# Proof of Bloch's Theorem

I'm reading the book Solid State Physics by Ashcroft and Mermin. In its second proof of Bloch's Theorem on p.137, the periodic potential $$U(\mathbf{r})$$ and the wavefunction $$\psi(\mathbf{r})$$ both have been decomposed into their Fourier components $$U(\mathbf{r})=\sum_{\mathbf{K}}U_{\mathbf{K}}e^{i\mathbf{K}\cdot \mathbf{r}}$$ and $$\psi(\mathbf{r})=\sum_{\mathbf{q}}c_{\mathbf{q}}e^{i\mathbf{q}\cdot \mathbf{r}}$$. By putting those two into the Schroedinger Equation, one arrives at,

$$\sum_{\mathbf{q}}e^{i\mathbf{q}\cdot\mathbf{r}}[(\frac{\hbar^2}{2m}q^2-\varepsilon)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}]=0,$$

where $$\mathbf{K}$$ is the reciprocal lattice vector and $$\mathbf{K}$$ has been renamed as $$\mathbf{K}'$$. Now, since the basis functions are orthogonal, their linear combination is zero if and only if the coefficients are zero themselves, so

$$(\frac{\hbar^2}{2m}q^2-\varepsilon)c_{\mathbf{q}}+\sum_{\mathbf{K'}}U_{\mathbf{K'}}c_{\mathbf{q}-\mathbf{K'}}=0.$$

My question is: why it is necessary to write $$\mathbf{q}=\mathbf{k}-\mathbf{K}$$, where $$\mathbf{K}$$ is a reciprocal lattice vector chosen so that $$\mathbf{k}$$ lies in the first Brillouin zone. Why is $$\mathbf{k}$$ in the first Brillouin zone rather than $$\mathbf{q}$$ if it is written in such way? What's the physical meaning/intuition behind such renaming process?

The book continues to state that we need to make another change of variable from $$\mathbf{K'}$$ to $$\mathbf{K'}-\mathbf{K}$$. Why is that? What's the physical meaning / intuition behind it? Sorry I'm totally confused here in this part.