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Berry curvature vanishes in TRS system
In spin 1/2 system with TR symmetry , the Berry curvature must vanish. Because Berry curvature is odd. How to prove it?
\begin{equation}
\langle\partial_{-k_x}u^{I}(-k)|\partial_{-k_y}u^{I}(-k)\rangle-...
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What is the logic connection between these two statements?
What is the connection between these two statements:
the berry curvature change sign under time-reversal operation
If the system has the time-reversal symmetry, then berry curvature is odd in k.
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