Question

Find an expectation value in the n th state of the harmonic oscillator.Find an expectation value in the n th state of the harmonic oscillator.

Answer 1

The classical motion for an oscillator that starts from rest at location x₀ is

                                           x(t) = x₀ cos(ωt)

The probability that the particle is at a particular x at a particular time t

is given by ρ(x, t) = δ(x − x(t)), and we can perform the temporal average

to get the spatial density. Our natural time scale for the averaging is a half

cycle, take t = 0 → π/ ω

Thus,

ρ =   $\frac{1}{\pi / w} \int\limits^\pi_0 {d(x - x_o cos(wt))} \, dt$

Limit is 0 to π/ω

We perform the change of variables to allow access to the δ, let y = x₀ cos(ωt)  so that

ρ(x) = $-\frac{w}{\pi } \int\limits^x_x {\frac{d ( x - y)}{x_ow sin(wt)} } \, dy$

Limit is x₀ to -x₀

$\frac{1}{\pi } \int\limits^x_x {\frac{d (x-y)}{x_o\sqrt{1 - cos^2(wt)} } } \, dy$

Limit is -x₀ to x₀

$= \frac{1}{\pi } \int\limits^x_x {\frac{d(x-y)}{\sqrt{x_o^2 - y^2} } } \, dy\\ \\= \frac{1}{\pi\sqrt{x_o^2 - x^2} }$

This has $\int\limits^x_x {p(x)} \, dx = 1$ as expected. Here the limit is -x₀ to x₀

The expectation value is 0 when the ρ(x) is symmetric, x ρ(x) is asymmetric and the limits of integration are asymmetric.