liquid helium) the classical condition 3 VN λ−1 >> 1 (right column in table) is no longer fulfilled. In the interpretation of section 3.4.1, the de Broglie wavelength, λ, becomes about the size of the inter-atomic distance, d = 3 VN . Hence, we must include interference effects of the particle wavefunctions. As a result of this, the occupation numbers of different energy states can no longer be chosen arbitrarily. (for example, see section 3.4.3, Pauli exclusion-principle for half integer spins particles) The following distinctions are results of relativistic quantum theory, which only holds when the symmetry of the (spin-) wavefunctions obeys the following restrictions (Wolfgang Pauli ca. 1945).
© Dr. Peter Blümler
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University of Canterbury
page:60
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
Principally we have to deal with two very different classes of substance. One has to obey Pauli’s exclusion principle and the other has. This results in following classification of matter and consequences due to spin quantum number and the symmetry of the total wavefunction. occupation number, Ni, of an energy spin-quantum state i no. no restrictions Ni = 0, 1, 2, ....
integer I = 0, 1, 2, ...
Pauli-exclusion principle: At least one in a particular state Ni = 0, 1
half-integer
Examples: for Bosons : for Fermions:
I=
1, 3,K 2 2
symmetry of wavefunction
classification
statistics
symmetric
BOSON
BoseEinstein
anti-symmetric
FERMION
FermiDirac
photons (I = 1), phonons (I = 0), π-mesons (I = 0) protons (I = ½), electrons (I = ½), neutrons (I = ½)
Similar arguments hold for interacting or combined particles (interacting atoms or combined atoms in molecules). To see which statistics they obey we have to combine the angular momentum (see section 3.4.3, net angular momentum of nuclear and electron spin plus orbital momentum). Such considerations can be summarised in the following rules for combined particles: number of fermions
number of bosons
composite spin
odd
any
I = half integer ⇒ Fermion
even
any
I = integer ⇒ Boson
For example: 4 He: 2p, 2n, 2e 3 He: 2p, 1n, 2e
even (6) no. of fermions odd (5) no. of fermions
⇒ Boson ⇒ Fermion
A somewhat qualitative understanding about the symmetry associated with bosons/fermions can be deduced from looking at linear combinations of wavefunctions (cf. excursus in section 3.4.3, page 42).
© Dr. Peter Blümler
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PH 605: Thermal & Statistical Physics
Bosons:
page:61
4. Quantum Statistics
supposed to have symmetric wavefunctions, hence symmetric combination: Boson ψ12 = φ1 ( a ) φ 2 (b ) + φ 2 (a ) φ1 (b)
single state wavefunction (e.g. nuclear spin) at boson 1 with quantum number a.
If there is no restriction in the occupation number associated with this state, we may just set a = b (violation of Pauli’s exclusion principle, which doesn’t apply here) and Boson see that the total wavefunction is then: ψ12 = 2 φ1 ( a) φ 2 (a ) ≠ 0 (the particle can exists in this state!)
Fermions:
supposed to have anti-symmetric wavefunctions, hence anti-symmetric combination: Fermion ψ12 = φ1 ( a) φ 2 (b) − φ 2 ( a ) φ1 (b)
A violation of Pauli’s exclusion principle (two particles in the same state, i.e. same Fermion quantum number), e.g. by letting a = b, causes ψ12 = 0 . The wavefunction collapses and the particle cannot exist in this state!)
Plausible, but no proof! As a consequence, the particles in quantum statistics have to be indistinguishable! This is because the energy levels of a system of identical particles cannot depend on which particle is where (whether or not they are experimentally distinguishable!). Hence, every micro-state can be interpreted as a macro-state (this doesn’t make sense in the classical treatment!) In other words: The Hamiltonian describing the entire system (in an extreme view, we have total interference of all wavefunctions of all particles) must obey its symmetry under any permutation (swapping of particle co-ordinates). This leads to the following POSTULATE: The wavefunction for a set of identical particles is either totally symmetric or anti-symmetric under any permutation (of particle co-ordinates).
© Dr. Peter Blümler
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
4.2.1
Bose-Einstein Statistics
Satyendra N. Bose and Albert Einstein (1924) [Recap the derivation of Boltzmann statistics in section 3.1.]
From the previous introduction we know that we have to deal with indistinguishable particles with no limitations in the occupation number. First we have to derive an expression for the statistical weight, Ω. For this purpose, imagine a line of Ni particles and a set of ( Ai − 1) barriers, which divide the particles in Ai groups (representing the energy-states). As shown in the following sketch.
state particle
1
2
3
...
1 2 3 4 5 6 7 8
barrier
1
2
Ai -1
Ai
... ... ...
Ni -1 Ni
Ai -2
Ai -1
So how many ways exist to arrange these ( Ni + Ai − 1) objects? The total number of permutations is given by ( Ni + Ai − 1)! . However, this also includes the physically meaningless permutations of the identical and indistinguishable particles and barriers. Hence we have to correct the total number of permutations by that of the permutation of the Ni particles and ( Ai − 1) barriers. Hence, the total number of different arrangements (micro-states) is: Ωi =
(N i + Ai − 1)! Ni ! ( Ai − 1)!
(4.2.1)
example: Ai = 3 (two barriers) and N i = 2 indistinguishable particles (dots!) no. 1 2 3 4 5 6 © Dr. Peter Blümler
Ai
I
II
l
l l
l
III l l
ll
Ω=
4! 24 = = 63 2! 2! 4
ll ll
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4. Quantum Statistics
Now we are ready to derive the Bose-Einstein (abbreviated by BE) distribution. The procedures we are going to apply are completely analogous to those in the derivation of the Boltzmann statistic (see section 3.1). Like there we start with evaluating the statistical weight of the micro-states. Ω =
∏
[ 4 .2 .1]
Ωi
=
i
(N
+ A − 1)!
∏ N ii! ( Aii − 1)! i
However, in order to simplify the following maths, we can assume that there are enough particles and energy levels to skip the ‘-1’ in the expression. We can of course (without lack in generalisation, see above) simply request ( Ai + 1) different levels. Hence, we get the following conditions for our BE-system: Ω BE
a)
( Ni
[4.2.2]
∑ Ni = N
[4.2.3]
∑ εi N i = E
[4.2.4]
=
∏
+ Ai )! Ni ! Ai !
i
b)
i
c)
i
Next step is finding the maximal Ω or calculating δlnΩ = 0. So we start with calculating lnΩ: ln Ω
BE
[ 4 .2 .2 ]
=
ln i
[ 3. 1. 6]
≅
∏
( N i + Ai )! N i! Ai !
∑ln ( Ni + Ai )!
∑ln ( Ni !) − ∑ln ( Ai !)
−
i
i
i
∑( N i + Ai ) ln ( Ni + Ai ) − ∑( Ni + Ai ) − ∑ Ni ln N i + ∑ N i i
−
i
∑ Ai ln Ai + ∑ Ai i
ln Ω BE =
=
i
i
∑ (N i + Ai )ln ( Ni + Ai ) − ∑ Ni ln Ni i
−
i
∑ Ai ln Ai
=
(N
=
)
∑ ( Nii + Aii ) δNi + ∑ ln ( Ni + Ai ) δN i − ∑ Nii δNi i
δ ln Ω BE
+A
N
i
i
∑ [ 1 + ln ( N i + Ai ) − 1 − ln N i ] δN i
[4.2.5]
i
∂ ln Ω BE Now, we have to find the most probable statistical weight via ∂N i this using a small change (mathematical rigour). δ ln Ω BE
i
= 0 . Again we calculate
−
∑ ln Ni δN i
= 0
i
= 0
i
δ ln Ω BE
=
N i + Ai A δN i = ln i + 1 δN i = 0 Ni Ni i
∑ln i
© Dr. Peter Blümler
∑
School of Physical Sciences
[4.2.6]
University of Canterbury
page:64
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
From eqs. [4.2.6], [4.2.3] and [4.2.4] we get the following three boundary conditions: δ ln Ω BE
i)
A
∑ln Nii + 1 δN i
=
= 0
[4.2.7]
i
∑ δN i
ii)
=0
[4.2.8]
i
∑ ε i δN i
iii)
=0
[4.2.9]
i
We combine them using the method of undetermined multiplies (λ and β): A
∑ δN i ln Nii + 1 + λ + βεi i
⇒
= 0
A ln i + 1 = − λ − βεi Ni Ai +1 Ni
=
e − λ e − βε i ≡ α e −βε i
[4.2.10]
1 k BT (the minus sign occurs because Ni is in the denominator of eq. [4.2.10]) β = −
Similar treatment to that on page 6 gives:
We will investigate the significance of α later (we just memorise that it is associated with boundary condition ii). Rearranging eq. [4.2.10] then gives:
Bose-Einstein distribution Ni Ai
4.2.2
=
1 ε α exp + i k BT
− 1
[4.2.11]
Fermi-Dirac Statistics
Enrico Fermi and Paul. A. Dirac (1926) From the introduction to this section we know that the particles are indistinguishable and can only realise occupation numbers of either 0 or 1 (particles) per available state. Hence, there simply must be more (or equal) states than particles.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:65
4. Quantum Statistics
To establish an expression for the statistical weight of Fermi-Dirac (abbreviated by FD) statistics, we wonder how many ways exist to pick Ni (occupied by one particle) cells from Ai totally available cells, while ( Ai − Ni ) remain unoccupied? The total permutation of all cells is Ai ! . But again we must correct for the permutation of the occupied N i ! and unoccupied ( Ai − Ni )! cells (their permutation has no physical significance). Hence, the total number of different arrangements (micro-states) is: Ai ! Ni ! ( Ai − N i )!
Ωi =
[4.2.12]
example: Ai = 3 cells and N i = 2 indistinguishable particles (dots!) no.
Ai
1 2 3
I
II
l
l
III Ω=
l
l
l
3! 6 = = 33 2! 1! 2
l
The treatment is completely analogous to the previous section (hence a bit abbreviated). We get the following conditions for our FD-system: Ω FD =
a)
A!
∏ Ni ! ( Aii − Ni )!
