Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section C / C8 / b

To preserve the convention of part (a), first count one spin species. Define $g_d^{(1)}(ϵ)dϵ$ as the number of spatial states in the whole specimen in that energy interval. PutIn one dimension there are intervals near both $+k$ and $-k$, so the one-dimensional free-electron density of states isThus $g_1^{(1)}\propto {ϵ}^{-1/2}$. 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 $g_1=2g_1^{(1)}=L\sqrt{2m}{ϵ}^{-1/2}/(\pi \hslash )$. These expressions are continuum approximations away from individual discrete levels.

In two dimensions the density of states in $k$ space per spin species is $(L/2\pi {)}^2$. A annulus of radius $k$ and thickness $dk$ has area $2\pi kdk$. ThereforeThe two-dimensional free-electron density of states is constant in energy, $g_2^{(1)}\propto {ϵ}^0$, above the band bottom. Including both spin states gives $g_2=L^{2}m/(\pi {\hslash }^2)$. Dividing by $L^2$ gives the density per unit physical area, rather than the total specimen's density.

In three dimensions use the $k$-space density of states $(L/2\pi {)}^3$ per spin species and the spherical shell volume $4\pi k^{2}dk$:Thus the three-dimensional free-electron density of states grows as ${ϵ}^{1/2}$. For physical Electrons, including both spin states givesAll three total density of states functions have units ${\mathrm{J}}^{-1}$ when energy is in joules; they count states per energy interval. The paper explicitly suppresses spin in (a) but does not repeat that convention in (b), so giving both conventions avoids a hidden factor-of-two ambiguity. Parts (c)--(f) below use the physical two-spin-state electron gas.