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The extended operator can be treated along similar lines as the $c=0$ case. One merely has to modify the algebra a little. Again, 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, $A$ takes the form $$ A=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & a^{\dagger } -a \\ -a^{\dagger } +a & -a^{\dagger } -a \end{pmatrix} + \begin{pmatrix} 0 & c \\ c & 0 \end{pmatrix} $$ One has the following eigenvectors/eigenvalues: $$ \begin{pmatrix} \psi_{0} \\ -\psi_{0} \end{pmatrix} \mbox{ with eigenvalue } \ \ \ -c $$ $$ \left[ \frac{c\pm \sqrt{c^2 +2n+2} }{\sqrt{2n+2} } \begin{pmatrix} \psi_{n} \\ \psi_{n} \end{pmatrix} + \begin{pmatrix} \psi_{n+1} \\ -\psi_{n+1} \end{pmatrix} \right] \ \ \ \mbox{with eigenvalue} \ \ \ \pm \sqrt{c^2 + 2n+2} $$ For $c=0$, the spectrum reduces to the one given in the answer to the previous question.