Diagonalizing Van der Waals Hamiltonian

Question

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?

Answer 1

We consider any potential of the form \begin{align} V( x) = K_{11}x_1^2 + 2K_{12} x_1x_2 + K_{22}x_2^2. \end{align} The key to "diagonalization" of this potential is to note that such a potential is quadratic in the positions $x_1$ and $x_2$, and can therefore be written in matrix notation as follows: \begin{align} V( x) = \underbrace{(x_1\, x_2)}_{ x^t}\underbrace{\begin{pmatrix} K_{11} & K_{12} \\ K_{12} & K_{22} \\ \end{pmatrix}}_{K} \underbrace{\begin{pmatrix} x_1 \\ x_2 \\ \end{pmatrix}}_{ x} \end{align} where a superscript $t$ denotes transpose, and $ x = (x_1, x_2)^t$ is considered a column vector. Now, when we want to "diagonalize" the potential, we simply need to diagonalize the matrix $K$ (assuming that it is diagonalizable). In other words, we find a matrix $Q$ for which \begin{align} K = Q^{-1}DQ \end{align} where $D$ is a diagonal matrix. In the case at hand, notice that $K$ is a real, symmetric matrix. It is a mathematical fact that any such matrix admits an orthogonal diagonalization, namely that the matrix $Q$ can be chosen orthogonal which means $Q^{-1} = Q^t$. It follows that we can write \begin{align} K = Q^tDQ \end{align} where $D$ is diagonal with the eigenvalues $\lambda_1, \lambda_2$ of $K$ on its diagonal so that \begin{align} V( x) = x^t Q^t D Q x = (Q x)^tD(Q x). \end{align} Now we simply note that if we define a new vector \begin{align} \xi =Q x, \end{align} then the potential becomes \begin{align} V(x) = \xi^tD\xi = \lambda_1\xi_1^2 + \lambda_2\xi_2^2, \end{align} as desired.