Diagonalization of a specific Dirac operator

Question

A few hours ago, a question was posed asking for the eigenvalues and eigenvectors of the Dirac operator $$ H=\begin{pmatrix} x & 0 & -i\partial_{x} & \bar{z} \\ 0 & x & z & i\partial_{x} \\ -i\partial_{x} & \bar{z} & -x & 0 \\ z & i\partial_{x} & 0 & -x \end{pmatrix} $$ The question was deleted just as I was about to post the complete answer. Since it seems a shame to waste the effort, I record the question along with the answer here.

Answer 1

Denote the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\lambda_{n} =2n+1$) as $\psi_{n} (x)$. Introduce also the standard raising and lowering operators $a^{\dagger } $, $a$, in terms of which $x=(a^{\dagger } +a)/\sqrt{2} $ and $-\partial_{x} =(a^{\dagger } -a)/\sqrt{2} $. Acting on the $\psi_{n} $, these act as $a\psi_{n} = \sqrt{n} \psi_{n-1} $, $a^{\dagger } \psi_{n} = \sqrt{n+1} \psi_{n+1} $. Now, $H$ takes the form $$ H=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & 0 & i(a^{\dagger } -a) &\sqrt{2} \bar{z} \\ 0 & a^{\dagger } +a & \sqrt{2} z & -i(a^{\dagger } -a) \\ i(a^{\dagger } -a) &\sqrt{2} \bar{z} & -(a^{\dagger } +a ) & 0 \\ \sqrt{2} z & -i(a^{\dagger } -a) & 0 & -(a^{\dagger } +a) \end{pmatrix} $$ One can readily verify that, in the following basis, which couples the $n$-th and $n+1$-th harmonic oscillator eigenfunctions, $$ e_1^n = \begin{pmatrix} \psi_{n} + \psi_{n+1} \\ 0 \\ -i(\psi_{n} - \psi_{n+1} )\\ 0 \end{pmatrix} \ \ \ \ \ e_2^n = \begin{pmatrix} \psi_{n} - \psi_{n+1} \\ 0 \\ -i(\psi_{n} + \psi_{n+1} )\\ 0 \end{pmatrix} $$ $$ e_3^n = \begin{pmatrix} 0 \\ \psi_{n} + \psi_{n+1} \\ 0 \\ i(\psi_{n} - \psi_{n+1} ) \end{pmatrix} \ \ \ \ \ e_4^n = \begin{pmatrix} 0 \\ \psi_{n} - \psi_{n+1} \\ 0 \\ i(\psi_{n} + \psi_{n+1} ) \end{pmatrix} $$ $H$ becomes block-diagonal, with a separate $4\times 4$ block for each integer $n\ge 0$ (in addition, at the lower end of the spectrum, this is supplemented by one more two-dimensional subspace involving only $\psi_{0} $, given further below); these $4\times 4$ blocks have the form $$ H_n = \begin{pmatrix} \sqrt{2n+2} & 0 & 0 & i\bar{z} \\ 0 & -\sqrt{2n+2} & i\bar{z} & 0 \\0 & -iz & \sqrt{2n+2} & 0 \\ -iz & 0 & 0 & -\sqrt{2n+2} \end{pmatrix} $$ They are readily diagonalized, yielding two degenerate doublets, $$ i (-\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_1^n + \bar{z} e_4^n \ \ \ \ \mbox{with eigenvalue} \ \ \ \ -\sqrt{2n+2+\bar{z}z} $$ $$ i (\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_2^n + \bar{z} e_3^n \ \ \ \ \mbox{with eigenvalue} \ \ \ \ -\sqrt{2n+2+\bar{z}z} $$ $$ -i (\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_1^n + \bar{z} e_4^n \ \ \ \ \mbox{with eigenvalue} \ \ \ \ \sqrt{2n+2+\bar{z}z} $$ $$ -i (-\sqrt{2n+2} + \sqrt{2n+2+\bar{z}z}) e_2^n + \bar{z} e_3^n \ \ \ \ \mbox{with eigenvalue} \ \ \ \ \sqrt{2n+2+\bar{z}z} $$ Finally, as mentioned above, one has the additional two eigenvectors $$ \begin{pmatrix} i\sqrt{\bar{z}z} \ \psi_{0} \\ z \ \psi_{0} \\ -\sqrt{\bar{z}z} \ \psi_{0} \\ -iz \ \psi_{0} \end{pmatrix} \ \ \ \ \mbox{with eigenvalue} \ \ \ \ -\sqrt{\bar{z} z} $$ $$ \begin{pmatrix} -i\sqrt{\bar{z}z} \ \psi_{0} \\ z \ \psi_{0} \\ \sqrt{\bar{z}z} \ \psi_{0} \\ -iz \ \psi_{0} \end{pmatrix} \ \ \ \ \mbox{with eigenvalue} \ \ \ \ \sqrt{\bar{z} z} $$ that become zero modes for $z=0$ (rescaling of the vectors needed in that limit).