D.83 Expectation powers of r for hydrogen

This note derives the expectation values of the powers of

The trickiest to derive is the expectation value of

Its energy eigenfunctions of given square and

where the

When this Hamiltonian is applied to an eigenfunction
dirty trick Hamiltonian

in which the angular
derivatives have been replaced by

The reason is that the angular derivatives are essentially the square angular momentum operator of chapter 4.2.3. Now, while in the hydrogen Hamiltonian the quantum number

(Yes, the eigenfunctions

So, if you can figure out how the dirty trick energy changes with

All other expectation values of

(D.61) |

Substituting

Note that for a sufficiently negative powers of

The remainder of this note derives the Kramers-Pasternack relation.
First note that the expectation values are defined as

When this integral is written in spherical coordinates, the integration of the square spherical harmonic over the angular coordinates produces one. So, the expectation value simplifies to

To simplify the notations, a nondimensional radial coordinate

To further shorten the notations, from now on the limits of integration and

Also note that the integrals are improper. It is to be assumed that the integrations are from a very small value of

According to derivation {D.15}, the function

where primes indicate derivatives with respect to

Since this makes

The idea is now to apply integration by parts on

Before embarking on this, first note that since

the latter from integration by parts, it follows that

This result will be used routinely in the manipulations below to reduce integrals of that form.

Now an obvious first integration by parts on

The first of the two integrals reduces to an expectation value of

In the final integral, according to the differential equation (D.62), the factor

and each of the terms is of the form (D.64), so you get

Plugging this into (D.65) and then equating that to
(D.63) produces the Kramers-Pasternack relation. It
also gives an additional right hand side

but that term becomes zero when the integration limits take their final values zero and infinity. In particular, the upper limit values always become zero in the limit of the upper bound going to infinity;

The above analysis is not valid when