Time reversal of Bloch Hamiltonian (at TRIM points)

Question

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: $H(textbf{k})=e^{-itextbf{k}cdottextbf{r}}H e^{itextbf{k}cdottextbf{r}}$. 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!

Shuangyuan Lu · 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
$$
H=sum_{kin R^3} c_k^dagger frac{k^2}{2m} c_k
$$
You can see this as a system with only discrete translational symmetry
$$
H=sum_{kin BZ} (sum_{iin Z^3}c_{k+G_i}^dagger frac{(k+G_i)^2}{2m} c_{k+G_i})
$$
and define for $k in BZ$
$$
H_k=sum_{iin Z^3}c_{k+G_i}^dagger frac{(k+G_i)^2}{2m} c_{k+G_i}
$$
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.

Nguyen D. H. Minh · Answer

In my opinion, you should see the problem in a different angle instead of following the paper's logical flow. Here, one only considers the first Brillouin zone and has that $Theta_kH(k)Theta^{-1}_k=H(-k)$ for all $k$. TRIM are points that satisfy $[H(k),Theta_k]=0$, which yields $-Gamma_i=Gamma_i$. However, since the Brillouin zone is defined in a torus, the author generalizes the result that $-Gamma_i=Gamma_i+mathbf{G}$.
Not only TRIM, one can also find inversion invariant momenta in the same manner.

Time reversal of Bloch Hamiltonian (at TRIM points)

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