Question

In a spinless system with time reversal symmetry, is $E_n(k)=E_n(-k)$ always true?

Answer 1

In general, we know that the time reversal operator ΘΘ acts on a Bloch state with band index nn (which is omitted in the above reference) and wave-vector kk (although here I am only treating the 1D case), ψnkψnk, as follows

Θψnk=ψ∗nkΘψnk=ψnk∗

Now, the translation operator T^RT^R acts as follows

T^Rψnk=eikRψnkT^Rψnk=eikRψnk

T^Rψ∗nk=ei(−k)Rψ∗nkT^Rψnk∗=ei(−k)Rψnk∗

Therefore, ψ∗nkψnk∗ corresponds to −k−k.

Now, the author says that, for nondegerate levels, we should identify ψ∗nk=ψn−kψnk∗=ψn−k. As I understand this, this is because we already know that for the nn-th band and at −k−k in the first Brillouin zone there is already the state ψn−kψn−k, so in order to not have more than one state there, we have to make the above identification.

The author then proceeds on by saying that for an (at least) doubly degenerate level, we cannot make the above identification and we have at least two states at (n,−k)(n,−k), namely ψn−kψn−k and ψ∗nkψnk∗ (again, correct me if I am not getting something right here).

Then the author says that if the system has time reversal symmetry, by [H^,Θ^]=0[H^,Θ^]=0, we have that

H^ψ∗nk=En(k)ψ∗nk.H^ψnk∗=En(k)ψnk∗.

So, in the nondegenerate case, this translates as

H^ψn−k=En(k)ψn−kH^ψn−k=En(k)ψn−k

so that

En(k)=En(−k).En(k)=En(−k).

While, for the doubly degenerate case we have that

ψ∗nk≠ψn−kψnk∗≠ψn−k

and they are distinct states. So, we just have H^ψ∗nk=En(k)ψ∗nkH^ψnk∗=En(k)ψnk∗ which implies that for each (n,k)(n,k) we have (at least) two states, namely we have ψnkψnk and ψ∗nkψnk∗ for each En(k)En(k).

I have three questions about this:

Is my above understanding correct? If so, then what restrictions does time reversal symmetry impose on the states? Because as I understand it, it only imposes restrictions on the band structure.

For the doubly degenerate case, even if ψ∗nk≠ψn−kψnk∗≠ψn−k, the state ψ∗nkψnk∗ still corresponds to −k−k and energy En(k)En(k), so could we again conclude that En(k)=En(−k)En(k)=En(−k)?

For the nondegenerate case, we can conclude that the Berry curvature is equal to zero. But what can we conclude about the Berry curvature in the non-degenerate case

Answer 2

yes
it's
true
becoz
the down
mentioned
will help you.....

ψ∗nk≠ψn−kψnk∗≠ψn−k, En(k)=En(−k)En(k)=En(−k)?