= Solution
The occupation probability of an individual one-electron state is the <Fermi-Dirac distribution>,
$$
\boxed{f(\epsilon,T)=\frac1{e^{(\epsilon-\mu(T))/(k_BT)}+1}}.
$$
At $T=0$, it is a step: one below $E_F$ and zero above it. Heating rounds the step over an <energy> range of order a few $k_BT$, creating empty states below and occupied states above the <chemical potential>. At every positive <temperature>, $f(\mu,T)=1/2$.
For fixed $N,V$ in the three-dimensional <free electron model>, the <chemical potential> is determined by $N=\int g(\epsilon)f(\epsilon,T)d\epsilon$; it is not exactly constant. In the degenerate regime $T\ll T_F$, the <Sommerfeld expansion> gives
$$
\mu(T)=E_F\left[1-\frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2+\cdots\right].
$$
It is therefore reasonable in a simple low-temperature sketch to set $\mu\simeq E_F$, provided this approximation is stated. The figure includes the leading small shift. At much higher <temperature> the fixed-$E_F$ approximation fails and the distribution approaches the classical dilute limit. The plotted occupation is $f$, rather than the <energy> distribution $g f$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-c10-fermi-distribution.png]
{title=Thermal rounding of Fermi-Dirac occupations at fixed particle number, with the leading chemical-potential shift}
Back to article page
