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Physics 127a: Class Notes
Lecture 17: Ideal Fermi Gas
General Properties
For an ideal Fermi gas
 = −kT ln Q = −kT
N=
X
hns i =
s
X
s
X
ln 1 + ze−βεs
(1)
s
1
z−1 eβεs + 1
(2)
where the sums run over single particle states s which are usually labelled by the wave vector k and the spin.
to −∞ <
The fugacity z = eβµ may run over the whole range 0 < z < ∞, correspondingP
R µ < ∞.
For a nonrelativistic gas in 3-dimensional free space εk = h̄2 k 2 /2m and s → dε ρ(ε) with the
density of states
2m 3/2 1/2
V
ε ,
(3)
ρ(ε) = g 2
4π
h̄2
with g the spin degeneracy, usually 2s + 1. Proceeding as in the Boson case leads to expressions
P
N
g
g
(4)
= 3 f5/2 (z)
= 3 f3/2 (z)
kT
λ
V
λ
√
with λ the thermal length h/ 2π mkT and fν (z) Fermi functions (their properties are discussed in Pathtria
appendix E)
Z ∞
1
y ν−1
dy.
(5)
fν (z) =
(ν − 1)! 0 z−1 ey + 1
High temperature expansions may be developed by expanding the integrands in small z as in the Bose Case.
“Zero Temperature” Behavior
The Fermi occupation function
1 1
1
nF (ε) = β(ε−µ)
= + tanh β (ε − µ)
(6)
e
+1
2 2
2
tells us the occupation number of a single particle state at energy ε. The function is 1 for ε many kT less than
µ, becomes 21 at ε = µ, and goes exponentially to zero for ε many kT greater than µ. At low temperatures
the chemical potential is the dividing energy between occupied and empty states. The second form in Eq.
(6) shows the symmetry
nF (µ + δ) = 1 − nF (µ − δ).
(7)
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At zero temperature the states are filled up to an energy called the Fermi energy F = µ(T = 0) and
empty above. The value of εF is fixed by demanding the total number of states with ε < εF be equal to the
number of Fermions N. For free particles, ε = εF defines a spherical surface in k-space |k| = kF with kF the
Fermi wave vector. Counting the states inside the Fermi surface (volume (2π )3 /V per wave vector) gives
2 1/3
g 43 π kF3
6π N
=N
i.e.
kF =
.
(8)
(2π )3 /V
g V
The order of magnitude of kF is the reciprocal of the interparticle spacing. The corresponding Fermi energy
εF = h̄2 kF2 /2m is of order the zero point energy of a particle confined to this scale. This is typically a
large energy, e.g. 104 K for electrons in a metal. Note that εF ∝ (N/V )2/3 so that ρ(εF ) = ∂N/∂εF |V =
(3N/2εF ), a result we will use below in the low temperature calculations.
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Thermodynamic Properties The total internal energy at zero temperature (all kinetic!) is given by integrating over the sphere up to radius kF . Clearly U = cN εF with c a constant of order unity. This shows us
that U (T = 0) scales with the thermodynamic variables as
U ∝ N 5/3 V −2/3 .
So we have
µ(T = 0) = εF =
(fixing the constant c = 3/5), and
∂U
∂N
V
5U
3N
=
∂U
P =−
∂V
or
N
(9)
=
3
U = N εF
5
2U
3V
(10)
(11)
(as we know from general considerations). This degeneracy pressure is important in diverse areas of physics,
such as solid state physics and astrophysics.
Degeneracy pressure and white dwarf stars White dwarf stars are old stars of mass similar to our sun,
that have used up most of their nuclear fuel. A simple model of these stars supposes that they consist of fully
ionized Helium, i.e. a mixture of alpha-particles and electrons. For simplicity we consider the density to be
uniform, so that there are N electrons and N/2 alpha particles in the volume of the star of radius R. The
electrons from a low temperature (kT εF ) degenerate electron gas, and the degeneracy pressure balances
the inward gravitational force. The Fermi energy turns out to be comparable to the electron rest mass, and
so we must consider a relativistic Fermi gas. The alpha particles provide charge neutrality, and dominate the
mass and so the gravitational attraction. Their kinetic energy (and contribution to the degeneracy pressure)
is small by the mass ratio me /4mp .
