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.

  • 3
    $\begingroup$ Two words: Virial theorem. $\endgroup$ Sep 25, 2016 at 1:34
  • $\begingroup$ After trying it, this is most certainly the easiest way to do this $\endgroup$
    – Andrew H.
    Sep 25, 2016 at 3:51

3 Answers 3


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:

$$H + V = \frac{p^2}{2m} + \frac{e'^2}{r}$$


$$e'^2 = e^2 + g$$

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:

$$\left<nlm\right|\frac{1}{r}\left|nlm\right> = \frac{me^2}{\hbar^2 n^2}=\frac{1}{a_0 n^2}$$

  • $\begingroup$ Nice derivation. You should add the centrifugal term $L(L+1)/ (2mr^2)$ to the Hamiltonian. $\endgroup$
    – wcc
    Sep 30, 2018 at 3:29

That sum in your radial wavefunction is proportional to a generalized Laguerre polynomial $L_m^{(\alpha)}(x)$, and the more usual way to write the radial wavefunction is




is a dimensionless radial coordinate and


is a wavefunction normalization factor that makes

$$\int_0^\infty r^2 R_{nl}^*(r)R_{nl}(r)dr=1.$$

The expectation value you want is thus

$$\begin{align} \left\langle\frac1r\right\rangle&=\int_0^\infty rR_{nl}^*(r)R_{nl}(r)dr\\ &=N_{nl}^2\left(\frac{na_0}{2Z}\right)^2\int_0^\infty\rho^{2l+1}[L_{n-l-1}^{(2l+1)}(\rho)]^2e^{-\rho}d\rho\\ &=\frac{2Z}{na_0}\frac{(n-l-1)!}{2n(n+l)!}\int_0^\infty\rho^{2l+1}[L_{n-l-1}^{(2l+1)}(\rho)]^2e^{-\rho}d\rho\\ &=\frac{Z}{n^2a_0}\frac{(n-l-1)!}{(n+l)!}\int_0^\infty\rho^{2l+1}[L_{n-l-1}^{(2l+1)}(\rho)]^2e^{-\rho}d\rho. \end{align}$$

Generalized Laguerre polynomials have been studied for about 150 years, and the integral above is very well known:

$$\int_0^\infty x^\alpha\,e^{-x}\,L_n^{(\alpha)}(x)\,L_m^{(\alpha)}(x)\,dx=\frac{(n+\alpha)!}{n!}\delta_{nm}.$$

Taking $n=m$ and substituting $\alpha\to 2l+1$ and $m\to n-l-1$, we find

$$\begin{align} \left\langle\frac1r\right\rangle&=\frac{Z}{n^2a_0}\frac{(n-l-1)!}{(n+l)!}\int_0^\infty\rho^{2l+1}[L_{n-l-1}^{(2l+1)}(\rho)]^2e^{-\rho}d\rho\\ &=\frac{Z}{n^2a_0}\frac{(n-l-1)!}{(n+l)!}\frac{(n+l)!}{(n-l-1)!}\\ &=\frac{Z}{n^2a_0}. \end{align}$$


Not a full answer but I think you can switch sigma and integral signs around.

Also look at Sums and Integrals

That is:

$${\displaystyle \int \sum _{r=a}^{b}f\left(r,x\right)\,dx=\sum _{r=a}^{b}\int f\left(r,x\right)\,dx}$$

This is simply because: \begin{align} \int \sum _{r=a}^{b}f(r,x)\,dx &=\int f\left(a,x\right)+f((a+1),x)+f((a+2),x)+\dots +f((b-1),x)+f(b,x)\,dx\\ &=\int f(a,x)\,dx+\int f((a+1),x)\,dx+\dots +\int f((b-1),x)\,dx+\int f(b,x)\,dx\\ &=\sum _{r=a}^{b}\int f(r,x)\,dx \end{align}


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