- #1

- 135

- 6

## Main Question or Discussion Point

Hello!

When computing the density of states of electrons in a lattice, a material with dimensions [itex]L_x[/itex], [itex]L_y[/itex], [itex]L_z[/itex] can be considered. The allowed [itex]\mathbf{k}[/itex] vectors will have components

[itex]k_x = \displaystyle \frac{\pi}{L_x}p[/itex]

[itex]k_y = \displaystyle \frac{\pi}{L_y}q[/itex]

[itex]k_z = \displaystyle \frac{\pi}{L_z}r[/itex]

with [itex]p, q, r \in \mathbb{Z}[/itex].

The only values of the wavevector [itex]\mathbf{k}[/itex] to be considered belong to just

Thank you anyway,

Emily

When computing the density of states of electrons in a lattice, a material with dimensions [itex]L_x[/itex], [itex]L_y[/itex], [itex]L_z[/itex] can be considered. The allowed [itex]\mathbf{k}[/itex] vectors will have components

[itex]k_x = \displaystyle \frac{\pi}{L_x}p[/itex]

[itex]k_y = \displaystyle \frac{\pi}{L_y}q[/itex]

[itex]k_z = \displaystyle \frac{\pi}{L_z}r[/itex]

with [itex]p, q, r \in \mathbb{Z}[/itex].

The only values of the wavevector [itex]\mathbf{k}[/itex] to be considered belong to just

*one*out of the 8 octants of the Fermi sphere. This is because the values of [itex]\mathbf{k}[/itex] in the remaining 7 octants are equivalent to the ones in the chosen octant. Why?Thank you anyway,

Emily