Principles of Quantum Mechanics · Problem 8.6.3

The Heisenberg Uncertainty Relations — R. Shankar

Physics — Quantum Mechanics · harmonic-oscillator-eigenfunctions-propagator

Problem

We know that given the eigenfunctions and the eigenvalues we can construct the propagator: U(x,t;x′,t′)=∑nψn(x)ψn∗(x′) e−iEn(t−t′)/ℏ(8.6.15)U(x, t; x', t') = \sum_n \psi_n(x) \psi_n^*(x')\, e^{-iE_n(t-t')/\hbar} \tag{8.6.15} Consider the reverse process (since the path integral approach gives UU directly), for the case of the oscillator. (1) Set x=x′=t′=0x = x' = t' = 0. Assume that A(t)=(mω/2πiℏsin⁡ωt)1/2A(t) = (m\omega/2\pi i\hbar \sin\omega t)^{1/2} for the oscillator. By expanding both sides of Eq.\ (8.6.15), you should find that E=ℏω/2, 5ℏω/2, 9ℏω/2,…E = \hbar\omega/2,\, 5\hbar\omega/2,\, 9\hbar\omega/2, \ldots, etc. What happened to the levels in between? (2) (Optional). Now consider the extraction of the eigenfunctions. Let x=x′x = x' and t′=0t' = 0. Find E0E_0, E1E_1, ∣ψ0(x)∣2|\psi_0(x)|^2, and ∣ψ1(x)∣2|\psi_1(x)|^2 by expanding in powers of α=exp⁡(iωt)\alpha = \exp(i\omega t).

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