Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Fourier Transforming a HgTe 2D Hamiltonian

  1. Jun 2, 2017 #1
    Hi! I am currently trying to derive the Fourier transform of a 2D HgTe Hamiltonian, with [itex] k_x [/itex] PBC and vanishing boundary conditions in the y direction at 0 and L. Here is the Hamiltonian:

    [itex]
    H = \sum_{k}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_x} + \cos{k_y}])\sigma_z]\tilde{c_k}
    [/itex]

    Where the [itex]\sigma_i [/itex] are the Pauli matrices, and A,M, and B are material dependent parameters, and the [itex]\tilde{c_k}^{\dagger}[/itex] and [itex]\tilde{c_k}[/itex] are the Nambu spinors ([itex]\tilde{c_k} = (c_k,c^{\dagger}_k))[/itex] and the c's are the fermion creation/annihilation operators. So what I want to do is Fourier transform in the [itex] y[/itex] direction to real space, and leave the [itex] x[/itex] direction in momentum space. My attempt was to split the [itex] H[/itex] into two parts,

    [itex]
    H_x = \sum_{k_x}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + (M-4B+2B[\cos{k_x})\sigma_z]\tilde{c_k}
    [/itex]

    [itex]
    H_y = \sum_{k_y}\tilde{c_k}^{\dagger}[ A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_y}])\sigma_z]\tilde{c_k}
    [/itex]

    I took the Fourier transform into real space for [itex]H_y[/itex] and got this:

    [itex]
    H_y = \sum_i^L\tilde{c_i}^{\dagger}D\tilde{c_i} + \sum_i^{L-1}(\tilde{c_{i+1}}^{\dagger}T\tilde{c_i} + \tilde{c_i}^{\dagger}T^{\dagger}\tilde{c_{i+1}})
    [/itex]

    With the matrices being equal to
    [itex]
    D = (M-4B)\sigma_z, \quad\quad T = \left(\frac{B}{a^2}\sigma_z - \frac{A}{2ia}\sigma_y\right)
    [/itex]

    But now it doesn't make sense to add the two parts together...I was thinking that maybe I fourier transform the whole thing into real space, and then write it in Nambu notation and include boundary conditions on the matrix like for y direction j and x direction i [itex]c_i = c_L+1 [/itex] and [itex] c_{j-1} = 0, c_{L+1} = 0[/itex]. Could someone help point me in the right direction? I've worked on 1D matrices in PBC before but I'm not sure how it should look in 2D. Would the Nambu spinors be 4N length? And if I were to FT into frequency space in the y direction, would the same solution method apply?
     
    Last edited: Jun 2, 2017
  2. jcsd
  3. Jun 5, 2017 #2
    I figured it out in case anyone is curious. The trick was to Fourier transform all of the [itex] c_k[/itex] operators into position space, giving 2-D Nambu spinors (length 4N) of [itex] c_{i,j}[/itex] and Fourier transforming the [itex] k_y [/itex] into position space (the fourier transforms lead to Delta functions, which can be rewritten as c operators). Because the [itex] k_x[/itex] was PBC, I left that in momentum space, so the final product came out to

    [itex]
    H = \sum_{i,j}^{L}\sum_{k_x}[c_{i,j}^{\dagger}Dc_{i,j} + c_{i,j}^{\dagger}(A\sin{k_x}\sigma_x + 2B\cos{k_x}\sigma_z)c_{i,j}] + \sum_i^{L-1}(c_{i,j}^{\dagger}T_ic_{i,j} + c_{i,j}T^{\dagger}c_{i,j}^{\dagger})
    [/itex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted