Eigenvalues of Hamiltonian with on-diagonal coordinate

A bit abstract, but if I take the standard graphene Hamiltonian (around the Dirac point) and introduce an on-diagonal term proportional to the coordinate $\hat{y}$, how would I find the eigenstates and eigenvalues of this Hamiltonian?

$$\hat{\mathcal{H}} = \begin{pmatrix} \hat{y} & \hat{p}_x-i\hat{p}_y\\ \hat{p}_x+i\hat{p}_y & \hat{y} \end{pmatrix}$$

where $\hat{y}$ and $\hat{p}_i$ are the normal quantum operators.

Multiplying out with the eigenstates $\psi = (\psi_A,\psi_B)^T$ we have two equations

$$\hat{y}\psi_A+(\hat{p}_x - i\hat{p}_y)\psi_B = \epsilon\psi_A$$ $$\hat{y}\psi_B+(\hat{p}_x + i\hat{p}_y)\psi_A = \epsilon\psi_B$$

How can I manipulate these to get an expression for $\epsilon$, the eigenfrequency? Under the influence of a magnetic field, or a strain field that resembles a psuedomagnetic field, off-diagonal terms proportional to coordinates ($x$ or $y$) are introduced, and this leads to Landau Levels etc. I'm wondering what happens when terms proportional to coordinates are on the on-diagonals.

• Concerning eigenvalues, $\lambda_\pm = y \pm \sqrt{p_x^2+p_y^2}$ since $\lambda_++\lambda_- = tr({\cal H}) = 2y$ and $\lambda_+\lambda_- = \det{\cal H} = y^2 -(p_x^2+p_y^2)$. – Valter Moretti Feb 16 '16 at 12:00
• @ValterMoretti On the right hand side you have operators. Do you mean the eigenvalues of these operators? – Urgje Feb 16 '16 at 12:13
• What do I mean... good question... well under the influence of a strain the graphene Hamiltonian above can be modified such that there is a 'gauge field' which introduces terms on the off-diagonal which are of the form $x$ or $y$. This leads to Landau levels etc. I'm wondering what happens when the coordinate dependent terms are of the diagonals. – Tom Feb 16 '16 at 12:17
• Sorry, I took a mistake: I interpreted $p_x$ and $p_y$ as numbers! – Valter Moretti Feb 16 '16 at 12:28