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?

## Best Answer

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.