[4.2.13]
i
∑ Ni = N
b)
[4.2.14]
i
∑ εi N i
c)
=E
[4.2.15]
i
∑ ln ( Ai !) − ∑ ln ( Ni !) − ∑ ln( Ai − Ni )!
ln Ω FD =
From eq. [4.2.13]:
i
i
i
Using Stirling’s approximation (eq. [3.1.6]): ln Ω FD ≅
∑ Ai ln Ai − ∑ Ai −∑ Ni ln Ni + ∑ N i − ∑( Ai − N i ) ln ( Ai − N i ) + ∑( Ai − N i ) i
=
i
i
i
i
i
∑ Ai ln Ai − ∑ Ni ln N i − ∑( Ai − Ni )ln ( Ai − Ni ) i
i
i
and maximising Ω FD via δ ln Ω FD =
−
∑ N ii δNi − ∑ ln Ni δNi N
i
© Dr. Peter Blümler
i
+
A −N
∑ Aii − Nii δNi + ∑ ln ( Ai − N i ) δNi i
School of Physical Sciences
= 0
i
University of Canterbury
page:66
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
From the previous equation, eqs. [4.2.14] and [4.2.15] we get the following three boundary conditions: δ ln Ω FD
i)
A
∑ ln Nii − 1 δNi
=
= 0
[4.2.16]
i
∑ δN i
ii)
=0
[4.2.17]
i
∑ ε i δN i
iii)
=0
[4.2.18]
i
which we again solve using undetermined multipliers: A δNi ln i − 1 + λ + βεi = 0 Ni i
∑
⇒
A ln i − 1 = − λ − βεi Ni Ai −1 = Ni
e − λ e − βε i ≡ αe − βε i
Similar treatment to that on page 6 gives here also:
β = −
[4.2.19]
1 k BT
Hence, we get
Fermi-Dirac distribution Ni Ai
4.2.3
=
1 ε α exp + i k BT
[4.2.20]
+ 1
Comparison of Boltzmann, BE and FD Statistics
In section 3.1 we derived Boltzmann’s distribution without referencing the Ai states (with energies between ε i and ε i + dεi ) over which the Ni particles can be distributed. We want to match up with this lack in section 3.1, in order to directly compare the three different statistics. In the (classical) Boltzmann statistics, the distinguishable (!) particles can be distributed over the Ai cells without any restrictions or limitations, except that swapping particles occupying the same cell doesn’t generate a new micro-state. Hence, the first particle can be distributed in Ai different ways over the Ai cells, and the same is true for the second and all other particles. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:67
4. Quantum Statistics
Because each assignment of a particle to a cell (energy state) can be combined with each other to
give a different micro-state, we have to consider ( Ai )N i different possibilities to distribute Ni particles in Ai cells. Next we have to ask how many ways exist to group the N particles in ensembles of N 0 , N1 ,K, N i ? The answer is eq. [3.1.1]. Hence, we get for the total number of realisations Ω = N!
N
Ai i Ni ! i
∏
[4.2.21]
Applying the same treatment as in section 3.1, we get ln Ω = N ln N + N i ln Ai − N i ln Ni
∑
∑
i
δ ln Ω =
∑ln Ai δN i
i
−
i
=
∑ ln N i δN i
= 0
i
A
∑ln Nii δN i
=0
i
and together with the usual boundary conditions A
∑ ln Nii + i
∑ δNi = 0 and ∑ εi δN i = 0
λ + βεi δN i = 0
Ni 1 1 = ≡ − λ − βε Ai e e i α e − βε i completely analogous procedures to those on page 5-6 gives: ⇒
α =
N
∑
Ai e − βε i
and β = −
1 k BT
[4.2.22]
i
and the usual form for Boltzmann's distribution (cf. eq. [3.1.17]) ε Ai exp − i Ni kBT = r −1 N ε Ai exp − i k BT i =0
[4.2.23]
∑
This equation is identical to eq. [3.1.17] except for the Ai . In the first derivation of Boltzmann's statistic (section 3.1) we had assumed that each energy level ε i is defined by exactly one quantum state. Now we have extended the distribution in such a way that each energy ε i is represented by Ai degenerate states (with Ai as the degree of degeneracy). This is of course the more general treatment. N The average number of particles ni = i in the ith state is then given by Ai © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:68
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
The distribution functions: niB =
Boltzmann:
Ni = Ai
Bose-Einstein:
niBE =
Ni = Ai
Fermi-Dirac:
niFD =
Ni = Ai
1 ε α exp + i k BT
[4.2.24]
1 ε α exp + i − 1 k BT 1 ε α exp + i + 1 kBT
[4.2.25]
[4.2.26]
Careful: The exponent is now positive (or in the denominator)! ε For α exp k iT >> 1 the two quantum distributions (BE and FD) are becoming identical to the B Boltzmann distribution. This is the same as ni being very small, which is identical to the discussion in section 3.2.1 and 3.4.1; or yet another formulation of the classical limit.
4.2.4
Determination of α
Before the derivation of the "grand partition function" in the next section, we want to try to relate α to some macroscopic properties. We remember that α was introduced as an undetermined multiplier scaling the boundary condition which stated the constant number of particles in the system. Hence it must be related to the number of particles in the system (same as β relates to the energy). We start with the free Helmholtz energy F = E − TS , where E = Hence,
F=
∑ Fi i
=
∑ [Ni ε i
∑ Ni ε i and S = k B ln Ω .
− T k B ln Ωi ( Ni )]
[4.2.27]
i
Next step is to minimise F with respect to Ni . But the Ni are not independent! We pick one, say N j and Nj=N−
∑ Ni
[4.2.28]
i≠ j
Now we can rewrite eq. [4.2.27]:
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:69
4. Quantum Statistics
F = N j ε j − T k B ln Ω j ( N j ) +
∑ [Ni ε i
− T k B ln Ω i ( Ni ) ]
[4.2.29]
i≠ j
Now we minimise F with respect to Ni but NOT N j , or ∂F ∂N i
= 0 N j
for i = 1, 2, 3, K , j − 1, j + 1, K ↑ not j
inserting eq. [4.2.29] gives ∂ ln Ω j ( N j ) ∂N j ε j − k BT ∂N j ∂N i From eq. [4.2.28] we get
∂N j ∂N i
∑ ε i
+
i≠ j
∂ ln Ωi ( Ni ) = 0 ∂N i
= − 1 , and if we look at each individual state the equation above
simplifies to εi − k BT
− k BT
∂ ln Ω j ( N j ) ε j − k BT ∂N j 144442444 4 3
∂ ln Ω i ( N i ) = ∂N i
this expression has some finite value for all groups (because j was chosen arbitraril y)
For the moment we define: ζ ≡ ε i − k BT
[4.2.30]
∂ ln Ω i ( Ni ) 1 =− (ζ − ε i ) ∂N i k BT
and
and with β = −
∂ ln Ω i ( Ni ) ∂N i
1 : k BT
∂ ln Ωi ( N i ) = β (ζ − ε i ) ∂N i
Now we can compare this with the distribution functions derived in eqs. [3.1.11]* or [4.2.22], [4.2.10] and [4.2.19] and see (cf. for instance to the derivation of eq. [4.2.19] ) that **) ∂ ln Ωi ( N i ) = β(ζ − εi ) = − λ − βεi ∂N i
hence,
λ = − βζ and
α
e.g. [ 4. 2. 19 ]
≡
e − λ = eβζ
[4.2.31]
*) but careful with the different definition of β in this chapter. Here β = -(k BT)-1 **) for all three statistics! © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:70
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
We now claim that ζ is nothing but the chemical potential, µ, (see page 22 Dr. Mallett's script) which is a constant for a system in equilibrium. [ 4. 2. 27 ] ∂F µ≡ = ∂N V , T [ 4. 2. 30 ]
=
∑ ε i
− T kB
i
∂N
∑[ ζ] ∂Ni i
= ζ
∂ ln Ωi ( Ni ) ∂Ni ∂N ∂N i ∂ ∂N Ni = ζ ∂N i ∂N
∑
⇒ζ = µ µ α = e βµ = exp − kBT
hence,
4.2.5
[4.2.32]
Systems with Variable Particle Numbers
(Recap grand canonical ensemble - Dr. Mallett's lecture/script) So far we have only investigated isolated systems. However, in some occasions particles will leave the system (e.g. black-body radiation, see later). In such situations we can either ignore the parameter α in the associated distribution function or we have to derive a different statistic which allows particles to leave the system. In the previous section we have already found an expression for α (cf. eq. [4.2.32]) with µ as the chemical potential: ∂F ∂S µ ≡ = − T ∂N V ,T ∂N V , T
[4.2.33]
We didn't derive α via the distributions (cf. section 3.1) for BE- and FD-statistics due to the fact that this involves extended maths and it is not easily possible to link it directly to macroscopic properties. However, there is a more general approach (introduced by Gibbs) by attaching a particle-reservoir to the observed system (like a heat-bath), which greatly simplifies the maths and allows then the variation of particle numbers in the system (e.g. due to reactions). The following figure shows a schematic representation of the different systems. Originally we have studied isolated systems (Figure a: with E, V and N fixed), which can be directly linked to macroscopic properties via the Boltzmann equation S = k B ln Ω (see treatment in section 3.1). This corresponds to a micro-canonical ensemble. Next we introduce exchange of energy with a heat bath (Figure b) which under isothermal conditions has T, V and N fixed. For small energy fluctuations we can evaluate the energy via F = E − TS = − k BT ln Z . This corresponds to a canonical ensemble.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:71
4. Quantum Statistics
Now we are going to consider a region (I) in exchange of energy and particles (Figure c) with a bath or reservoir (II), where N, E, V, T for the reservoir are fixed. The region which defines the system (I) is arbitrarily chosen (with T and V fixed). This corresponds to the grand canonical ensemble. We also introduce a parameter, which relates the change of energy and entropy due to the exchange of particles and name it µ (but we do not assume to know more about it!) The trick Gibbs used is to introduce the variation of particle numbers of a system via being a subsystem of a describable, large and isolated system, which we can determine.
4.2.6
The Grand Partition Function, Z
After the introduction in the previous chapter (nomenclature follows part c) of the previous figure), we are now ready to determine a system with variable particle numbers. The isolated heat- and particle-reservoir (II) has: N 0 , V0 and E0 The subsystem (I) of volume V can contain particles N = 0, 1, 2, K For each N we define: ε N1 ≤ ε N 2 ≤ ε N 3 ≤ K ≤ ε N r where ε N r is the energy of the rth state of the N particles. When the system is in a state associated with ε N r or N r we have to "cut" the subsystem (I) away from the reservoir system (II), and analyse system II: EII = E0 − ε N r , N II = N 0 − N r and VII = V0 − V The macro-state of the reservoir-system has then a statistical weight: Ω II ( E0 − ε N r , N 0 − N r , V0 − V ) The next question is then to determine the probability of finding the system in the state N r .
