= Solution
{c}
Let $A$ be the real-space area and use $\sum_{\mathbf p}\to C\int d^2p$, with $C=A/(2\pi\hbar)^2$. There is one state per <momentum> cell, since the <fermions> are spinless. An annulus between $p$ and $p+dp$ contains $C\,2\pi p\,dp$ states. Near the <Fermi surface>, the <quasiparticle> <dispersion relation> has $d\epsilon/dp=v_F^*=p_F/m^*$, so the total <density of states> is
$$
\boxed{D^*(\mu)=C\frac{2\pi p_F}{v_F^*}=\frac{Am^*}{2\pi\hbar^2}.}
$$
The <density of states> per unit area and the occupied-state count are, respectively,
$$
\boxed{g^*(\mu)=\frac{D^*(\mu)}A=\frac{m^*}{2\pi\hbar^2},\qquad
N=C\pi p_F^2=\frac{Ap_F^2}{4\pi\hbar^2}.}
$$
In particular, $D^*(\mu)=2Nm^*/p_F^2$. No factor of two for <spin degeneracy> is present. This counts <Landau quasiparticle> levels, so it is not multiplied by the <quasiparticle residue> of the microscopic particle <Green's function>.
Back to article page
