Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.2.1.
[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]
The energy levels of hydrogen, when calculated from the Coulomb potential alone (ignoring various perturbations) depend only on the principal quantum number according to
The quantization arises entirely from the requirement that the radial function remain finite for large , and makes no mention of the angular quantum numbers and . Thus each energy level (each value of ) has a degeneracy of , with degenerate states for each . Each symmetry is associated with the conservation of some dynamical quantity, with the degeneracy in due to conservation of angular momentum.
Shankar points out that, in classical mechanics, any potential with a dependence conserves the Runge-Lenz vector, defined for the hydregen atom potential as
where I’ve used for the electron mass to avoid confusion with the quantum number .
Although it doesn’t make sense to talk about the orbit of the electron in quantum mechanics, classically we can see that the conservation of implies that the orbit is closed. We can see this as follows.
In the second line, we used the vector identity
Since we’re dealing with a bound state, must always remain finite, so it must have a maximum value. At this point , which means that there is no radial motion, which in turn means that all motion at that point must be perpendicular to . In other words
Also, from conservation of energy, we have
so at we have
Plugging these into 6, we get
Exactly the same argument applies to the case where is a minimum: again so and we end up with
If is conserved (constant), then it must be parallel or anti-parallel to both and , and the latter two vectors must therefore always have the same direction. In other words, the particle reaches its maximum (and minimum) distance always at the same point in its orbit, meaning that the orbit is closed.
In a general (elliptical) orbit, so . Since for a bound orbit, we therefore must have
This in turn implies that is anti-parallel to and parallel to .
For a circular orbit, both and are constant, so both the kinetic and potential energies are also constant. From the virial theorem, we know that, for
Thus from 12, we see that for a circular orbit.