quantum mechanics - Orthonormality of Radial Wave Function

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.

Answer

No, the radial parts of the wavefunctions are not orthogonal, at least not quite to that extent.

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 $1$.

You can recover the full orthogonality you expect, but only by adding on the angular dependence given by the spherical harmonics for the full wavefunction.