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

$$ \psi_{n\textbf{k}}(\textbf{r}) = e^{i \textbf{k} \cdot \textbf{r}}u(\textbf{r}), $$


$$ u (\textbf{r} + \textbf{R}) = u(\textbf{r}). $$

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:

  1. 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 $\psi_{n\textbf{k}+\textbf{K}}?$

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

  3. 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?

  • $\begingroup$ of the top of my head, $K$ seems like the kernel/null space of momentums, so $k+K$ momentum acting on a state should have same effect as $k$ momentum acting on a state (provided what is stated that $k \notin \{K\}$) $\endgroup$
    – Nikos M.
    Aug 9, 2014 at 22:44

1 Answer 1


(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.

  • $\begingroup$ Ah, thank you for clearing up my confusion! It makes sense that you must expand $\psi$ in k-space to get the equality. Could you possibly point me to a proof of how it follows that $\textbf{k}$ is the conserved quantity? I assume it probably consists of expressing the Hamiltonian in k-space as well, and showing that [H, k] = 0, but I could be wrong. $\endgroup$
    – Ultima
    Aug 10, 2014 at 5:19
  • $\begingroup$ Well it is not a conserved quantity per se (only $\textbf k mod \textbf K$). Anyway, Ashcroft and Mermin chapter 8 is pretty good on this front. $\endgroup$
    – Xcheckr
    Aug 10, 2014 at 6:22
  • 1
    $\begingroup$ That's what I thought. It's not a conserved quantity in time, just a quantity that's invariant under transformations of the form: $$ \textbf{k} \to \textbf{k} + \textbf{K} $$ $\endgroup$
    – Ultima
    Aug 10, 2014 at 6:32
  • $\begingroup$ @Ultima Why not consider the commutation relation $[H,T_{\textbf{R}}]=0$ plus the Heisenberg equation of motion to claim that $\textbf{k}$ is conserved in time? We can also assume $\textbf{k}$ to be the generator of $T_{\textbf{R}}$. $\endgroup$
    – SRS
    Aug 7, 2017 at 5:49

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