First lets suppose the electrons are strongly relativistic, so the spectrum is εp = cp. The total kinetic
energy is of order Ukin ∼ N εF ∼ N h̄c(N/V )1/3 ∼ h̄cN 4/3 /R. This provides the repulsive degeneracy
pressure. The gravitational potential energy is Upot ∼ −G(Nmp )2 /R. These expressions imply that for a
star mass M ∼ N mp satisfying
(h̄c/G)3/2
(12)
M&
mp
the attractive forces overcome the repulsive one, so that so solution can be found—the star collapses in
this model. This limit is known as the Chandrasekhar limit, and indeed more massive stars are believed to
collapse to neutron stars or black holes.
If we instead suppose the classical limit εp = p 2 /2m the kinetic energy scales as Ukin ∼ N 5/3 (h̄2 /2m)/R 2
and so collapse never occurs. Clearly we need to consider the full expression for the kinetic energy
p
(13)
εp = m2 c4 + p 2 c2 − mc2 .
The pressure is calculated from the grand potential
Z ∞
−3
P = gkT h
ln[1 + e−β(εp −µ) ]4πp 2 dp
(14)
0
which becomes on integrating by parts
g
P = 3
3h
Z
0
∞
1
eβ(εp −µ)
We can now take the T → 0 limit
4πg
P =
3h3
Z
pF
p
0
2
dεp
4πp 2 dp.
+ 1 dp
(15)
dεp
4πp 2 dp,
dp
(16)
p
and finally put in the form for εp , use the spin degeneracy g = 2, and introduce the scaled integration variable
y = p/mc to get
π m4 c 5
A(pF /mc)
(17)
P =
3 h3
where the function A(x) is defined by
Z x
y4
A(x) = 8
dy.
(18)
2 1/2
0 (1 + y )
The function A is rather boring. For small x it varies as A(x) ' 85 x 5 (and then c drops out of P and P ∝ R −5 ,
the classical limit) and for for large x as A(x) ' 2x 4 (so that P ∝ cR −4 , the strongly relativistic limit.
To determine the equilibrium radius of the star we minimize the total energy with respect to the radius
dE/dR = 0. For the kinetic energy
and for the potential energy
dEkin
3V dE
3V
=
=−
P = −4π R 2 P
dR
R dV
R
(19)
dEpot
d
αGM 2
αGM 2
=
−
=
dR
dR
R
R2
(20)
where α is a fudge factor that depends on the density distribution. Minimizing the total energy gives
4π R 2 P =
αGM 2
R2
(21)
balancing the outward pressure force on the surface with the inward gravitational force. Now it simply
remains to collect all the factors, using pF = h̄(3π 2 n)1/3 , n = N/( 43 π R 3 ), M = 4mp × N2 . Defining the
mass scale
1/2
3π
9
(Planck mass)3/2
(h̄c/G)3/2
M0 =
∼
(22)
64
α3
m2p
(Proton mass)2
and the length scale
3
R0 =
8
3π
α
1/2
gives the final expression
(h̄c/G)1/2 h̄
(Planck mass)
∼
× (Compton wavelength of electron)
mp
mc
(Proton mass)
(23)
1
M 2/3
A(x) 1/2 =
.
R0
2x 4
M0
x= M
M0
R
You can find a figure of the resulting solution for R/R0 as a function of M/M0 in Pathria §8.4. Since the left
hand side of this equation approaches 1 for small R/R0 this shows there is a maximum M = M0 for which
solutions can be found, the Chandrasekhar limit.
Pauli spin susceptibility In a magnetic field, for charged particles such as the electron, the single particle
energies depends on the spin. For an electron, a spin- 21 particle
εk,σ = εk − σ µB B,
(24)
where σ = ±1 for the ↑ and ↓ spin respectively, and µB = eh̄/2mc
isP
the Bohr magneton. The statistical
P
mechanics is now easily generalized to this case by replacing 2 k by k,σ .
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The number of up or down spin particles is then
X
Nσ (T , µ, B) =
k
1
eβ(εk −σ µB B−µ)
+1
.
If we define N0 as the number of particles of one spin in zero field
X
1
N0 (T , µ) =
β(ε
−µ)
k
e
+1
k
(25)
(26)
then by inspection
Nσ (T , µ, B) = N0 (T , µ + σ µB B).
