Evaluate the integrals (Equation5.108 and 5.109) for the case of identical fermions at absolute zero. Compare your results with equations 5.43 and5.45. (Note for electrons there is an extra factor of 2 in Equations 5.108 and 5.109. to account for the spin degeneracy.)

A notion in quantum physics is called Fermi Energy. The Fermi energy is the value of the Fermi level at absolute zero temperature. The sea of fermions, in which no particle can live, is a clear indicator of it.

Equation 5.108 $\Rightarrow N=\frac{v}{2{\pi }^2}\underset{0}{\overset{\infty }{\int }}{k}^{2}n(\in )dk,$, where $n(\in )$is given as $(T\to 0)$by Eq. 5.104. So $N=\frac{V}{2{\pi }^2}\underset{0}{\overset{k_{\max }}{\int }}{k}^{2}dk=\frac{V}{2{\pi }^2}\frac{{k^3}_{\max }}{3},$where$k_{\max }$ is given by$\begin{array}{rcl}\frac{h^2{k^2}_{\max }}{2m}& =& \mu \left(0\right)\\ & =& E_F\Rightarrow k_{\max }\\ & =& \frac{\sqrt{2m{E}_F}}{h}\\ & & \end{array}$

Compare Eq. 5.43, which says

$\begin{array}{rcl}{E}_F& =& \frac{h^2}{2m}(3{\pi }^2\frac{Nq}{V}{)}^{2/3},or\frac{{\left(2m{E}_F\right)}^{3/2}}{h^3}\\ & =& 3{\pi }^2\frac{Nq}{V}, N=\frac{V}{3{\pi }^{2}q{h}^3}\left(2m{E}_F\right)3/2.\\ & & \end{array}$

Here q = 1, and Eq. 5.108 needs an extra factor of 2 on the right, to account for spin, so the two formulas agree.

Equation 5.109

$$\begin{gathered} E_{tot}=\frac{V{h}^2}{4{\pi }^{2}m}\underset{0}{\overset{k_{\max }}{\int }}{k}^{4}dk=\frac{V{h}^2}{4{\pi }^{2}m}\frac{{k^5}_{\max }}{5} \\ {E}_{tot}=\frac{V}{20{\pi }^{2}m{h}^3}\left(2m{E}_F\right)5/2 \end{gathered}$$

Compare Eq. 5.45, which says $E_{tot}=\frac{V{h}^2}{10{\pi }^{2}m}{k^5}_{\max }.$Again, Eq. 5.109 for electrons has an extra factor of 2, so the two agree.