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:72
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
For this purpose we use the Principle of Equal 'A-Priori' Probability (Dr. Mallett's lecture), which is another word for the intuitive assumption that the probability of something occurring is proportional to the number of ways in which it can occur. Or P( n) ∝ Ω( n) . Hence, we get the probability of finding the system in the state N r via: p ( N r ) = c Ω II( E0 − ε N r , N 0 − N , V0 − V ) where c is a normalisation constant (proportionality factor). Thus, p (N r )
[ 3. 1. 2 ]
=
S II ( E0 − ε N r , N 0 − N , V0 − V ) c exp kB
[4.2.34]
However, the subsystem is small compared to the reservoir (II) and the exchange of E and N is then also small. Hence we are allowed to expand the entropy in eq. [4.2.34] in a Taylor series. S II ( E0 − ε N r , N 0 − N , V0 − V ) = S II ( E0 , N0 ,V0 ) −
∂S II ( E0 , N0 ,V0 ) ∂V0
V−
∂ S II (E 0 , N 0 ,V0 ) ∂E 144 420444 3
ε Nr
=1 / T
−
1 ∂ 2 S (E , N , V ) II 0 0 0 N + V 2 + ... 2 ∂ N0 2 ∂V0 144 42444 3 ∂S II ( E0 , N 0 ,V0 ) = µ / T (eq. [4.2 .33])
S II ( E0 − ε N r , N 0 − N , V0 − V )
εN r µ ∂S II ( E0 , N 0 ,V0 ) S II ( E0 , N 0 ,V0 ) − V− − N ∂V T T 144444442444404443
=
constant
The first two terms are constant, because the reservoir is assumed to be isolated (no exchange of entropy can be introduced in c in eq. [4.2.34]) and the volume is also fixed. Hence, eq. [4.2.34] becomes: Nµ − ε N r p ( N r ) = c exp kB T The normalisation constant is given by
[4.2.35]
∑ p (N r ) ≡ 1 and so we get N, r
c =
1 Nµ − ε N r exp kB T N ,r
∑
≡
1
Z
where Z is the grand (canonical) partition function.
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:73
4. Quantum Statistics
Grand partition function: µN i − ε N i r kB T
∞
N
Z = ∑ ∑ exp N i r =0
[4.2.36]
where ε N i is a single particle state (of particle i) in energy state r. r We want to reformulate this expression in order to get rid of the double sum. Therefore, we refer back to the occupation number of an energy state. εN = r
∞
∑ ni εi i =1
with ni as the occupation number of the ith energy state. Then the double sum in eq. [4.2.36] becomes a sum of products and we only have to sum over all occupation numbers to cover all states and particles: ∞
∑
ni εi µni − i =1 = exp kB T ni
∑
Z
=
∑{exp[β(µ − ε1 ) n1] ⋅ exp [β(µ − ε 2 ) n2 ]⋅ K ⋅ exp [β(µ − ε i ) ni ]} ni
with β =
1
[4.2.36a]
kB T
where all the factors are independent from each other, hence
Z = ∑ exp [β(µ − ε1 ) n1 ] ⋅ n1
=
∑ exp [β(µ − ε 2 ) n2 ] ⋅ K ⋅ ∑exp [β(µ − ε i ) ni ] n2
ni
∏∑exp [β(µ − ε i )ni ] ≡ ∏ Z i i
where
ni
i
Z i ≡ ∑ exp [β(µ − ε i ) ni ]
[4.2.37]
ni
Due to eq. [4.2.35] the probability of finding n1 in the single state 1, n 2 in the single state 2, etc. is then exp[β(µ − ε1 ) n1 ]⋅ exp [β(µ − ε 2 ) n2 ]⋅K p (n1 , n2 , n3 , K) =
Z
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
page:74
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
p (n1 , n2 , n3 , K) =
∞
∏ pi (ni ) = ∏ i
exp [β(µ − ε i ) ni ]
i
[4.2.38]
Zi
Now we can calculate Z i for the different distributions. Starting with Fermi-Dirac, hence ni = 0 or 1 from [4.2.37]:
ZiFD
=
∑exp [β(µ − ε i ) ni ] =
exp[0] + exp [β(µ − ε i ) ]
ni
ZiFD
= 1 + exp [β(µ − ε i ) ]
[4.2.39]
and for Bose-Einstein ( ni = 0, 1, 2,K )
ZiBE
=
∞
∑ exp [β(µ − εi ) ni ] = 1 + exp [β(µ − ε i ) ] + exp [2β(µ − ε i ) ] + K
ni = 0
This is a geometric series, which converges only for exp [β(µ − ε i ) ] < 1 ⇒ µ < ε i This condition must hold for all single particle states. It does so, if it holds for the ground-state ε1 (because all other ε i > ε1 ), hence µ < ε1
[4.2.40]
We can speculate ε1 ≈ 0 or deliberately set it to 0 to define a suitable energy scale). Thus, for a converging geometric series we get
ZiBE
=
1 1 − exp [β(µ − ε i ) ]
[4.2.41]
Grand partition functions of the i th state: Fermi-Dirac:
ZiFD
= 1 + exp [β(µ − ε i ) ]
[4.2.39]
Bose-Einstein:
ZiBE
=
1 1 − exp [β(µ − ε i ) ]
[4.2.41]
or
© Dr. Peter Blümler
Zi±
= {1 ± exp[β(µ − ε i ) ]}
±1
⊕ for FD and $ for BE
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
page:75
4. Quantum Statistics
We now want to check this result by calculating the mean occupation number (as before in sections 4.2.1 and 4.2.2) of the ith single particle state: ni =
[ 4 .2 .38 ] Ni exp [β(µ − ε i ) ni ] = ni pi (ni ) = ni Ai Zi ni ni
∑
∑
we use a little trick by calculating ∂Z i ∂µ
[ 4 .2 .37 ] ∂ = exp [β(µ − ε i ) ni ] = β ni exp [β(µ − ε i ) ni ] ∂ µ T ,V ni ni
∑
∑
∂Z i ∂µ T ,V 1 ∂ ln Z i 1 ni = = β Zi β ∂µ T ,V
Fermi-Dirac: niFD =
∑ ni
[4.2.42]
exp [β(µ − ε i ) ni ]
ZiFD
ni
(
[ 4 .2 .39 ] ∂ ln ZiFD ∂ β exp [β(µ − εi ) ] β ZiFD −1 = {ln (1 + exp [β(µ − εi ) ] )} = = ∂µ ∂µ 1 + exp[β(µ − ε i ) ] ZiFD T ,V
and using eq. [4.2.42]:
niFD
( ZiFD − 1) = = ZiFD
re-substituting eq. [4.2.36a] niFD =
exp [β(µ − εi ) ] 1 = 1 + exp[β(µ − ε i ) ] exp [β(εi − µ ) ] + 1
1 = eq. [4.2.26] with eq. [4.2.32]3 εi − µ + 1 exp k BT
Bose-Einstein: niBE =
∑ ni ni
∂ ln ZiBE ∂µ
)
exp [β(µ − ε i ) ni ]
ZiBE
[ 4 .2 .41 ] ∂ = {− ln (1 − exp [β(µ − ε i ) ] )} = − − β exp [β(µ − ε i ) ] = β exp [β(µ − ε i ) ]ZiBE ∂µ 1 − exp [β(µ − ε i ) ] T ,V
exp [β(µ − ε i ) ] 1 = 1 − exp [β(µ − εi ) ] exp [β(ε i − µ ) ] − 1 1 re-substituting eq. [4.2.36a] niBE = = eq. [4.2.25] with eq. [4.2.32]3 εi − µ −1 exp k BT and using eq. [4.2.42]:
© Dr. Peter Blümler
niBE =
School of Physical Sciences
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page:76
4.3 4.3.1
PH 605: Thermal & Statistical Physics
4. Quantum Statistics
Application to Fermion/Boson-Systems Free electrons in metals
We consider a metal as a regular three-dimensional lattice (cf. section 4.1) of ions and containing a large number N of (valence) electrons that are free to move throughout the entire metal. Metals are characterised by these weakly bound electrons, which can float around. Experimentally one finds about 1-4 of these conductive electrons per metal-atom depending on the material and its structure. In the absence of an electric field the electrons move about the metal randomly, very much in the way like gas molecules in a container. Therefore, they are considered as an "electron gas". In a classical treatment we would expect ~ 3R CV = 3R + = 4.5 R 2 4 123 1 42 3 Dulong − Petit for lattice vibrations
translatio n of the electrons (equipartition theorem)
~ However, the experiments give completely different values, which are much lower (e.g. CV =3.02R for Au at room temperature) and the classical limit of 4.5R is not even reached at temperatures where metals evaporate at normal pressure (2000-6000K). Because the electrons are Fermions we have to obey the Pauli-exclusion principle when describing the electron gas. We want to model them as a perfect gas obeying Fermi-Dirac statistics (cf. section 4.2.2). Perfect gas means introducing the following simplifications: • we neglect interactions with the lattice (free propagating electron-waves) • we assume that the lattice shields the electrons sufficiently from each other, hence collisions can be neglected. From eq. [4.2.26] with eq. [4.2.32] we obtain the Fermi-Dirac distribution of the average occupation number per available energy level niFD =
1 1 1 = with β ≡ ε −µ exp (β[ε i − µ ]) + 1 k BT + 1 exp i k BT
The energy of the electrons in the "electron gas" is considered to be purely translational. Hence, it can be expressed by the density of states (cf. eq. [3.2.17] and eq. [3.2.14]) with momentum and energy as the variables: N ( k ) dk
=
k 2 dk 2π 2
N ( p )dp
© Dr. Peter Blümler
=
V and with k = 4πVp 2 h3
2π p h
dp
School of Physical Sciences
University of Canterbury
PH 605: Thermal & Statistical Physics
for non-relativistic velocities the energy is given by ε = Hence p = 2me ε and dp = N (ε) dε
=
me 2ε
p2 where me is the mass of the electron. 2me
dε :
4πV 2meε me 2ε
h3
page:77
4. Quantum Statistics
dε =
2 πV h3
3
(2me ) 2
ε dε (4.3.1a)
However, this is not correct, because we have neglected the spin of the electrons (I = ± ½) which results in the same energy for different quantum numbers. Hence two electrons can occupy the same energy level without violating Pauli's exclusion principle (see graph on the right).