(27)
N = N↑ + N↓ = N0 (T , µ + µB B) + N0 (T , µ − µB B) = 2N0 (T , µ) + O(B 2 )
∂N0
M = µB (N↑ − N↓ ) = N0 (T , µ + µB B) − N0 (T , µ − µB B) ' 2
µ2 B
∂µ T ,B=0 B
(28a)
Then for small B we have
(28b)
where the second equalities are true for small B. Equation (28a) is the same as in zero field, and shows that
the chemical potential for fixed N is unchanged at O(B). Equation (28b) then gives us the susceptibility
1 ∂N
χ = µ2B
(29)
V ∂µ T
(using N = 2N0 for the total number of electrons). At zero temperature µ = εF , so
3 µ2B
(30)
n
2 εF
where the first expression in quite general, and the second evaluated for nonrelativistic electrons N ∝
2/3
kf3 ∝ εF . This is the Pauli susceptibility, and is independent of temperature. At high temperatures
µ = kT ln(N λ3 /V ) and so ∂N/∂µ = N/kT and
χ = µ2B ρ(εF ) =
µ2B
kT
the usual Curie susceptibility of free spins. These susceptibilities are positive, i.e. paramagnetic.
Charged particles also couple to a magnetic field via their charge, given by the Hamiltonian
X 1
e
H =
[pi + A(xi )]2
2m
c
i
χ =n
(31)
(32)
with B = ∇ × A. This turns out to lead to a diamagnetic contribution to the magnetic susceptibility coming
from circulating currents (diamagnetic because of Lenz’s law): at high temperatures
1 µ2
χD → − n B ,
3 kB T
and at low temperatures (known as Landau diamagnetism)
(33)
1 µ2
(34)
χD →= − n B .
2 εF
For both high and low temperatures, the diamagnetic susceptibility is a factor of 3 smaller, but of opposite
sign, than the paramagnetic susceptibilities Eqs. (30) and (31) coming from the coupling of the field to the
spins. The calculation of the diamagnetic component of the susceptibility involves the rather tricky solution
of the quantum mechanics of the Hamiltonian Eq. (32) giving “Landau levels”, which I will not describe
here.
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Low-T Fermi Integrals
We are often interested in the low temperature thermodynamic properties at temperature kT F . Only
particle near within a few kT of the Fermi surface may be thermally excited, because of the Pauli exclusion
principle (cf. the form of nF (ε)). This means that the temperature dependence of thermodynamic properties
for kT εF should only depend on properties near the Fermi surface, and in particular only on ρ(ε ' εF ),
not on the whole density of states. This is important because in some applications of the results, e.g. to
electrons in metals, the function ρ(ε) might be quite complicated or even unknown in full. The following
method to evaluate low temperature Fermi integrals stresses this result. (The result is equivalent to Eq. (16)
of Appendix E of Pathria, although his derivation is rather obscure.)
Thermodynamic sums over the single particle (momentum) states lead to integrals over the single particle
energy ε of the form
Z
∞
I=
dε φ(ε) nF (ε)
(35)
0
where nF (ε) = (eβ(ε−µ) + 1)−1 is the Fermi occupation function, and φ will depend on the quantity being
calculated, e.g. φ = ρ(ε) for I = N , φ = ερ(ε) for I = U (the internal energy), with ρ(ε) the density of
states. For the low temperature expansion of these expression, since all the action is for an energy range of
order kT about the Fermi energy εF , we try to isolate this region of the energy integral.
First split up the integral
Z µ
Z µ
Z ∞
I=
dε φ(ε) −
dε φ(ε) [1 − nF (ε)] +
dε φ(ε) nF (ε).
(36)
0
µ
0
In the second two integrals, shift the integration variable ε = µ + δ
Z 0
Z
Z µ
dε φ(ε) −
dδ φ(µ + δ) [1 − nF (µ + δ)] +
I=
−µ
0
∞
dδ φ(µ + δ) nF (µ + δ).
(37)
0
For low temperatures the lower limit in the second integral can be replaced by −∞, with an error of order
ε−βµ (and µ ' εF ). Then in this integral put δ → −δ, and use the symmetry of the nF function
1 − nF (µ − δ) = nF (µ + δ)
to give
Z
I=
0
µ
Z
∞
dε φ(ε) +
dδ [φ(µ + δ) − φ(µ − δ)] nF (µ + δ) + O(ε −βµ ).
