Crystal Momentum in a Periodic Potential I'm working through some basic theory on periodic potentials, and I would appreciate help in understanding the crystal momentum. Suppose we have a Bravais lattice with lattice vectors $\textbf{R}$. There is an associated reciprocal lattice with lattice vectors $\textbf{K}$ such that $\textbf{K} \cdot \textbf{R} = 2\pi n$ for $n \in \mathbb{Z}$. The relationship between these two lattices ensures that plane waves of the form $e^{i \textbf{K} \cdot \textbf{r}}$ are periodic in the direct lattice. A consequence of Bloch's theorem is that the states $
\langle x|\psi \rangle$ of a particle assume the form
$$
\psi_{n\textbf{k}}(\textbf{r}) = e^{i \textbf{k} \cdot \textbf{r}}u(\textbf{r}),
$$
where
$$
u (\textbf{r} + \textbf{R}) = u(\textbf{r}).
$$
For these wavefunctions, $\textbf{p} \equiv \hbar \textbf{k}$ is defined to be the crystal momentum. Canonical momentum is ill-defined for this problem since the crystal breaks translation symmetry. However, for any translation $T_{\textbf{R}}$ within a lattice vector, $[H, T_{\textbf{R}}] = 0$. My questions are:


*

*In the first equation, I currently believe that $\textbf{k}$ can be any vector, and is not necessarily in the set of reciprocal wave vectors (i.e., $\textbf{k} \notin \{\textbf{K}\}$ necessarily). Since this is true, what is $\psi_{n\textbf{k}+\textbf{K}}?$

*Suppose a particle has crystal momentum $\textbf{p} = \hbar \textbf{k}$. How do we interpret $\textbf{p}' = \hbar (\textbf{k} + \textbf{K})$?

*Although there is no continuous symmetry in the lattice, there is a discrete symmetry of the potential $U(\textbf{r} + \textbf{R}) = U(\textbf{r})$, and therefore of the Hamiltonian. If Noether's theorem does not apply here, what quantity is "conserved" in time, and how do we justify such a conservation in general?
 A: (1) Since $u(\textbf{r}) = u(\textbf{r}+\textbf{R})$, we can expand this part in terms of reciprocal lattice vectors, $u_k(\textbf{r}) = \sum_\textbf{G}{e^{i\textbf{G}\cdot \textbf{r}}u_\textbf{k-G}}$. We can therefore write:
\begin{equation}
\psi_{\textbf k+\textbf K} = e^{i(\textbf k + \textbf K)\cdot \textbf r}\sum_\textbf{G'}{e^{i\textbf{G'}\cdot \textbf{r}}u_{\textbf k-\textbf K- \textbf G'}} = e^{i\textbf k \cdot \textbf r}\sum_\textbf{G'}{e^{i(\textbf{G'}+\textbf K)\cdot \textbf{r}}u_{\textbf k-\textbf K- \textbf G'}}=e^{i\textbf k \cdot \textbf r}\sum_\textbf{G}{e^{i\textbf{G}\cdot \textbf{r}}u_{\textbf k-\textbf G}} = \psi_\textbf k
\end{equation}
where $\textbf G = \textbf K+\textbf G'$.
(2) You can interpret $\textbf p'$ as being equal to $\textbf p$. This is true because the real space lattice is periodic; $\textbf k$ is always equal to $\textbf k + \textbf K$.
(3) The conserved quantity is $\textbf k$ $\textit mod$ $\textbf K$. You can see that I used this fact in the answer to (2).
You can read just about any solid state physics textbooks for complete justification though my personal favorite is Ziman's Theory of Solids.
