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Amplitude for a free particle

  1. Feb 15, 2009 #1
    I hope this is the correct place for my question. I posted it here, because it`s from Peskin & Schroeder:
    "Consider the amplitude for a free particle to propagate from [tex] \mathbf{x}_{0} [/tex] to [tex] \mathbf{x} [/tex] :

    [tex]U(t)=\left\langle \mathbf{x}\right|e^{-iHt}\left|\mathbf{x_{0}}\right\rangle[/tex]

    In nonrelativistic quantum mechanics we have E=p^2/2m, so

    [tex] U(t)&=&\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{x}_{0}\right\rangle [/tex]

    =\int\frac{d^{3}p}{(2\pi)^{3}}\left\langle \mathbf{x}\right|e^{-i(\mathbf{p}^{2}/2m)t}\left|\mathbf{p}\right\rangle \left\langle \mathbf{p}\right.\left|\mathbf{x_{0}}\right\rangle [/tex]

    =\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i(\mathbf{p}^{2}/2m)t}\cdot e^{i\mathbf{p}\cdot(\mathbf{x}-\mathbf{x}_{0})}

    =\left(\frac{m}{2\pi it}\right)^{3/2}\, e^{im(\mathbf{x}-\mathbf{x}_{0})/2t}}

    I don't understand the last equation.
    Why I can't use the fourier-transformation:

    =\frac{1}{(2\pi)^{3}}\int d^{3}p\, e^{-i\mathbf{p}\cdot(\mathbf{x}_{0}-\mathbf{x})}\widetilde{f}(\mathbf{p})


  2. jcsd
  3. Feb 15, 2009 #2
    but is what you have don actually ;)

  4. Feb 16, 2009 #3
    Sry, I don't understand. My solution is a different one as from Peskin & Schroeder. But why?
  5. Feb 16, 2009 #4
    The last step is made by completing the square and evaluating the gaussian integral (although with complex coefficient). I suspect the expression should be ~exp[im(x-x_0)^2/2t] though...

    The last step is precisely the evaluation of a fourier transform. In what you write next:

    It looks like you have simply taken [tex]\widetilde{f}=f[/tex] and replaced the [itex]\mathbf{p}[/itex] argument with [itex]\mathbf{x}-\mathbf{x}_0[/itex].. If so then this is wrong. The fourier transform should be evaluated as in the last step of your first equation (i.e. completing the square as I wrote)
  6. Feb 16, 2009 #5
    You are right.

    Thanks a lot.
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