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Harmonic Oscillator violating Heisenberg's Uncertainity

  • #1

Homework Statement


Does the n = 2 state of a quantum harmonic oscillator violate the Heisenberg Uncertainty Principle?

Homework Equations



$$\sigma_x\sigma_p = \frac{\hbar}{2}$$

The Attempt at a Solution


[/B]
I worked out the solution for the second state of the harmonic oscillator.
$$\frac{1}{\sqrt{2}}(\frac{mw}{\pi\hbar})^{1/4}(\frac{1}{\hbar})(e^{\frac{-mwx^2}{2\hbar}}(2mwx^2 - \hbar)$$

Should I be solving for standard deviation of position and momentum? Or is there another way to do this problem. I'm not sure what the next step is. Thank you for the guidance.
 

Answers and Replies

  • #2
Orodruin
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Should I be solving for standard deviation of position and momentum?
Yes.

Or is there another way to do this problem.
If you work in terms of creation and annihilation operators you never need to compute the wave function at all.
 
  • #3
Yes.

If you work in terms of creation and annihilation operators you never need to compute the wave function at all.
You mean by using the following and the momentum operator in terms of creation/annihilation operators?
$$x = \sqrt{\frac{\hbar}{2mw}}(a_+ + a_{-})$$
 
  • #4
Orodruin
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Along with the properly normalised definitions of the energy eigenstates ##|n\rangle##, yes.
 
  • #5
Along with the properly normalised definitions of the energy eigenstates ##|n\rangle##, yes.
Edit: I figured it out I got standard deviation of x * standard deviation of p = $$2\hbar$$. Which works out.
 
  • #6
Orodruin
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I just worked through it and after evaluating the limit over all space I getting that the standard deviation of the expectation value of position is 0. Does this make sense?
No, it does not. However, it is impossible to see where you went wrong without seeing your maths.
 
  • #7
No, it does not. However, it is impossible to see where you went wrong without seeing your maths.
I realized I messed up the <x^2> calculation. Thank you for your help!
 
  • #8
Orodruin
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So the challenge is now to generalise the problem to the state ##|n\rangle## and then to an arbitrary linear combination of the ##|n\rangle##.
 
  • #9
So the challenge is now to generalise the problem to the state ##|n\rangle## and then to an arbitrary linear combination of the ##|n\rangle##.
I was actually working through Griffiths problems just now and I've seen the nth generalization, very cool stuff.
 

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