A Modern Approach to Quantum Mechanics · Problem 11.18

Identical Particles — John S. Townsend

Physics — Quantum Mechanics · hydrogen-expectation-1-over-r-squared

Problem

In order to evaluate ⟨1/r2⟩\langle 1/r^2 \rangle take γ/r2\gamma/r^2 as a perturbation for the hydrogen atom. Here again we can obtain the exact solution since the perturbation modifies the centrifugal potential in (10.12) as follows: l(l+1)ℏ22μr2+γr2=l′(l′+1)ℏ22μr2\frac{l(l+1)\hbar^2}{2\mu r^2} + \frac{\gamma}{r^2} = \frac{l'(l'+1)\hbar^2}{2\mu r^2} Thus the exact energy is given by E=−μc2Z2α22(nr+l′+1)2E = -\frac{\mu c^2 Z^2 \alpha^2}{2(n_r + l' + 1)^2} Show that ⟨γr2⟩n,l,m=En,l,m(1)=γ(dEdγ)γ=0=γ(dEdl′)l′=l(dl′dγ)l′=l=γZ2n3a02(l+12)\left\langle \frac{\gamma}{r^2} \right\rangle_{n,l,m} = E_{n,l,m}^{(1)} = \gamma\left(\frac{dE}{d\gamma}\right)_{\gamma=0} = \gamma\left(\frac{dE}{dl'}\right)_{l'=l}\left(\frac{dl'}{d\gamma}\right)_{l'=l} = \frac{\gamma Z^2}{n^3 a_0^2 (l+\frac{1}{2})}

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