Question about the radial hydrogen eigenfunctions

Question

When calculating the selection rules for electronic transition in the hydrogen atom in dipole approximation, we always focus on the angular integrals. But why the integral
$$
int_{0}^{infty}[rR_{nl}(r)]R_{n'l'}(r)r^{2}dr
$$
(where $nneq n'$ and $lneq l'$)
always gives a non-zero result, so we don't have any selection rule for the $n$ quantum number? Or, in other words, why can we say that $rR_{nl}(r)$ has always a non-zero component on the subspace generated by $R_{n'l'}(r)$?

MEisebitt · Answer

I would take a different pint of view to explain this:
Normally to calculate rates/probabilities for dipole transitions (things like spontaneous emission rates) we have to evaluate $langlepsi_b|textbf{r}|psi_arangle$. If you start to calculate those for e.g. hydrogen you will notice that many integrals evaluate to zero. Of course, it would be helpful to somewhat shorten that process so that we could know in advance whether an integral will be zero because then we don't 'waste' time on solving that integral.
We can use the commutator relations for the angular momentum and that the angular momentum operators are hermitian to identify some cases where the integral will definitely be zero. In the end, this will lead us to selection rules for $m_S$ and $l$. But this is only a mean to identify some cases, that we know will be zero. This doesn't mean that there are no other cases that will be zero. And for $n$ there are just no known selection rules (or at least I don't know them).
So I would say that regarding the selection rules your integral could be zero in some cases if there are no other constraints that make it always non zero.
In case you are asking about these additional constraints, I probably missed the point of your question.
Source: Griffiths 2nd Edition Chapter 9.3.3

Question about the radial hydrogen eigenfunctions

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