Deriving Bloch's Theorem 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!
 A: 
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. 
Assume
$$\psi(\mathbf{r})=\sum_{\mathbf{q}}c_{\mathbf{q}}e^{i\mathbf{q}\cdot\mathbf{r}},$$
with the $\mathbf{q}$'s arbitrarily chosen. Substituting this ansatz in the Schrödinger equation and rearranging the terms yields
$$\sum_{\mathbf{q}}\left(\frac{\hbar^2}{2m}q^2-\epsilon\right)c_{\mathbf{q}}\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}}+\sum_{\mathbf{q},\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}}\mathrm{e}^{\mathrm{i}(\mathbf{q}+\mathbf{K'})\cdot\mathbf{r}}=0.$$
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
$$\sum_{\mathbf{q},\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}},$$
and then collect the exponential $\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}}$ between the two terms to obtain
$$\sum_{\mathbf{q}}\left[\left(\frac{\hbar^2}{2m}q^2-\epsilon\right)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}\right]\mathrm{e}^{\mathrm{i}\mathbf{q}\cdot\mathbf{r}}=0.$$
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.
A: I completely understand your confusion, because that was exactly mine too.
The logic is that, generally by the Schrodinger equation $\hat{H} \psi(x) = \epsilon \psi(x)$, the energy $\epsilon$ is determined by the full wave function $\psi(x)$. Similarly, if the potential is not periodic, in Fourier space we will need the full set of $c_\mathbf{q}$ (with $\mathbf{q}$ running through all the allowed values) to determine $\epsilon$.
However in the periodic potential case, the Schrodinger equation in terms of the Fourier coefficients
$$
\left(\frac{\hbar^2}{2m}q^2-\epsilon\right)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}=0
$$
tells us that knowing just the set of $c_\mathbf{q}$ with $\mathbf{q} = \ldots, \mathbf{k}, \mathbf{k} + \mathbf{K}, \ldots$ instead of the full set is enough to solve for $\epsilon$ (just imagine when the coefficients are all known, the only unknown is $\epsilon$). In other words, all the information in $\psi(x)$ is contained in this subset of $c_\mathbf{q}$, i.e. we can safely say that $c_\mathbf{q} = 0$ for any other values of $\mathbf{q}$. Every time we pick a different $\mathbf{k}$, the set $c_\mathbf{q}$ represents a different energy eigenfunction, so the eigenstates are labelled by $\mathbf{k}$
$$
\psi_{\mathbf{k}}=\sum_{\mathbf{K}}c_{\mathbf{k}-\mathbf{K}}e^{i(\mathbf{k}-\mathbf{K})\cdot\mathbf{r}}.
$$
This might be a late answer but I still hope it is helpful, and possibly to any other readers that are having the same confusion.
A: 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.  
