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Gaussian wave packet in position space

  1. Aug 29, 2015 #1

    KFC

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    Hi all,
    The Gaussian wave packet is widely discussed in the text. I got the following expression for wave packet in momentum space

    ##\psi(p, 0) = A \exp\left[-(p-p_0)^2/ (2\sigma_p^2)\right]##
    with ##A=\sqrt{2\sigma_p/\sqrt{2\pi}}##

    As my understanding, the corresponding wave packet in position space should be inverse Fourier transformation of ##\psi(p, 0)##. I plot the ##\psi(p, 0)## in matlab, I saw the Gaussian profile. But when I take the inverse Fourier transformation ifftshift(ifft(psi)), I saw something of the order ##10^{-6}##. I have no idea why is it.
     
  2. jcsd
  3. Aug 30, 2015 #2

    blue_leaf77

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    So long as it's normalized, what's the problem? It depends on how you set the parameter values anyway.
     
  4. Aug 30, 2015 #3

    KFC

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    The problem is after the numerical transformation, all samples for the packet is about ##10^{-6}## instead of ranged from 0 to ##10^{-6}## that means it is almost zero everywhere. Let me restate my question with some code and data. I think I made an mistake in the first post, here is the Guassian in momentum

    ##
    \psi(p) = \sqrt{\dfrac{1}{\sqrt{2\pi}\sigma_p}}\exp\left[-\dfrac{(p-p_0)^2}{4\sigma_p^2}\right]
    ##

    If you integrate this in mathematica, ##\int_{-\infty}^{+\infty} |\psi(p)|^2dp = 1##. I also tried to find the wave packet in position by inverse Fourier transformation, I got

    ##
    \phi(x) = \sqrt{\dfrac{2\sigma_p}{\sqrt{2\pi}}} \exp\left[-\sigma_p x^2\right]\exp(ip_0x)
    ##

    and ##\int_{-\infty}^{+\infty} |\phi(x)|^2dx = 1##. Plotting those functions with ##p_0=0, \sigma_p=1##, we observe that ##|\psi(p)|^2## with minimum and maximum as 0 and about 0.9; the minimum and maximum for ##|\phi(x)|^2## is about 0 to 1.2

    Now doing numerical evaluation by matlab, I try

    p0 = 0;
    sigma_beta = 1;
    psi = sqrt(1/sqrt(2*pi)/sigma_beta);*exp(-((p-p0).^2)/(4*sigma_beta^2));
    phi = ifftshift(ifft(psi));

    plot(abs(phi).^2);

    The range for the plot is about ##10^{-5}## and is we sum over the ##|\phi|^2##, I've got 5.9921e-04 while the sum of ##|\psi|^2## gives 1. See the attachment (don't know why it upload twice and seems I cannot remove one of them)
     

    Attached Files:

  5. Aug 30, 2015 #4

    blue_leaf77

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    I haven't checked your calculation, but I just want to confirm, have you considered that ##p=\hbar k##? Forgetting this may also affect the scaling of the wavepacket in position space, and hence the numerical integration.
     
  6. Aug 30, 2015 #5

    KFC

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    Thanks for your reply. I rescale the system with ##\hbar=1##. Moreover, in both analytical calculation (mathematica) and numerical one (matlab), I have the same scale (##\hbar=1##). so the result should be identical. But they are not. My question is why the numerical one doesn't give the same one from analytical expression.
     
  7. Aug 30, 2015 #6

    Avodyne

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    Your analytic formulas seem to be correct, provided that you set ##\hbar=1##. So you have a matlab problem, not a physics problem.
     
    Last edited: Aug 31, 2015
  8. Aug 31, 2015 #7

    DrClaude

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    Do you simply sum up the components, or do you also multiply by ##dx## or ##dp##?

    While ifft(fft(psi)) should recover the original psi, I don't know what normalization of the single action of fft or ifft should be. From what I can gather from the manual, there is a ##1/N## factor in ifft, with nothing for the forward fft, so that might be the origin of the scaling you see.
     
  9. Aug 31, 2015 #8

    KFC

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    Thanks for reply. Yes, I sum up all components and multiply it with dx or dp to test normalization.
    For fft part, yes, it has the 1/N in front of series, I've already taking that factor in consideration, but still no help.
     
  10. Aug 31, 2015 #9

    DrClaude

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    Staff: Mentor

    How?

    I don't see anything in there that involves N.
     
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