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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):

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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?

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2 Answers 2

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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 $\hbar^2\ell(\ell+1)$.

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Just consider the radial Hamiltonian as a hamiltonian on its own. So you want to compute an observable that depends only on the radial behaviour, vaguely speaking.

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 $\hat{H}_\ell$

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  • $\begingroup$ 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"? $\endgroup$
    – Arbiter
    May 18, 2018 at 15:44
  • $\begingroup$ Define it (per value of $\ell$ if you wish) as it is written in the Wikipedia. The proof basically holds step by step. $\endgroup$
    – ohneVal
    May 18, 2018 at 16:25

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