I've been having difficulty calculating this quantity.
We have an electron in the 2p state of a hydrogen atom (Coulomb potential). As a preliminary step to finding the energy difference between the $2p_{1/2}$ and $2p_{3/2}$ states, I am attempting to find expectation values for $\vec S \cdot \vec L$ and $\frac 1{r^3}$.
However, when I attempt to calculate $<\frac 1{r^3}>$, I run into a non-convergent integral.
If I understand the rules with expectation values correctly, the operator for $<\frac 1{r^3}>$ would be $\frac 1{r^3}$; thus,
$$<\frac 1{r^3}>=\int \Psi^* \frac 1{r^3} \Psi dV $$
over all space. However, in the 2p state, the radial wave function is of a form
$\Psi = C r e^{-r/2a}$ where $a$ is the Bohr radius and C is a mess of constants. Inserting this into the above equation gives me
$$<\frac 1{r^3}> = 4 \pi^2 C^2\int_0^\infty \frac 1r e^{-r/2a} dr $$
which does not converge at $0$.
Any help would be appreciated.
Best Answer
If your wave function is $\Psi = Cre^{-r/2a}$, then you mustn't forget the additional $r^2$ you get from the Jacobi determinant (actually $r^2\sin\theta$, but the $\sin\theta$ is accounted for with the $4\pi$) when changing coordinates: $$ \left<\frac{1}{r^3}\right> = 4\pi C^2\int_0^\infty\mathrm{d}r \;r^2\cdot\frac{1}{r^3}\cdot(re^{-r/2a})^2 = 4\pi C^2\int_0^\infty\mathrm{d}r \;r\cdot e^{-r/a} = 4\pi C^2a^2 $$