How to prove Bloch function is periodic in reciprocal lattice?

I saw in some textbooks this formula: $$ \Psi_{\mathbf{k}} (\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k}+\mathbf{G}}e^{i(\mathbf{k}+\mathbf{G})\cdot \mathbf{r}} $$ which makes the statement of this question obvious. ($\mathbf{G}$ is reciprocal lattice vectors)

But I don't understand this formula. I know $$ \Psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(\mathbf{r}) $$ and $u_{\mathbf{k}}(\mathbf{r})$ is periodic function of lattice, therefore can be written in Fourier series: $$ u_{\mathbf{k}}(\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k},\mathbf{G}}e^{i\mathbf{G}\cdot\mathbf{r}} $$ Now I don't understand why $c_{\mathbf{k},\mathbf{G}}$ can be written as $c_{\mathbf{k}+\mathbf{G}}$ ?


Because the reciprocal lattice $G$-periodic, the state with a wave vector $k+G$ describes the same state as that of wave vector $k$. You can therefore reduce your study to the first Brillouin Zone ($-\pi<k\leq\pi$). This means that the coefficient in your Fourier expansion will only depend on where you are within this zone. You can add or subtract as many times $G$ as you like from your $k$ vector, and the result will stay the same. At least in simple descriptions where no further corrections make the Bloch theorem only an approximation.


because the index of summation only relates to G,you can forget about "k",and also k=G+k(that shows the transnational symmetry). and look here.

  • $\begingroup$ G=G+k ? Do you mean k = k+G ? $\endgroup$ – Tim Jan 26 '15 at 20:51
  • $\begingroup$ Also your argument is not true, a function of two arguments $c_{k,G}$ is not necessary to only depend on $k+G$ $\endgroup$ – Tim Jan 26 '15 at 20:54
  • $\begingroup$ yes,corrected my statement! $\endgroup$ – user71065 Jan 26 '15 at 20:54

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.