Degenerate state, Fermi energy and density of states

Degenerate state

Two or more different quantum states (different sets of quantum numbers, i.e. different wave functions) that have the same energy are called degenerate states. For an electron in a cubic box of side $L$,

$$E = \frac{h^2}{8mL^2}(n_1^2 + n_2^2 + n_3^2)$$

The states $(n_1, n_2, n_3) = (1,1,2), (1,2,1), (2,1,1)$ all have $n_1^2 + n_2^2 + n_3^2 = 6$, so they have the same energy $6h^2/8mL^2$. This level is three-fold degenerate. With spin, each state is also two-fold spin degenerate.

Fermi energy

The Fermi energy $E_F$ is the energy of the highest occupied level in a metal at absolute zero (0 K). At 0 K all states below $E_F$ are filled and all above are empty. At any temperature it is the energy at which the probability of occupation is exactly 1/2, $f(E_F) = 1/2$. For copper $E_F \approx 7$ eV.

Density of states versus energy

Density of states $g(E)$ is the number of states per unit energy per unit volume at energy $E$.

Let $n^2 = n_1^2 + n_2^2 + n_3^2 = \dfrac{8mL^2E}{h^2}$. Each state is a point with unit volume in n-space. All states with energy up to $E$ lie inside a sphere of radius $n$, but only the positive octant is allowed ($n_i > 0$).

   n3
   |
   |  . . .
   |  . . . .      states with energy <= E
   |  . . . . .    fill 1/8 of a sphere
   +-----------n2  of radius n
  /
 n1

Number of orbital states $= \frac{1}{8}\cdot\frac{4}{3}\pi n^3$. With 2 spin states each:

$$\begin{aligned} S(E) &= 2\times\frac{1}{8}\times\frac{4}{3}\pi n^3 = \frac{\pi}{3}\left(\frac{8mL^2E}{h^2}\right)^{3/2} \end{aligned}$$

Per unit volume (divide by $L^3$):

$$S_v(E) = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E^{3/2}$$

Differentiating:

$$\begin{aligned} g(E) &= \frac{dS_v}{dE} = \frac{\pi}{2}\left(\frac{8m}{h^2}\right)^{3/2}E^{1/2} \\ &= \frac{8\sqrt{2}\,\pi\,m^{3/2}}{h^3}\,E^{1/2} \end{aligned}$$

So $g(E) \propto \sqrt{E}$: the number of available states per unit energy increases as the square root of energy.

 g(E)
  ^                      ___..
  |                __..--
  |  filled   _.-''|
  |  at 0 K .'     |  empty
  |       /        |
  |     /          |
  +----------------+--------> E
                  E_F

Fermi energy from $g(E)$

At 0 K all states up to $E_F$ are full, so the electron concentration is

$$n = \int_0^{E_F} g(E)\,dE = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E_F^{3/2} \quad\Rightarrow\quad E_F = \frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{2/3}$$