Integral of (psi_0 * d/dx psi_1)

  • Thread starter Aziza
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  • #1
Aziza
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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!
 

Answers and Replies

  • #2
PhysicsGente
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3
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.
 
  • #3
Aziza
190
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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.

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
 
  • #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
Aziza
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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
 
  • #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
$$\int \frac{1}{\sin\theta}\,d(\sin\theta) = \frac{1}{\sin\theta}{\sin\theta}=1.$$
 
  • #8
Aziza
190
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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
$$\int \frac{1}{\sin\theta}\,d(\sin\theta) = \frac{1}{\sin\theta}{\sin\theta}=1.$$

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?
 
  • #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##.
 
  • #10
Aziza
190
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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##.

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
 
  • #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.
 
  • #12
PhysicsGente
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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

Yes, I'm sorry I overlooked the [itex]^2[/itex]. My bad.
 

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