Therefore, the correct density of states in terms of energies is given by : N (ε) dε
4πV
=
h3
3
(2 me ) 2
ε dε
[4.3.1]
We can now derive the fraction, dN(ε), of electrons with energies between ε and ε+dε dN (ε) = n (ε ) N (ε ) dε =
4 πV h3
3
(2me ) 2
ε dε exp (β[ε − µ ]) + 1
[4.3.2]
where the mean occupation number, niFD , is simply expressed in terms of their energy, n (ε) . We further know that normalisation of this expression must give the total number, N, of electrons in the metal: ∞
∫ dN (ε) = 0
4πV h3
(
3 2 me 2
)
∞
ε
∫ exp (β[ε − µ]) + 1 dε =
N
[4.3.3]
0
Before we can continue, we have to inquire µ. Other than for the derivation of the Bose-Einstein distribution (cf. page 74 eq. [4.2.40]) we didn't have to make any assumptions about sign and value of the chemical potential to conclude to eq. [4.2.39]) for Fermi-Dirac systems. Generally µ = µ(T ,V , N ) with V and N fixed it only depends on the temperature. Therefore, it is convenient to define a lower boundary, or an energy (or characteristic temperature) defined by the chemical potential of the substance at absolute zero temperature. Fermi-energy: © Dr. Peter Blümler
ε F ≡ µ (0K ) School of Physical Sciences
[4.3.4] University of Canterbury
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
We still don't know the sign of µ in eq. [4.3.3]. Therefore, we substitute x ≡ βε = ε
k BT
4πV h
3
(2me k BT )
3 2
µ exp k T B
∞
x dx = N µ x 0 e + exp k T B
∫
[4.3.5]
Now we check eq. [4.3.5] in the low limit µ→ε F for different signs of ε F ε ε lim F = − ∞ and lim exp F = 0 T →0 k B T T →0 k BT and hence lim {eq. [ 4.3.5]} = 0 ≠ N this is clearly nonsense!
case ε F < 0: then
T →0
ε ε lim F = ∞ and lim exp F T → 0 k BT T →0 k BT this makes sense!
case ε F > 0: then
= ∞ (or a very big number, like N )
Now we have a second look at eq. [4.3.5] in checking the denominator in the integral: ε − ε F ∞ for ε > ε F lim exp = T →0 k BT 0 for ε < ε F 0 for ε > ε F and lim n ( ε) ∝ lim = T →0 T → 0 exp ε − ε F + 1 k T 1 for ε < ε F B 1
[4.3.6]
Equation [4.3.8] is illustrated in part a) of the figure above. This can be explained, because the system at 0K is seeking for the lowest possible energy state. However, due to the Pauli-exclusion principle it cannot condense in the (single) lowest state (cf. part b of the figure). Therefore, the lowest N/2 energy states are subsequently occupied with a cut-off at ε F (above they are empty).
© Dr. Peter Blümler
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University of Canterbury
PH 605: Thermal & Statistical Physics
page:79
4. Quantum Statistics
The energy distribution of the Fermi-"electron gas" at 0K is shown in part c) of the figure, which is an illustration of eq. [4.3.2] ( dN (ε, T = 0 K) ∝ ε multiplied with the distribution function n (ε) from fig. a). We recognise that a considerable amount of energy is stored in the system even at absolute zero! From these results we can readily derive an expression for ε F, because now we know that we only have to integrate up to ε F (no electrons are above this level at 0K) and the exponential term in the denominator of eq. [4.3.3] is negligible (close to zero, cf. eq. [4.3.6]) in the low limit. N =
4πV h3
(
εF
3 2 me 2
)
∫
4πV
ε dε =
h3
0
3h 3
3
ε F2 =
8π(
3 2me 2
)
2 32 ε 3 F
3
(2 me ) 2
[4.3.6a]
N V
Fermi-energy and Fermi-temperature 2
h2 3 N 3 εF = 2 me 8π V
and
TF ≡
εF kB
[4.3.7]
The total energy of the "electron gas" at 0K (area under figure c) is then given by: εF
E =
∫ 0
[ 4 .3 .7 ]
=
ε N (ε ) dε =
4πV h3
(
3 2me 2
)
εF
∫
3
ε 2 dε =
0
4 πV h3
3
(2me ) 2
2 52 ε 5 F
3
3h 2 3N 2 3 N = Nε F 40me πV 5 Average energy per electron at 0K: ε =
© Dr. Peter Blümler
E 3 = ε N 5 F
School of Physical Sciences
[4.3.8]
University of Canterbury
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
When we plug some typical values for N/V of metals (electrons per volume, see weekly question) into eqs. [4.3.7] and [4.3.8] we can estimate typical ranges (orders of magnitude for ε F and TF) ε F ≈ 10 −18 − 10 −19 J
and TF ≈ 7000 − 70000 K
The following table gives some specific values for different metals: Metal
N/V [cm3]
ε F [eV]
TF [K]
Li
4.7 · 1022
4.72
5.48 · 104
K
1.4 · 1022
2.12
2.46 · 104
Cu
8.5 · 1022
7.00
8.12 · 104
Au
5.9 · 1022
5.51
6.39 · 104
We see that TF is much higher than the "normal" temperature of use (actually at T = TF the metals in the table would no longer be metals, but plasma). We recognise that typically ("normal" temperatures) the electrons in metals are in a degenerate state which is not very different from the completely degenerate state at absolute zero temperature. (or: 300K is very cold for the electrons, almost as cold as 0K!) If we want to calculate the energy of the electron gas at higher temperatures (T ≈ TF) we cannot simply set N(ε) either to 0 or 1 as in eq. [4.3.6], but must solve: E (T ) =
4πV h3
(
3 2 me 2
)
∞
3
ε2
∫ exp ε − µ(T ) + 1 dε 0
[4.3.9]
k T B
This integral is not generally analytical solvable. However, using advanced models an estimate of µ(T) is possible: π2 T 2 for 0 < T << TF µ(T ) ≅ ε F 1 − 12 TF Then numerical approximations of eq. [4.3.9] can be generated as displayed in the next figure for T/TF = 0.2 (still very "cold" for the electrons but "hot" for us, e.g. for TF ≈ 104 K ⇒ T = 2000K). Hence the displayed distribution will correspond to higher temperatures of use! We also see that only those electrons close to ε F are excited to higher states. From that we can calculate a rough estimate: By assuming that changing the temperature from 0K to T will only excite those electrons in an energy range ∆ε = k BT centred at ε F.
© Dr. Peter Blümler
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PH 605: Thermal & Statistical Physics
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4. Quantum Statistics
Figure: a) The Fermi-Dirac average occupation number, n (ε) , at a temperature T = 0.2 µ/k B. b) The FermiDirac energy distribution at a temperature T = 0.2 µ/k B . Grey shaded area is the deviation to T=0K (cf. previous graph)
In this rough estimate the number of excited electrons, Nexc, can then be estimated via N exc ≈
N (ε F ) 123
density of states centred at ε F [ 4 .3 .1]
N (ε F ) with eq. [4.3.6a]:
and
N=
4πV h3
=
3
(2 me ) 2
N exc ≈
4πV
⋅
k{ BT
in an interval kB T around ε F
3
(2me ) 2
εF h3 2 32 2 3 N ε F = N (ε F ) ε F ⇒ N ( ε F ) = 3 3 2 εF
[ 4 .3 .7 ] 3N 3 T kBT = N 2 εF 2 TF
[4.3.10]
Usually T/TF represents only a very small fraction (e.g. for T=300K and TF = 104, Nexc ≈ 4%) which validates our previous treatment. In this rough estimate the excitation energy per electron can be estimated to be ≈ k BT
and
3 T2 E (T ) ≈ N exc k BT = N kB 2 TF
[4.3.11]
T ∂E (T ) CV (T ) = ≈ 3N kB TF ∂T V
[4.3.12]
an advanced theory gives a better estimate of CV (T ) ≅
© Dr. Peter Blümler
π2 T T N kB ≈ 4.9 N k B 2 TF TF
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
The figure below illustrates the deviation of the Fermi-theory from the classical treatment at low temperatures (as approximated by the T 2 dependence in eq. [4.3.11]).
Figure: E(T) as approximated by eq. [4.3.11]. FD= Fermi-Dirac model of the electron gas, E0 =
3 5
Nε F
Hence for a metal at high T:
low T:
lattice:
CV ∝ 3k B
electrons:
CV ≈ 4.9 N kB
T TF
lattice: electrons:
CV ∝ T 3 CV ∝ T
(Debye model, see eq. [4.1.18]) (which dominates at very low T, see figure)
(Dulong-Petit per atom) (usually negligible)
Figure: Illustration of the dominance of electron contribution(Cel) to CV at very low temperatures (lattice = Cvib, C= Cel+ Cvib). © Dr. Peter Blümler
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4.3.2
page:83
4. Quantum Statistics
Pauli-Paramagnetism Recap part from Dr. Mallett (page 43-45): For single, spatially fixed atoms (separation is large enough to neglect spin-spin interactions) with spin I = ½ , the magnetisation is given by the Langevin function: µ B M = N µ B tanh B k BT
Now we are going to consider conducting electrons in metals (application of Pauli exclusion principle and FD statistics) and want to consider all electrons and their spins simultaneously. The magnetic energy per electron is given by Em = −µ B B
[4.3.13]
(example: B = 2T ⇒ Em ≈ 10-4 eV << ε F ≈ 5eV) Hence the argument about the electron spins being in a degenerate state similar to T=0K at normal temperatures is even more true (than the translational energy of the electrons), because even at very high magnetic fields the electron spins will be close to their ground state. ⇒ Treatment at T = 0K is justified! We want to start with picturing the distribution of the spin-up / spin-down states for the electrons: a) T = 0 and B = 0
... ...
E εF
The spins are filling the lowest available states with up and down oriented moments (electron: I = ±½). The cut-off is the same as in the previous section (Fermi-energy: ε F) b) T = 0 and B ≠ 0 When an external magnetic field, B, is applied, the energy of each single particle state changes
© Dr. Peter Blümler
∆ε ↑ = -µBB
for an up movement (↑)
∆ε ↓ = +µBB
for an down movement (↓)
School of Physical Sciences
[4.3.14]
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
We have to shift some electrons from down to up due to supplied energy by the magnetic field and each of these shifts consumes ∆E = ∆ε ↑ - ∆ε ↓ = -2µBB. However, the Pauli-exclusion principle forces these up-shifted states to have energies higher than ε F , because all lower states are already occupied (as illustrated in the next figure).
... ...
