In Kittel's Solid State Physics, he attempts to find the energy exchange due to the van der Waals interaction. He starts by writing the hamiltonian: two oscillators with coordinates $x_1$ and $x_2$ $$H_0 = \frac{1}{2m}p_1^2 + \frac{1}{2}C x_1^2 + \frac{1}{2m}p_2^2 + \frac{1}{2}C x_2^2$$ and an approximate coulomb interaction $$ H_1 \approx -\frac{2e^2 x_1 x_2}{R^3}.$$ He then "diagonalizes by the normal mode transformation" $$x_s = \frac{1}{\sqrt{2}}(x_1 + x_2)$$ $$x_a = \frac{1}{\sqrt{2}}(x_1 - x_2)$$ Intuitively, I understand what this transformation is. You construct a new, orthonormal basis. $x_s$ and $x_a$ oscillate independently of one another.
I would like to repeat this but with differing spring constants, $K_1$ and $K_2$. I would like to diagonalize with matrices, but the hamiltonian cannot be represented as such (due to the coupling term $x_1 x_2$). I have tried writing an arbitrary basis $x_1 = c_1 x_a + c_2 x_b$ and $x_2 = c_3 x_a + c_b$ subject to the constraint that the coupling terms cancel: $$k_1 c_1 c_2 + k_2 c_3 c_4 - (c_1 c_4 + c_2 c_3) = 0$$ but this doesn't feel like the right way to do it.
How would I diagonalize the above hamiltonian where the spring constants differ?