12.3: Separation of Variables for a Central Potential; Cyclic Variables

Landau presents in some details the separation of variables method for a $1/r$ potential, interesting here because it results in equations you’ve met before—those arising in the standard quantum treatment of the hydrogen atom.

How do we make any progress with these formidable differential equations? One possibility is that some coordinates are cyclic, meaning that $q_{1}$ say, does not appear explicitly in the Hamiltonian—for example, an angle variable in a spherically symmetric field. Then we have immediately that the corresponding momentum, $p_{1}=\partial S / \partial q_{1}=\alpha_{1}$, a constant.

The Hamiltonian for a central potential is:

$$\begin{equation} H=\dfrac{1}{2 m}\left(p_{r}^{2}+\dfrac{p_{\theta}^{2}}{r^{2}}+\dfrac{p_{\phi}^{2}}{r^{2} \sin ^{2} \theta}\right)+V(r) \end{equation}$$

The Hamilton-Jacobi equation is therefore

$$\begin{equation} \dfrac{1}{2 m}\left(\dfrac{\partial S_{0}}{\partial r}\right)^{2}+V(r)+\dfrac{1}{2 m r^{2}}\left(\dfrac{\partial S_{0}}{\partial \theta}\right)^{2}+\dfrac{1}{2 m r^{2} \sin ^{2} \theta}\left(\dfrac{\partial S_{0}}{\partial \phi}\right)^{2}=E \end{equation}$$

The first thing to note is that $\phi$ is cyclic (it doesn't appear in the Hamiltonian), so we can immediately replace $\partial S_{0} / \partial \phi \text { with a constant } p_{\phi}$.

Then we have:

$$\begin{equation} \dfrac{1}{2 m}\left(\dfrac{\partial S_{0}}{\partial r}\right)^{2}+V(r)+\dfrac{1}{2 m r^{2}}\left[\left(\dfrac{\partial S_{0}}{\partial \theta}\right)^{2}+\dfrac{p_{\phi}^{2}}{\sin ^{2} \theta}\right]=E \end{equation}$$

Now we seek a solution of the form

$$\begin{equation} S_{0}(r, \theta, \phi)=S_{r}(r)+S_{\theta}(\theta)+p_{\phi} \phi \end{equation}$$

Substituting in the equation, notice that the expression in square brackets will become

$$\begin{equation} \left(\dfrac{\partial S_{\theta}}{\partial \theta}\right)^{2}+\dfrac{p_{\phi}^{2}}{\sin ^{2} \theta} \end{equation}$$

independent of $r$, but on multiplying the full equation by $r^{2}$, and staring at the result, we see that in fact it is purely a function of $r$. This means that it’s a constant, say

$$\begin{equation} \left(\dfrac{\partial S_{\theta}}{\partial \theta}\right)^{2}+\dfrac{p_{\phi}^{2}}{\sin ^{2} \theta}=\beta \end{equation}$$

and then

$$\begin{equation} \dfrac{1}{2 m}\left(\dfrac{\partial S_{r}}{\partial r}\right)^{2}+V(r)+\dfrac{\beta}{2 m r^{2}}=E \end{equation}$$

These first-order equations can then be solved, at least numerically (and of course exactly for some cases). Physically, $\beta=\ell^{2}, \quad \ell$ being the total angular momentum, and $E$ is the total energy.

Note: recall that in quantum mechanics, for example in solving the Schrödinger equation for the hydrogen atom, the separation of variables was achieved by writing the wave function as a product of functions belonging to the different variables. Here we use a sum—remember that the action corresponds closely to the phase of a quantum mechanical system, so a sum of actions is analogous to a product of wave functions.