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Integral of (psi_0 * d/dx psi_1)

  1. Oct 2, 2013 #1
    I need help determining if the following statement is mathematically valid, where psi_0 and psi_1 are the first two stationary states of the SHO:


    http://af10.mail.ru/cgi-bin/readmsg?id=13807553990000000798;0;1&mode=attachment&bs=2783&bl=281383&ct=image%2fjpeg&cn=mms_picture.jpeg&cte=base64

    Basically what I did was cancel the dx's as if it was algebraic equation, and then since my integration became wrt psi_1, I changed my limits from 0->0, since at +/- infinity, psi_1 must be zero. And of course if your limits of integration are same, result must be zero. Specifically, I would like to know where exactly (at which of my above steps) is the flaw? (there must be a flaw since I do not get correct solution when I use this to calculate other stuff...this integral should in fact not be zero).

    If I replace psi_1 by psi_0, for example, then I think this is correct, since I did it on my recent QM test and got full credit.

    Any help with this is appreciated!








    Any help is appreciated!
     
  2. jcsd
  3. Oct 2, 2013 #2
    Hello,

    [itex] \psi_{1}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n}n!}}2\xi e^{-\frac{\xi^{2}}{2}} [/itex]. And [itex]\xi = \sqrt{\frac{m\omega}{\hbar}}x[/itex]

    So it does not vanish at minus infinity.
     
  4. Oct 2, 2013 #3
    How does it not vanish at minus infinity? first, it is a wavefunction, so by the QM postulates it must vanish at +/- inifinity for it to be a physical solution.

    second, just looking at the formula above you can see it does vanish, since at -∞ you basically have the quantity (-∞)*exp(-∞)...this is equal to zero as you can prove with l'hopital rule
     
  5. Oct 3, 2013 #4

    Borek

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    Aziza, we can't see the image attached to an email hosted at mail.ru. Please post it in the thread.
     
  6. Oct 3, 2013 #5

    jtbell

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  7. Oct 3, 2013 #6
    sorry i did not realize that my picture could not be seen. Here is the equation:


    ∫ψ0[itex]\frac{d}{dx}[/itex]ψ1dx = ∫ψ01 = ψ0ψ1|[itex]^{0}_{0}[/itex] = 0

    Where the first integral runs from -inf to +inf, and the second from 0 to 0, since my integration is now wrt dψ1 and ψ1 is zero at +/- inf
     
  8. Oct 3, 2013 #7

    vela

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    Here's a similar specific example of what you did:
    $$\int \cot\theta\,d\theta = \int \frac{1}{\sin\theta}\cos \theta\,d\theta = \int \frac{1}{\sin\theta}\,\frac{d}{d\theta}\sin \theta\,d\theta = \int \frac{1}{\sin\theta}\,d(\sin\theta).$$ So far so good, but from here, you claimed that
    $$\int \frac{1}{\sin\theta}\,d(\sin\theta) = \frac{1}{\sin\theta}{\sin\theta}=1.$$
     
  9. Oct 3, 2013 #8
    no...then using my method above i would say sinθ=u and so integrate wrt u.....so the answer is ln(sinθ) by my method.


    the reason i just pulled out the ψ0 was because the integration is wrt ψ1....so ψ0 is treated as constant...is there an error here?
     
  10. Oct 3, 2013 #9

    vela

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    I did exactly what you did where ##\psi_0 = \frac{1}{\sin \theta}## and ##\psi_1 = \sin\theta## and where ##\theta## plays the role of ##x##.
     
  11. Oct 3, 2013 #10
    ohh im sorry yes i see what you mean...so basically if i dont know (or actually just dont plug in, in this case) the actual psi's and their dependence on x, then there is no way to do this integral
     
  12. Oct 3, 2013 #11

    vela

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    The Hermite polynomials have certain properties that might be helpful to evaluate the integral. You could try looking into that.
     
  13. Oct 3, 2013 #12
    Yes, I'm sorry I overlooked the [itex]^2[/itex]. My bad.
     
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