E εF
These excess shifts are small (Em << ε F), hence we can still assume the density of the down-states to be about the same as the up-states ≈ 12 N (ε F ) ; and the number of shifted electrons, ns, to be approximately: ns = 12 N ( ε F )µ B B
[4.3.15]
But each shifted electron increases the magnetic moment by 2µB. Thus, the total magnetic moment along B for N conducting electrons is: M = ns ⋅ 2µ B
[ 4. 3. 15 ]
=
N (ε F ) µ 2B B
[4.3.16]
The total number of electrons with moments parallel (+) or anti-parallel (-) to B is then:
∫
N ± = 12 N ( ε) n ( ε m µ B B ) dε
[4.3.17]
where n is the average number of electrons in these (±)states. Of course, this average number is given by a Fermi-Dirac distribution (cf. eq. [4.3.2]), and we are allowed to replace µ with ε F because the system is similar to equilibrium at 0K. 1 for ε m µ B B < ε F n (ε m µ B B) = = exp [β(ε m µ B B − ε F )] + 1 0 for ε m µ B B > ε F 1
Because the states are either full or empty n (ε m µ B B) is either 0 or 1, and we only have to integrate eq. [4.3.17] from 0 to ε F ± µ B B (the full states): N± =
1 2
ε F ±µ B B
∫
N± =
© Dr. Peter Blümler
ε ± µB B
3 2 3 F 1 4πV N ( ε) dε = ( 2 me ) 2 ε 2 2 h3 3 0 0
4πV 3h
3
3
3
(2me ) 2 (ε F ± µ B B ) 2
School of Physical Sciences
[4.2.18]
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4. Quantum Statistics
Because we know that µ B B << ε F we can approximate N ± in eq. [4.2.18] by expansion: N± ≈
4πV 3h3
3
(2 me ) 2
3 3 µ B B [ 4. 3. 7 ] N 3 µB B = 1 ± ε F2 1 ± 2 εF 2 2 ε F
The net magnetic moment (macroscopic magnetisation of the sample) is given by: M =
µB (N + − N − ) = µ B N 1 + 3 µ B B − 1 − 3 µ B B V V 2 2 εF 2 ε F Pauli-magnetisation 3 N µ 2B B M = 2 V εF
[4.3.19]
When we compare this with the Langevin-function for isolated spins in the low temperature limit: µ B M isolated = N µ B tanh B and lim M isolated = N µ B T →0 k BT which means that all the isolated spins would line up.
4.3.3
The perfect photon gas - Black-Body Radiation
We want to look at this well known problem from two different perspectives. Part a) will show the historical derivation by M. Planck (1900) in a somewhat classical treatment; while part b) will deal with the back-body as a perfect Boson gas (photon spin =0) obeying Bose-Einstein statistics. a) historical approach (M. Planck) We observe photons in thermal equilibrium emitted by oscillation of the atoms in the cavity. We know that from a quantum-mechanical treatment a single harmonic oscillator has (cf. eq. [3.2.38])
(
ε s = hν s + 12
)
A hole in a heated cavity approximates an ideal black-body. Entering radiation will be absorbed, emitted radiation hence corresponds to the temperature of the cavity.
hν with s = 0, 1, 2,K and zero point energy : ε 0 = 2
Planck (just inventing Quantum-Theory) had assumed: ε n = hνn + const . , where n ∈ ¥
© Dr. Peter Blümler
School of Physical Sciences
[4.3.20]
University of Canterbury
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
∂ ln Z ε n = k BT 2 ∂T V
from eq. [3.2.24] we know:
εn
d (ln Z ) =
hence:
k BT 2
ln Z = −
which gives:
dT
∫
εn + C, kBT
with C as the integration constant
This corresponds to the partition function of a single mode of the oscillating black-body (cf. Debye theory), which has then: ε [ 4 .3 .20 ] hνn Z1 = C′ exp − n = exp − k BT k BT
const . hν n ⋅ C ′ exp − = exp − k BT k BT 144 24 43 ≡1 Planck set all the constants to 1 (referencing to zero-point energy, can be calculated later via normalisation). Hence, the partition function of all modes in the black-body is given by: ∞
hνn Z all = exp − k BT n = 0
∑
n ∞ hνn 2 1 = = exp − k BT 1 − exp − hν n =0 k BT
∑
where the last step was calculated using a geometric series (hν << k BT). The energy can then be derived by re-substituting into eq. [3.2.24] ∂ ∂ ln Z all h ν Eall modes = k BT 2 = kBT 2 − ln 1 − exp − k B T ∂ T ∂ T V − exp − hν hν k T B k BT 2 hν = k BT 2 − = 1 − exp − h ν exp + h ν −1 kB T k BT when introducing the density of modes/states we get Planck's famous distribution (but this is the same for both treatments). b) treatment by Boson-statistics: Since photons do not possess a spin, they are Bosons and described by eq. [4.2.25]. For the determination of the density of states we have to realise that light possesses two directions of polarisation, which - of course- have the same energy, hence they are degenerate and we have to correct eq. [3.2.17] by a factor of 2: N (ν )dν = 2 © Dr. Peter Blümler
4πV c3
ν 2 dν =
8πV 2 ν dν c3
School of Physical Sciences
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4. Quantum Statistics
The average occupation number in a Bose-Einstein distribution is given by eq. [4.2.25]: n (ε ) =
[ 4 .2 .32 ]
1
1 hν − µ − 1 exp k BT
=
ε − 1 α exp + kBT
[4.3.22]
But we are dealing with an open system! The number of photons is not a constant (they are emitted as a function of temperature, the higher T the more photons are emitted). Hence the chemical µ ∂F = 1 . potential is "undefined" µ = → 0 or α = exp − ∂N V , T k BT Hence the number of photons with frequencies between ν and ν+dν is (cf. to part a): 8πV
n (ν )dν = N (ν ) n ( ν ) dν =
c3
ν2 dν hν − 1 exp k BT
[4.3.23]
and the spectral energy density (distribution): E (ν )dν = n(ν ) ε dν =
8πV c3
ν2 8πhV ν3 hνdν = hν hν c3 − 1 exp exp k BT k BT
− 1
dν
The spectral energy density (distribution) per volume is called
spectral radiation density of a black-body: ρr ( ν )dν
E (ν ) V
=
=
ν3 hν c3 exp k BT
8πh
− 1
dν
[4.3.24]
This function is illustrated in the next figure for three different temperatures. (see next page) The temperature dependence of the maximum of the distribution (ν max) can be analysed by rewriting eq. [4.3.24] according to: 3
ρr ( ν )dν = hν and substituting x ≡ k BT
© Dr. Peter Blümler
, hence
8πh k BT c3 h
3
hν k BT dν hν − 1 exp k BT
ρr ( x) dx = const.
x3 ex − 1
School of Physical Sciences
dx
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
∂ ρ ( x) The maximum is found for ∂rx = 0 and results in the following condition: exp ( xmax ) (3 − xmax ) = 3 . This is a transcendental equation with a solution (root) at
x max =
hν max ≅ 2.8214 or kBT
ν max = 5.877 ⋅1010 Hz K − 1 T
also known as Wien's displacement law! (try it yourself!) Integration of eq. [4.3.24] gives Stefan-Boltzmann's law! (see weekly question) a)
b)
Planck's law: a) Radiated power versus wavelength! b) Energy density, ρ r(ν), versus frequency (eq. [4.3.24])
4.3.4
Bose-Einstein condensation
Since there is NO Pauli-exclusion principle to obey for Bosons, they will all settle in the ground state for T → 0 (see graph).
The average occupation number is given by eq. [4.2.25] in combination with eqs. [4.2.32] and [4.2.40]: ni
© Dr. Peter Blümler
=
1 ε −µ − 1 exp i k BT
with eq. [ 4.2.40] : µ < ε1
School of Physical Sciences
[4.3.25]
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4. Quantum Statistics
and the density of states is given (in terms of energy) by eq. [4.3.1a] (me = m) N (ε) dε
2 πV
=
h3
3
(2m ) 2
ε dε
[4.3.26]
We are now going to observe what will happen to the BE system in the limits T → 0K and T > 0K. a) for T = 0K: We know that all Bosons will try to minimise their energy and settle in the ground state, ε 1 Hence: E (0 K) = N ε1 and
lim ni → n1 → N
T →0
1 ε −µ T →0 − 1 exp i kBT
or using eq. [4.3.25]: lim ni = lim T →0
= N
ε −µ ε −µ = exp 1 ‰ 1 or µ < ε1 to realise this limit lim exp i T →0 k BT k BT [4.2.40] in order to give a big number like N.
as requested by eq.
Hence the exponent must be small and we can expand it (see section 4.2.6) for T → 0K: N ≅
1 1+
ε1 − µ k BT
+K − 1
≈
lim µ
or
T →0
k BT ε1 − µ ≈
k BT N {
ε1 −
[4.3.27]
very small
So µ→ε 1 as T → 0K and all ni with i >1 are emptied! (we have just proved our initial assumptions, but found some new expressions for them!) b) for T > 0K (but close): For this purpose we combine eqs. [4.3.25] and [4.3.26]: N=
∑ N (ε)dε n (ε) ≅ i
© Dr. Peter Blümler
2πV h
3
(2m)
3 2
∞
ε dε ε−µ 0 exp k T − 1 B
∫
School of Physical Sciences
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
N = V
or
∞
C
ε dε ε −µ 0 exp k T − 1 B
∫
with C =
2π h
3
3
(2 m ) 2
[4.3.28]
This equation relates the particle density to an integral which contains the chemical potential and temperature as parameters. N and V are supposed to be constant, so the particle density N/V is a constant too, thus for varying the temperature the integral must also remain constant. This means that µ must increase as T decreases, because µ < ε (≈ε 1) (eq. [4.2.40]). However, for T→0 (ε, µ→ε 1) the integration must give a measurable density, this implies that µ is negative; or µ must decrease for decreasing T. So we find a negative µ at T > 0K and finally reach µ = ε 1 ≥ 0 at T = 0K. In order to make things more convenient, we can take the ground-state as the zero of the energy scale (ε 1 = 0 and µ ≤ 0). The integral in eq. [4.3.28] then defines a minimum temperature such that for T = Tc the chemical potential vanishes (µ = 0). Now we want to calculate Tc from eq. [4.3.28] N = V we substitute x ≡
N = V
∞
C
ε k BTc C(
3 k BTc 2
)
∞
ε dε ε 0 exp k T − 1 B c
∫
2πmk BTc d x = x e − 1 h2 0
∫
x
3
∞
2 2 x dx π 0 ex − 1 144 42444 3 ≅ 2.612...
