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I have a question regarding to how to construct an operator from k.p Hamiltonian. Maybe there are some problems in my understanding, I hope you can point me out and correct my description if I made something wrong below.

We know that the k.p Hamiltonian can be obtained by taking the inner product of the basis function that belongs to the same irreducible representation for a given point group. Another way is to write down the tight binding model explicitly and do the expansion. For example, the Dirac Hamiltonian obtained by expanding the tight binding element in the high symmetry point in graphene. But for the former, I don't how to construct an operator explicitly. As I understand, for a tight binding Hamiltonian, we can monitor how the basis was transformed so that we can write down the explicit matrix representation of a certain symmetry operation. I want to split this into 2 sub-questions.

Like I said above, if the effective k.p Hamiltonian can be obtained by doing an expansion around the high symmetry point in the tight binding element, does it mean that the basis of the k.p Hamiltonian matrix obtained from the point group approach must correspond to some basis in my tight binding matrix? If yes, that means I can use the symmetry operator in my tight binding form to act on the effective k.p form. If no, how can we construct it explicitly?

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  • $\begingroup$ I do not understand the question, maybe you can clarify. The crystal symmetries should be present in both your k dot p Hamiltonian and your tight-binding model. $\endgroup$
    – B. Brekke
    Sep 8 at 9:01
  • $\begingroup$ @B.Brekke May be I ask in this way. Yes you are right, both tight binding and k dot p Hamiltonian both satisfy the crystal point group symmetry. And in principle the k.p form can be derive from tight binding by doing the expansion around some high symmetry point. But I encounter a problem in writing a symmetry operator based on the k.p form. I have a cubic crystal, it has a reflection symmetry along 111 direction lattice plane. I want to used the Eigenstate of this symmetry operator to transform the Hamiltonian into block diagonal form. S I’m wondering whether I made any mistake in my concept. $\endgroup$
    – JensenPang
    Sep 9 at 3:55
  • $\begingroup$ What does it mean to "write a symmetry operator based on the k.p form"? That is, what do you mean by symmetry operator? Is it the unitary operator that transforms a radial vector under your reflection symmetry? How is this symmetry operator related to your k dot p theory? $\endgroup$
    – B. Brekke
    Sep 9 at 7:44

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