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.


1 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.

  • $\begingroup$ Thanks for your answer. I'm still a little confused trying to understand this mathematically. After eqn 1.17 in this reference, the author mentions that for a reciprocal lattice vector G, two Bloch states |n, k> and |n, k+G> can differ only by a phase, as you also pointed out. How could we prove that? We would need to show that |n, k> and |n, k+G> satisfy the same Schrodinger equation, but I don't understand how we can demonstrate that. I'm guessing there is a simple linear algebra result that I'm not realizing applies here. $\endgroup$
    – Jamin
    Feb 6, 2021 at 18:58
  • $\begingroup$ It is just the Bloch theorem I think. The form of wavefunction is $e^{ikx}u(X)$, where is translationally invariant (with respect to the lattice). $k$ is such that the exponential changes with a constant phase when $x$ varies in the lattices. This condition defines the reciprocal lattice. $\endgroup$ Feb 6, 2021 at 19:08
  • $\begingroup$ Right, I realize my mistake now. Fundamentally, each Bloch state is labelled by the quantum number $\mathbf{k}$, but it is only defined modulo a shift by reciprocal lattice vectors. I guess I should go back to Ashcroft & Mermin to review, but hopefully my understanding is correct now. In that case, shouldn't we really be saying that $K\psi_{n\mathbf{k}} = e^{i\phi}\psi_{n,-\mathbf{k}}$ for some unknown phase factor? $\endgroup$
    – Jamin
    Feb 6, 2021 at 20:10
  • $\begingroup$ I think so. An issue would be if that phase depends on the eigenvector or not... $\endgroup$ Feb 6, 2021 at 20:25

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