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Homework Help: Use of Fourier Transform in Quantum Mechanics

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    The solution of Schrodinger’s equation for a free particle can be written in the form:

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

    [Q1]: Explain why the function [tex] \phi(k) [/tex] is given by:

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

    [Q2]: A particle is described by a wavefunction such that:

    [tex] \psi(x,0) = Ne^{-\alpha|x|}[/tex] , where N is a normalisation constant.

    i) Find [tex] \phi(k) [/tex] and find [tex] | \phi(k) |^{2} [/tex]

    ii) By thinking of the value of k at which [tex] | \phi(k) |^{2} [/tex] begins to rapidly decrease with k, estimate the most probable range of momentum values for this state.

    2. Relevant equations

    (Within the first section and solution attempts)

    3. The attempt at a solution

    [Q1]: [tex] \psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-wt)}dk [/tex]

    t=0 --> [tex] \psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x,0)e^{ikx}dk [/tex] which is the inverse Fourier transform of [tex]\phi(k)[/tex]

    Therefore [tex][\phi(k)][/tex] is the Fourier transform of [tex]\psi(x,0)[/tex]

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

    .. correct method?!

    [Q2]:

    i)

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

    and:

    [tex] \psi(x,0) = Ne^{-\alpha|x|}[/tex]

    therefore:

    [tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}Ne^{-\alpha|x|}e^{-ikx}dx [/tex]

    [tex] \phi(k) = \frac{N}{\sqrt{2\pi}}\int_{-\infty}^{0}e^{\alpha x}e^{-ikx}dx + \int_{0}^{\infty}e^{- \alpha x}e^{-ikx}dx[/tex]

    .. after some calculations ..

    [tex] \phi(k) = \frac{2N}{\sqrt{2\pi}}\left( \frac{\alpha}{\alpha^{2} + k^{2}} \right)[/tex]

    Hence:


    [tex] | \phi(k) |^{2} = \frac{4N^{2}}{2\pi}}\left( \frac{\alpha}{\alpha^{2} + k^{2}} \right)^{2}[/tex]


    ii) I have no idea how to attempt this part of the question at the moment.
     
  2. jcsd
  3. Feb 27, 2010 #2

    vela

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    Looks good so far.

    Question ii is worded badly. If you plot [itex]|\phi(k)|^2[/itex], you'll see it's symmetric about k=0 and decreases rapidly initially as you move away from k=0. It flattens out as you move farther away from the origin, so the rate of decrease slows. What the problem should ask you to find is where [itex]|\phi(k)|^2[/itex] stops, not begins, rapidly decreasing. The last sentence makes clear what you're supposed to find, namely a characteristic width for the curve.
     
  4. Feb 27, 2010 #3
    I sort of get what to do, not really sure though. :frown:

    ..something like:

    set k=0 therefore:

    [tex] | \phi(0) |^{2} = \frac{4N^{2}}{2\pi}}\left( \frac{1}{\alpha} \right)^{2}[/tex]

    the graph looks like a Gaussian/normal distribution/'bell curve' type shape, and then most of the measurements will be in the region where [tex]|\phi(k)|^{2}[/tex] is largest, i.e. over a region of [tex]\pi[/tex]

    so:

    [tex]\frac{4N^{2}}{2\pi \alpha^{2}}} \simeq \pi[/tex]

    .. hopefully that's something along the right lines?!?
     
  5. Feb 27, 2010 #4

    vela

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    This particular function is called a Lorentzian.
    How did you come up with "over a region of π"?
     
  6. Feb 27, 2010 #5
    Just a quick thing about the graph..

    1. Obviously it is symmetrical.. (about [tex]\left(\frac{2N^{2}}{\pi \alpha^{2}}\right) = 0[/tex]) ??

    2. There is a central largest peak, which I assumed reaches the x-axis (both sides) at [tex]\pm \pi[/tex] ?? Then, does it just stop or is there then another smaller peak either side of the main one? Hopefully that makes sense!

