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 m in a harmonic oscillator potential V(x)=21mω2x2 is in its ground state. In the position representation the normalised ground-state wave function is ϕ0(x)=(πℏmω)1/4exp(−2ℏmωx2), In the energy representation the state is simply cn(0)=δn0, i.e. (1,0,0,…), with En=(n+21)ℏω. (a) Write ∣ϕ0(x)∣2 as a Gaussian ∝e−x2/2σx2 and read off ⟨x^2⟩=σx2 without doing an integral. (b) Fourier transform ϕ0, ϕ~0(p)=(2πℏ)−1/2∫−∞∞ϕ0(x)e−ipx/ℏdx, by completing the square in the exponent. Then do for ∣ϕ~0(p)∣2 what part (a) did for ∣ϕ0(x)∣2 to get ⟨p^2⟩=σp2, and combine it with ⟨x^2⟩ from part (a) to compute ⟨H^⟩=⟨p^2⟩/2m+21mω2⟨x^2⟩. Compare with E0 read directly off the energy-basis list. (c) Substitute ϕ0(x) into the time-independent Schrodinger equation −2mℏ2ϕ0′′(x)+21mω2x2ϕ0(x)=Eϕ0(x) and show by direct differentiation that it is satisfied with a specific value of E, confirming that this position-basis equation is the component form of the basis-free statement H^∣ψ⟩=E0∣ψ⟩. (d) The full time-dependent state is Ψ(x,t)=ϕ0(x)e−iE0t/ℏ. State, without further calculation, the value of P(E0) predicted from the energy representation and the value of ∫−∞∞∣Ψ(x,t)∣2dx predicted from the position representation, for every t, and say in one sentence why the two must agree.