∫
[4.3.29a]
Bose-Einstein condensation temperature, Tc 2
2
0.527 h 2 N 3 h2 N 3 Tc = = 0.167 π 2mk B V 2mkB V
[4.3.29]
However, there is something wrong with this argument. There is no point in saying "there is a minimum temperature, Tc , below which the Bosons cannot be cooled"... we had calculated lim ni = N . T →0
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
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page:91
4. Quantum Statistics
So what is wrong? The integral in eq. [4.3.28] scales with ε , which gives the ground-state the wrong weight ( namely = 0). We see that this causes the integral to look as if there is a minimum temperature. We must correct for this wrong weight. N =
Correction:
∑ ni
= n1 + nε> ε1
[4.3.30]
i
particles in ground-state
other states
∞
nε > ε1 = VC
with
ε dε ε−µ 0 exp k T − 1 B
∫
This is no problem, since the ground-state gets the wrong weight. Hence, we do not count the groundstate twice nε > ε1 = V
2π h
3
(2m)
3 2
∞
ε dε ε −µ 0 exp k T − 1 B
∫
[4.3.31]
Now we have to determine µ again. We know that for T < Tc and in the limit T → 0K µ ≈ ε 1 k T From eq. [4.3.27] we know that for T ≈ 0K: µ ≈ ε1 − B N Hence, for another energy level, ε :
ε − µ ε − ε1 1 = + k BT k BT N
The average occupation number of such a level is then n (ε )
1
=
ε − ε1 1 exp + − 1 k T N B with the condition ε − ε1 >> k BT = ε1 − µ . This is then true for all states but the ground-state, and we get a BE-distribution with µ ≈ ε 1. Let's look at an example: 4He in a box of volume V = L3
λ=
ground state:
h h = 2 L ⇒ p1 = p1 2L
L
© Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
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PH 605: Thermal & Statistical Physics
4. Quantum Statistics
λ=
first excited state
h h = L ⇒ p2 = p2 L
L p22 − p12 h2 1 1 3 h2 Hence, ∆ε = ε 2 − ε1 = = − = 2m 2m L2 4 L2 4 2 mL2 for L = 1 cm: ∆ε ≈ 10-18 eV k BT = 10-4 eV T (T is small!) N ≈ 1022 for 1 cm3 k T ⇒ B ≈ 10 − 24 T = ε1 − µ N ⇒ ∆ε = ε − ε1 >> ε1 − µ hence we can treat µ ≈ 0 for T < Tc With the assumption µ ≈ 0 for T < Tc we can then solve eq. [4.3.31]: 2π
nε > ε1 ≅ V 3 (2 m) h [ 4 .3 .29 ]
3 2
∞
ε ε 0 exp k T B
∫
− 1
dε
[ 4. 3. 29 a ]
=
2πmkBT V h2
3
2 2.612...
3
T 2 nε > ε1 = N Tc these are the particles which are not in the ground-state for 0 < T < Tc. 3
With eq. [4.3.30]:
T 2 N = n1 + nε > ε1 = n1 + N Tc
Fraction of particles in the Bose-Einstein condensate (BEC): 3
T 2 n1 = 1 − N Tc
[4.3.32]
This equation means, that below a certain temperature, Tc (given by eq. [4.3.29]), the Bosons start to condense in the ground-state, with dramatically increasing number as T → 0K (cf. following figure). When all or almost all particles are in the same state (with almost no but zero-point energy, for translational energy this is zero!), they must share the same wavefunction. Then N particles behave like a single particle (no inter-particle interactions!). If such a system can be realised it will have completely new properties, it is in fact a new material = Bose-Einstein condensate (BEC). © Dr. Peter Blümler
School of Physical Sciences
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PH 605: Thermal & Statistical Physics
page:93
4. Quantum Statistics
Figure: Fraction n1 / N of particles in the zero-energy ground-state, as a function of temperature. Tc is the Bose-Einstein condensation temperature.
Below Tc: The BE-ideal gas separates into a 'normal' gas (or fluid) and a 'BE-condensed' gas (or fluid). The BEC has no contribution to energy or pressure (we expect 'odd' behaviour). The mean energy in the regime 0 < T < Tc is then given by ∞
E = ε1 n1 +
[ 4 .3 .32 ]
E
=
ε N ( ε) dε with µ ≈ 0 ε −µ 0 exp k T − 1 B
∫
3 3 2 2 T T ε1 N 1 − + 0.77... NkBT Tc Tc 14442 444 3 1442443 5 negligible for T > 0 ∝T2
[4.3.33]
An approximate solution is (cf. page 81 eq. [4.3.10] ff.): 5
E ≈ nε > ε1 k BT ≈ N k B
T2 3 T2
for 0 < T < Tc
c
3
and
5 T 2 ~ CV ≈ R 2 Tc
for 0 < T < Tc
The exact solution (see weekly problem) is: 3
T 2 ~ CV ≅ 1.93... R Tc © Dr. Peter Blümler
for 0 < T < Tc
School of Physical Sciences
[4.3.34]
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4. Quantum Statistics
PH 605: Thermal & Statistical Physics
A numerical analysis of 0 for 0 < T < Tc ∞ 3 ε Nh3 µ = solution of dε = (2 m )− 2 ε−µ 2πV 0 exp − 1 k T B is shown in the following figure.
∫
Tc
for T > Tc
T Tc
µ
~ Once µ is determined E and CV can be calculated via eq. [4.3.33] as shown in the next figure. ~ CV
T Tc ~ The piecewise behaviour of µ results in a kink (→ phase transition) of the molar specific heat CV at ~ Tc. More detailed analysis reveals that CV (Tc ) is continuous, but not differentiable. ~ For much higher temperatures CV (T ) must approach the classical value of 32 R (dashed line in figure). The classical limit is typically reached at about 10 Tc. In the region Tc < T < Tclassical we must evaluate numerically.
© Dr. Peter Blümler
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4.3.5
page:95
4. Quantum Statistics
Superconductivity and superfluidity, BEC
In the previous section we have derived the theory for Bose-Einstein condensation. Unfortunately this derivation is only valid for an ideal gas. Under normal circumstances, no gas can be cooled to temperatures necessary to reach Tc. Most materials will be solid and hence are not subject to BEC, because they are already condensed and have no translational degrees of freedom. An exception is 4He, which is a Boson, and -under normal pressure- doesn't solidify even at very low temperatures. However, it is not a gas but condenses to a liquid below 4.2K (density = 0.178 g/cm3). But the interactions in a liquid are still weak and translational motion is present. When we treat liquid 4He as an ideal gas, we can use eq. [4.3.29] to calculate Tc = 3.2 K. Experiments indeed show a phase transition at Tλ = 2.2 K, where liquid 4HeI changes to 4HeII (see figure). CV
Tλ Experimental specific heat of liquid 4He (in coexistence with its vapour) versus T [K]
Because of it's shape (like the Greek letter λ) the point of phase transition is called "lambda-point", Tλ. This figure resembles the features shown in the theoretical curve in the previous chapter. This is reason why it is believed that the "lambda transition" happens because of Bose-Einstein condensation. (The deviation between Tc and Tλ can be explained because 4He is a liquid and not a gas). 4HeII also shows the expected "odd" behaviour. It has no heat capacity (the decrease in the above figure is due to the co-existence of 4HeI and 4HeII) and no viscosity (superfluid). 3
He (Fermion) doesn't show this behaviour until T < 2 mK. (see later)
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Superfluidity: HELIUM-4 4
He shows fascinating behaviour at low temperatures:
• Although it doesn't solidify (considerable pressure, 25 bar, has to be applied to get solid 4He), thermodynamic data show that below Tλ it is in an extremely ordered state (∆S = 0 for 4
II He liquid to 4 He solid ) • Below Tλ the heat conductivity of the liquid becomes so large that hot spots necessary for the formation of bubbles of boiling cannot occur. Under proper conditions the thermal conductivity of the superfluid can be as large as 2000 times that of copper at room temperature. A drop of temperature of only 1K in going from 4HeI to 4HeII causes an increase of the thermal conductivity by a factor of several million.
• Heat can flow in superfluids in form of a wave known as second sound. Under normal condition heat diffuses (via random molecule motion) or is transported via convection. However in superfluid He, a pulsed heater causes a temperature pulse to travel across the container. A heater that is cycled sinusoidally produces a sinusoidal temperature wave which travels through the liquid with a speed similar to that of sound. • Superfluid He shows negligible viscosity. In a container the vapour above the liquid coats the walls with a layer only a few atomic layers thick (this is not a speciality of superfluids). However, since the viscosity is so small, this layer is able to flow upwards towards the rim of the container and then over the edge. This process will empty an open container. This extraordinary flow properties give the substance the name and also allows it to flow through very narrow pores, which are too narrow for normal fluid He. Such channels are called superleaks. • The low viscosity also causes a superfluid flowing in a circle to continue this motion in principle to eternity (as long as the velocity stays below a critical velocity). Turbulence is easily achieved in superfluids due to low viscosity. 4
Fountain of superfluid He, photographed by Jack Allen in the 1970s at the University of St. Andrews in Scotland, who was the first to observe this spectacular phenomenon.
© Dr. Peter Blümler
• The thermomechanical or fountain effect in superfluids is a special arrangement of two containers of helium connected by a superleak. If one side is heated slightly the liquid level of this side rises at the expense of the liquid in the other container. It can be arranged in such a way, that the superfluid spurts out of container in form of a fountain (see figure on the left).
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HELIUM-3: 3He Like 4He, Helium-3 stays liquid (under normal pressures) down to absolute zero. 3He shows even more bizarre behaviour than 4He (many of it is still not completely understood). The following graph shows its phase diagram at low temperatures.
Phase diagram for 3He in the absence of an external magnetic field. The phases A and B are superfluid (hashed areas).
