Past exam of the thermal and statistical physics course of the University of Cambridge / 2012 / Paper 1 / 3 / Solution

Let $\beta =1/(k_{B}T)$. Each independent fermion orbital is either empty or occupied once, so its grand partition factor is $1+e^{-\beta ({ϵ}_j-\mu )}$. Multiplying factors and taking the logarithm givesSince $1-n_F=[1+e^{-\beta (ϵ-\mu )}{]}^{-1}$, this is the requested form. Spin or other orbital degeneracy is already included in $g$.

For the Fermi-window convolution of the grand potential, defineDirect differentiation gives $F^{\prime }=+n_F$. The supplied PDF hint has the opposite sign; using that sign would contradict its grand-potential formula. Integrating twice with $A(0)={\Phi }_0(0)=0$ and vanishing upper boundary terms,At zero temperature $-{\partial }_{ϵ}{n}_F\to \delta (ϵ-\mu )$, confirming ${\Phi }_0(\mu )=\Phi (0,\mu )$.

Use the source's implicit unit normalization of the constant density of states; restoring a constant baseline $g_0$ multiplies the following potentials by $g_0$. For the sinusoidal density of states, direct integration givesPhysical nonnegative density of states requires $|\alpha |\le 1$. Write $t=k_{B}T$ and $y=ϵ-\mu$. The Fermi windowis even, with full-line moments $\int wdy=1$, $\int ywdy=0$ and $\int y^{2}wdy={\pi }^2{t}^2/3$. The last integral follows, for example, by expanding $e^x/(1+e^x{)}^2$ on $x>0$ into its alternating exponential series. When $t\ll \mu$, extending the lower integration limit to $-\infty$ introduces only exponentially small band-edge corrections. The supplied Fourier transform givesThe odd sine term vanishes. Substitution provesThe thermal damping of a sinusoidal density of states is therefore $X/\sinh X$ in the full oscillatory potential. The requested ratio instead concerns the thermal change $\Phi (T,\mu )-{\Phi }_0(\mu )$. Its oscillatory thermal-change amplitude ratio isFor positive $X$ it decreases smoothly from the limit $|\alpha |$ to zero:At exactly $T=0$ both thermal changes vanish; the quoted value is their limiting ratio. The amplitude is the envelope in chemical potential. At a particular fixed $\mu$, the signed contribution ratio has $\alpha \cos (2\pi \mu /\Delta )$ in place of $|\alpha |$; at an oscillation node the actual oscillatory contribution is zero. The large-$X$ behaviour still requires $k_{B}T\ll \mu$, so the useful range is $1\ll X\ll 2{\pi }^2\mu /\Delta$.