- #1
EmilyRuck
- 136
- 6
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 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
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