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Quantum Mechanics, a free particle prepared as a gaussian wavepacket

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data
    The problem is given in the attached image. I'm currently trying to work out question one.

    2. Relevant equations

    [tex]\phi (k) = \dfrac{1}{2 \pi} \int_{- \infty}^{ \infty} \Psi (x,0) e^{-ikx} dx[/tex]

    3. The attempt at a solution

    Okay, so the first thing I did was to normalise it, but then I realised it was already normalised, so there's no need to do so there.

    Using the above relevant equation:

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-ax^2} e^{ik'x} e^{-ikx} dx[/tex]

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-ax^2 + ik'x -ikx} dx[/tex]

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{- \infty}^{ \infty} e^{-[ax^2 + (ik - ik')x]} dx[/tex]

    Using the hint, ## ax^2 + (ik - ik')x = u^2 - \dfrac{(ik - ik')^2}{4a} ## where ## u = \sqrt{a} \left( x+\dfrac{(ik-ik')}{2a} \right)##
    And therefore ## du = \sqrt{a} dx ##

    So then the equation becomes:

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{x = - \infty}^{ x = \infty} exp - \left( u^2 - \dfrac{(ik - ik')^2}{4a} \right) dx[/tex]

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \int_{x = - \infty}^{ x = \infty} e^{-u^2} e^{\dfrac{(ik - ik')^2}{4a}} \dfrac{du}{\sqrt{a}}[/tex]

    [tex]\phi (k) = \dfrac{1}{2 \pi} \left( \dfrac{2a}{\pi} \right)^{\frac{1}{4}} \dfrac{1}{\sqrt{a}} e^{\dfrac{(ik - ik')^2}{4a}} \int_{x = - \infty}^{ x = \infty} e^{-u^2} du [/tex]

    Now how do I change the bounds of integration? If there were still from negative infinity to positive infinity, I'd get ## \sqrt{ \pi} ## for the integral part, so is that right?

    I feel like it's not from negative infinity to infinity though.
  2. jcsd
  3. Apr 27, 2013 #2
    The bounds look correct, as your definition of u goes toward -∞ as x goes to -∞, and ∞ as x goes to ∞ (provided that [itex]\sqrt{a}[/itex] is taken to be positive, if it's not, the integral just becomes negative and the [itex]\sqrt{a}[/itex] in front of the integral will make it positive again).
    Last edited: Apr 28, 2013
  4. Apr 28, 2013 #3
    Okay thanks.

    I have done question 2 and this is what I got:

    [tex]\Psi (x,t) = \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} \right)}{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}}[/tex]

    where exp is the exponential function.
    I asked a friend and he said he also got something this messy for the answer.

    However, I run into trouble when I try to do the expectation values of ## x ## and ## x^2 ## for question 3.
    I remember my lecturer saying something about a way of solving this (if you know ##<x>## and ## \sigma_x ## then you can work out ## <x^2> ## without having to evaluate 2 more gaussian integrals.

    So to work out ##<x>## I use:

    [tex]<x> = \int^{\infty}_{- \infty} \Psi (x,t)^* (x) \Psi (x,t)dx[/tex]

    And of course I can extract out the constants so I get:

    [tex]<x> = \left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} \right) exp \left( \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)[/tex]

    [tex]<x> = \left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} + \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)[/tex]

    I don't really know what to do now because I have imaginary's on the top and bottom of the fraction. It doesn't really seem like a gaussian integral.

    Anyone know what I can do or if there's an easier/better way to do this (my lecturer said I can do all of question 3 on one page if I knew the shortcuts)? Doing the ## <x^2>## will be similar to this as well I presume.
  5. Apr 29, 2013 #4
    huh, how would I got about opening the attachment?
  6. Apr 29, 2013 #5
    I don't know, I'm sure I attached it before.

    Here it is again... hopefully it works

    Attached Files:

  7. Apr 29, 2013 #6
    I think your lecturer was talking about finding [itex]<x^2>[/itex] with the relation σ=[itex]\sqrt{<x^2> - <x>^2}[/itex]
    Last edited: Apr 29, 2013
  8. May 1, 2013 #7
    Last edited: May 1, 2013
  9. May 1, 2013 #8
    Yea I've already solved it... I just have to transform this:

    [tex]\left( \dfrac{\dfrac{1}{2 \sqrt{2a}} \left( \dfrac{2a}{\pi} \right) ^{\frac{1}{4}} }{\sqrt{\frac{1}{4a} + \frac{iht}{2m}}} \right)^2 \int^{\infty}_{- \infty} x \cdot exp \left( \dfrac{(ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} + \frac{iht}{2m}]} + \dfrac{(-ix - \frac{2k'}{4a})^2}{4[\frac{1}{4a} - \frac{iht}{2m}]} \right)[/tex]

    into the form [tex]\dfrac{1}{ \sigma \sqrt{2 \pi}} \int_{- \infty}^{\infty} e^{\frac{-(x- \mu)^2}{2 \sigma ^2}}[/tex]

    using a few algebra tricks and compare co-efficients. This is just the general equation for a normal distribution.

    The function is already normalised and so ## \mu ## is the expectation value of x and ## \sigma_x ^2 ## is the variance and I can use those 2 things to work out ## <x^2> ##
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