A particle of mass in a harmonic oscillator potential is in its ground state

PHYSICS · Subjective · Fermi University Physics · Quantum Mechanics · Wave mechanics in one dimension · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch The wave function is not the state

Problem

A particle of mass mm in a harmonic oscillator potential V(x)=12mω2x2V(x)=\frac12m\omega^2x^2 is in its ground state. In the position representation the normalised ground-state wave function is ϕ0(x)=(mωπℏ)1/4exp⁡ ⁣(−mωx22ℏ),\phi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\!\left(-\frac{m\omega x^2}{2\hbar}\right), In the energy representation the state is simply cn(0)=δn0c_n(0)=\delta_{n0}, i.e. (1,0,0,… )(1,0,0,\dots), with En=(n+12)ℏωE_n=(n+\tfrac12)\hbar\omega. (a) Write ∣ϕ0(x)∣2|\phi_0(x)|^2 as a Gaussian ∝e−x2/2σx2\propto e^{-x^2/2\sigma_x^2} and read off ⟨x^2⟩=σx2\langle \hat x^2\rangle=\sigma_x^2 without doing an integral. (b) Fourier transform ϕ0\phi_0, ϕ~0(p)=(2πℏ)−1/2∫−∞∞ϕ0(x)e−ipx/ℏ dx\tilde\phi_0(p)=(2\pi\hbar)^{-1/2}\int_{-\infty}^{\infty}\phi_0(x)e^{-ipx/\hbar}\,\mathrm{d}x, by completing the square in the exponent. Then do for ∣ϕ~0(p)∣2|\tilde\phi_0(p)|^2 what part (a) did for ∣ϕ0(x)∣2|\phi_0(x)|^2 to get ⟨p^2⟩=σp2\langle \hat p^2\rangle=\sigma_p^2, and combine it with ⟨x^2⟩\langle \hat x^2\rangle from part (a) to compute ⟨H^⟩=⟨p^2⟩/2m+12mω2⟨x^2⟩\langle\hat H\rangle = \langle\hat p^2\rangle/2m + \frac12m\omega^2\langle\hat x^2\rangle. Compare with E0E_0 read directly off the energy-basis list. (c) Substitute ϕ0(x)\phi_0(x) into the time-independent Schrodinger equation −ℏ22mϕ0′′(x)+12mω2x2ϕ0(x)=Eϕ0(x)-\frac{\hbar^2}{2m}\phi_0''(x)+\frac12m\omega^2x^2\phi_0(x) = E\phi_0(x) and show by direct differentiation that it is satisfied with a specific value of EE, confirming that this position-basis equation is the component form of the basis-free statement H^∣ψ⟩=E0∣ψ⟩\hat H|\psi\rangle=E_0|\psi\rangle. (d) The full time-dependent state is Ψ(x,t)=ϕ0(x)e−iE0t/ℏ\Psi(x,t)=\phi_0(x)e^{-iE_0t/\hbar}. State, without further calculation, the value of P(E0)P(E_0) predicted from the energy representation and the value of ∫−∞∞∣Ψ(x,t)∣2dx\int_{-\infty}^{\infty}|\Psi(x,t)|^2dx predicted from the position representation, for every tt, and say in one sentence why the two must agree.

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