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What's the kinetic energy uncertainty for gaussian wave packet?

  1. Oct 5, 2011 #1
    What's the kinetic energy uncertainty for gaussian wave packet ψ(x)=((α/pi)^(1/4))exp(-αx^2/2)?
     
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  3. Oct 5, 2011 #2

    Matterwave

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    Can you compute the uncertainty in p? If you know the uncertainty in p and you know that K=p^2/2m can you compute the uncertainty in K?
     
  4. Oct 5, 2011 #3
    I guess you can get the EXPECTATION VALUE of K, not the uncertainty in K, by squaring the uncertainty in p and dividing it by 2m
     
  5. Oct 5, 2011 #4

    Matterwave

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    [tex]K=\frac{p^2}{2m}[/tex]

    [tex]<K>=<\frac{p^2}{2m}>=\frac{<p^2>}{2m}[/tex]

    [tex]\sigma_K=\frac{\partial K}{\partial p}\sigma_p[/tex]

    Does that look reasonable?
     
  6. Oct 5, 2011 #5
    σK=(p/m)σp

    what is p? does p= <p>?
     
  7. Oct 5, 2011 #6

    Matterwave

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    Yes, as usual, you put the expectation values in there.
     
  8. Oct 5, 2011 #7
    Then <p> should be zero..
     
  9. Oct 5, 2011 #8

    Matterwave

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    Ah, well I think there may be something subtle happening here that prevents my formula from working. In the mean time then, you can compute this by brute force:

    [tex]\sigma_K=\sqrt{\langle \psi | K^2|\psi\rangle-(\langle \psi | K|\psi\rangle)^2}=\frac{1}{m}\sqrt{\langle \psi |\frac{p^4}{4}|\psi\rangle-(\langle \psi | \frac{p^2}{2}|\psi\rangle)^2}[/tex]

    This involves the 4th moment of the Gaussian though...which may be kind of hard...maybe someone more enlightened can fill us in then...XD
     
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