Principles of Quantum Mechanics · Problem 13.2.1

The Hydrogen Atom — R. Shankar

Physics — Quantum Mechanics · runge-lenz-vector

Problem

Let us see why the conservation of the Runge--Lenz vector

\mathbf{n}

implies closed orbits. (1) Express

\mathbf{n}

in terms of

\mathbf{r}

and

\mathbf{p}

alone (get rid of

\mathbf{l}

). (2) Since the particle is bound, it cannot escape to infinity. So, as we follow it from some arbitrary time onward, it must reach a point

\mathbf{r}_{\max}

where its distance from the origin stops growing. Show that

\mathbf{n} = \mathbf{r}_{\max}\left(2E + \frac{e^2}{r_{\max}}\right)

at this point. (Use the law of conservation of energy to eliminate

p^2

.) Show that, for similar reasons, if we wait some more, it will come to

\mathbf{r}_{\min}

, where

\mathbf{n} = \mathbf{r}_{\min}\left(2E + \frac{e^2}{r_{\min}}\right)

Thus

\mathbf{r}_{\max}

and

\mathbf{r}_{\min}

are parallel to each other and to

\mathbf{n}

. The conservation or constancy of

\mathbf{n}

implies that the maximum (minimum) separation is always reached at the same point

\mathbf{r}_{\max}(\mathbf{r}_{\min})

, i.e., the orbit is closed. In fact, all three vectors

\mathbf{r}_{\max}

\mathbf{r}_{\min}

, and

\mathbf{n}

are aligned with the major axis of the ellipse along which the particle moves;

\mathbf{n}

and

\mathbf{r}_{\min}

are parallel, while

\mathbf{n}

and

\mathbf{r}_{\max}

are antiparallel. (Why?) Convince yourself that for a circular orbit,

\mathbf{n}

must and does vanish.