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Problem 6.33 in Griffiths (typing out only the relevant parts) is stated as follows:

The Feynman-Hellmann theorem can be used to determine the expectation values of $1/r$ and $1/r^2$ for hydrogen. The effective Hamiltonian for the radial wave functions is $$H=-\frac{\hbar^2}{2m}\frac{d^2}{dr^2} + \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}.$$

(a) Use $\lambda=e$ in the Feynman-Hellmann theorem to obtain $\langle 1/r\rangle$.

I understand that the given Hamiltonian above is that for the radial wavefunction $u(r)$ where $u(r) = r\cdot R_{n\ell}(r)$ and $\Psi=R_{n\ell}(r)\cdot Y_{\ell m}(\theta,\phi)$. In using the Feynman-Hellmann to calculate $\langle 1/r\rangle$ with respect to the state $\Psi$, I appreciated the careful treatment as explained by this post to remove the angular parts. Following the argument, it seems like I should be computing $\frac{dH_\lambda}{d\lambda}$ where

$$H_\lambda = -\frac{\hbar^2}{2mr^2}\frac{d}{dr}r^2 \frac{d}{dr} + \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}-\frac{e^2}{4\pi\epsilon_0}\frac{1}{r}.$$

(i.e. the "full" Hamiltonian for $R_{n \ell}(r)$) rather than the effective Hamiltonian $H$ given in Griffith.

Of course both $\langle \frac{dH}{d\lambda}\rangle = \langle\frac{dH_\lambda}{d\lambda}\rangle = \langle \frac{1}{r} \rangle$ as the first term in $H$ and $H_\lambda$ both disappear when we differentiate with respect to $\lambda$. However, conceptually, I would like to check if the correct application of Feynman-Hellmann should be involving $H_\lambda$ instead of $H$.

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  • $\begingroup$ Which edition do you have? It's problem 6.27 in the 1995 printing. $\endgroup$
    – Qmechanic
    Nov 22, 2019 at 17:22
  • $\begingroup$ Second edition, the 2005 printing $\endgroup$
    – suncup224
    Nov 23, 2019 at 6:31

1 Answer 1

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To be clear, one can use both radial wavefunctions $$u_{\lambda}(r)~\equiv~r R_{\lambda}(r) $$ in the radial Feynman-Hellmann theorem $$\frac{\partial E_{\lambda}}{\partial\lambda^j} ~=~\langle R_{\lambda} | \frac{\partial \hat{H}_{\lambda}}{\partial\lambda^j}| R_{\lambda} \rangle .$$ Of course one should be aware that the norms are weighted differently $$1~=~ \langle R_{\lambda} | R_{\lambda} \rangle~=~\int_{\mathbb{R}_+}\!r^2\mathrm{d}r~|R_{\lambda}(r)|^2 ~=~\int_{\mathbb{R}_+}\!\mathrm{d}r~|u_{\lambda}(r)|^2, $$ and that the effective Hamiltonians take different forms as listed by OP.

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