What are the dimensions of the radial function $R_{n,l}(r)$ in the three dimensional wave function $\psi_{n,l,m}(r,\theta,\phi)=R_{n,l}(r)Y_{l,m}(\theta,\phi)$? I'm trying to confirm that the integral $\int r^2 R_{n,l}(r)R_{n',l'}(r)dr$ is dimensionless.
1 Answer
One way to answer this is to look at the form of the radial functions e.g.
$$ R_{1s} = \frac{2}{a_0^{3/2}}e^{-r/a_0} $$
and you immediately find that the dimensions are $[L]^{-3/2}$. A more interesting way is to note that the probability of finding a particle in a volume element $v$ is:
$$ P = \psi^*\psi\,dV $$
Since $P$ is a dimensionless number the dimensions of $\psi$ must be $[L]^{-3/2}$. Since the angular part of the wavefunction is dimensionless that means the dimensions of the radial part must also be $[L]^{-3/2}$
The dimensions of your expression $r^2 R_{n,l}(r)R_{n',l'}(r)dr$ are therefore:
$$ [L]^2 \left([L]^{-3/2}\right)^2 [L] $$
which is dimensionless.