I know that time-reversal symmetry requires the Bloch Hamiltonian $H(\textbf{k})$ to transform as:
$$
\Theta H(\textbf{k}) \Theta^{-1}=H(-\textbf{k})
$$
where $\Theta$ is the time-reversal operator.
Now, let me consider the Bloch Hamiltonian at some time-reversal invariant momenta (TRIM) $\Gamma_i$ (i.e. $-\Gamma_i=\Gamma_i + \textbf{G}$ for some reciprocal lattice vector $\textbf{G}$). Then, as I have seen in many places (including this paper by Fu & Kane, just below Eq.(2.2)), people would claim that
$$
\Theta H(\Gamma_i) \Theta^{-1}=H(-\Gamma_i)=H(\Gamma_i)
$$
and then the Kramers theorem can be applied to argue that the band is degenerate at TRIM points.
What confuses me here is that I don't know why $H(-\Gamma_i)=H(\Gamma_i)$. I am tempted to think that $H(\textbf{k}+\textbf{G})=H(\textbf{k})$, which would then imply $H(-\Gamma_i)=H(\Gamma_i+\textbf{G})=H(\Gamma_i)$, but I cannot prove this statement either.
People may say that $\textbf{k}$ and $\textbf{k+G}$ are the same point in the 1st BZ, so it is "obvious" that $H(\textbf{k}+\textbf{G})=H(\textbf{k})$. But I am skeptical. Well, I can show that $E(\textbf{k}+\textbf{G})=E(\textbf{k})$ for the folded band dispersion, but the band dispersion and the Bloch Hamiltonian are really two things.
Ultimately, if $H(\textbf{k})=H(\textbf{k}+\textbf{G})$, I want to see this from the basic definition of the Bloch Hamiltonian: $H(\textbf{k})=e^{-i\textbf{k}\cdot\textbf{r}}H e^{i\textbf{k}\cdot\textbf{r}}$. When I naively put $\textbf{k}+\textbf{G}$ into the definition, I don't see an immediate way to equate $H(\textbf{k})=H(\textbf{k}+\textbf{G})$.
It seems to be a trivial question but somehow I am stuck in it. Any comments on this issue would be greatly appreciated!
Best Answer
You seems to treat $k$ as a good quantum number, which is not the case for system with lattice (discrete translational symmetry). $k$ out of BZ makes no sense but artificially defined as something the same as their conterpart $k+G$ in BZ.
Consider free particle $$ H=\sum_{k\in R^3} c_k^\dagger \frac{k^2}{2m} c_k $$ You can see this as a system with only discrete translational symmetry $$ H=\sum_{k\in BZ} (\sum_{i\in Z^3}c_{k+G_i}^\dagger \frac{(k+G_i)^2}{2m} c_{k+G_i}) $$ and define for $k \in BZ$ $$ H_k=\sum_{i\in Z^3}c_{k+G_i}^\dagger \frac{(k+G_i)^2}{2m} c_{k+G_i} $$ define for $k \notin BZ$ $$ H_k=H_{k+G_i} $$ So, you fold the bands in BZ and it seems natural that $H_k$ is periodic with k.