Is crystal momentum an operator? My teacher has for Bloch waves the notation $\langle \vec{r}|\vec{k} \rangle = e^{i\vec{k}\cdot \vec{r}}u_{\vec{k}}(r)$ and uses it consistently. However, does this not assume that there is an operator that has eigenstates $|\vec{k} \rangle$? If so, how would such an operator be defined?
 A: Sure, it is certainly possible to define a crystal momentum operator, although I haven't heard of people doing this.
You define it by saying that the eigenstates of this operator are Bloch states, and the eigenvalue of each Bloch state is its crystal momentum (translated into the first Brillouin zone). There is a unique linear operator that satisfies these specifications.
A: It turns out the Bloch states are eigenstates of the translational operator, $T(\vec{R}_{j})$, namely, $T(\vec{R}_{j})\left\vert\vec{k}\right\rangle=e^{i\vec{R}_{j}\cdot\vec{k}}\left\vert \vec{k}\right\rangle$, where $\vec{R}_{j}$'s are lattice vectors. The translation group element $T(\vec{R}_{j})$ has a unitary representation, say, $T(\vec{R}_{j})=e^{i\hat{\vec{K}}\cdot\vec{R}_{j}}$ with $\hat{\vec{K}}$ being hermitian. If we have $\hat{\vec{K}}\left\vert\vec{k}\right\rangle=\vec{k}\left\vert\vec{k}\right\rangle$, then this leads to $e^{i\hat{\vec{K}}\cdot\vec{R}_{j}}\left\vert\vec{k}\right\rangle=e^{i\vec{R}_{j}\cdot\vec{k}}\left\vert \vec{k}\right\rangle$ consistent with $T(\vec{R}_{j})$, namely, $T(\vec{R}_{j})\left\vert\vec{k}\right\rangle=e^{i\vec{R}_{j}\cdot\vec{k}}\left\vert \vec{k}\right\rangle$. Therefore, it seems that the crystal momentum operator is the generator $\hat{\vec{K}}$ of the translational group. Unfortunaltely, I don't know how to write $\hat{\vec{K}}$ in terms of more familiar expressions.
A: I was confused when I wrote that question... The answer, trivially, is that $|\vec{k} \rangle$ is just a state, not necessarily an eigenstate of any operator.
