0
$\begingroup$

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 $k'$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.