# Hydrogen atom 1/r^2 expectation value

1. Sep 12, 2009

### Unkraut

1. The problem statement, all variables and given/known data
Using the Feynman-Hellman theorem, determine the expectation values of 1/r and 1/r^2 for the hydrogen atom.

2. Relevant equations
Hamiltonian: $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}$$
energy eigenvalues: $$E_n=-\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2(N+l+1)^2}$$
N is the largest power of the Laguerre polynomial associated with the solution of $u_{nl}$ (and I have actually no idea what that is and I hope I don't need it), l is the azimuthal quantum number and n the principal quantum number.
n=N+l+1

Feynman-Gellman theorem: $$\frac{\partial E_n}{\partial \lambda}=\langle \psi_n | \frac{\partial H}{\partial \lambda} | \psi_n \rangle$$ for a Hamiltonian which depends on a parameter $$\lambda$$

3. The attempt at a solution
For <1/r^2>:
Take the derivative of the Hamiltonian with respect to l:
$$\frac{\partial H}{\partial l}=\frac{\hbar^2}{2m}\frac{2l+1}{r^2}$$
$$\Rightarrow \frac{1}{r^2}=\frac{2m}{\hbar^2(2l+1)}\frac{\partial H}{\partial l}$$
$$\Rightarrow \langle\frac{1}{r^2}\rangle=\frac{2m}{\hbar^2(2l+1)}\langle\frac{\partial H}{\partial l}\rangle$$
$$=\frac{2m}{\hbar^2(2l+1)}\frac{\partial E_n}{\partial l}$$ (Feynman-Gellman theorem used here)
$$=...$$ ... straightforward direct calculation which is missing now because texing is so tedious and I think maybe I have made an error in my notes which I must check later.

So, now I have the expectation value of 1/r^2.

But I can't do the same for 1/r. I cannot express 1/r in terms of the derivative of the Hamiltonian with respect to some parameter because in the term $$\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}$$ there is no parameter.
Seems I need the expectation value of $$\frac{d^2}{dr^2}$$, then I can calculate the expectation value of 1/r from the Hamiltonian. My uneducated intuition tells me that maybe that is zero, but that is not a good argument and often my intuition is wrong. I have no idea. Any hint appreciated.

2. Sep 12, 2009

### xepma

You're seeing too many boundaries ;)

Why wouldn't the electric charge serve as a parameter..?

3. Sep 12, 2009

### Unkraut

Oh, now that you say it, that seems reasonable. Thank you!
In quantum mechanics I always have problems determining what kind of reasoning is actually reasonable :)