Expectation values of the quantum harmonic oscillator

AI Thread Summary
The discussion focuses on deriving the mean position and momentum of a particle in a quantum harmonic oscillator (QHO) state, specifically ψγ. The expected values are given as <x> = sqrt(2ħ/mω) Re(γ) and <p> = sqrt(2ħmω) Im(γ), with γ being a complex parameter related to coherent states. Participants express confusion over the definition of ψγ and its relation to the lowering operator, suggesting that it is an eigenstate corresponding to eigenvalue γ. They recommend using linear algebra techniques to express position and momentum operators instead of integration for a clearer solution.
Dean Navels
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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



##\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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Dean Navels said:

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.
 
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.
 
PeroK said:
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
 
Dean Navels said:
##\psi_{\gamma} (x) = Dexp((-\frac{mw(x-<x>)^2}{2\hbar})+\frac{i<p>(x-<x>)}{ħ})##

What's ##\gamma##?
 
PeroK said:
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.
PeroK said:
What's ##\gamma##?
i haven't got that information, ψγ represents coherent states.
 
Dean Navels said:
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.
 
PeroK said:
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 realized 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
 
Dean Navels said:
Just realized 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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