Explicit eigenvalues of matrix?

Question

Consider the matrix-valued operator

$$A = \begin{pmatrix} x & -\partial_x \\ \partial_x & -x \end{pmatrix}.$$

I am wondering if one can explicitly compute the eigenfunctions of that object on the space $L^2(\mathbb R)$?

Answer 1

First some heuristics, before constructing the complete answer - this looks a bit more transparent if one considers $$ A^2 = \begin{pmatrix} -\partial_{x}^{2} +x^2 & 1 \\ 1 & -\partial_{x}^{2} +x^2 \end{pmatrix} $$ Then, denoting the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial_{x}^{2} +x^2 $ with eigenvalues $\lambda_{n} =2n+1$) as $\psi_{n} (x)$, $A^2 $ has the eigenvectors $$ \begin{pmatrix} \psi_{n} (x) \\ \psi_{n} (x) \end{pmatrix} \ \ \ \mbox{and} \ \ \ \begin{pmatrix} \psi_{n} (x) \\ -\psi_{n} (x) \end{pmatrix} $$ with eigenvalues $\lambda_{n} +1$ and $\lambda_{n} -1$, respectively. The eigenvalues of $A$ are the square roots of the aforementioned, but this doesn't yet directly yield the eigenvectors of $A$ - some more algebra is needed.

However, with these preliminaries, it now becomes apparent how the eigenvectors of $A$ are structured: Introduce 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} $. $A$ takes the form $$ A=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & a^{\dagger } -a \\ -a^{\dagger } +a & -a^{\dagger } -a \end{pmatrix} $$ One immediately has the state with zero eigenvalue, $(\psi_{0} , -\psi_{0} )$. In addition, the action of $A$ on the other eigenstates of $A^2 $ constructed previously above is $$ A\begin{pmatrix} \psi_{n} \\ \psi_{n} \end{pmatrix} =\sqrt{2n+2} \begin{pmatrix} \psi_{n+1} \\ -\psi_{n+1} \end{pmatrix} \ \ \ \mbox{and} \ \ \ A\begin{pmatrix} \psi_{n+1} \\ -\psi_{n+1} \end{pmatrix} =\sqrt{2n+2} \begin{pmatrix} \psi_{n} \\ \psi_{n} \end{pmatrix} $$ and therefore all that remains is to form the right linear combinations of these doublets: $$ A\begin{pmatrix} \psi_{n} + \psi_{n+1} \\ \psi_{n} - \psi_{n+1} \end{pmatrix} =\sqrt{2n+2} \begin{pmatrix} \psi_{n} + \psi_{n+1} \\ \psi_{n} - \psi_{n+1} \end{pmatrix} $$ and $$ A\begin{pmatrix} \psi_{n} - \psi_{n+1} \\ \psi_{n} + \psi_{n+1} \end{pmatrix} =-\sqrt{2n+2} \begin{pmatrix} \psi_{n} - \psi_{n+1} \\ \psi_{n} + \psi_{n+1} \end{pmatrix} $$ So the doubly degenerate eigenvalues $2n+2$ of $A^2 $ split up into separate eigenvalues $\pm \sqrt{2n+2} $ of $A$.