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Homework Help: Expectation Value/ Evaluating Gaussian Integral

  1. Oct 21, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm re-hashing a problem from my notes; given the wave function

    [tex]\psi(x)=Ne^{-(x-x_0)/2k^2}[/tex]

    Find the expectation value <x>.

    2. Relevant equations

    The normalization constant N for this is in my notes as [tex]N^2=1/\sqrt{2 \pi k^2}[/tex] [tex]N=1/(2\pi k^2)^{(1/4)}[/tex] It should be solved through a u substitution.

    3. The attempt at a solution

    [tex]<x>=\int \psi(x)^{*}x \psi(x) dx=\int x |\psi(x)|^{2} dx[/tex]
    [tex]=\int x N^2 e^{-(x-x_0)^2/k^2} dx[/tex]
    [tex]=N^2 \int x e^{-(x-x_0)^2/k^2} dx[/tex]
    [tex]u=x-x_0, du=x dx[/tex]
    [tex]<x>=N^2 \int e^{-u^2/k^2}du[/tex]
    This is where I stumble. Assuming I wrote the problem down correctly the expectation value should come out to <x>=x0. But I'm just not seeing it which leads me to believe that I'm not solving the final integral correctly.
     
    Last edited: Oct 21, 2011
  2. jcsd
  3. Oct 21, 2011 #2
  4. Oct 21, 2011 #3
    Woops, type-o on my part. I've edited the original post...the question should be correct now.
     
  5. Oct 21, 2011 #4
    where did that x go? you subbed u = x - xo but you didnt set x = to u + xo
     
  6. Oct 21, 2011 #5

    D H

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    This is obviously wrong.

    Try using u=(x-x0)2, or use the substitution you used (but do it correctly).
     
  7. Oct 21, 2011 #6
    If I set [tex]u=x-x_0[/tex] and say [tex]x=u+x_0[/tex] I then get [tex]xdx=du[/tex] don't I?
     
  8. Oct 21, 2011 #7
    Yeah, moment of clarity here... let me re-do it with [tex]u=(x-x_0)^2[/tex].
     
  9. Oct 21, 2011 #8
    no, just du = dx
     
  10. Oct 24, 2011 #9
    So in a fit of mathematical confusion I decided to take the problem all the way back to the beginning. And I might be closer to having done it properly...I hope.

    Given my wave function [tex]\psi(x)=Ne^{-(x-x_0)/2k^{2}}[/tex] I needed to determine the normalization constant N which should be found via [tex]\int|\psi(x)|^{2}dx=1[/tex] so I found

    [tex]\int|Ne^{-(x-x_0)/2k^{2}}|^{2}dx=1[/tex]
    [tex]\int|N^{2}e^{-(x-x_0)^2/k^2}|dx=N^{2}\int e^{-(x-x_0)^2/k^2}dx[/tex]
    My substitution was [tex]u=\frac{x-x_0}{k}, kdu=dx[/tex] and then
    [tex]N^{2}\int ke^{-u^2}du=N^{2}k\int e^{-u^2}du=N^{2}k\sqrt{pi}=1[/tex]
    [tex]N=\frac{1}{\pi^{1/4}\sqrt{k}}[/tex] and [tex]N^2=\frac{1}{\sqrt{\pi}k}[/tex]
    Similarly then, the expectation value <x> is
    [tex]<x>=\int \psi(x)^{*}x \psi(x) dx=\int x |\psi(x)|^{2} dx[/tex]
    [tex]=\int x N^{2} e^{-(x-x_0)^2/k^{2}} dx[/tex]
    [tex]=N^{2} \int x e^{-(x-x_0)^2/k^{2}} dx[/tex]
    [tex]u=\frac{x-x_0}{k}, kdu=dx[/tex]
    [tex]<x>=N^{2} \int xk e^{-u^{2}}du[/tex]
    [tex]<x>=N^{2} xk \int e^{-u^{2}}du=N^{2} xk \sqrt{pi}=\frac{kx\sqrt{\pi}}{\sqrt{\pi}k}=x[/tex]

    Success?
     
    Last edited: Oct 24, 2011
  11. Oct 24, 2011 #10

    vela

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    No, you made several errors. You can't pull x out of the integral because x is a function of u.

    What's the initial wave function? In one step, you essentially said (ex)2 = ex2, but that's incorrect since (ex)2 = e2x.
     
  12. Oct 24, 2011 #11
    It's clearly been a long time since I've done much with u-subs. How would I handle the x in the integrand? (Though that may be immaterial to the greater error with the exponent)

    The wave function
    [tex]\psi(x)=Ne^{-(x-x_0)/2k^{2}}[/tex]
    [tex]\psi(x)^{2}=N^{2}e^{-(x-x_0)/k^{2}}[/tex]
    This then brings me back to my original stumbling point of choosing an appropriate substitution.
     
  13. Oct 24, 2011 #12

    vela

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    The problem with that wave function is that it diverges as x goes to negative infinity. It's not normalizable. Are you sure you don't mean
    [tex]\psi(x) = Ne^{-(x-x_0)^2/2k^2}[/tex]
    If so, after you make the substitution, you find the normalization constant the way you did above.

    When finding <x>, you do the substitution, but you have to now write x in terms of u.
     
  14. Oct 24, 2011 #13
    Oh, wow, yes. I did mean [tex]\psi(x) = Ne^{-(x-x_0)^2/2k^2}[/tex] (clearly I've been looking at this too long) This would then square to [tex]\psi(x)^2 = N^2e^{-(x-x_0)^2/k^2}[/tex]. Is the normalization constant N I found in the first part correct (success despite myself...although I'm suspecting not)?
     
  15. Oct 24, 2011 #14

    vela

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    Yes, the two mistakes canceled each other out. :smile:
     
  16. Oct 24, 2011 #15
    Now that's refreshing, so wrong I was right! :smile:

    Ok, going back to the expectation value then; I have the dangling x in the integrand post substitution [tex]N^{2}k\int x e^{-u^{2}}du[/tex] how do I handle the x? Solve for x in terms of u? And since it's been about 7 years since I took Calc II I may need the basic lesson here.
     
    Last edited: Oct 24, 2011
  17. Oct 25, 2011 #16
    Try setting x in terms of u. I didn't follow the whole thread, but I believe u=(x-xo)/k. So solve for x in terms of u, put that in and the integral should be more manageable... though you'll still need integration by parts, no way around that.
     
  18. Oct 25, 2011 #17
    Right on,
    [tex]u=\frac{x-x_0}{k}[/tex]
    [tex]x=ku+x_0[/tex]
    And so,
    [tex]N^{2}k\int ku+x_0 e^{-u^{2}}du[/tex]
     
  19. Oct 25, 2011 #18

    vela

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    Use parentheses!
    [tex]N^2 k\int_{-\infty}^\infty (ku+x_0) e^{-u^{2}}\,du = N^2 k^2\int_{-\infty}^\infty ue^{-u^{2}}\,du + N^2 kx_0\int_{-\infty}^\infty e^{-u^{2}}\,du[/tex]Can you see by inspection why the first integral is equal to 0?
     
  20. Oct 25, 2011 #19
    Oh wow, that is fantastically elegant!
     
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