Although a Fermion (TF ≈ 6K), 3He can become superfluid (at ca. 2mK). (D. D. Osheroff, R. C. Richardson, and D. M. Lee discovered this in 1972, Nobel prize 1996) How can that be? When the conducting electrons from Cooper-pairs (see below) -in order to become a Boson- their spins are oriented oppositely (hence no net spin nor magnetic moment). 3He is different. The atoms in the superfluid phases (A and B, see figure) also form pairs, which are held together by weak interatomic forces, and their nuclear spins are aligned parallel or anti-parallel, hence the (Boson)-pair has a net spin of h and a net magnetic moment or zero (phases A with I = 1 and B with I =0 in figure). These superfluid A-phase responds surprisingly strong to external magnetic fields (creating yet another phase), and the properties of the superfluid are then becoming anisotropic relative to the orientation of the magnetic field (e.g. speed of sound is different parallel or normal to the M-field). 3
He is used for creation of low temperatures in two different types of refrigerators: a) Dilution refrigerator and b) Pomeranchuk refrigerator. a) Dilution refrigerator: Below 0.8 K the two helium isotopes (3He and 4He) separate spontaneously from a mixture. The separation is not complete and the lower phase (heavier 4He) still has ca. 6% of 3He in it, while the 3 He-phase floating on top is practically pure (see figure). The temperature is too high for 3He to be a superfluid (hence having a nonzero heat capacity), but 4He is and hence has a heat capacity close to zero.
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We can explain the function of this refrigerator by looking at it as a boiling ideal gas of 3He (right part of figure). The upper phase is pure 3He, while the lower phase has only few 3He atoms in an inert atmosphere of superfluid 4He, which directly cools the gas due to its enormous heat conduction and almost non-existing heat capacity. So the upper layer boils off 3He, when the equilibrium is disturbed by removing stuff via a pump. The heat of evaporation is removed from the upper phase, which is cooled by this process. The main advantage of a dilution refrigerator is that it can maintain very low temperatures for a long time.
3 He
4He
+ 6% 3He
3 He to pump
to pump
b) Pomeranchuk refrigerator
liq u id
3
He
There is another abnormality in the behaviour of 3He, which was predicted by I. Y. Pomeranchuk (1950). It results in the fact, that solid 3He has a higher entropy than the liquid. Hence the solid is more disordered than the liquid at low temperatures (T < 0.32 K). In the solid the atoms and their nuclei are located at specific lattice points and can be understood as acting approximately independent from each other. They form solid 3He distinguishable lattice points and are not really subject S to the Pauli exclusion principle, they are said to be quasi-independent. While in the liquid, the indistinguishable atoms are forced by the exclusion principle to correlate their spins and the spin contribution to the entropy is decreased. A schematic entropy-temperature diagram is shown in the figure on the right. The liquid is in equilibrium with its vapour 2 1 and the solid is under pressure. If we take liquid 3He (point 1) and compress it adiabatically we can reach T point 2, where it solidifies with the shown decrease in temperature. Or from a different perspective, when solid 3He melts the volume per atom increases. Hence, when the liquid is compressed, V decreases and the solid forms, but entropy increases which requires heat which is removed from the system, which is therefore cooled. This bizarre behaviour can be summarised in the paradox statement: "To solidify 3He, heat it!"
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Superconductivity: Because of their lower mass, conducting electrons show similar behaviour to BEC at much higher temperatures (T > 7K, see table). Below such critical temperatures the electrons become superconducting (zero electrical resistance, see graph), which we can now understand as another term for "superfluid" (the resistance of a fluid to a external force potential is viscosity, while that of conducting electrons is called electrical resistance). Element Ag Cu Au Mo Ga Al Sn Ta Pb Nb
Tc [K] 0.9 1.1 1.2 3.7 4.4 7.2 9.2
Plot by K. Onnes, who discovered superconductivity, of the resistance of mercury versus temperature, showing a sudden decrease (jump-temperature) at T = 4.2K.
Prior to the discovery of very-high-temperature superconducting oxides (high-Tc superconductors) the highest Tc observed was that of a Nb-alloy with Tc = 23K. The conversion of Fermions to "Boson-behaviour" is explained by forming pairs, which are hold together by very weak interactions (phonons). Due to this pairing (Cooper-pairs, see figure) their 'molecular' or pair spin becomes integer (→Bosons) thus they can avoid Pauli's exclusion principle and reach zero-point energy. (This so-called BCS-theory was accomplished by J. Bardeen, L. N. Cooper, and J. R. Schrieffer in 1956). This theory is beyond the scope of Simultaneously acquired topographic (blue on top) and this lecture. However, the reason why spectroscopic (grey on bottom) images of three gadolinium atoms electrons form pairs at low on top of a superconducting niobium surface. In the region near the gadolinium atoms the magnetic properties of these individual temperatures can be illustrated by two defects break up Cooper electron pairs (dark rings), therefore marbles in a container with a soft modifying the superconductivity of the niobium bottom (e.g. a drum). If the container is shaken violently (high T) the marbles move independent of each other. If it is moved gently (low T) the marbles move in pairs, each in the small trough created in the soft cover.
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The properties of superconductors are well known. A current once started will persist even after the applied potential difference has been removed (e.g. a shorted coil). This allows to "store" currents over a long time without a significant decrease and allows to build super-stable magnets, which do not have to be cooled because the resistance of the coil generates heat, but to achieve a superconducting state (e.g. MRI magnets, magentic propelled trains, etc.). Another field of application uses the Josephson effect for detecting very small magnetic fields with SQUIDs (Superconducting QUantum Interference Device), which measures the phase difference of the highly coherent (electrons in the BEC) tunneling through a junction. Finally superconductors are perfect diamagnets, which means that an external magnetic flux is completely repelled from them, which makes them to hover over a magnet (Meissner effect, see figures below).
Meissner-effect: Left: A superconducting pendulum is repelled from a horse-shoe magnet. Right: A small permanent magnet levitates over a disk of a high-Tc superconductor (Yttrium-Barium Copper Oxide)
A real Bose-Einstein condensate! Bose's and Einstein's prediction of the existence and properties of a BEC date back to 1924. However, until 1995 there was no real example (analogies of superfluids and superconducters exist, but their either not gas or real Bosons). In 1995 E. Cornell and C. Wieman were the first to demonstrate BEC in a dilute gas of 87Rb atoms (87Rb has even number of neutrons, I = 0, 1, 2 ; hence is a Boson). How did they achieve this new state of matter? 2
a) from eq. [4.3.29]:
0.527 h 2 N 3 T < Tc = π 2 mkB V
b) from eq. [3.4.7]:
V 3 λ ≈ 1.4 N
⇒ decrease T
1
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⇒ increase density (N/V)
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The successful approach by Cornell and Wieman uses both ways (a and b):
2) The vapour is cooled by LASERs (from room temperature to ca. 20 mK). This LASER-cooling is achieved by setting the LASER frequency slightly too low for resonant absorption by the Rb-atoms at rest. Hence only the moving (hot) Rb atoms are affected, and from them only those who are Doppler blue-shifted (those who move towards the LASER, see figure). They are the only one who can absorb a photon by a head-on collision, which slows them down (cools them), because they then emit light with no preferred direction, so on average there is no recoil.
LASER
1) The Rb-atoms are polarised by circularly polarised light to a spin 2h . This trick prevents the vapour to reach thermal equilibrium at low temperatures (cf. adiabatic demagnetisation) to form a liquid or solid. In this way they can condense a super-saturated vapour.
3) To confine the atoms magnetic fields are applied (see figure below). After the LASER-cooling of step 2, the gradient of the magnetic trap is increased and the vapour is compressed by a factor of ca. 10. The heat generated by this process is removed by further LASER-cooling until a temperature of ca. 20 µK is reached.
In a conventional trap the field falls to zero at the centre (left), allowing the atoms to leak out if they are cool enough. The TOP trap modulates the position of the trap by additional rotation (centre), which results in an average potential which is closed (right).
4) Eventually the LASER-cooling is stopped and the characteristics of the magnetic trap changed (TOP trap, see figure above). Now evaporative cooling is used to achieve even lower temperatures. The more energy the atoms possess the further they travel from the centre of the trap with the lowest potential. The maximum potential is now lowered to let the hot (energetic) atoms escape, which cools the rest. Since the 87Rb atoms were spin polarised in step 1, radio-frequency (NMR-experiment) can be used to selectively flip the spin of the trap, hence raise their energy over a critical level and © Dr. Peter Blümler
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remove them from the trap. The more energetic the atoms are higher is the local field they experience in the trap. Due to the Larmor condition ωrf ∝ B the more energetic atoms can be selectively removed without touching the cool stuff. By ramping down the rf-frequency ultracold temperatures can be achieved, while the material stay in a gaseous state. The first BEC of ca. 20000 atoms appeared at T = 170 nK. Further evaporative cooling caused almost all atoms to settle in the ground-state, forming a new state of completely coherent matter (ca. 2000 atoms in a volume of ca. (10 µm)3 all in the same quantum state). The BEC can be observed by another trick: The LASERs are off during evaporative cooling (step 4). Once the state is reached, where the system should be detected, the magnetic traps are removed. Then the atoms are given some time (milli-seconds) to fly apart to allow the measurement of their individual velocities. This cloud is then illuminated by a LASER at resonance frequency (maximum scattering from the atoms) and its intensity measured by a CCD-chip. This procedure gave the following images. [M. H. Anderson, J. R. Ensher, M. R. Matthews, C. E. Wiema, and E. A. Cornell, Science 269 (1995) 198.]
Observation of BEC: The images represent velocity distributions of a super-cool cloud of 87Rb atoms. a) Before the condensation the distribution is isotropic, as expected from a gas in thermal equilibrium. b) The condensate appears in the centre by atoms having zero velocity. c) Further cooling gives almost pure BEC. The distribution is slightly elliptical because an elliptical trap was used. Each image is 200×500 µm and is derived from LASER illumination after a period of 60 ms of free expansion. © Dr. Peter Blümler
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Possible applications....who knows?! BECs could be used for atomic lasers, where the atoms are pumped into a single particle state with one de Broglie wavelength (see figure below). Hence they emit coherent matter rather than just coherent light. This might have applications for high precision lithography. They could also be used for a new type of interferometer, because their de Broglie wavelength can be made much shorter than that of light? Any other ideas?
Atomic LASER: Density distribution of two superimposed waves of coherent BEC-matter of ultra-cold rubidium gas. a) no interference for normal Rb-gas, b) interference appears when some Rb-atoms enter BEC-state, c) for almost all Rb-atoms forming a BEC. The scale on the right is distance in mm! The strong contrast of the interference pattern is another proof that the Rb-atoms form a BEC, because the atoms share one state with one de Broglie wavelength. [I. Bloch, T. W. Hänsch and T. Esslinger, Spektrum der Wissenschaft, 7 (2000) 23.]
4.3.6
Thermodynamics of Stars
Some basics first! How does a star form? From a gas cloud from clusters which are pulled together by gravitation. Hence gravitation is the attractive force.
