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Free particle has a Gaussian wave packet wave function.

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    This is problem 2.22 from D.J. Griffiths Introduction to Quantum Mechanics

    A free particle has the initial wave function:

    [itex]\Psi(x,0)[/itex]=A[itex]e^{-ax^{2}}[/itex]

    Find [itex]\Psi(x,t)[/itex]. Hint Integrals of the form:
    [itex]\int_{-\infty}^{\infty}[/itex][itex]e^{-(ax^{2}+bx)}dx[/itex]

    can be handled by completing the square: Let [itex]y\equiv \sqrt{a}[x+(b/2a)][/itex], and note that [itex](ax^{2}+bx)=y^{2}-(b^{2}/4a)[/itex].


    2. Relevant equations

    [itex]\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k)e^{i(kx-\omega t)}dk[/itex]

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

    [itex]\omega=\frac{\hbar k^{2}}{2m}[/itex]

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    So I found [itex]\phi(k)=\left(\frac{1}{2\pi a}\right)^{1/4}e^{-k^{2}/4a}[/itex].

    Plugging this into my eq for [itex]\Psi(x,t)[/itex] I get the following:

    [itex]\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty} e^{-k^{2}/4a}e^{i(kx-(\hbar k^{2}/2m)t)}dk[/itex]

    [itex]=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty}exp[-\left(\left(\frac{i\hbar t}{2m}+\frac{1}{4a}\right)k^{2}-ikx\right)]dk[/itex]

    Now here is where I get stuck. I feel like I need to do another completing the square manipulation to argument of the exponential,but I am having trouble seeing how the obtained the following solution:

    [itex]\Psi(x,t)=\left(\frac{2a}{\pi}\right)^{(1/4)}\frac{e^{-ax^{2}}/[1+(i2\hbar at/m}{\sqrt{1+(i2\hbar at/m)}}[/itex]

    Any help would be greatly appreciated. Seems as though Professor Griffiths has some real cute tricks up his sleeve. Thanks in advance.

    Joe
     
    Last edited: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2

    vela

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    If you pull out a factor of 1/4a in the coefficient of the quadratic term, the exponent will be equal to
    $$-\left[\left(1+\frac{i2\hbar a t}{m}\right) \frac{k^2}{4a} - ikx\right]$$ To save you some writing, it would convenient to define ##\beta = 1+ i2\hbar at/m## since that quantity appears in the expression you're trying to derive. So you want to complete the square on
    $$-\left(\frac{\beta}{4a}k^2 - ikx\right)$$
     
  4. Feb 26, 2012 #3
    Ah ha! Thanks a lot Vela.
     
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