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Quantum Question

  1. Mar 8, 2009 #1
    Just reading through my notes and found a step I can't follow:

    If you look at p3 of the follwing document, one of the lines in the proof of the Generalised Uncertainty Principle has a (2) next to it. I can't get from the line before it to that line.

    Can anyone help me out?

    http://www.ph.ed.ac.uk/teaching/course-notes/documents/64/786-lecture5.pdf


    thanks
     
  2. jcsd
  3. Mar 8, 2009 #2

    gabbagabbahey

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    Use, equation (1) on that page to show that [tex][\hat{X},\hat{Y}]=[\hat{A},\hat{B}][/tex], then use the definition of expectation value
     
  4. Mar 8, 2009 #3
    hey. thanks. i have another question about it though.

    in the line before (2), how does the [itex](\Delta \hat{A}_t)^2[/itex] term work? Surely when we square [itex]\hat{X}[/itex], we get [itex](\Delta \hat{A}_t)^2[/itex] as well as other stuff arising from the cross terms???

    also in the very last line, where does he get the [itex]i[/itex] in the RHS from - i'm assuming it's so that we end up with a [itex]\geq[/itex] not a [itex]\leq[/itex] but i don't follow it...

    thanks
     
  5. Mar 8, 2009 #4

    gabbagabbahey

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    By definition, [tex](\Delta \hat{A}_t)^2=(\hat{A}-\langle\hat{A}\rangle_t)^2=\hat{X}^2[/tex]

    And the [itex]i[/itex] is just a way to account for the negative sign since [tex]\langle i [\hat{A},\hat{B}]\rangle_t^2=i^2\langle [\hat{A},\hat{B}]\rangle_t^2=-\langle[\hat{A},\hat{B}]\rangle_t^2[/tex]
     
  6. Mar 8, 2009 #5
    but [itex]\Delta \hat{A}_t = \sqrt{\langle \hat{A^2}_t \rangle - \langle \hat{A}_t \rangle^2}[/itex]

    so why do you get what you've written?
     
  7. Mar 8, 2009 #6

    gabbagabbahey

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    The two definitions are equivalent:

    [tex](\hat{A}-\langle\hat{A}\rangle_t)^2=\hat{A}^2-2\hat{A}\langle\hat{A}\rangle_t+\langle\hat{A}\rangle_t^2[/tex]

    [tex]\implies \int_{-\infty}^{\infty} \Psi^{*}(x,t)(\hat{A}-\langle\hat{A}\rangle_t)^2\Psi(x,t)dx=\int_{-\infty}^{\infty} \Psi^{*}(x,t)\hat{A}^2\Psi(x,t)dx-2\langle\hat{A}\rangle_t \int_{-\infty}^{\infty} \Psi^{*}(x,t)\hat{A}\Psi(x,t)dx+\langle\hat{A}\rangle_t^2\int_{-\infty}^{\infty} \Psi^{*}(x,t)\Psi(x,t)dx[/tex]

    [tex]=\langle\hat{A}^2\rangle_t-2\langle\hat{A}\rangle_t^2+\langle\hat{A}\rangle_t^2=\langle\hat{A}^2\rangle_t-\langle\hat{A}\rangle_t^2=(\Delta\hat{A}_t)^2[/tex]

    [tex]\implies (\Delta\hat{A}_t)^2=(\hat{A}-\langle\hat{A}\rangle_t)^2[/tex]

    since they both integrate to the same thing
     
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