Is the radial component $R_{n\ell}$ of the hydrogen wavefunction orthonormal? Doing out one of the integrals, I find that

$$\int_0^{\infty} R_{10}R_{21}~r^2dr ~\neq~0$$

However, the link below says that these wave functions should be orthonormal (go to the top of page 3):

http://www.phys.spbu.ru/content/File/Library/studentlectures/schlippe/qm07-05.pdf

Am I doing something wrong? Are the radial components orthogonal, or aren't they? Are there some kind of special condition on $n$ and $\ell$ that make $R_{n\ell}$ orthogonal? Any help on this problem would be appreciated.

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The radial components are built out of Laguerre polynomials, whose orthogonality only holds when leaving the secondary index fixed (the $\ell$ or $2\ell+1$ or whatever depending on your convention). That is, $$\langle R_{n'\ell} \vert R_{n\ell} \rangle \equiv \int_0^\infty R_{n'\ell}^*(r) R_{n\ell}(r) r^2 \, \mathrm{d}r = \delta_{nn'}.$$ You can check this yourself, using some of the lower-order functions, e.g. \begin{align} R_{10}(r) & = \frac{2}{a_0^{3/2}} \mathrm{e}^{-r/a_0}, \\ R_{21}(r) & = \frac{1}{\sqrt{3} (2a_0)^{3/2}} \left(\frac{r}{a_0}\right) \mathrm{e}^{-r/2a_0}, \\ R_{31}(r) & = \frac{4\sqrt{2}}{9 (3a_0)^{3/2}} \left(\frac{r}{a_0}\right) \left(1 - \frac{r}{6a_0}\right) \mathrm{e}^{-r/3a_0}. \end{align} (Note that $R_{10}$ and $R_{21}$ are in fact both strictly positive, so they can't integrate to $0$.) You should find $$\langle R_{10} \vert R_{10} \rangle = \langle R_{21} \vert R_{21} \rangle = \langle R_{31} \vert R_{31} \rangle = 1$$ and $$\langle R_{21} \vert R_{31} \rangle = \langle R_{31} \vert R_{21} \rangle = 0,$$ as expected. However, $\langle R_{10} \vert R_{21} \rangle = \langle R_{21} \vert R_{10} \rangle$ and $\langle R_{10} \vert R_{31} \rangle = \langle R_{31} \vert R_{10} \rangle$ are very much neither $0$ nor $1$.