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A How to get Peierls substitution in edge state?

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  1. Aug 13, 2017 #1

    haw

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    In paper PRL 101, 246807 (2008), authors use "Peierls substitution", that is ky -> -i y. As we know, ky is eigenvalue of translation operator in period potential, while -i y is momentum operator, it seems they are huge different. So I wonder how to get ""Peierls substitution" in strict math way?
     
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  3. Aug 15, 2017 #2

    king vitamin

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    No, [itex]k_y[/itex] is the eigenvalue of momentum [itex]p_y = -i \partial_y[/itex].

    The translation operator in the y direction is given by [itex]T_y(a) = e^{-i p_y a}[/itex].
     
  4. Aug 15, 2017 #3
    Peierls substitution is a way to couple a tight binding Hamiltonian to a external magnetic field within the lattice approximation. I see what you are referring to in the paper; they say that they use that substitution to say [itex] k_y \rightarrow \partial_y[/itex]. I think they might be misusing the term; When they move from PBC's to finite BC's along the y direction, [itex] k_y[/itex] is no longer a good quantum number. And in moving from a lattice model with momentum [itex] k_y[/itex] to a continuum model with crystal momentum [itex]\hbar k [/itex] they make the substitution [itex] k_y \rightarrow \hbar k_y[/itex] or [itex]\partial_y [/itex]. I'm not sure why they call it a Peierls substitution, it looks more like a substitution like lattice to continuum model. Here is a nice forum post about the math behind the Peierls substitution: https://physics.stackexchange.com/questions/178003/tight-binding-model-in-a-magnetic-field

    I've actually solved this model before, and the way to do it is to take the momentum space Hamiltonian and do a partial fourier transform along the y direction, so that in the final product you have a Hamiltonian that is PBC in the x direction but lattice model in the y direction.
     
  5. Aug 15, 2017 #4

    haw

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    Thanks for your help! Actually helpful.
     
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