In Bloch bands, when we consider the energy degeneracy caused by translation symmetry and spatial inversion symmetry, we use the standard procedure like in Bloch waves under space inversion (parity) operator to find the degenerate Bloch state.

Now we consider a different line of argument using a different version of the original Hamiltonian, which is the k-dependent version of the original Hamiltonian, written as

$ H_k = \frac{(p + \hbar k)^2}{2m} + U(r) $

In this representation, the Schrödinger equation can be written as

$ H_k | u_{nk}\rangle = \epsilon_{nk} | u_{nk} \rangle $

where the Bloch wave $| \psi_{nk} \rangle= e^{ikr} | u_{nk} \rangle $

But unlike the original Hamiltonian, when we perform spatial inversion on this Hamiltonian, namely

$ I H_k I^{-1} $

it actually change from $H_k$ to $H_{-k}$. But the eigenvalue of $H_{-k}$ is the same with $H_k$. (Just simple calculations) Indeed, by this line of argument it seems that we could still get the conclusion that there is a two-level degeneracy at opposite crystal momentum $k$. The question that I want to ask is that:

In general, when we say that the Hamiltonian has some kind of “symmetry”, we usually mean that the symmetry operator commutes with the Hamiltonian. Or written as:

$ U H U^{\dagger} = H $

I wonder when we say that the Hamiltonian is not changed under this transformation, do we really mean that the $form$ of the Hamiltonian is not changed, or the eigenvalue of the Hamiltonian is not changed? (As the above example) If it is the former, how to interpret the above example?


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