Explicit eigenvalues of matrix? 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 $L^2(\mathbb R)$?
 A: 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 $A$.
