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Expectation values of the quantum harmonic oscillator

  • #1

Homework Statement



Show the mean position and momentum of a particle in a QHO in the state ψγ to be:

<x> = sqrt(2ħ/mω) Re(γ)
<p> = sqrt (2ħmω) Im(γ)

Homework Equations



##\psi_{\gamma} (x) = Dexp((-\frac{mw(x-<x>)^2}{2\hbar})+\frac{i<p>(x-<x>)}{ħ})##


The Attempt at a Solution



I put ψγ into the equation

<p> = ∫[ψ(x,t)]* (-iħψ'(x)) dx

Which gave me

(-(mω(x-<x>)/ħ)+((<p>)(i/ħ))*ψγ which I can't help but feel is taking me further away from what I'm looking for. Are there easier alternative route to doing this?

Thank you in advance
 
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Answers and Replies

  • #2
PeroK
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Homework Statement



Show the mean position and momentum of a particle in a QHO in the state ψγ to be:

<x> = sqrt(2ħ/mω) Re(γ)
<p> = sqrt (2ħmω) Im(γ)

Homework Equations



ψγ = Dexp(-((mω(x-<x>)^2)/(2ħ))+ ((<p>)*(i/ħ)(x-<x>)))
How is ##\psi_{\gamma}## defined? That expression has no ##\gamma## in it, but it does have the expected values of ##x## and ##p##.

It's hard to read your latex. I'll post some you can copy in a minute.
 
  • #3
PeroK
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Some latex:

##\psi_0 = (\frac{m \omega}{\pi \hbar})^{\frac14}exp(-\frac{mwx^2}{2\hbar})##

##\psi_{\gamma} = ?##

If you reply to this post, it will give you something to cut and paste.
 
  • #4
Some latex:

##\psi_0 = (\frac{m \omega}{\pi \hbar})^{\frac14}exp(-\frac{mwx^2}{2\hbar})##

##\psi_{\gamma} = ?##

If you reply to this post, it will give you something to cut and paste.
I've edited it accordingly
 
  • #5
PeroK
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##\psi_{\gamma} (x) = Dexp((-\frac{mw(x-<x>)^2}{2\hbar})+\frac{i<p>(x-<x>)}{ħ})##
What's ##\gamma##?
 
  • #6
Some latex:

##\psi_0 = (\frac{m \omega}{\pi \hbar})^{\frac14}exp(-\frac{mwx^2}{2\hbar})##

##\psi_{\gamma} = ?##

If you reply to this post, it will give you something to cut and paste.
What's ##\gamma##?
i haven't got that information, ψγ represents coherent states.
 
  • #7
PeroK
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i haven't got that information, ψγ represents coherent states.
Given that the required answer (for the expectation values) depends on ##\gamma##, it must be a parameter in the state. You need to check the question.
 
  • #8
Given that the required answer (for the expectation values) depends on ##\gamma##, it must be a parameter in the state. You need to check the question.
Just realised I have missed a little bit out,

γ is a complex parameter and
a_ψγ(x) = γψγ(x)

That's 100% all the information I have now
 
  • #9
PeroK
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Just realised I have missed a little bit out,

γ is a complex parameter and
a_ψγ(x) = γψγ(x)

That's 100% all the information I have now
That's entirely different. That means that ##\psi_{\gamma}## is an eigenstate of the lowering operator, corresponding to eigenvalue ##\gamma##.

The expression you quoted for ##\psi_{\gamma}## before makes no sense.

Hint: can you express the position and momentum operators in terms of the raising and lowering operators?

HInt #2: I suspect you can do this using Linear Algebra, without resorting to integration.
 
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