# Diagonalization of a hamiltonian for a quantum wire

1. Aug 7, 2013

### 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.

2. Aug 7, 2013

### 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 $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, 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.

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