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Homework Help: Find Wave Function given <x>, sigmax, <p>

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Come up with a wave function Psi[x] that satisfies the given known values:
    sigma x = 1
    <p> = h bar

    2. Relevant equations

    3. The attempt at a solution
    So far I have this equation, which satisfies <x>, <p>, but not sigma x.
    1/[Pi]^(1/4) E^(i (x + 2)) E^(-(1/2) (x + 1)^2)
  2. jcsd
  3. Sep 18, 2008 #2
    What does sigma x refer to?
  4. Sep 18, 2008 #3
    Sigma x is the standard deviation of x.
  5. Sep 18, 2008 #4
    Remember that the standard deviation of a gaussian is controlled by dividing the x^2 term by sigma^2.
  6. Sep 18, 2008 #5
    Try looking at the below wikipedia article on Gaussian or Normal distributions for a good way to create a wavepacket with a given uncertainty ([tex]\sigma_x[/tex]) and position [tex]\langle x\rangle[/tex].

    This wikipedia link should give an adequate definition:
  7. Sep 18, 2008 #6
    For some reason, I just can't find the function.

    I don't know how I could make a wave function that could possibly have a standard distribution of 1 and an expected value of -1, graphically.
  8. Sep 18, 2008 #7
    A Gaussian wavefunction has the form
    [tex] A\exp\left(-\frac{\left(x - x_0\right)^2}{2\sigma^2}\right)[/tex]
    where A is the normalization, [tex]x_0[/tex] is the center of the Gaussian, and [tex]\sigma[/tex] is the standard deviation.

    Try calculating
    [tex]\sigma = \sqrt{\langle x^2\rangle - \langle x\rangle^2}[/tex] to convince yourself of this property of Gaussians and then see if you can put the momentum part into the total wavefunction.
  9. Sep 18, 2008 #8
    I have tried this, the only problem is that the function must also have the momentum, -h bar, which requires another exponential term, exp(-i x).

    Also, sigma^2 would be 1, and that would have the form A * exp^(-i x)*exp^((-(x+1)^2)/2) where A = 1/ pi^(1/4)

    Because of the first exponential function, exp^(-i x), the standard deviation becomes 1/rad(2).

    How can I get around this?
  10. Sep 18, 2008 #9
    I'm unsure how you're getting the standard deviation to be [tex]1/\sqrt{2}[/tex] because of the complex exponential part. I'm assuming rad(2) is the square root.

    Don't forget that to calculate the expectation value you use the complex conjugate on the left, e.g.
    [tex]\langle x \rangle = \int \psi^{*}\left(x\right)x\psi\left(x\right)\,dx[/tex].
    So the complex exponential times its complex conjugate should just give 1 inside the integral and not affect the expectation value of [tex]x[/tex] and [tex]x^2[/tex].

    One more thing to consider, when you multiply two Gaussians together you get another Gaussian:

    [tex]A\exp\left(-\frac{\left(x-x_0\right)^2}{2\sigma^2}\right) * A\exp\left(-\frac{\left(x-x_0\right^2}{2\sigma^2}\right) = A^2\exp\left(-\frac{\left(x - x_0\right)^2}{\sigma^2}\right)[/tex]
    This about this relates the standard deviation of [tex]\psi\left(x\right)[/tex] to [tex]\psi^{*}\left(x\right)\psi\left(x\right)[/tex].
  11. Sep 18, 2008 #10
    Yeah, that's how I've been getting <x>.

    It must the be exp^(i x) that's causing the problems. The x in the exponent must be messing something up.

    How can I get around this?
  12. Sep 18, 2008 #11
    I'm pretty sure the complex exponential should cancel in the calculation of the x and x^2 expectation values.

    Remember what I mentioned about the product of Gaussians. This means that while the standard deviation of [tex]\psi\left(x\right)[/tex] might be [tex]\sigma[/tex] the standard deviation of [tex]|\psi\left(x\right)|^2[/tex] will be [tex]\sigma/\sqrt{2}[/tex]...
  13. Sep 18, 2008 #12
    Well, sigma x comes out to 1/rad(2) for the sqrt((integral from -inf to +inf of Psi* x^2 Psi) - (integral from -inf to +inf of Psi* x Psi)^2).

    How would I change the wave function to make sigma x = 1 then?
  14. Sep 18, 2008 #13
    Try changing the value of [tex]sigma[/tex] in the Gaussian part of the wavefunction and see how that changes what you calculate [tex]\sigma_x[/tex]. See if can figure it out own your own exactly what you need to change then.
    Also, remember the correct normalization of a Gaussian involves [tex]\sigma[/tex] as well. It should be in the wikipedia article.
  15. Sep 18, 2008 #14
    Oh that's awesome. Thanks, I changed 1/2 to 1/4 and then renormalized the wave function, and everything works perfectly.

    Thanks a lot.
  16. Sep 18, 2008 #15
    Glad I could help :)
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