Ground State Energy and Bulk Modulus of Electron Gas

The total number of electrons $N = \sum\limits_{\textstyle \vec{k}\sigma} n_{\textstyle \vec{k}\sigma} = ... ...\limits^{\textstyle \varepsilon _F}_{\textstyle 0}g(\varepsilon ) d\varepsilon$. We could check our expression for $g(\varepsilon )$ by plugging it in to see if we get $N$ back. The total energy is

$$E = \sum\limits_{\vec{k}\sigma} \varepsilon _{k} n_{k\sigma} = \int\limits^{\varepsilon _F}_{0} \varepsilon g(\varepsilon ) d\varepsilon$$

Now $g(\varepsilon ) = A\varepsilon ^{1/2}$ (It won't matter what $A$ is.)

$$\Rightarrow N = \frac{2}{3} A \varepsilon ^{3/2}_{F}\quad {\rm and}\quad E = \frac{2}{5} A\varepsilon ^{5/2}_{F}.$$

Thus $\qquad$ (no dependence on $A$)
(Contrast w/ classical gas $\frac{\textstyle E}{\textstyle N} = \frac{\textstyle 3}{\textstyle 2} kT$)
Pressure: Note $\varepsilon _F$ is proportional to $(\frac{\textstyle N}{\textstyle V})^{2/3}$, i.e., it depends on $V$. If V is changed, the electron gas should exert a pressure. At $T = 0$,

$$P = - \left( \frac{\partial E}{\partial V}\right)_N = \frac{2}{3}... ...}{V}\right)\varepsilon _F\qquad ({\rm use}\quad E = \frac{3}{5}N\varepsilon _F)$$

Aside: Dimensions or units dictate $-(\partial E/\partial V)_N=2E/3V$:

$$\varepsilon _F \sim k^{2}_{F}\sim n^{2/3}\sim \left(\frac{N}{V}\right)^{2/3}$$

$$E = A\varepsilon _F \quad\quad\quad \left(A=\frac{3}{5}N\right)$$

$$= {\tilde A}\left(\frac{N}{V}\right)^{2/3}$$

$$\frac{dE}{dV} = \frac{2}{3}{\tilde A}N^{2/3}V^{-5/3}$$

$$= \frac{2}{3}\frac{E}{V}$$

where ${\tilde A}$ is a constant. The inverse compressibility (or bulk modulus $B$) is defined as $-V(\frac{\textstyle dP}{\textstyle dV})$. (At $T = 0$, the derivative is taken with $N = const.)$ Since

$$P \propto \frac{\varepsilon _F}{V} \propto V^{-5/3}, \quad B = - V \frac{\partial P}{\partial V} = \frac{5}{3} P$$

Using $P = \frac{\textstyle 2}{\textstyle 3}\frac{\textstyle E}{\textstyle V}$, we obtain $B = \frac{\textstyle 10}{\textstyle 9} \frac{\textstyle E}{\textstyle V} = \f... ...textstyle 2}{\textstyle 3} (\frac{\textstyle N}{\textstyle V})\varepsilon _F .$

We would not expect $B$ to represent the total bulk modulus (ions contribute, too!), but the fact that this gives the right order of magnitude means that the electronic contribution is substantial.