I understand that the wave function $\psi$ in a lattice with periodic boundary conditions satisfies $$ \psi(\bf{r}) = \sum_{\bf{k}}c_{\bf{k}}e^{i\bf{k}\cdot\bf{r}}$$ After solving the Schrodinger equation for a periodic potential, we arrive at $$ \left ( \frac{\hbar k^2}{2m} - E \right )c_{\bf{k}} + \sum_{\bf{G}}U_{\bf{G}}c_{\bf{k} - \bf{G}} = 0 $$ where $U_{\bf{G}}$ is the coefficient in the Fourier expansion of the periodic potential in terms of reciprocal lattice vectors $\bf{G}$. Hence for every $k$ in the Brillouin zone, the $c_{\bf{k}}$ are determined only by the $c_{\bf{k} - \bf{G}}$ summing over the $\bf{G}$. Apparently this implies that the wavefunction contains non-vanishing terms with these coefficients and therefore can be written as : $$ \psi_\bf{k}(\bf{r}) = \sum_{\bf{G}}c_{\bf{k} - \bf{G}}e^{i(\bf{k}-\bf{G})\cdot\bf{r}}$$ I cannot understand how $\psi_{\bf{k}} = c_{\bf{k}}e^{i\bf{k}\cdot\bf{r}}$ could be equal to the above under those conditions.

  • $\begingroup$ Have you followed the derivation all the way through? Particularly, do you know how you get the resulting wave function from that form of the SE (second in your question)? How about how you go from the first equation to the second? Do you understand why you can separate each part of the summation over $\mathbf{k}$ to give $N$ independent equations? $\endgroup$ Sep 15, 2016 at 15:34
  • $\begingroup$ Ashcroft and Mermin have a nice derivation (albeit a little hard to follow what with all the changes of variable - 4!), pp. 137-139 I think. $\endgroup$ Sep 15, 2016 at 15:40
  • $\begingroup$ Thank you for your response. Yes, I have read the whole derivation and thought about it for some time. I have also googled quite a bit and now I think I have everything sorted. Thank you for your suggestion. If something seems amiss, I'll check it out. $\endgroup$
    – Kaitou1412
    Sep 15, 2016 at 21:03
  • 2
    $\begingroup$ Does this answer your question? How can one write the electron wavefunction in a periodic lattice as $\psi(x)=\sum_GC_{k-G}e^{i(k-G)x}?$ $\endgroup$
    – Rishi
    May 19, 2021 at 18:15

1 Answer 1


Due to the periodic nature of the lattice any field that exists as a solution of the Schrödinger equation can be written as a superposition of Bloch modes. These Bloch modes (your last expression) would have a kind of periodicity that is related to the periodic lattice. If one would shift such a mode by an amount given by integer multiples of the lattice spacings one would end up with the same function times some phase. If one only had the original field being strictly periodic, it would be possible to express it simply as a Fourier series with frequency components denoted by the $\mathbf{G}$s, but due to this extra phase factor one can have an arbitrary uniform shift in the components of the Fourier series. This shift is represented by the Bloch vector $\mathbf{k}$ in your last expression.

  • $\begingroup$ Thank you for your response. I still do not exactly understand why the relation given between the $c_{\bf{k}}$ and $c_{\bf{k-G}}$ would imply that the wavefunction needs to be periodic in the reciprocal lattice vectors although I can grasp the intuition behind it. $\endgroup$
    – Kaitou1412
    Sep 14, 2016 at 19:28
  • $\begingroup$ It comes down to the Floquet theory (see: en.wikipedia.org/wiki/Floquet_theory), which tells us that the field within such a periodic structure also needs to be periodic apart from some phase shift. $\endgroup$ Sep 15, 2016 at 4:05

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