In every solid state physics book it says that the wave vectors appearing in Bloch's theorem can be confined to the first Brillouin zone and provide a hint on how to show this. Most of the times this hint is something like:

Assume $$k'= k+G$$ where $k$ belongs to the 1st BZ and $G$ is a reciprocal lattice vector. Use $$ \psi_k(r+R)= e^{ik \cdot R} \psi_k(r) \tag{1}$$ to show that if $(1)$ holds for $k'$ it also holds for $k$.

My attempt at a "solution/proof"

$$ \psi_{k'} (r+R)= e^{ik'(r+R)}=e^{i(k+G)(r+R)}= \psi_{k'}(r)e^{ik \cdot R}$$

But the last equation should be equal to (by Bloch's thm):

$$ \psi_{k'}(r) e^{ik' \cdot R}$$

And this should imply $$ k'=k$$?

Also what is a good way of understanding this geometrically? Does this mean that the $G$ vector that "connects" $k'$ and $k$ introduces a phase shift and that is the only difference between $k$ and $k'$?


Starting from $$ \psi_k(r+R) = e^{ikR}\psi_k(r) \quad (1)$$ you evaluate $$ \psi_{k'}(r+R) = \psi_{k+G}(r+R) = e^{i(k+G)R}\psi_{k+G}(r) = e^{iGR}e^{ikR}\psi_{k+G}(r).$$ Since $$ e^{iGR} = 1,$$ you find $$ \psi_{k'}(r+R) = e^{ikR}\psi_{k'}(r). \quad (2)$$ If you compare this result with eq. (1), you see that both $\psi_{k'}(r)$ and $\psi_k(r)$ are eigenfunctions of translations by $R$ with the same eigenvalue. We regard $k$ as a label for the class of wave functions solving Schrödinger's equation for a crystal with the translational property (1). The result (2) shows that the label $k$ is not unique, because also wave functions labeled with $k'=k+G$ have the same translational property. In order to have unique labels for classes of wave functions with the same translational property (1), we therefore choose $k$ by convention to be within the first Brillouin zone.

  • $\begingroup$ So this shows only that they are equivalent and we voluntarily choose the vectors in the first bz. Thanks! $\endgroup$ – Pablo Bähler Feb 13 at 9:26

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