Question 5.39: In Section 5.3.1 we put the electrons in a box with impenetr...

(a)

\psi\left(x+l_{x}, y, z\right)=\frac{1}{\sqrt{l_{x} l_{y} l_{z}}} e^{i\left[k_{x}\left(x+l_{x}\right)+k_{y} y+k_{z} z\right]}=\psi(x, y, z)=\frac{1}{\sqrt{l_{x} l_{y} l_{z}}} e^{i\left(k_{x} x+k_{y} y+k_{z} z\right)} \Rightarrow e^{i k_{x} l_{x}}=1 ,

\text { so } k_{x} l_{x}=0, \pm 2 \pi, \pm 4 \pi, \ldots=2 n_{x} \pi ; \text { likewise } k_{y} l_{y}=2 n_{y} \pi, k_{z} l_{z}=2 n_{z} \pi .

\Delta k_{x} \Delta k_{y} \Delta k_{z}=\frac{2 \pi}{l_{x}} \frac{2 \pi}{l_{y}} \frac{2 \pi}{l_{z}}=\frac{8 \pi^{3}}{V} .

(b) Because n_{x}, n_{y}, \text { and } n_{z} run negative as well as positive, we want the whole sphere, not just one octant:

\frac{4}{3} \pi k_{F}^{3}=\frac{8 \pi^{3}}{V} \frac{N d}{2} \Rightarrow k_{F}=\left(\frac{3 \pi^{2} N d}{V}\right)^{1 / 3}=\left(3 \rho \pi^{2}\right)^{1 / 3} .

where \rho \equiv N d / V (same as Equation 5.52).

k_{F}=\left(3 \rho \pi^{2}\right)^{1 / 3}              (5.52).

E_{F}=\frac{\hbar^{2}}{2 m} k_{F}^{2}=\frac{\hbar^{2}}{2 m}\left(3 \rho \pi^{2}\right)^{2 / 3}   (same as Equation 5.54).

E_{F}=\frac{\hbar^{2}}{2 m}\left(3 \rho \pi^{2}\right)^{2 / 3}      (5.54).

A shell of thickness dk contains a volume 4 \pi k^{2} d k of k-space, so the number of electron states in the shell is 2 \frac{4 \pi k^{2} d k}{8 \pi^{3} / V}=\frac{V}{\pi^{2}} k^{2} d k   (same as in the text), so E_{\text {tot }}=\frac{\hbar^{2}\left(3 \pi^{2} N d\right)^{5 / 3}}{10 \pi^{2} m} V^{-2 / 3}   (same as Equation 5.56).

E_{ tot }=\frac{\hbar^{2} V}{2 \pi^{2} m} \int_{0}^{k_{F}} k^{4} d k=\frac{\hbar^{2} k_{F}^{5} V}{10 \pi^{2} m}=\frac{\hbar^{2}\left(3 \pi^{2} N d\right)^{5 / 3}}{10 \pi^{2} m} V^{-2 / 3}            (5.56).

Each state occupies an 8-times larger volume of k-space (because of the 2’s in part (a)), but the k-space itself is 8 times larger (because negative n’s are allowed, so with periodic boundary conditions all 8 octants are accessible).