Introduction to Quantum Mechanics · Problem 7.43

Time-Independent Perturbation Theory — David J. Griffiths

Physics — Quantum Mechanics · kramers_relation

Problem

Prove Kramers' relation: s+1n2⟨rs⟩−(2s+1)a⟨rs−1⟩+s4[(2ℓ+1)2−s2]a2⟨rs−2⟩=0,(7.113)\frac{s+1}{n^2}\langle r^s \rangle - (2s+1)a\langle r^{s-1} \rangle + \frac{s}{4}\left[(2\ell+1)^2 - s^2\right]a^2\langle r^{s-2} \rangle = 0, \qquad (7.113) which relates the expectation values of rr to three different powers (ss, s−1s-1, and s−2s-2), for an electron in the state ψnℓm\psi_{n\ell m} of hydrogen. *Hint:* Rewrite the radial equation (Equation 4.53) in the form u′′=[ℓ(ℓ+1)r2−2ar+1n2a2]u,u'' = \left[\frac{\ell(\ell+1)}{r^2} - \frac{2}{ar} + \frac{1}{n^2 a^2}\right]u, and use it to express ∫(ursu′′) dr\int (ur^s u'')\,dr in terms of ⟨rs⟩\langle r^s \rangle, ⟨rs−1⟩\langle r^{s-1} \rangle, and ⟨rs−2⟩\langle r^{s-2} \rangle. Then use integration by parts to reduce the second derivative. Show that ∫(ursu′) dr=−(s/2)⟨rs−1⟩\int (ur^s u')\,dr = -(s/2)\langle r^{s-1} \rangle, and ∫(u′rsu′) dr=−[2/(s+1)]∫(u′′rs+1u′) dr\int (u'r^s u')\,dr = -\left[2/(s+1)\right]\int (u''r^{s+1}u')\,dr. Take it from there.

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