Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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|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]

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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
    This then brings me back to my original stumbling point of choosing an appropriate substitution.
  13. Oct 24, 2011 #12


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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,
    And so,
    [tex]N^{2}k\int ku+x_0 e^{-u^{2}}du[/tex]
  19. Oct 25, 2011 #18


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook