Empty lattice approximation for a body centered tetragonal crystal

\(\Large \frac{E}{\frac{\hbar^2}{2ma^2}}\)

a/c:   

$\large \vec{k}=u\vec{b}_1+v\vec{b}_2+w\vec{b}_3\,:\,(u,v,w)$

 Symmetry points $(u,v,w)$ $[k_x,k_y,k_z]$
 $\Gamma:\,(0,0,0)$  $[0,0,0]$
 $X:\,(\frac{1}{2},0,0)$  $[\frac{\pi}{a},\frac{\pi}{a},0]$ 
 $Z:\, (\frac{1}{2},\frac{1}{2},-\frac{1}{2})$  $[\frac{2\pi}{a},0,0]$ 
 $N:\, (0,\frac{1}{2},0)$  $[\frac{\pi}{a},0,\frac{\pi}{c}]$ 
 $P:\, (\frac{1}{4},\frac{1}{4},\frac{1}{4})$  $[\frac{\pi}{a},\frac{\pi}{a},\frac{\pi}{c}]$ 
 

$\overline{\Gamma X} = \frac{\sqrt{2}\pi}{a}$

$\overline{\Gamma Z} = \frac{2\pi}{a}$

$\overline{\Gamma N} = \frac{\pi}{ac}\sqrt{a^2+c^2}$

$\overline{\Gamma P} = \frac{\pi}{ac}\sqrt{a^2+2c^2}$

The real space and reciprocal space primitive translation vectors are:

$\large \vec{a}_1 = \frac{a}{2}(\hat{x}+\hat{y})-\frac{c}{2}\hat{z}$  $\large \vec{a}_2 = \frac{a}{2}(\hat{x}-\hat{y})+\frac{c}{2}\hat{z}$  $\large \vec{a}_3 = \frac{a}{2}(-\hat{x}+\hat{y})+\frac{c}{2}\hat{z}$,

$\large \vec{b}_1 = \frac{2\pi}{a}(\hat{k_x}+\hat{k_y})$  $\large \vec{b}_2 =\frac{2\pi}{a}\hat{k_x}+\frac{2\pi}{c}\hat{k_z}$  $\large \vec{b}_3 = \frac{2\pi}{a}\hat{k_y}+\frac{2\pi}{c}\hat{k_z}$.

The colors correspond to the bands. Each band has 2 electron states per unit cell of the crystal. If the number of valence electrons per unit cell is known, the bands can be filled to determine the Fermi energy. The colors also correspond to the Brillouin zones. Electron states in the same Brillouin zone have the same color. Where the bands have the same energy, they have been offset slightly for visibility.