A particle in a three-dimensional isotropic harmonic oscillator of frequency has…

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · The formalism of quantum mechanics · Density matrices, open systems, entanglement and interpretation

The Structure of Quantum Mechanics, Ch The postulates, measurement and collapse

Problem

A particle in a three-dimensional isotropic harmonic oscillator of frequency ω\omega has energy levels EN=(N+3/2)ℏωE_N = (N+3/2)\hbar\omega, with simultaneous eigenstates ∣N,l,m⟩|N,l,m\rangle of H^\hat H, L^2\hat L^2 and L^z\hat L_z; for N=0N=0 the only state is l=0,m=0l=0,m=0, while for N=2N=2 the allowed values are l=0l=0 (m=0m=0) and l=2l=2 (m=−2,…,2m=-2,\dots,2). The particle is prepared in the normalised state∣ϕ⟩=0.36 ∣0,0,0⟩+0.16 ∣2,0,0⟩+0.48 ∣2,2,0⟩.|\phi\rangle = \sqrt{0.36}\,|0,0,0\rangle + \sqrt{0.16}\,|2,0,0\rangle + \sqrt{0.48}\,|2,2,0\rangle. ((a)) The total energy is measured. Find the probability of obtaining E0E_0 and the probability of obtaining E2E_2, using the projector onto each degenerate energy eigenspace. ((b)) The energy is measured non-selectively: its result is recorded for every run, but no run is discarded. Compute the probability that a subsequent measurement of L^2\hat L^2 then gives the eigenvalue for l=0l=0, by weighting each branch's conditional l=0l=0 probability by its energy probability from part (a). ((c)) Compute the probability of l=0l=0 if L^2\hat L^2 is instead measured directly on ∣ϕ⟩|\phi\rangle, with no prior energy measurement, and state briefly why the two answers in (b) and (c) can agree numerically even though the object produced by the non-selective measurement in (b) is not a single ket.

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