# Feynman-Hellmann theorem problem in Griffiths - somewhat inaccurate?

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 $$H$$.

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

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.