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Number of states and density of states

Short answer

For free electrons in a 3D box, the number of states per unit volume up to energy E is S_v(E) = (π/3)(8m/h²)^(3/2) E^(3/2). The density of states is g(E) = dS_v/dE = (π/2)(8m/h²)^(3/2) E^(1/2) = (8√2 π m^(3/2)/h³) √E, proportional to √E.

What do you understand by number of states and density of states in quantum mechanics? Derive appropriate expressions for them.

  • The number of states S(E)S(E) is the total number of allowed electron states (including spin) with energy from 0 up to EE (usually per unit volume).
  • The density of states g(E)g(E) is the number of states per unit energy per unit volume at energy EE: g(E)=dSv/dEg(E) = dS_v/dE. So g(E) dEg(E)\,dE is the number of states between EE and E+dEE + dE per unit volume.

Electron in a three-dimensional box

Treat the free electrons of a metal as particles in a cubic box of side LL (infinite well in 3D). The energies are

E=h28mL2(n12+n22+n32),n1,n2,n3=1,2,3,…E = \frac{h^2}{8mL^2}\left(n_1^2 + n_2^2 + n_3^2\right), \qquad n_1, n_2, n_3 = 1, 2, 3, \ldots

Each set (n1,n2,n3)(n_1, n_2, n_3) is one orbital state, i.e. one point in "n-space" with unit volume per point. Define

n2=n12+n22+n32=8mL2Eh2n^2 = n_1^2 + n_2^2 + n_3^2 = \frac{8mL^2E}{h^2}

All states with energy ≤E\le E lie inside a sphere of radius nn. Since n1,n2,n3n_1, n_2, n_3 are positive, only one octant (1/8) of the sphere counts.

   n2
   ^
   | . . . .
   | . . . . .    one octant of a sphere
   | . . . . . .  of radius n; each dot
   | . . . . . .  = one orbital state
   +--------------> n1

Number of states

Each orbital state holds 2 electrons (spin), so

S(E)=2×18×43πn3=π3n3=π3(8mL2Eh2)3/2\begin{aligned} S(E) &= 2\times\frac{1}{8}\times\frac{4}{3}\pi n^3 = \frac{\pi}{3}n^3 \\ &= \frac{\pi}{3}\left(\frac{8mL^2E}{h^2}\right)^{3/2} \end{aligned}

Dividing by the volume L3L^3:

Sv(E)=π3(8mh2)3/2E3/2S_v(E) = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}E^{3/2}

Density of states

g(E)=dSvdE=π3(8mh2)3/2⋅32E1/2=π2(8mh2)3/2E1/2=82 π m3/2h3E1/2\begin{aligned} g(E) &= \frac{dS_v}{dE} = \frac{\pi}{3}\left(\frac{8m}{h^2}\right)^{3/2}\cdot\frac{3}{2}E^{1/2} \\ &= \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)∝Eg(E) \propto \sqrt{E}.

 g(E)
  ^                 __..--
  |           _.--''
  |       _.-'      g(E) ~ sqrt(E)
  |    .-'
  |  /
  | /
  +-------------------------> E

Use

Multiplying g(E)g(E) by the Fermi–Dirac function f(E)f(E) gives the number of electrons per unit energy, nE=g(E)f(E)n_E = g(E)f(E). Integrating up to EFE_F at 0 K gives the Fermi energy EF=h28m(3nπ)2/3E_F = \frac{h^2}{8m}\left(\frac{3n}{\pi}\right)^{2/3}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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