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I'm following the Topological Condensed Matter online course by TU Delft. For systems with spin 1/2, time-reversal symmetry as $$ \mathcal T = i\sigma_y\mathcal K, $$ where $\sigma_y$ is the second Pauli matrix. A Hamiltonian with this kind of symmetry satisfies the equation $$ \mathcal H = \sigma_y \mathcal H^* \sigma_y, $$

Based on the authors, the following is an example of such Hamiltonian

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where all the entries are real and $i$ is imaginary. I tried to compute the above equation (in Mathematica) by redefining $\sigma_y$ to be $\sigma_y\otimes\text{I}$ and calculating $(\sigma_y\otimes\text{I}) \mathcal H^* (\sigma_y\otimes\text{I})$. I obtained

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which clearly is a different Hamiltonian from the original. I uploaded the Mathematica sheet to GitHub if someone wants to check it: https://github.com/ManuelGmBH/Time-Reversal-Hamiltonians.

The calculation should be straightforward, but I don't know why I don't obtain the desired result. Any help or comment is appreciated.

Manuel E
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  • Physically what degrees of freedom does $\sigma_y$ act on and what is the basis the Hamiltonian is written in (including the order of the basis vectors)? – By Symmetry Nov 09 '23 at 17:44
  • $\sigma_y$ is acting on the spin degree of freedom. I don't know on which basis the Hamiltonian is written; it is not mentioned in the course notes: https://topocondmat.org/w1_topointro/0d.html#Time-reversal-symmetry. They give an example with specific numbers, I just changed them by variables, i.e. a,b,c,etc. I don't see how the basis will change anything in the discussion, though. – Manuel E Nov 09 '23 at 19:51

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