Expectation Value $\langle \frac{1}{r^2} \rangle$ using Hellmann–Feynman theorem

Suppose we have the hydrogen atom$$H ~=~ \frac{p_r ^2}{2m} + \frac{L^2}{2mr^2} -\frac{e^2}{r} \,.$$And have solved the Schrodinger equation finding $$E_n = - \frac{me^4}{2 \hbar ^2 n^2}$$ and $$Ψ_{nlm}~=~R_{E,l}\left(r\right) Y_{lm}\left(φ,\, θ\right)\,.$$ Using the charge $e$ as a parameter it's quite easy to use the Hellmann–Feynman theorem:$$\frac{\mathrm{d}E_λ}{\mathrm{d}e}~=~\left< Ψ_{nlm} \, \middle| \, \frac{\mathrm{d}H}{\mathrm{d}λ} \, \middle| \, Ψ_{nlm} \right> \,,$$to find $\left< \frac{1}{r} \right>$. Now I'm trying to find a proper way of doing the "same" to find $\left< \frac{1}{r^2} \right>$.

I found the following solution (in wikipedia):

What I don't understand is why we can use the radial part $H_l$ the way it's written in the solution, instead of the whole hamiltonian? I understand that $L^2$ acts only on $Y_{lm}$ giving $\hbar^2 l(l+1) Y_{lm}$, but I can't see how we can get this simplification in$$\left< R_{E,l}\left(r\right) Y_{lm} \left(φ,\, θ\right) \, \middle| \, \frac{\mathrm{d}H}{\mathrm{d}l} \, \middle| \, R_{E,l} \left(r\right) Y_{lm} \left(φ, \, θ \right) \right> \,.$$ Doesn't the hamiltonian first have to act on $Y_{lm}$ to be dependent on $l$ in the first place?

• – Qmechanic May 18 '18 at 16:21

Once we have decided to look at states with a certain azimuthal quantum number $\ell$, i.e. an irreducible representation of the 3D rotation group $SO(3)$, then the $SO(3)$ Casimir operator $\vec{L}^2$ acts as an eigenvalue $\hbar^2\ell(\ell+1)$.
Alternatively you can always start from the full Hamiltonian, take partial derivatives on any parameter you wish and you arrive to the same answer, since the Hellman-Feynman theorem still applies for $\hat{H}$ and $\hat{H}_\ell$
• Taking the radial Hamiltonian we can substitute $L^2$ with it's eigenvalues then? Doesn't seem that trivial to me, how do we exactly define the "radial Hamiltonian"? – Dimitris May 18 '18 at 15:44
• Define it (per value of $\ell$ if you wish) as it is written in the Wikipedia. The proof basically holds step by step. – ohneVal May 18 '18 at 16:25