10.4: Simple Example
( \newcommand{\kernel}{\mathrm{null}\,}\)
The best way to clarify this abstract discussion is to consider the quantum mechanics of the Harmonic oscillator of mass m and frequency ω,
ˆH=−ℏ22md2dx2+12mω2x2
If we assume that the wave function at time t=0 is a linear superposition of the first two eigenfunctions,
ψ(x,t=0)=√12ϕ0(x)−√12ϕ1(x)ϕ0(x)=(mωπℏ)1/4exp(−mω2ℏx2)ϕ1(x)=(mωπℏ)1/4exp(−mω2ℏx2)√mωℏx
(The functions ϕ0 and ϕ1 are the normalised first and second states of the harmonic oscillator, with energies E0=12ℏω and E1=32ℏω.) Thus we now kow the wave function for all time:
ψ(x,t)=√12ϕ0(x)e−12iωt−√12ϕ1(x)e−32iωt.
In figure 10.4.1 we plot this quantity for a few times.
The best way to visualize what is happening is to look at the probability density,
P(x,t)=ψ(x,t)∗ψ(x,t)=12(ϕ0(x)2+ϕ1(x)2−2ϕ0(x)ϕ1(x)cosωt)
This clearly oscillates with frequency ω.
Question: Show that ∫∞−∞P(x,t)dx=1.

Another way to look at that is to calculate the expectation value of ˆx :
⟨ˆx⟩=∫∞−∞P(x,t)dx=12∫∞−∞ϕ0(x)2xdx⏟=0+12∫∞−∞ϕ1(x)2xdx⏟=0−cosωt∫∞−∞ϕ0(x)ϕ1(x)xdx=−cosωt√ℏmω1√π∫∞−∞y2e−y2=−12√ℏmωcosωt.
This once again exhibits oscillatory behaviour!