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Normalizing wavefunction of (x^2)e^(-x^2)?

  • #1
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Homework Statement



ψ(x,t)=Axe[itex]^{-cx^2}[/itex]e[itex]^{-iωt}[/itex]

Homework Equations



1=∫ψ*ψdx


The Attempt at a Solution



1=[itex]\int[/itex]A[itex]^{2}[/itex]x[itex]^{2}[/itex]e[itex]^{-2cx^2}[/itex]

I think I multiplied ψ*ψ correctly... I'm surprised a little that there is an x^2 with this problem. We have not discussed these integrals in class. I watched a video on how to integrate e[itex]^{-x^2}[/itex], but don't know what to do with the product x^2 and e^(-2cx^2)
 

Answers and Replies

  • #2
dextercioby
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Did you take statistical mechanics ? You learn about integrating gaussians there. If not, then at least a course in mathematical methods of physics should have explained you this.

Anyways, think I(c) = int_R exp(-cx2) dx. Compute it then differentiate the result wrt c. (c is a parameter).
 
  • #3
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I'll give you a hint.
I presume you know that: [itex]\int_{- \infty}^{+\infty} e^{-\lambda x^2}dx = \sqrt{\pi/ \lambda}[/itex]. Consider the function:
[tex]I(\lambda )=\int_{- \infty}^{+\infty}e^{-\lambda x^2}dx[/tex]
Then:
[tex]\frac{d}{d\lambda }I(\lambda )=-\int_{- \infty}^{+\infty}x^2e^{-\lambda x^2}dx[/tex]
 
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  • #4
vela
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You could also integrate by parts.
 
  • #5
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Thanks for the help, guys! Didn't take statistical mechanics, unfortunately...

Does this mean that the derivative of [itex]\sqrt{\frac{\pi}{\lambda}}[/itex] is my answer?
 
  • #6
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vela is correct, but kostas230's trick is one of the most beautiful in mathematics and one of the most important in physics. The technique is referred to as using a "generating functional."
 
  • #7
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I don't know if you ever had to evaluate the Gaussian integral, but here's an elegant way to do it. Consider the integral:

[tex]I= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-(x^2+y^2)}dxdy[/tex]

By changing from cartesian coordinates to polar coordinates, the integral above becomes:
[tex]I= \int_{0}^{+\infty}\int_{0}^{2\pi}re^{-r^2}drd\theta[/tex]

I trust that you are able to evaluate the integral. Can you find the gaussian integral [itex]I'=\int_{-\infty}^{+\infty}e^{-x^2}dx[/itex] using the above results? ;)
 
  • #8
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Thanks kostas230, appreciate your help:smile:... The result should be in the form of a Hermite polynomial, is that correct? I think the derivative of re[itex]^{-r^2}[/itex] is

e[itex]^{-r^2}[/itex](1-2r[itex]^{2}[/itex])

Not quite sure what to do next
 
  • #9
96
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No, I said: compute the integral, not find the derivative of the integrated function
 
  • #10
186
2
If you are asking me to find the integral of re[itex]^{-r^2}[/itex]drd[itex]\theta[/itex] from 0[itex]\leq[/itex]r[itex]\leq[/itex][itex]\infty[/itex] and 0[itex]\leq\theta[/itex][itex]\leq[/itex]2[itex]\pi[/itex] I believe it is [itex]\pi[/itex]. And then the integral with a coefficient λ in the power would be [itex]\frac{\pi}{\lambda}[/itex]
 
  • #11
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That is right, and it allows you to find the value of the original Gaussian integral, if you find the relation between that and the formula kostas230 used.
 
  • #12
186
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Thank you, mfb.

So is it that I[itex]^{2}[/itex]= [itex]\frac{\pi}{\lambda}[/itex]

So the derivative of [itex]\frac{\pi}{\lambda}[/itex] is -[itex]\frac{\pi}{\lambda^2}[/itex]?

Since the derivative would be 2I for I[itex]^{2}[/itex], should I divide -[itex]\frac{\pi}{\lambda^2}[/itex] by 2, so that I=-[itex]\frac{\pi}{2\lambda^2}[/itex] ?

Although A is a positive value.. would the negative sign in front of kostas' integral cancel this one, so that I =[itex]\frac{\pi}{2\lambda^2}[/itex] ?
 
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  • #13
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So the derivative of [itex]\frac{\pi}{\lambda}[/itex] is -[itex]\frac{\pi}{\lambda^2}[/itex]?
I think you are mixing two different approaches here.
 
  • #14
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My first post was on an algorithmic process in finding integrals of the form: [itex]I_n=\int_{-\infty}^{+\infty}x^n e^{-\lambda x^2}dx[/itex] presuming that you know the integral [itex]I_0=\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}[/itex]. My other posts where to show you a way to compute the integral [itex]\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}[/itex].
 

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