Diagonalization of a hamiltonian for a quantum wire

[Lars Milz]

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)

H=\eta(k)τz+Bσ_x+αkσ_yτ_z+Δτ_x

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

Now the correct result is: 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}
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.

 

[krome]

I believe what you have here is tensor notation but with the tensor sign left out to reduce clutter. By the tensor A \otimes B of two matrices A,B, we mean at every entry of A, insert a copy of B multiplied by that entry in A. For example, i'll work out the first term in that Hamiltonian. I'll take the convention that the \sigma matrices come first followed by the \tau matrices and 1 as a matrix is the 2 \times 2 identity matrix:

\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}.

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

Using the standard Pauli matrices \sigma_x = \begin{pmatrix} 0 &amp; 1 \\ 1 &amp; 0 \end{pmatrix},<br /> \sigma_y = \begin{pmatrix} 0 &amp; -i \\ i &amp; 0 \end{pmatrix}, \sigma_z = \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; -1 \end{pmatrix}, i don't actually agree with the expression for energy you wrote down. When I plug the 4 \times 4 matrix H 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 \pm \sqrt{\phantom{a}} term in E^2.

I know you're getting this from a paper. After a quick scan, the most similar expression for H I can find is equation 3 on page 4. The expression for E^2 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.