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The free particle

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\psi(x,0)=Ae^{-ax^2}[/tex]
    Normalize and find:
    [tex]\psi(x,t)[/tex]


    2. Relevant equations
    [tex]1=\int_{-\infty}^\infty\psi^*\psi dx[/tex]


    3. The attempt at a solution
    [tex]1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx[/tex]
    let:
    [tex]u=x\sqrt{2a}[/tex]

    [tex]1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}[/tex]
    Therefore:
    [tex] A=(\frac{2a}{\pi})^{1/4}[/tex]

    [tex]\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}[/tex]
    This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.
     
  2. jcsd
  3. Oct 5, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Your value of [itex]A[/itex] is correct; which leaves you with

    [tex]\psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}[/tex]

    How did you go from that, to your equation for [itex]\psi(x,t)[/itex]?
     
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