Introduction to Quantum Mechanics · Problem 7.18

Time-Independent Perturbation Theory — David J. Griffiths

Physics — Quantum Mechanics · hermiticity_momentum_operators

Problem

Show that p2p^2 is hermitian, for hydrogen states with ℓ=0\ell = 0. Hint: For such states ψ\psi is independent of θ\theta and ϕ\phi, so p2=−ℏ2r2ddr(r2ddr)p^2 = -\frac{\hbar^2}{r^2} \frac{d}{dr}\left(r^2 \frac{d}{dr}\right) (Equation 4.13). Using integration by parts, show that \left\langle f \left| p^2 g \right\rangle = -4\pi\hbar^2 \left(r^2 f \frac{dg}{dr} - r^2 g \frac{df}{dr}\right)\bigg|_0^\infty + \left\langle p^2 f \left| g \right\rangle.\right. Check that the boundary term vanishes for ψn00\psi_{n00}, which goes like ψn00∼1π(na)3/2exp⁡(−r/na)\psi_{n00} \sim \frac{1}{\sqrt{\pi}(na)^{3/2}} \exp(-r/na) near the origin. The case of p4p^4 is more subtle. The Laplacian of 1/r1/r picks up a delta function (see, for example, D. J. Griffiths, Introduction to Electrodynamics, 4th edn, Eq. 1.102). Show that ∇4[e−kr]=(−4k3r+k4)e−kr+8πk δ3(r),\nabla^4\left[e^{-kr}\right] = \left(-\frac{4k^3}{r} + k^4\right)e^{-kr} + 8\pi k\,\delta^3(\mathbf{r}), and confirm that p4p^4 is hermitian.

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