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Statistical Physics.
Solutions Sheet 5.
HS 2014
Prof. Manfred Sigrist
Exercise 1. Ideal fermionic quantum gas in a harmonic trap
In this exercise we study the fermionic spinless ideal gas confined in a three-dimensional harmonic
potential and compare it with the classical case (see Exercise Sheet 3). The eigenenergies of the
gas are given by
εa = ℏω(ax + ay + az ) , (1)
where a = (ax, ay, az ), with ai ∈ { 0 , 1 , 2 , ...}, labels the states and the zero point energy ε 0 =
3 ℏω/2 was omitted. The occupation number corresponding to state a is given by na.
(a) Consider the high-temperature, low-density limit (z 1). Derive the grand canonical
partition function Zf of this system and compute the grand potential Ωf. Show that
Ωf ∝ f 4 (z) , (2)
where the function fs(z) is defined as
fs(z) = −
∑^ ∞
l=
(−1)l^
zl
ls^
Solution. We begin with the general definition of the grand canonical partition function within the occupation number formalism (chapter 3 of the lecture notes) and find Zf =
a
na
ze−βεa
)na
a
(1 + ze−βεa^ ). (S.1)
In order to compute the grand potential Ω = − 1 /β log Z, we use the series expansion
log(1 + x) = −
∑^ ∞
l=
(−x)l l for − 1 < x ≤ 1. (S.2)
This expansion is applicable to the logarithm of the partition function in (S.1) if ze−βεa^ ≤ 1 (it is always positive) in the high-temperature, low density limit z 1. We find
log Zf =
a
log(1 + ze−βεa^ ) = −
a
∑^ ∞
l=
(−1)l^ zl l e−lβεa^ = −
∑^ ∞
l=
(−1)l^ zl l
a=
e−lβℏωa
∑^ ∞
l=
(−1)l^ zl l
1 − e−lβℏω
l=1(−1) l zl l 1 (lβℏω)^3 if^ β^ →^0
−
l=1(−1) l zl l if^ β^ → ∞
=
1 (βℏω)^3 f^4 (z)^ if^ β^ →^0 f 1 (z) if β → ∞
, (S.3)
and obtained both the high and the low temperature limits (in either case z 1 must be given). Alternatively, for the high temperature limit, with the help of the Euler-Maclaurin formula (see, e.g., Exercise Sheet 4), we can approximate the sum over the oscillator modes by an integral and find in leading order
log Zf = −
∑^ ∞
l=
(−1)l^ zl l
a=
e−lβℏωa
(S.4)
∑^ ∞
l=
(−1)l^ zl l
0
da e−lβℏωa
∑^ ∞
l=
(−1)l^ zl l
0
da e−lβℏωa
(βℏω)^3
∑^ ∞
l=
[
(−1)l^ zl l^4
]
(βℏω)^3 f 4 (z) (S.5)
Either way, the high temperature expansion results in the grand potential
Ωf = − 1 β
(βℏω)^3
f 4 (z). (S.6)
(b) Calculate the particle number 〈N 〉 and the internal energy U as a function of N. In order
to get U in terms of N (instead of dealing with the chemical potential), introduce the
parameter
ℏω〈N 〉^1 /^3
kB T
and relate it to z using the high-temperature, low-density expansion of 〈N 〉 (up to O(z^2 )).
Interpret the condition ρ 1.
Finally, expand U up to second order in ρ, relating it to N.
Solution. First, we compute the internal energy of the system,
Uf = ∂(β Ωf ) ∂β
z
, (S.7)
where the derivative has to be taken at constant fugacity z = eβμ. Starting from (S.6) we find
Uf =^3 β
(βℏω)^3
f 4 (z), (S.8)
which shows that the internal energy is proportional to the grand potential, Uf = −3 Ωf. The average particle number can be computed in a similar way,
〈Nf 〉 = z
∂z log Zf. (S.9)
We have
〈Nf 〉 = z ∂ ∂z
(βℏω)^3 f 4 (z) = 1 (βℏω)^3 f 3 (z), (S.10)
where we used
z ∂ ∂z
f 4 (z) = f 3 (z). (S.11)
In order to relate the internal energy to the particle number, we start with the high-temperature, low- density expansion of the total particle number,
〈Nf 〉 =
(βℏω)^3 f 3 (z) ≈
(βℏω)^3
z − z^2 8
. (S.12)
Rewriting this equation using the parameter ρ leads to
ρ = z − z^2 8
. (S.13)
The condition z 1 therefore implies ρ 1. Solving this equation for z, we obtain z = 4 ± 2
4 − 2 ρ. Choosing the relevant solution and expanding
(1 + x) ≈ 1 + x 2 − x 2 8 we find
z = ρ + ρ
2 8
. (S.14)
Expanding in ρ allows us to deal with the particle number instead of the chemical potential. To interpret the condition ρ 1 we first note that for this system, the Fermi energy follows F = 3ℏωamax, while the number of occupied states is proportional to a^3 max. The characteristic energy scale is thus given by 3ℏωN 1 /^3. Therefore, this condition requires that the characteristic energy scale is much smaller than
kB T
U
@ k
T B
DêH
3N
L
kB T
C
N^
@ k
T B
DêH
3N
L
kB T
Κ T
@ k
T B
DêH
N
êV
L
Figure 1: Thermodynamics of the fermionic gas (dahsed, blue) compared to the classical gas (solid, black). Note that these quantities are computed within the high-temperature, low-density approximation and are therefore not exact results. Still, they can be used to observe trends. We set N (ℏω)^3 = 100.
