Past exam of the condensed matter physics course of the University of Oxford / 2016 / 2 / b / Solution

Use periodic boundaries in a box of side $L$ and include two spin states for each wavevector. One orbital state occupies $(2\pi /L{)}^d$ in wavevector space. Let $v_d$ be the volume of the unit ball in dimension $d$, namely $2,\pi ,4\pi /3$. Filling the ball of radius $k_F$ givesSince $E_F={\hslash }^2{k}_F^2/(2m)$,The density $n$ means number per length, area or volume as appropriate; it is not assigned three-dimensional units in every case.

For the same spin-summed states, the cumulative number per $d$-dimensional volume below energy $E$ is $n(E)=2v_d(2mE/{\hslash }^2{)}^{d/2}/(2\pi {)}^d$. Differentiating yields $g(E)=(d/2)n(E)/E$. Evaluating at the filling energy,In one, two and three dimensions the density of states varies respectively as $E^{-1/2}$, a constant and $E^{1/2}$. The normalization counts both signs of momentum and both spin states exactly once. A single-spin density of states or extra valley degeneracy would require a different counting convention, not an extra unexplained factor in these constants.