I'm trying to derive $\left\langle \frac{1}{r} \right\rangle = \frac{1}{n^2 a_0}$ (where $a_0$ is the Bohr radius) for the $| nlm\rangle$ state of hydrogen.
I've separated the radial and angular parts of the hydrogen wavefunction and split up the integration to yield
\begin{align}\left\langle \frac{1}{r} \right\rangle &= \int R^*_{nl}(r) {Y^{m}_{l}}^*(\theta, \phi) \frac{1}{r} R_{nl}(r) Y_{l}^{m}(\theta, \phi) r^2 \sin \theta ~\mathrm dr ~\mathrm d\theta~\mathrm d\phi\\ &= \int {Y^{m}_{l}}^*(\theta, \phi) Y_{l}^{m}(\theta, \phi) \sin \theta ~\mathrm d\theta ~\mathrm d\phi \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\\ &= \int \limits_{0}^{\infty} r R^*_{nl}(r) R_{nl}(r) ~\mathrm dr\end{align}
But from here, I'm not sure how to continue. Since $R_{nl}(r)$ is defined as
$$R_{nl}(r) = \left( \frac{2 Z}{n a_0} r \right)^l \sum \limits_{k = 0}^{n – l – 1} a_k \left( \frac{2 Z}{n a_0} r \right)^k e^{-Zr/n a_0}$$
where
$$a_{k + 1} = \frac{k + l + 1 – n}{(k + 1)(k + 2l + 2)}a_k$$
How do I deal with the sum inside the integral (especially a sum that requires recursion to compute coefficients)?
Have looked this source so far.
Best Answer
While you should learn to compute the integrals, let me point out in this additional answer that one can compute this without evaluating any integrals.
The Hamiltonian is:
$$H = \frac{p^2}{2m} + \frac{e^2}{r}$$
the energy eigenvalue of the $\left|n,l,m\right>$ eigenstate is
$$E_{nlm} = \frac{m e^4}{2\hbar^2 n^2}\tag{1}$$
here I've written this explicitly in terms of the constants that appear in the Hamiltonian. If we add the term $$V = \frac{g}{r}$$ as a perturbation to the Hamiltonian, then we know from first order perturbation theory that the perturbed energy eigenvalues are given to first order in $g$ by:
$$E(g) = E_{nlm} + g \left<nlm\right|\frac{1}{r}\left|nlm\right>+\mathcal{O}(g^2)\tag{2}$$
We can also compute the perturbed energy directly by modifying the charge $e$. We can write:
$$H + V = \frac{p^2}{2m} + \frac{e'^2}{r}$$
with:
$$e'^2 = e^2 + g$$
From Eq. (1) it follows that:
$$E(g) = \frac{me'^4}{2\hbar^2} = E_{nlm} + g\frac{me^2}{\hbar^2 n^2} + \mathcal{O}(g^2)$$
Comparing this to Eq. (2) yields:
$$\left<nlm\right|\frac{1}{r}\left|nlm\right> = \frac{me^2}{\hbar^2 n^2}=\frac{1}{a_0 n^2}$$