Find the expectation value of momentum squared for a simple harmonic oscillator

  • #1
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Find the expectation value of (px)2, keeping in mind that ψ0(x) = A0e−ax2
where A0 = (2mω0/h)^1/4, and
<x2> = ∫x2|ψ|2dx = h_bar / 2mω0


<ψ(x)|px2|ψ(x)> = ∫ψ(x)(pop2)ψ(x) dx

pop = [hbar / i] ([itex]\delta[/itex]/[itex]\delta[/itex]x)


I'm not going to attempt to type out me solving the integral because it would be extremely messy, as I am unfamiliar with syntax on this site.
I take the partial twice of ψ(x), and plug in the value they give for ∫x2|ψ|2:

-4hbar3a2 / 2mω0

This isn't correct, and I'm not too sure what I'm doing wrong. The negative doesn't make sense to me, but that's what the math is giving (the negative comes from squaring i in the momentum operator).

Any help is greatly appreciated.
 

Answers and Replies

  • #2
The syntax on this site is LaTex which is quite easy to use. Related to your problem, can you use the Virial Theorem ? Or you could go for an algebraic approach, because the average is computed in the ground state...
 
  • #3
Find the expectation value of (px)2, keeping in mind that ψ0(x) = A0e−ax2
where A0 = (2mω0/h)^1/4, and
<x2> = ∫x2|ψ|2dx = h_bar / 2mω0


<ψ(x)|px2|ψ(x)> = ∫ψ(x)(pop2)ψ(x) dx

pop = [hbar / i] ([itex]\delta[/itex]/[itex]\delta[/itex]x)


I'm not going to attempt to type out me solving the integral because it would be extremely messy, as I am unfamiliar with syntax on this site.
I take the partial twice of ψ(x), and plug in the value they give for ∫x2|ψ|2:

-4hbar3a2 / 2mω0

This isn't correct, and I'm not too sure what I'm doing wrong. The negative doesn't make sense to me, but that's what the math is giving (the negative comes from squaring i in the momentum operator).

Any help is greatly appreciated.

Can you explain to yourself why you are taking the partial twice? Are you missing any factors for the momentum operator?
 
  • #4
The momentum operator squared is
-hbar2 δ2/δx2
 
  • #5
-4hbar3a2 / 2mω0

Except for the minus sign, this looks correct. Don't forget to substitute the expression for [itex]a[/itex].
 
  • #6
Uh oh, I didn't substitute for a. Is this a common constant? They give what A0 is, but it goes away with the identity for <x2>.
 
  • #7
Why compute tough integrals and derivatives, why not just convert p^2 into creation and annihilation operators?
 

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