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Finding the wave function for a Gaussian wave packet

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose a Gaussian wave packet ψ(x,0) is built out of plane waves according to the amplitude distribution function

    [itex]A_{k} = \frac{Ca}{\sqrt{\pi}}e^{(-a^2(k-k_{0} )^2)}[/itex]

    Calculate ψ(x,t) for this packet and describe its evolution.

    2. Relevant equations

    [itex]ψ(x,t) = ∫^{∞}_{-∞} A_{k} e^{i(kx - wt)}dk[/itex]

    Gaussian integrals:

    [itex]∫^{∞}_{-∞} e^{-Au^2}du = \sqrt{\frac{\pi}{A}}[/itex]

    [itex]∫^{∞}_{-∞} u^2 e^{-Au^2}du = \frac{\sqrt{\pi}}{2A^{3/2}}[/itex]

    3. The attempt at a solution

    I'm fairly certain I'm supposed to put the equation for ψ(x,t) into the form of one of the above Gaussian integrals and use those identities to integrate it. But I'm really stuck as to how to do this. This is what I have to begin with:

    [itex]ψ(x,t) = ∫^{∞}_{-∞} \frac{Ca}{\sqrt{\pi}}e^{(-a^2(k-k_{0} )^2)} e^{i(kx - wt)}dk[/itex]
     
  2. jcsd
  3. Mar 31, 2013 #2
    Any Gaussian integral can be solved by pretty much the same procedure. You should start by expanding out the exponents, and combining the exponentials to form a single polynomial. Then see if you can find a way to simplify the resulting integral.
     
  4. Apr 2, 2013 #3
    Thanks, I think I got it after doing an *insane* amount of algebra.
     
  5. Apr 2, 2013 #4
    Shouldn't have been all that insane...once you've assembled the polynomial, you just complete the square and shift the integration variable, and then use your Gaussian identity to evaluate the integral. Is that what you ended up doing, or was it something more complicated?
     
  6. Apr 5, 2013 #5
    Basically what I did, but used the first 3 or so terms of the Taylor expansion of w around k0 as a sub for w, I guess part of it was just that it was my first time doing something like this, so I made a few wrong turns :p
     
  7. Apr 5, 2013 #6
    Eesh. I think I may have mislead you with some poor wording choices--when I said "expand out the exponents", I didn't mean to do a Taylor expansion, I just meant to multiply out the quadratic term so that you could combine it with the other exponential.

    [tex]e^{-a^2(k-k_0)^2}e^{i(kx-\omega t)} = e^{-a^2k^2 + 2a^2kk_0 - a^2k_0^2}e^{ikx-i\omega t} = e^{-a^2k^2 + (2a^2k_0+ix)k - a^2k_0^2 -i\omega t}[/tex]

    Once you have it in that form, all Gaussian integrals can be solved by the same procedure:

    [tex]\int_{-\infty}^\infty dx\: e^{-\frac{A}{2}x^2 + Bx + C}[/tex]

    Remove constants, and factor in terms:
    [tex] = e^C \int_{-\infty}^\infty dx\: e^{-\frac{A}{2}(x^2 - \frac{2B}{A}x})[/tex]

    Complete the square:
    [tex] = e^C \int_{-\infty}^\infty dx\: e^{-\frac{A}{2}(x^2 - \frac{2B}{A}x + \frac{B^2}{A^2} - \frac{B^2}{A^2})} = e^C \int_{-\infty}^\infty dx\: e^{-\frac{A}{2}(x^2 - \frac{2B}{A}x + \frac{B^2}{A^2})}e^{\frac{B^2}{2A}} = e^{C+\frac{B^2}{2A}}\int_{-\infty}^\infty dx\: e^{-\frac{A}{2}(x - \frac{B}{A})^2}[/tex]

    Now shift integration variables to [itex]u=x - \frac{B}{A}[/itex]:
    [tex] = e^{C+\frac{B^2}{2A}} \int_{-\infty}^\infty du\: e^{-\frac{A}{2}u^2}[/tex]

    And finally, use your Gaussian integral identity [itex]\int_{-\infty}^\infty du\: e^{-\frac{A}{2}u^2} = \sqrt{\frac{2\pi}{A}}[/itex]:

    [tex] = \sqrt{\frac{2\pi}{A}}e^{C+\frac{B^2}{2A}}[/tex]

    Just plug in the right constants for [itex]A[/itex], [itex]B[/itex], and [itex]C[/itex], and you're done. You're going to be doing a lot of Gaussian integrals throughout your study of QM, so it's important to get comfortable with this procedure.

    Sorry again if I mislead you, hopefully this helps!
     
  8. Apr 5, 2013 #7
    Oh no, it wasn't because of you, it's just something the lecturer did in a similar problem. Since w is a function of k, but we aren't assuming that we know what w is, I approximated it with the first few terms of the Taylor expansion around k0. My answer was the same form as yours though, so I think I got the right answer. Thanks for your help :)
     
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