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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 $c_{\mathbf{k}+\mathbf{G}}$ ?

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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.

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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.

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  • $\begingroup$ 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

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