I'm currently working numerically with wave equations and I was wondering if one can always decouple two wave equations, with potentials, which are linearly coupled. The system I'm talking about is the following: $$ \Box \psi - V_{11}\psi = V_{12}\phi \\ \Box \phi - V_{22}\phi = V_{21}\psi $$ or more compactly $$ \Box\left(\begin{array}{c} \psi \\ \phi \end{array}\right) = \left(\begin{array}{c} V_{11} V_{12} \\ V_{21} V_{22} \end{array}\right) \cdot \left(\begin{array}{c} \psi \\ \phi \end{array}\right) \equiv V \left(\begin{array}{c} \psi \\ \phi \end{array}\right) $$ with $V_{ij} = V_{ij}(x)$.
If I try to diagonalize the RHS matrix, the eigenvalues and as a result the eigenvectors, are x-dependent, so its diagonalized form is : $$ D(x)=P(x)\cdot V\cdot P(x)^{-1} $$ where D is the eigenvalue matrix and P the corresponding eigenvector matrix. If we try to apply this transformation to the initial equation, we have $$ P(x)^{-1}\Box\left(\begin{array}{c} \psi \\ \phi \end{array}\right)=\underbrace{P(x)^{-1}\cdot V\cdot P(x)}_{D(x)}\cdot P(x)^{-1}\left(\begin{array}{c} \psi \\ \phi \end{array}\right)=D(x) P(x)^{-1}\left(\begin{array}{c} \psi \\ \phi \end{array}\right) $$
So, the RHS is just fine. Diagonal to the ''new functions'' $\left(\begin{array}{c} u \\ v \end{array}\right)=P(x)^{-1}\left(\begin{array}{c} \psi \\ \phi \end{array}\right)$. However, the LHS cannot be expressed directly into the terms of $P(x)^{-1}\left(\begin{array}{c} \psi \\ \phi \end{array}\right)$.
Also, if we did something like inserting a $P(x)P(x)^{-1}$ next to the $\Box$, then the operator is gone.
Is there anything that comes in mind to help?
