This is a question about the 'Second Proof of Bloch's Theorem' which can be found in chapter 8 of Solid State Physics by Ashcroft and Mermin. Alternatively a similar (one dimensional) version of the proof can be found at http://ph.qmul.ac.uk/~anthony/spfm/21.html

The Schrodinger equation is given by $$\left[-\frac{\hbar^2}{2m}\mathbf{\nabla}^2+U(\mathbf{r})\right]\psi=\epsilon\psi$$ We write the periodic potential as the Fourier series $$U(\mathbf{r})=\sum_{\mathbf{K}}U_{\mathbf{K}}e^{i\mathbf{K}.\mathbf{r}}$$ where $\mathbf{K}$ is a reciprocal lattice vector, whilst the eigenfunctions may be written as the plane wave expansion $$\psi(\mathbf{r})=\sum_{\mathbf{q}}c_{\mathbf{q}}e^{i\mathbf{q}\cdot\mathbf{r}}$$ Substitution yields the condition on the coefficients $$\left(\frac{\hbar^2}{2m}q^2-\epsilon\right)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}=0$$ This couples wavevectors that differ only by a reciprocal lattice vector. My question is why does this imply that $c_{\mathbf{q}}=0$ unless $\mathbf{q}=\mathbf{k},\mathbf{k}+\mathbf{K},...$ (with $\mathbf{k}$ in the first Brillouin zone) so that the eigenfunctions are given by $$\psi_{\mathbf{k}}=\sum_{\mathbf{K}}c_{\mathbf{k}-\mathbf{K}}e^{i(\mathbf{k}-\mathbf{K})\cdot\mathbf{r}}$$ I see no reason to suggest that the eigenfunctions aren't given by $$\psi(\mathbf{r})=\sum_{\mathbf{k}}\sum_{\mathbf{K}}c_{\mathbf{k}-\mathbf{K}}e^{i(\mathbf{k}-\mathbf{K})\cdot\mathbf{r}}$$ but this doesn't seem to explained in much detail in either of the above sources. Thanks!

  • $\begingroup$ Your last expression is not even well-defined. You have a fixed $\boldsymbol k$ on the left hand side, so how can you sum over it on the right hand side? $\endgroup$ – Ruben Verresen Jan 14 '17 at 19:30
  • $\begingroup$ True - corrected. $\endgroup$ – Watw Jan 14 '17 at 19:42
  • $\begingroup$ You can always take sums of eigenfunctions, and if these eigenfunctions had the same energy then the sum is still an energy eigenfunction. So the question is not 'why do eigenfunctions have to be of the form $\psi_{\mathbf k}$?' (i.e. they don't) but 'how can we choose a basis of eigenfunctions of the form $\psi_{\mathbf k}$?'. The answer for the latter question is given by the above (fourth) equation, which shows that the equation to be solved for a given $c_{\mathbf q}$ only involves components $c_\mathbf{q'}$ with $\mathbf{q-q'} = \mathbf{K}$. $\endgroup$ – Ruben Verresen Jan 14 '17 at 21:50

Substitution yields the condition on the coefficients $$\left(\frac{\hbar^2}{2m}q^2-\epsilon\right)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}=0$$

To understand the condition on the $\mathbf{q}$'s you have to make a step back in the derivation of that equation.



with the $\mathbf{q}$'s arbitrarily chosen. Substituting this ansatz in the Schrödinger equation and rearranging the terms yields


If the sets $\{\mathbf{q}\}$ and $\{\mathbf{q}+\mathbf{K'}\}$ do not coincide, the exponential functions in the two terms on the left hand side of the above equation are linearly independent, and the only way in which that equation can be satisfied is to have $c_\mathbf{q}=0$, that is, $\psi(\mathbf{r}) = 0$.

If, instead, the sets $\{\mathbf{q}\}$ and $\{\mathbf{q}+\mathbf{K'}\}$ are equal, that is, $\{\mathbf{q}\}$ is invariant for translations of reciprocal lattice vectors, you can relabel $\mathbf{q}+\mathbf{K'}$ as $\mathbf{q}$ and rewrite the second term as


and then collect the exponential $\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}}$ between the two terms to obtain


The above equation holds if and only if the coefficients of the exponentials $\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}}$ are zero, and this yields the required condition on the coefficients.

  • $\begingroup$ Unless I am missing something, this proves the condition on the coefficients but doesn't seem to answer the question... $\endgroup$ – Watw Jan 15 '17 at 16:37
  • $\begingroup$ @Watw You are missing the part "If the sets [...] do not coincide" ;-) $\endgroup$ – Massimo Ortolano Jan 15 '17 at 16:39
  • $\begingroup$ But don't they always coincide? $\endgroup$ – Watw Jan 15 '17 at 23:23
  • $\begingroup$ @Watw No, if you choose the q's arbitrarily. But if you choose the q's so that the set is not invariant under translations of reciprocal lattice vectors, there is no non-zero solution in that form, as I've shown in the answer. $\endgroup$ – Massimo Ortolano Jan 15 '17 at 23:29
  • $\begingroup$ But the set $\{\mathbf{q}\}$ is the set of wave vectors obeying periodic boundary conditions - there isn't a question of 'choosing' them is there? $\endgroup$ – Watw Jan 16 '17 at 11:03

Final equations are eigenvalue equations in the form $H'c= \epsilon c$ where $c$ is the column vector of $c_q$ and $H'$ is the matrix of the coefficients. Now the point is that $H'$ is block diagonal and each block corresponds to one $k$. That is why you can construct a wave function by only using plane waves in one block.


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.