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The material system is described by the Hamiltonian and overlap matrice, with the Gaussian basis set.

At Gamma point in the reciprocal space, the upper triangle block and the lower triangle block in the Hamiltonian matrix are symmetric to each other.

What about other k point? At other k points in the reciprocal space, are the upper and lower triangle block in the Hamiltonian still symmetric to each other or not?

Would anyone give me some suggestions?

Thank you in advance.

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    $\begingroup$ +1 but please proofread your post. For example even your first sentence has errors which don't look good when trying to attract the best scientists to answer our questions here. A more descriptive title would also be appreciated! $\endgroup$ Aug 17, 2023 at 2:42
  • $\begingroup$ @Nike Thank you for your suggestion. I modified my title. Is it fine now? If not, would you please recommend some explanation files to show how to write the post title in the proper way? $\endgroup$
    – Kieran
    Aug 17, 2023 at 2:52
  • $\begingroup$ Please read my comment again. I would appreciate that. $\endgroup$ Aug 17, 2023 at 2:55
  • $\begingroup$ @Nike Sorry for the error. I have corrected it. $\endgroup$
    – Kieran
    Aug 17, 2023 at 3:08
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    $\begingroup$ At all k-points the Hamiltonian is Hermitian. This reduces to symmetric when the Hamiltonian is purely real, such as happens at the gamma point (assuming Ising spins) $\endgroup$
    – Ian Bush
    Aug 17, 2023 at 6:23

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The simplest way of putting this is to note that the Hamiltonian is an operator corresponding to an observable (the energy) and hence by the postulates of quantum mechanics must be Hermitian. In the case of a real Hamiltonian matrix, such as typically occurs at the $\Gamma$ point, this reduces to (real) symmetric. However at other k points this is not the usual case, and the matrix is (complex) Hermitian.

More strictly one should note that the full reciprocal space representation of the Hamiltonian doesn't contain merely one k point, but in fact is an infinite sized matrix that contains all the k points. However as the whole point of the introduction of k points is to symmetry adapt the basis to take advantage of the translational symmetry of the periodic system, this infinite matrix is block diagonal, with each block containing 1 k point. Thus we can consider each k point separately, and the argument in the first paragraph holds for each k point individually.

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  • $\begingroup$ Thank you for the explanation with detials. I understand now. $\endgroup$
    – Kieran
    Aug 23, 2023 at 1:51

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