(38)
(39)
0
Now expanding φ(µ+δ)−φ(µ−δ) as a Taylor expansion in δ and integrating over δ gives a series expansion
in (kT /εF )2 . We will just go up to quadratic order
Z µ
π2
2
dε φ(ε) +
(40)
I'
(kT ) (dφ/dε)
6
0
ε=µ
where we have used
Z
∞
π2
δ
=
(kT )2 .
(41)
βδ + 1
e
12
0
In the second term, which is already of quadratic order we can replace µ by the zero temperature value
µ ' εF , and we split up the first term into two pieces
Z εF
Z µ
π2
2 dφ I'
(kT )
dε φ(ε) +
dε φ(ε) +
.
(42)
6
dε ε=εF
0
εF
dδ
Note that the first term is temperature independent. The second two terms give the temperature dependence
through µ(T ) and the explicit temperature dependence.
5
Number of Particles This fixes µ(T , N ). We use φ = ρ(ε):
Z µ
Z εF
π2
2 dρ dε ρ(ε) +
dε ρ(ε) +
.
(kT )
N'
6
dε ε=εF
0
εF
(43)
Subtracting the zero temperature expression
Z
N'
εF
dε ρ(ε)
(44)
0
gives
Z
µ
εF
so that
(45)
π2
2
(µ − εF )ρ(εF ) ' −
(kT ) (dρ/dε)
6
ε=εF
(46)
π2
2
µ ' εF −
(kT ) (d ln ρ/dε)
6
ε=εF
(47)
or
Specific Heat
π2
2
dε ρ(ε) ' −
(kT ) (dρ/dε)
6
ε=εF
Use CV = ∂U/∂T |N,V , and for the internal energy U we use φ = ερ(ε):
Z
U'
0
µ
π2
2 d(ερ) (kT )
dε ερ(ε) +
.
6
dε ε=εF
(48)
Differentiating with respect to T
dµ
π 2 2 d(ερ) .
CV ' µρ(µ)
+
k T
dT
3
dε ε=εF
(49)
To evaluate dµ/dT we differentiate Eq. (43) using dN/dT = 0
dµ
π 2 2 dρ 0 ' ρ(µ)
+
k T
dT
3
dε ε=εF
(50)
and substituting into Eq. (49) and then putting µ ' εF gives
CV '
π2 2
k T ρ(εF ).
3
For the free particle spectrum ε = h̄2 k 2 /2m and then ρ(εF ) = 3N/2εF , so that
1
2 kT
.
CV ' N kπ
2
εF
6
(51)
(52)
Internal Energy Another way to calculate the temperature dependence of the internal energy is to use
φ(ε) = (ε − εF )ρ(ε). This gives
Z εF
Z µ
π2
2 d[(ε − εF )ρ] (kT )
(ε − εF )ρ(ε) +
(ε − εF )ρ(ε) +
.
(53)
U − N εF =
6
dε
0
εF
ε=εF
The second term is of order (µ − εF )2 and so is negligible at order T 2 , and similarly in the ε derivative we
can neglect the term (ε − εF )dρ/dε, so that
Z εF
π2
(kT )2 ρ(εF )
(ε − εF )ρ(ε) +
(54)
U = N εF +
6
0
and all the temperature dependence is in the second term. For the free, nonrelativistic gas
3
5π 2
U = N εF [1 +
5
12
Other Results
kT
εF
2
+ · · · ].
(55)
Some other results are
PV =
2
5π 2
2
U = N εF [1 +
3
5
12
kT
εF
2
+ · · · ].
(56)
The Helmholtz free energy A is given by A = N µ − P V
3
5π 2
A = NεF [1 −
5
12
kT
εF
2
+ · · · ].
(57)
The entropy is derived as ∂A/∂T |N,V or from CV = T ∂S/∂T |N,V
S = Nk
π 2 kT
+ ··· .
2 εF
(58)
Note that the free energy spectrum ε = h̄2 k 2 /2m has been used in these expressions: the quadratic temperature
dependences could more generally be written in terms of ρ(εF ) as in Eq. (51).
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