- In first order in ρ the results for the classical (Boltzmann) gas in a harmonic trap are recovered (cf. Exercise Sheet 3).
- Due to quantum corrections, the internal energy U for fermions is higher than the classical ideal gas. This can be understood by taking quantum statistics into account. Fermions are not allowed to occupy the same state multiply (Pauli). Lowering the temperature, the system tends to occupy low energy states with growing probability. While the classical system does not care about multiple occupancies, in the fermionic system the double occupancy is forbidden and occupation of low-energy states is thus reduced, increasing the inner energy Uf compared to the classical gas.
Exercise 2. Sommerfeld expansion and density of states
Consider a thermally equilibrated system of non-interacting fermions with single particle states
labeled by the quantum numbers ν and corresponding energies εν.
(a) Work in the grand canonical ensemble and write the particle and energy densities in the
form
n =
V
ν
f (εν ) =
dε g(ε)f (ε) , (5)
u =
V
ν
εν f (εν ) =
dε εg(ε)f (ε) , (6)
where f (ε) is the Fermi-Dirac distribution function. What is g(ε)?
Solution. In thermal equilibrium, the occupation probability of a particular state with energy ε is given by the Fermi-Dirac distribution
f (ε) =
e(ε−μ)/kB^ T^ + 1
, (S.22)
and we find from the definition of n and u
n =
V
ν
f (εν ) =
dεf (ε)
V
ν
δ(ε − εν ) =
dε g(ε)f (ε) , (S.23)
u =
V
ν
εν f (εν ) =
dε εf (ε)
V
ν
δ(ε − εν ) =
dε εg(ε)f (ε) , (S.24)
with
g(ε) =
V
ν
δ(ε − εν ) =
V
ω(ε) , (S.25)
and ω(ε) as in the lecture notes. That is, g(ε) is the density of energy levels divided by the volume.
(b) The above expressions for n and u are of the form
−∞
dε H(ε)f (ε). (7)
For temperatures T ε kFB (which is typically the case for metals), H(ε) is slowly varying
in the region where dfdε 6 = 0 significantly and the Sommerfeld expansion^1
−∞
dε H(ε)f (ε) =
−∞
dε H(ε)+
π^2
(kB T )^2 H′(μ)+
7 π^4
(kB T )^4 H′′′(μ)+O
kB T
becomes handy. Make use of this expansion up to O
( k
B T μ
to expand n and u in T.
Hint: Use (in a self-consistent way) that μ − εF ∝ T 2 in leading order in T and expand
−∞
dε H(ε) ≈
∫ εF
−∞
dε H(ε) + (μ − εF )H(εF ). (9)
(^1) For a reference on the Sommerfeld expansion see, e.g., Ashcroft, N. W. and Mermin N. D., Solid State Physics,
Holt, Rinehart and Winston, 1976.
Using this result we obtain (as for the low-temperature, high-density limit in the lecture)
μ = εf
[
π^2 12
kB T εF
) 2 ]
, (S.37)
cv = π^2 2
kB T εF
nkB. (S.38)
For a classical ideal gas (Maxwell-Boltzmann distribution) we find
cv =^3 2
nkB , (S.39)
which means that in the fermionic case the specific heat is surpressed by a factor π^2 2
kB T εF
. (S.40)
The origin of this surpression lies in the Fermi-Dirac distribution and the Pauli principle. For low tem- peratures kB T εF (note that this can easily be several hunderd Kelvin for electrons in metals) only a fraction of all the fermions, namely the ones around the Fermi energy, get thermally excited and contribute to the heat capacity, whereas for a classical gas all the particles can contribute.
(e) For the free Fermi gas g′(εF ) > 0. This does not need to be the true in more complex
systems such as solids (cf., e.g., semiconductors). What are the consequences of g′(εF ) ≤ 0?
Solution. The sign of the density of states determines whether the chemical potential increases, decreases or stays constant with respect to the temperature. A negative sign would lead to an increase in the chemical potential by increasing temperature.
