Population density and temperature-independent Fermi energy

Population density

The population density $n_E(E)$ is the number of electrons per unit volume per unit energy at energy $E$. It is the product of the number of available states and the probability that they are occupied:

$$n_E(E) = g(E)\, f(E), \qquad g(E) = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} E^{1/2}, \quad f(E) = \frac{1}{1 + e^{(E-E_F)/kT}}$$

The area under the $n_E$ versus $E$ curve gives the total electron concentration $n$.

 nE(E)
  ^        T=0: sharp cut at EF
  |      ..---.
  |   .-'      |\  <- T>0: small tail
  |  /         | \
  | /          |  '..
  |/___________|______> E
  0            EF

Fermi energy depends only on n

At $T = 0$ K, $f(E) = 1$ for $E < E_{F0}$ and $0$ above it. Hence

$$n = \int_0^{E_{F0}} g(E)\, dE = \frac{8\pi\sqrt{2}\, m_e^{3/2}}{h^3} \cdot \frac{2}{3} E_{F0}^{3/2} = \frac{8\pi}{3}\left(\frac{2m_e E_{F0}}{h^2}\right)^{3/2}$$

Solving,

$$E_{F0} = \frac{h^2}{8m_e}\left(\frac{3n}{\pi}\right)^{2/3}$$

Thus $E_{F0}$ contains only constants and $n$, the free electron concentration. A metal with more free electrons per volume has a higher Fermi energy (Cu, $n = 8.5 \times 10^{28}$ m⁻³: 7.0 eV; Na, $2.5 \times 10^{28}$: 3.1 eV).

Temperature has almost no effect

At $T > 0$, the electron number must stay the same:

$$n = \int_0^{\infty} g(E) f(E)\, dE$$

Evaluating this integral (Sommerfeld expansion) gives

$$E_F(T) = E_{F0}\left[1 - \frac{\pi^2}{12}\left(\frac{kT}{E_{F0}}\right)^2\right]$$

For copper at 300 K: $kT = 0.0259$ eV, $E_{F0} = 7.0$ eV, so $kT/E_{F0} = 3.7 \times 10^{-3}$ and the correction is $\frac{\pi^2}{12}(3.7 \times 10^{-3})^2 \approx 1.1 \times 10^{-5}$, a change of only about 0.001%.

Physical reason

Only electrons within a few $kT$ of $E_F$ can be thermally excited, because states deeper down have all their neighbouring states already filled (Pauli principle). Since $kT \ll E_F$, only a tiny fraction of electrons are disturbed, and the electrons moved above $E_F$ match the holes left below $E_F$ (the Fermi function is symmetric about $E_F$). Also $n$ in a metal does not change with $T$. Hence the Fermi energy of a metal is practically independent of temperature and depends only on its electron concentration.