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Some references mentioned that the calculation of $Z_2$ topological invariant of a crystal can be greatly simplified if the crystal contains inversion symmetry. But it involves the calculation of the parity of the occupied band at a specific TRIM point. I have a very basic question about this:

If I have a tight binding Hamiltonian obtained by fitting the data from a first principles calculation, how can we calculate the parity of the occupied band at particular TRIM point?

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    $\begingroup$ +1 But I've made some edits which I think you should look over so that no one has to do that next time. Also the question of how "parity of a band" is defined ought to be a separate question, but you can add a link to that second question here if you wish. $\endgroup$ – Nike Dattani Mar 11 at 19:24
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    $\begingroup$ @NikeDattani Thank you for your help. I will open a new post to ask the second question. $\endgroup$ – JensenPang Mar 12 at 2:47
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Consider the energy eigenstate $|\psi_{n\mathbf{k}}\rangle$, where $n$ is the band you are interested in and $\mathbf{k}$ is wave vector of the TRIM point. Then, if the system has inversion symmetry, these energy eigenstates are also eigenstates of the parity operator $\hat{\pi}$. This means that to determine the parity of that state, you can calculate the expectation value of the parity operator with respect to the energy eigenstate:

$$ \langle\psi_{n\mathbf{k}}|\hat{\pi}|\psi_{n\mathbf{k}}\rangle $$

This can only be $\pm1$, and the parity is "even" for $+1$ and "odd" for $-1$.

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    $\begingroup$ Thank you for your answer. But given a Hamiltonian in terms of orbital basis, how can we construct this parity operator and calculate the parity ? $\endgroup$ – JensenPang Mar 12 at 13:35
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    $\begingroup$ The parity operator can be defined by its action on position eigenkets: $\hat{\pi}|\mathbf{r}\rangle=|-\mathbf{r}\rangle$. So I imagine that a way forward would be to expand the energy eigenstates in the position basis, and then apply the operator? $\endgroup$ – ProfM Mar 12 at 15:53
  • $\begingroup$ Thank you! noticed that $\endgroup$ – JensenPang Mar 13 at 15:47

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