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-x\leftrightarrow y=-\langle\partial_{k_x}u^{II}(k)|\partial_{k_y}u^{II}(k)\rangle-x\leftrightarrow y \end{equation} \begin{equation} \langle\partial_{-k_x}u^{II}(-k)|\partial_{-k_y}u^{II}(-k)\rangle-x\leftrightarrow y=-\langle\partial_{k_x}u^{I}(k)|\partial_{k_y}u^{I}(k)\rangle-x\leftrightarrow y \end{equation} The equation above means $F(-k_x,-k_y)=-F(k_x,k_y) $ How do we prove the above equation ?
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$\begingroup$ See answer already given here: physics.stackexchange.com/questions/422788/… $\endgroup$– PPRCommented Aug 19, 2020 at 4:02
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1$\begingroup$ No, this is a different argument: the presence of time-reversal symmetry is enough to make the Chern number/relative index vanish, but it does not imply that the Berry curvature vanishes. In fact, the statement in the title is false: the presence of time-reversal symmetry makes Berry curvature an odd function. It does not imply that it vanishes pointwise. You need e. g. inversion symmetry in combination with TRS: the presence of inversion symmetry implies $F(-k) = + F(+k)$ and the only function that is odd and even at the same time is $F(k) = 0$. $\endgroup$– Max LeinCommented Sep 6, 2020 at 3:12
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