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Calculating an integral (QM / probability)

  1. Mar 27, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hi guys, I've an integral that poped up in QM, it should be different from 0 (else the particle's position is known with 100% certainty while it should not with the wave function I was given). However I get 0 and I don't see where my mistake(s) is/are.
    ##<(x-a)^2>=\int _{-\infty } ^\infty (x-a)^2N^2 \exp \{ - \frac{(x-a)^2}{2 \sigma ^2} \}dx##.


    2. The attempt at a solution
    I make the change of variables ##\alpha = (x-a)^2## so that ##d\alpha = 2(x-a)dx##. So as x goes from - infinity to + infinity, alpha goes from + infinity to + infinity and therefore the integral becomes ##\int _{\infty} ^\infty \frac{\sqrt \alpha }{2}N^2 \exp \{ - \frac{\alpha }{2 \sigma ^2} \}d\alpha =0## because of the limits of integration (they are the same).
    I've done the exact same change of variables with another integral that yielded 0 this way and it was correct. However here I must not get 0, but I do.
    Where did I go wrong?
     
  2. jcsd
  3. Mar 27, 2013 #2

    Ray Vickson

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    ##x = a \pm \sqrt{\alpha}##, so the integrations for ##x < a## and ##x > a## need to be kept separate. So, no, you do not get 0. However, I don't know why you want to make that change of variable. It would be much, much easier to let ##y = (x-a) / \sigma##.
     
  4. Mar 28, 2013 #3

    fluidistic

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    I see. So if I understand well, even though alpha goes from +infinity to +infinity, it "passes by" 0 (in which case x=a) so I should take this into account in the limits of integration.
    Ok I'll follow your tip for the change of variables.
     
  5. Mar 28, 2013 #4

    fluidistic

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    You're a boss Ray Vickson! I reached a result that makes sense (variance is worth ##\sigma ^2##).
    I chose the change of variables ##\alpha = \frac{x-a}{\sqrt 2 \sigma}##. In order to compute ##\int _{-\infty}^\infty \alpha ^2 e^{-\alpha ^2 }d\alpha## I used a trick: I thought the integral as ##\int _{-\infty}^\infty \alpha ^2 e^{-\alpha ^2 \gamma}d\alpha## as if it was dependent on gamma (despite it being worth 1). Then I rewrote the integral as ##-\int _{-\infty}^{\infty} \frac{\partial }{\partial \gamma} (e^{-\alpha ^2 \gamma}) d\alpha##.
    After a few algebra I had to use the fact that ##\int _{-\infty}^\infty e^{-\alpha ^2 \gamma} d\alpha=\sqrt \frac{\pi}{\gamma}## (that I had demonstrated in a previous exercise).
    All in all I solved the original integral, it's worth ##\sqrt{2\pi} \sigma ^3N^2##.
     
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