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Diagonalization of a hamiltonian for a quantum wire

  1. Aug 7, 2013 #1
    I try diagonalize the Hamiltonian for a 1D wire with proximity-induced superconductivity. In the case without a superconductor is all fine. However, with a superconductor I don't get the correct result for the energy spectrum of the Hamiltonian (arxiv:1302.5433)

    [itex]H=\eta(k)τz+Bσ_x+αkσ_yτ_z+Δτ_x[/itex]

    Here σ and τ are the Pauli matrices for the spin and particle-hole space.

    Now the correct result is: [itex]E^2(k)=Δ^2+η^2(k)+B^2+(αk)^2 ± \sqrt{B^2Δ^2+η^2(k)B2+η^2(k)(αk)^2}[/itex]
    My problem is now that I don't know how I bring the Hamiltonian in the correct matrix form for the calculation of the eigenvalues. If i try it with the upper Hamiltonian I have completely wrong results for the energy spectrum. I believe my mistake is the interpretation of the Pauli matrices τ but I don't know how I can write the Hamiltonian in the form to get the correct eigenvalues.
     
  2. jcsd
  3. Aug 7, 2013 #2
    I believe what you have here is tensor notation but with the tensor sign left out to reduce clutter. By the tensor [itex]A \otimes B[/itex] of two matrices [itex]A,B[/itex], we mean at every entry of [itex]A[/itex], insert a copy of [itex]B[/itex] multiplied by that entry in [itex]A[/itex]. For example, i'll work out the first term in that Hamiltonian. I'll take the convention that the [itex]\sigma[/itex] matrices come first followed by the [itex]\tau[/itex] matrices and 1 as a matrix is the [itex]2 \times 2[/itex] identity matrix:

    [itex]\displaystyle \tau_z = 1 \otimes \tau_z = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} [/itex].

    Similarly, the second term is really [itex]\sigma_x = \sigma_x \otimes 1[/itex], the third is [itex]\sigma_y \otimes \tau_z[/itex] and so on.

    Using the standard Pauli matrices [itex]\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},
    \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}[/itex], i don't actually agree with the expression for energy you wrote down. When I plug the [itex]4 \times 4[/itex] matrix [itex]H[/itex] into Mathematica and ask for the eigenvalues I get the same expression you wrote down except with a factor of 2 in front of the [itex]\pm \sqrt{\phantom{a}}[/itex] term in [itex]E^2[/itex].

    I know you're getting this from a paper. After a quick scan, the most similar expression for [itex]H[/itex] I can find is equation 3 on page 4. The expression for [itex]E^2[/itex] in equation 4 on page 5 has the factor of 2 in front of the square root. However, admittedly, I haven't followed that calculation closely and I'm not entirely sure what all the symbols in equation 4 mean.
     
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