# $T$-invariant Hamiltonians

If $T$ is time-reversal transformation $t\mapsto -t$, Why do $T$-invariant Bloch Hamiltonians obey $$H(-k) = T H(k) T^{-1}$$ and not $$H(k) = T H(k) T^{-1}$$ Somehow I understand the word "invariant" as being unchanged under a transformation. As the way an operator $O$ on a Hilbert space transforms as $O\mapsto P O P^{-1}$ where $P$ is the transformation, I don't see why it is usually the former and not the latter which you find in literature (for example Hasan and Kane 2010).

• Time-reversal is an anti-unitary operator (not linear). Does it answer your question? – Prof. Legolasov Jan 13 '15 at 4:47
• Hi, thanks for your comment. It doesn't answer my question yet, at least not explicitly so. What kind of condition is more important, from which the first equation in my original question stems? Is it the invariance of the absolute value of a transition amplitude? If so, why is the Hamiltonian called invariant? – PPR Jan 13 '15 at 10:53
• it is the same thing... – Phoenix87 Jan 16 '15 at 23:21
• Found a discussion that shows why this is true in Chapter 4 of a book by Bernevig: press.princeton.edu/titles/10039.html – PPR Jan 16 '15 at 23:49

It is the total $H = \sum_k H_k$ invariant under time reversal