We want to estimate the potential energy due to gravitation. Acceleration of a mass element located at a distance r from the centre:
The matter on a spherical shell of radius r has a mass m(r) (see graph): r
m( r ) =
∫ ρ(r′) 4πr′ dr′ 2
[4.3.35]
0
where ρ is the density. The gravitational acceleration is then:
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g (r) =
Gm( r ) r2
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[4.3.36]
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The particles get closer. There has to be some counteracting pressure to evolve (we do not specify at this moment by what means) to prevent the system from collapse.
To reach a stable state, the forces have to be in equilibrium: PdA − g (r )dm − ( P + dP)dA = 0 PdA − g ( r ) ρ( r ) dAdr − PdA − dPdA = 0 − g (r ) ρ(r ) dAdr − dPdA = 0 g ( r ) ρ(r ) =
−
dP [ 4 .3 .36 ] G m(r ) = ρ( r ) dr r2
[4.3.37]
The system is in equilibrium when eq. [4.3.37] holds for all r, hence we have to integrate over the volume to establish this condition (assuming a sphere of radius R and mass M): R
∫
dP dr = − dr
R
3
dP dr = − dr
∫ 4πr
0
R
∫ 4πr 0
= −
∫
0 R 0 R
G m( r ) r2 3
ρ(r )dr
G m( r ) r2
ρ(r )dr
∫ 4πr G m(r) ρ(r) dr 0
substitute from eq. [4.3.35]: dm = ρ( r ) 4 πr 2 d r (mass between r and r+dr) R
∫
0
dP 4πr dr = − dr 3
m=M
R
∫ 4 πr G m(r ) ρ(r) dr 0
=
∫
m =0
Gm( r ) dm ≡ EG r
[4.3.38]
where EG denotes the gravitational potential energy.
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The left side of eq. [4.3.38] can be obtained via integration by parts: R
∫
4πr 3
0
[
]
R R r =R dP dr = 4πr 3 P( r ) dr = P( r ) 4πr 3 r = 0 − 3 P( r ) 41 π2 r 23 dr dr dV 0 0
[
∫
]
r=R
where P( r ) 4πr 3 r = 0 =
⋅ 4πR 3 − 0
P ( R) {
∫
≈0
surface pressure ≈ 0
combination with eq. [4.3.38] gives: R
∫
− 3 P( r ) 41 π2 r 23 dr = − 3 P V = EG where P is the volume averaged pressure dV 0 volume averaged pressure:
P
=
−
1 EG 3 V
[4.3.39]
(cf. Virial theorem. However the particles might become so energetic and fast that they become relativistic!) The main physical properties of the sun: Property
Value
Mass
M• = 1.99 ⋅ 1030 kg
Radius
R• = 6.96 ⋅ 108 m
Photon luminosity
L• = 3.86 ⋅ 1026 W
Surface temperature
TS = 5780 K
Central temperature
TC = 1.56 ⋅ 107 K
Central pressure
PC = 2.29 ⋅ 1016 Pa
Central density
ρC = 1.48 ⋅ 105 kg m-3
Age
t • ≈ 4.55 ⋅ 109 years
The question which interests us in the context of this lecture is: What can produce the necessary pressure to avoid gravitational collapse? 1) Photons: Black-body radiation pressure: For the sun (see table above): First we can calculate the average pressure (due to gravitation) inside the sun: 2 2 GM e 1 EG 1 GM e 3 P = ≈ = ≈ 1014 Pa 3 4 3 V 3 R e 4 πRe 4πRe
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[4.3.40]
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If we treat the sun as an ideal gas, we can estimate the average temperature (or get it from the table): PV = N k BT ρ P = kB T m M e 3M e = is the average density and m the average mass per sun particle (e.g. 3 Ve 4πRe ionised H⊕, which is the mass of a proton, mp). where ρ =
T=
2 3 G Me 1 m [ 4. 3. 40 ] 1 GM e 4πRe G M e P = m = m = mp ≈ 7 ⋅10 6 K 4 kB ρ k B 4πRe 3M e 3k B Re 3k B Re
In a weekly set problem we had calculated (from Planck's law, see eq. [4.3.24]) the radiation pressure of a black-body to be: b Pr = T 4 3
with
b=
8π5 k B4
[4.3.42]
15c 3 h 3
For the sun: (b = 7.5 ⋅ 106 Pa K-4) Surface: Our estimate: Centre:
TS = 5780 K T = 7 ⋅ 106 K TC = 1.56 ⋅ 107 K
Pr = 0.28 Pa Pr = 6 ⋅ 1011 Pa Pr =1.5 ⋅ 1013 Pa
This is not sufficient: Our estimate for the average pressure was 1014 Pa (table gives 1016 Pa in the centre). Hence radiation contributes only ca. 0.1 % to the necessary pressure that stabilises the sun. 4
4 b bG M M However, because of Pr = T 4 = m ∝ 3 3 3k B R R the radiation pressure plays a significant role for small and massive stars (e.g. white dwarfs)!
The gas pressure is Pg = Hence:
N M M M2 kB T ∝ = V R3 R R 4
Pr ∝ M 2 which illustrates that radiation pressure dominates for dense stars (≥ 50 M•) Pg
2) Pressure of the degenerate Fermi-electron gas: Because stars consist of plasma, the electrons can move around free. Although we might have to alter the description from section 4.3.1 due to collisions. The treatment of the electron gas close to absolute zero temperature is still valid, because of TF ∝ ρ 2 / 3 (cf. eq. [4.3.7]) and stars are much denser (see table next page) than metals which already gave TF of ca. 10000 K.
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So what pressure is generated by the electron gas? N 1 N P = px vx = pv V 3 V
From kinetic gas theory (see first week problem):
or
1 p2 N 2 N [4.3.8] P= = ε = 3 me V 3 V
2
2 N [ 4 .3 .7 ] h 2 ε = 5 FV 5me
5
3 3 N 3 8π V
[4.3.43]
This is the pressure of the degenerate electron-gas at 0K. We see that, P =C
2
h2 3 3 with C = 8 8π 3 5m
5 ρ3
e
5 3
M P ∝ 3 R
or
If we compare this -in form of proportionalities- with the gravitational pressure from eq. [4.3.39] or eq. [4.3.40]: 5
M 3 C ′ = R3
5
M3
C′ 5 1 42R4 3
GM 2
=
pressure of the electron gas
4 4π2 R3 1
gravitatio nal pressure
We realise that the left side grows in the fifth power as the radius decreases, while the left side onyl grows in the fourth power. Hence, the pressure of the electron gas is able to counter-act gravitational compression. − 13 4 πC ′ We also find: R = ∝ M 1 GM 3 Hence, the larger the mass the smaller the star despite the electron pressure. A more detailed analysis gives R ∝ M −0 .57 for M ˆ M•. The following table comprises some relevant (non-relativistic) data for this discussion for white dwarfs: M/ M•
R/R•
d = 3 V / N [m]
ε F [eV]
TF [K]
pF/(mec)
v F/c
40 Eri B
0.447
0.013
2.3 ⋅ 10-12
6.9 ⋅ 104
8.9 ⋅ 108
0.52
0.46
Sirius B
1.05
0.0073
9.6 ⋅ 10-13
4.0 ⋅ 105
4.6 ⋅ 109
1.2
0.78
Name
The last column shows the classical assumptions about electron velocity and hence mass is not correct, because the electrons in stars reach speeds of 50-80% of the speed of light. Hence we have to treat them relativistically.
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1 N pv is relativistically correct. However, we must avoid introducing me, 3 V which is the mass of the electron at rest (p ≠ mev in the relativistic domain). In this new regime the electrons have a speed close to c (hence to simplify things -sufficient in this context- we say their speed is that of light). If we assume the electrons still as completely degenerate, we can simply equate (avoiding relativistic corrections). The equation P =
P≅
pressure of relativistic electrons:
1 N pF c 3 V
[4.3.44]
The momentum of an electron in a Fermi-gas at 0K is: p F = 2
pF =
h2 3 N 3 2me 2me 8π V 1
and eq. [4.3.44] gives:
2 mε F with ε F given by eq. [4.3.7]. 1
3 N 3 = h 8π V
[4.3.45]
4
c 3 3 N 3 P ≅ h 3 8π V
Hence in the extreme relativistic regime (electrons have speed of light), the pressure of the degenerate Fermi-gas gives a mass/radius ratio of P ∝
4 ρ3
4
4
M 3 M3 = 3 = 4 R R
Comparison with the gravitational pressure from eq. [4.3.39] or eq. [4.3.40] then gives: 4
C ′′
M3 R4
=
GM 2
[4.3.46]
4 πR 4
Once the extreme relativistic regime (maximum speed of electrons) is reached further contraction doesn't increase pressure faster than gravitation. (The balance doesn't depend on R anymore!) However, further information can be gathered from eq. [4.3.46]: 1
GM 2
4
hc 3 3 M 3 = 4 4 8π 4πR 4 3m 3 R e
M
2 4
M3
=M
2 3
1 3
π2 hc 3 4π = = 2 9 3 8π G 1
3
1
3 hc G 3
8π 2 2 hc 2 = 2 2π hc 2 M Ch = 9 G 3 G
[4.3.47]
If the mass of a star doesn't satisfy this equation, it cannot adjust its radius to provide an equality between electron and gravitational pressure. The route to stellar equilibrium is blocked. © Dr. Peter Blümler
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In 1930 S. Chandrasekhar (Noble prize 1983) calculated eq. [4.3.47]. The critical (limiting) mass MCh is also called Chandrasekhar-limit. Further analysis shows that stars with masses up to M ≈ M• may evolve to white dwarfs. More massive stars cannot be supported by the pressure of the degenerate electron gas and must search for other routes to establish equilibrium or collapse. Such other routes can include for instance inverted β-decay (i.e. e- + p → n + ν e). This process removes electrons, which drops the pressure even further causing further contraction and more inverted β-decay. This chain stops when the pressure of a degenerate neutron gas holds up against gravity (neutron stars).
Schematic Hertzsprung-Russell diagram. Snapshot of the luminosity (L) and surface temperature of stars at different stages in their evolution. Most of the observed stars are grouped along the main-sequence; these are hydrogen burning stars like the sun (yellow circle). As stars evolve the contraction of the central core is accompanied by an expansion of the outer layers of the star to form luminous stars with low surface temperature (e.g. red giants). The end-point of stellar evolution of a star with mass comparable to the sun is a compact object supported by degenerate electrons, a white dwarf. The evolution of more massive stars (M > MCh) can lead to the formation of neutron stars or blck holes. © Dr. Peter Blümler
School of Physical Sciences
University of Canterbury
