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

Limit is x₀ to -x₀

`(1)/(\pi ) \int\limits^x_x {(d (x-y))/(x_o√(1 - cos^2(wt)) ) } \, dy`

Limit is -x₀ to x₀

`= (1)/(\pi ) \int\limits^x_x {(d(x-y))/(√(x_o^2 - y^2) ) } \, dy\\ \\= (1)/(\pi√(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.