Ideal Fermi Gas at Low Temperatures

H.2 Euler-Maclaur in Sum Formula

25.1 Ideal Fermi Gas at Low Temperatures

For an ideal Fermi gas the average occupancyfFD(ε)(see Eq. (21.88)) of an orbitalεis given by the bounded quantity

0≤ 1

λ−1eβε+1 = 1

exp[β(ε−μ)] +1 ≤1. (25.1)

For an ideal Bose gas, the corresponding average occupancy becomes infinite forε = 0 asλ → 1. However, for fermions,λ =eβμcan be any positive number, so 0≤λ ≤ ∞. In particular, one does not have to take the ground state into account explicitly, so conversion from a sum to an integral presents no problem. Therefore, for fermions, there is no critical temperature, such as the condensation temperatureTcfor bosons.

Thermal Physics.http://dx.doi.org/10.1016/B978-0-12-803304-3.00025-9 425

426 THERMAL PHYSICS

At all temperatures (see Section 23.1 and Eq. (23.13) withfν(λ)=hν(λ, 1)), the particle densityn=N/Vcan be written in the form

n=g0nQ(T)f3/2(λ), (25.2)

which can be regarded as an implicit equation for μ(n,T), with n specified. In particular, the function f3/2(λ) is not bounded as λ→ ∞. As we shall see later, f3/2(λ)→(lnλ)3/2/ (5/2)=(βμ)3/2/ (5/2) as λ→ ∞, so the product nQ(T)f3/2(λ) becomes independent ofTand proportional toμ3/2asT→0. This leads to an equation for μ(n, 0), the chemical potential at zero temperature, which is known as theFermi energy, εF≡μ(n, 0).

This sameT=0 limit may be explored in an elementary way by returning to the sum (see Eq. (21.89)) that led to Eq. (25.2), namely

N =g0

exp[β(ε−μ)] +1, (25.3)

whereg0=2s+1 is the degeneracy due to spinsthat is half integral for fermions. Thus, the prime on the sum means that one should exclude the degeneracyg0due to spin. As T→0,β→ ∞, soμ→εFthat depends only onn. Thus,fFD(ε)becomes a step function of the form

β→∞lim

exp[β(ε−μ)] +1 =

1 ifε < εF

0 ifε > εF. (25.4)

In other words, all of the states forε < εFare full and all of the states forε > εFare empty.

So forT=0, Eq. (25.3) takes the simple form N =g0

ε<εF

1. (25.5)

For the free particle and periodic boundary conditions, we know thatε=h¯2k2/2mand that the quantum states are distributed uniformly inkspace with a densityV/(2π)3. Therefore, we only need to compute the volume inkspace for whichε < εF, known as the volume of theFermi sphere. Specifically,

N=g0

V (2π)3

4πk2dk=g0

V (2π)3

3πkF3, (25.6)

where theFermi wavenumberkFsatisfiesεF=¯h2k2F/2m. We therefore obtain

kF=(6π2n/g0)1/3 (25.7)

and

εF= ¯h2

2m(6π2n/g0)2/3. (25.8)

AtT=0 the energy is also easy to calculate because one can include a factor ofεin the sum in Eq. (25.5) to obtain

U0=g0

ε<εF

ε=g0 ¯h2 2m

V (2π)3

kF 0

4πk4dk=3

5NεF. (25.9)

According to Eq. (23.18), the pressure is two-thirds of the energy density at all temperatures, so the pressure atT =0 is given by

p0=2

5nεF. (25.10)

In summary, atT = 0, the fermions are forced by the Pauli exclusion principle to fill the states of lowest energy that can accommodate all of them. Thus, all states up to the Fermi energyεFare occupied and all states above that energy are unoccupied. This forced occupation of high energy states results in a cumulative energy given by Eq. (25.9) and a corresponding pressure given by Eq. (25.10).

One can also define aFermi temperature TF:= εF

kB = h¯2

2mkB(6π2n/g0)2/3. (25.11) This may be rewritten in the form

n=g0

mkBTF

2πh¯2 3/2

3π1/2, (25.12)

which greatly resembles Eq. (24.5) for the critical temperatureTcof an ideal Bose gas. We emphasize, however, thatTF isnot a critical temperature but rather a temperature that characterizes the degree to which fermions atT = 0 are forced into excited states by the Pauli exclusion principle.

A word about the relative magnitudes ofTFandTcis relevant. If we consider fermions or bosons that have comparable number densities and masses, say the masses of He3 (a fermion with half integral spin) and He4(a boson with integral spin), the magnitudes of TFandTcwill be comparable. As we saw previously,Tcwas typically a few K degrees at the density of He4near the lambda transition. But electrons are fermions and the electron mass is about 1836 times smaller than the mass of a proton. Therefore, for free electron gases in metals at their usual densities,TFis typically 50,000 K degrees. In such cases, one hasT TFfor any temperature of interest. We shall see that a Fermi gas at temperature T > 0 butT TFdisplays characteristics very similar to a Fermi gas at T = 0 except a small fraction∼ T/TF of electrons is now in excited states. Consequently, a Fermi gas atT TFis usually referred to as adegenerate Fermi gas. Equivalent conditions for a degenerate Fermi gas are thereforeβμ 1,λ 1, orn/nQ(T) 1.

Before leaving this section, it is worth pointing out that the integrals in Eqs. (25.6) and (25.9) could equally well have been written as integrals overεby expressingk=

2mε/¯h2 and then using dk=(∂k/∂ε)dε. This leads to an intensive density of states of the form

g(ε):=G(ε) V :=g0

2 m

¯ h2π2

¯ h2

1/2

ε1/2=3 2

n εF

ε εF

1/2

. (25.13)

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Here,G(ε)dεis the number of states, including spin, with energy betweenεandε+dεand g(ε)dεis the number of states per unit volume in that same interval. Then

n= εF

g(ε)dε (25.14)

and the energy density

uV(T=0)= εF

g(ε)εdε=3

5nεF. (25.15)

Work Due to Expansion of an Ideal Gas

Statement of the Second Law