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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?

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  • $\begingroup$ Have a look at this answer, which might help you understand more of it. $\endgroup$ – taper Dec 7 '15 at 5:58
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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.

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    $\begingroup$ This is correct, with the caveat that crystal momentum isn't uniquely defined; you can break that degeneracy by demanding that the expansion only uses momenta in a given Brillouin zone. The choice of the first Brillouin zone is relatively natural, but it is still an artificial imposition. $\endgroup$ – Emilio Pisanty Sep 21 '18 at 13:09
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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.

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  • $\begingroup$ "$|\psi⟩$ is an eigenstate of $e^{iR\hat p}$" does not imply that $|\psi⟩$ is an eigenstate of $\hat p$ if all you have is a discrete set of $R$s for which it holds: if you have a continuum then you can differentiate with respect to $R$, but that is not applicable here. The generator $\hat{\vec K}$ of the translational group is simply the canonical momentum $\hat{\vec p}$, but Bloch states are not eigenstates of it. Bloch states are indeed eigenvectors of $e^{i\hat{\vec K}\cdot\vec R_j}$, but you can't take a unique logarithm of that operator. $\endgroup$ – Emilio Pisanty Sep 21 '18 at 13:06
  • $\begingroup$ (Though not to be discouraging: this answer is mostly on the right track.) $\endgroup$ – Emilio Pisanty Sep 21 '18 at 15:17
  • $\begingroup$ Good point. That's why the logic order of the above statement is : "If" $\left\vert\vec{k}\right\rangle$ is already an eigenvector of $\hat{\vec{K}}$, "then" such state is also an eigenvector of $e^{i\hat{\vec{K}}\cdot\vec{R}_{j}}$. But I think you really point to the key point, the discreteness of $\vec{R}_{j}$. $\endgroup$ – wayer tue Sep 24 '18 at 2:58
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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.

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