Time reversal of scalar Bloch states

Question

There's a well-known result about time reversal symmetry in crystals that's important when studying the physics of topological insulators, but I am having trouble proving it. I'll start with notation. Recall that a Bloch state is a simultaneous eigenfunction of $\hat{H}$ and $\hat{D}_{\mathbf{R}}$, where $\hat{H}=H(\mathbf{r})=\frac{\hat{p}^2}{2m} + V_L(\mathbf{r})$ with $V_L(\mathbf{r})=V_L(\mathbf{r}+\mathbf{R})$ and $\hat{D}_{R}:\mathbf{r}\rightarrow\mathbf{r}+\mathbf{R}$. Specifically, $$\hat{H}\psi_{n\mathbf{k}}(\mathbf{r}) = \varepsilon_{n\mathbf{k}}\psi_{n\mathbf{k}}(\mathbf{r})$$ $$\hat{D}_{\mathbf{R}}\psi_{n\mathbf{k}}(\mathbf{r}) = \lambda_{\mathbf{k}}\psi_{n\mathbf{k}}(\mathbf{r})$$ $$\lambda_{\mathbf{k}}=e^{i\mathbf{k}\cdot{\mathbf{R}}}$$

In reality, electrons are spinors, hence the time reversal operator should be $\hat{\mathcal{T}}=c\sigma_yK$ with $|c|=1$ and $K$ denoting complex conjugation as shown by Wigner. But if we consider them as scalar particles, then instead we use $\hat{\mathcal{T}}=K$.

Now, here's what I haven't been able to prove: if $H$ is time-reversal symmetric (i.e., $\hat{\mathcal{T}}H\hat{\mathcal{T}}{}^{-1}=KHK=H$), then for all non-degenerate levels, $K\psi_{n\mathbf{k}}=\psi_{n,-\mathbf{k}}$. I'll share what I have so far.

Since $\hat{H}K\psi_{n\mathbf{k}} = K\hat{H}\psi_{n\mathbf{k}} = \varepsilon_{n\mathbf{k}}K\psi_{n\mathbf{k}}$, it follows that $K\psi_{n\mathbf{k}}$ is itself a Bloch state and can hence be labelled as $K\psi_{n\mathbf{k}}=\psi_{n'\mathbf{k}'}$ for some $n'$ and $\mathbf{k}'$. Furthermore, non-degeneracy implies that $n'=n$, or else a different band would have the same energy as the $n^{\text{th}}$ band at some other point in reciprocal space (I'm assuming this is what is meant by "degeneracy" in the context of Bloch states, but if not, please correct me). We now just need to show that $\mathbf{k}' = -\mathbf{k}$.

The usual approach I've seen in references (e.g., page 3 here and section 16.3 here) is to observe that $$\hat{D}_{\mathbf{R}}\psi_{n\mathbf{k}'}(\mathbf{r}) = \psi_{n\mathbf{k}'}(\mathbf{r}+\mathbf{R})= K\psi_{n\mathbf{k}}(\mathbf{r}+\mathbf{R})= K(\lambda_{\mathbf{k}}\psi_{n\mathbf{k}}(\mathbf{r}))=\lambda_{-\mathbf{k}}\psi_{n\mathbf{k}'}$$

This eigenvalue equation tells us that $\lambda_{\mathbf{k}'}=\lambda_{\mathbf{-k}}$. But how does this imply that $\mathbf{k}' = -\mathbf{k}$? The function $\lambda(\mathbf{k})=\lambda_{\mathbf{k}}$ is clearly not one-to-one; notice that for all $m \in \mathbb{Z}$,

$$\lambda\left(\mathbf{k}+\frac{2\pi m\mathbf{R}}{\mathbf{R}\cdot\mathbf{R}}\right)=\lambda(\mathbf{k})e^{i2\pi m} = \lambda(\mathbf{k})$$

Any help understanding this last step would be much appreciated. As a final note, I see that a similar post has been made before. However, I don't believe this is a duplicate, as I am specifically asking about the proof for the non-degenerate case.

Answer 1

Notice that the vectors $k$ belong to the reciprocal lattice and the label $k$ is by definition ambiguous: if you change $k$ by adding a generator of the reciprocal lattice, the wave function changes just due to a constant phase. Hence, the quantum state associated to the wavefunction is not affected and remains the same, since pure states are normalized vectors up to phases. As a matter of fact only a portion of the reciprocal lattice is sufficient to label the wavefunctions without redundancies: the Brillouin cell. All $k$ giving rise to the same eigenvalue of $D$ are here physically equivalent. In summary, the ambiguity you pointed out does not change the quantum state, so that you are authorised to assume that $k’=-k$. This "ambiguity" gives rise to well known physical consequences as the so-called umklapp processes.