422 Quantum ideal gases
AsT→0,ζ→∞and only the first term in the above expansion survives:ρλ^3 =ρ(
2 π ̄h^2
mkT) 3 / 2
≈
4 g
3√
π(lnζ)^3 /^2 =
4 g
3√
π(μ
kT) 3 / 2
. (11.5.26)
According to the procedure of the grand canonical ensemble, we need to solve forζas
a function ofρor equivalently forμas a function ofρ. From eqn. (11.5.26), we find
μ=
̄h^2
2 m(
6 π^2 ρ
g) 2 / 3
≡μ 0 =εF, (11.5.27)which is independent ofT. The special value of the chemical potentialμ 0 =μ(T=
0) is known as theFermi energy,εF. The Fermi energy plays an important role in
characterizing many-fermion systems such as metals and semiconductors. In order to
shed more light on the physical significance of the Fermi energy, consider the expression
for the average number of particles:
〈N〉=
∑
m∑
nζe−βεn
1 +ζe−βεn. (11.5.28)
However, recall that the occupation numbers must sum to the total number of particles
in the system: ∑
m∑
nfnm=N. (11.5.29)Thus, taking an average of both sides over the grand canonical ensemble, we obtain
〈N〉=∑
m∑
n〈fnm〉. (11.5.30)Comparing eqns. (11.5.28) and (11.5.30), we can deduce that the average occupation
number of a given state with quantum numbersnandmis
〈fnm〉=e−β(εn−μ)
1 + e−β(εn−μ)=
1
1 + eβ(εn−μ). (11.5.31)
Eqn. (11.5.31) gives the average occupancy of each quantum state in the ideal fermion
gas and is known as theFermi–Dirac distribution function. AsT→0,β→ ∞, and
eβ(εn−μ^0 )→ ∞ifεn> μ 0 , and eβ(εn−μ^0 )→0 ifεn< μ 0. Recognizing thatμ 0 =εF,
we have theT= 0 result
〈fnm〉=
0 εn> εF1 εn< εF. (11.5.32)
That is, at zero temperature, the Fermi–Dirac distribution becomes a simple step
function:
〈fnm〉=θ(εF−εn). (11.5.33)
A plot of the average occupation number versusεnatT= 0 is shown in Fig. 11.1.