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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 is the total number of allowed electron states (including spin) with energy from 0 up to (usually per unit volume).
- The density of states is the number of states per unit energy per unit volume at energy : . So is the number of states between and per unit volume.
Electron in a three-dimensional box
Treat the free electrons of a metal as particles in a cubic box of side (infinite well in 3D). The energies are
Each set is one orbital state, i.e. one point in "n-space" with unit volume per point. Define
All states with energy lie inside a sphere of radius . Since 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
Dividing by the volume :
Density of states
So .
g(E)
^ __..--
| _.--''
| _.-' g(E) ~ sqrt(E)
| .-'
| /
| /
+-------------------------> E
Use
Multiplying by the Fermi–Dirac function gives the number of electrons per unit energy, . Integrating up to at 0 K gives the Fermi energy .
✓ Answer checked by Claude Opus 5.5 (high effort)
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