Question

Show that for a square lattice in two dimensions the kinetic energy of a free electron at a
corner of the first Brillouin zone is higher than that of an electron at the midpoint of a
side face of the zone by a factor of 2.

Answer 1

Answer on Question #51240, Physics, Solid State Physics

Show that for a square lattice in two dimensions the kinetic energy of a free electron at a corner of the first Brillouin zone is higher than that of an electron at the midpoint of a side face of the zone by a factor of 2.

Solution:

The term “first zone” must be referring to the first Brillouin zone in $k$-space. The reciprocal lattice of a square space lattice is also a square lattice. For a real space (i.e., crystal) lattice of side $a$, the reciprocal lattice points are separated by steps of $2\pi / a$ in the $k_X$ and $k_Y$ directions. The Wigner-Seitz cells in reciprocal space are squares also. These have boundaries that bisect the segments connecting reciprocal lattice points. The first Brillouin zone is therefore the square $\left[-\frac{\pi}{a}; \frac{\pi}{a}\right] \times \left[-\frac{\pi}{a}; -\frac{\pi}{a}\right]$. A corner of the zone is $(k_X, k_Y) = \left(\pm \frac{\pi}{a}, \pm \frac{\pi}{a}\right)$. A free electron with one of these four wavevectors has kinetic energy $E_{CORNER} = \frac{\hbar^2}{2m} \vec{k}^2 = \frac{\hbar^2 \pi^2}{ma^2}$. The faces of the wave zone have one of $k_X, k_Y$ with magnitude $\pi / a$ and the other component zero, giving $E_{FACE} = \frac{\hbar^2}{2m} (\pi / a)^2 = E_{CORNER} / 2$.

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