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Particle Released From Narrow Potential - Fourier Transform

  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data

    A free particle moving in one dimension is initially bound by a very narrow potential well at the origin. At time ##t = 0## the potential is switched off and the particle is released; its wave function is:

    ##\psi (x,0) = N e^{-\frac{|x|}{\lambda}}##

    where λ is a positive real constant giving the decay length of the initial wave function away from the origin.

    (a) Calculate N.

    (b) The wave function ##\psi(x, 0)## can be written as a superposition of stationary states, which for a free particle are planes waves,

    ##\psi (x,0) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{\infty} dk c(k) e^{ikx}##

    Calculate c(k) by applying the inverse Fourier transformation.

    (c) How does ##c(k)## behave (i) if the initial spread of the wave function is very small (that is: if ##\lambda## is very small), and (ii) if ##\lambda## is very large? Bearing in mind that for a free particle ##k## is ##\frac{p}{\hbar}##, can you see a connection between the dependence of ##c(k)## on ##λ##, the width of the initial distribution, and the Heisenberg uncertainty principle?

    2. Relevant equations


    3. The attempt at a solution

    I am stuck with part c of this question.

    I find that ##N = \frac{1}{\sqrt{\lambda}}##, so that ##\psi (x,0) = \frac{1}{\sqrt{\lambda}} e^{-\frac{|x|}{\lambda}}##

    ##c(k) = \frac{1}{\sqrt{2 \pi} \sqrt{\lambda}} (\frac{2 \lambda}{1 + \lambda^2 k^2})##

    I think that for very small ##\lambda##, ##(1 + \lambda^2 k^2) \approx 1##, so ##c(k) \approx \frac{1}{\sqrt{2 \pi} \sqrt{\lambda}} 2 \lambda##. So for small ##\lambda##, ##c(k)## is also small.

    For large ##\lambda##, ##(1 + \lambda^2 k^2) \approx \lambda^2 k^2## whick seems to suggest that ##c(k)## is also small for large ##\lambda##.

    I don't think i'm seeing the physics of what's going on here, and would really appreciate a conversation about the situation.

    Thank you!
     
  2. jcsd
  3. Mar 2, 2016 #2

    blue_leaf77

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    I think this part of the problem asks you to investigate the behavior of the width of ##|c(k)|^2## for the two cases pertaining ##\lambda## mentioned above.
     
  4. Mar 2, 2016 #3
    Ok, that makes sense.

    Using the ideas of Full Width Half Maximum, I find that ##c(k)_{max} = \sqrt{\frac{2}{\pi}} \sqrt{\lambda}## and thus ##\frac{1}{2} c(k)_{max} = \frac{\sqrt{\lambda}}{\sqrt{2 \pi}}##.

    the values of k that give me the half maximum values are ##k = \pm \frac{1}{\lambda}##, making the width of the function at FWHM ##\frac{2}{\lambda}## which corresponds to a broader peak for small values of ##\lambda## and a narrower peak for large values of ##\lambda##.

    Regarding the relation to Heisenberg's uncertainty principle, when the question says "the width of the initial distribution" is it talking about the wave function or the inverse Fourier transform?
     
  5. Mar 2, 2016 #4

    blue_leaf77

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    When mentioning the "width", the question refers to ##\lambda## which has a unit of length, therefore it must be talking about ##\psi(x,0)##. By the way, when calculating the width, you should be using the probability distribution instead of the wavefunction.
     
  6. Mar 2, 2016 #5
    Earlier you said that I should calculate ##|c(k)|^2##. Is that equivalent to both calculating the inverse Fourier transform of ##|\psi (x,0)|^2## and taking the modulus squared of the fourier transform of ##|\psi (x,0)|##?

    i.e, is it safe to take the modulus squared of what I calculated, or should I have started with the modulus squared of my wave function?
     
  7. Mar 2, 2016 #6

    blue_leaf77

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    It's not "safe", that way is indeed one of the correct ways to obtain ##|c(k)|^2## - calculate ##c(k)## first, which you have done, and then take its modulus square.
     
  8. Mar 2, 2016 #7
    Ok, thank you.

    I find the width of ##|c(k)|^2 = \frac{2}{\lambda}## at FWHM.

    I find the width of ##|\psi (x,0)|^2 = \frac{1}{\lambda}## at FWHM.

    I can't see the link between what I have done and the Heisenberg uncertainty principle. Has the fourier transform told me what the function looks like in momentum space?
     
  9. Mar 2, 2016 #8

    blue_leaf77

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    That cannot be true, the width of ##\psi(x,0)## should have a dimension of length. In the Heisenberg uncertainty principle, ##\Delta x## and ##\Delta p## are actually the standard deviations of the probability functions in position and momentum spaces. So, if you want to be exactly following the Heisenberg principle, you should calculate the standard deviations instead of the FWHM widths. But I think part c) just want you to verify that the widths, whatever formula used to calculate them, of the probability function in the two spaces are reciprocal of each other.
     
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