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 Incoming links: One-dimensional free-electron density of states

Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section C / C8 / b / i / Solution 2026-10-18

To preserve the convention of part (a), first count one spin species. Define

g_{d}^{(1)} (ϵ) d ϵ

as the number of spatial states in the whole specimen in that energy interval. Put

k = \frac{2 m ϵ}{ℏ}, \frac{d k}{d ϵ} = \frac{m}{ℏ ^{2} k} .

In one dimension there are intervals near both

+ k

and

- k

, so the one-dimensional free-electron density of states is

g_{1}^{(1)} (ϵ) = \frac{L}{2 π} 2 \frac{d k}{d ϵ} = \frac{L 2 m}{2 π ℏ} ϵ^{- 1/2} .

Thus

g_{1}^{(1)} \propto ϵ^{- 1/2}

. The factor of two here comes from opposite momenta, not spin. For physical Electrons with two spin states, the spin-summed density of states is

g_{1} = 2 g_{1}^{(1)} = L 2 m ϵ^{- 1/2} / (π ℏ)

. These expressions are continuum approximations away from individual discrete levels.

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