    I wasn't sure what the region would be, so I guessed [tex]\pi[/tex] as in between [tex]
    -\frac{\pi}{2}[/tex] and [tex]\frac{\pi}{2}[/tex]

    I don't really get what the question is asking, it's not worded brilliantly I don't think. Could do with some help on this last part, seeing how the rest seems to be ok then.
     
  7. Feb 27, 2010 #6

    vela

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    Why would you assume this? What do you get when you plug k=pi into the function? It has to be 0 if it hits the x-axis there. Are there terms or factors that would cause the function to equal 0 when k=pi?

    Plot the function and see for yourself what it looks like. You have a parameter alpha that you can play around with. Try starting with alpha=1. See what happens to the curve when alpha=2 and alpha=1/2. Figure out why what happens makes sense given the function you have.
     
  8. Feb 27, 2010 #7
    Right, Ok.. hopefully know what you're getting at..

    Firstly as there is no k in the expression for [tex]|\phi(x)|^{2}[/tex], I assume that can just replace k with [tex]\alpha [/tex]? So then..

    Just plotted a little graph of [tex]|\phi(x)|^{2}[/tex] against [tex]\alpha[/tex], and got the result of a lorentzian curve, with a central peak at [tex]\alpha = 0[/tex], which rapidly decreases down towards [tex]|\phi(x)|^{2} = 0 [/tex] in terms of [tex]\frac{1}{\alpha^{2}}[/tex] dependance.

    The only way I can see that the function would reach [tex]|\phi(x)|^{2} = 0[/tex] is when [tex]\alpha \implies \infty [/tex] such that can then say effectively [tex]|\phi(x)|^{2} = \frac{1}{\infty} = 0[/tex]

    I don't see how there would be any secondary peaks.

    .. but then need to find the point at which the function stops rapidly decreasing, and I don't see how any of this has helped to solve that.
     
  9. Feb 27, 2010 #8

    vela

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    Where did you get [itex]\phi(x)[/itex] from? It's [itex]\phi(k)[/itex]. And no, you can't just randomly replace one variable with another.
    I have no idea what this means.
    Why should there be?
    You need to learn how to analyze these types of questions and that requires an understanding of how functions behave. A good way to do this is to experiment by varying parameters to see how they affect the graph and then to understand why it makes sense.
     
  10. Feb 27, 2010 #9
    1. It should indeed be [tex]|\phi(k)|^2}[/tex] .. I too have no idea where [tex]|\phi(x)|^2}[/tex] came from!

    2. The dependance thing was just my way of saying that the function [tex]|\phi(k)|^2}[/tex] decreases at a rate of [tex]\frac{1}{\alpha^{2}}[/tex] since none of the other values/variables in the equation change, so for example:

    [tex]\alpha = 2 \implies |\phi(k)|^2} = \frac{2N^{2}}{\pi}\left(\frac{1}{4}\right)[/tex]

    [tex]\alpha = 3 \implies |\phi(k)|^2} = \frac{2N^{2}}{\pi}\left(\frac{1}{9}\right)[/tex]

    [tex]\alpha = n \implies |\phi(k)|^2} = \frac{2N^{2}}{\pi}\left(\frac{1}{n^{2}}\right)[/tex]

    Hopefully that's a bit clearer now as to what I meant, not that it matters that much though anyway.

    3. I was just pointing out that after plotting the function with varying [tex]\alpha[/tex], I now see there is no secondary peaks and just the one, which is symmetrical about the point [tex]|\phi(k)|^2} = 0[/tex]

    4. I know that I need to find the point at which the function stops rapidly decreasing.. but I don't know how to find this value without just making it wherever I determine it to be.. so I assume there is a method for this, that I don't know of :|
     
  11. Feb 27, 2010 #10

    vela

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    There's actually another factor of alpha floating around in there: if you find N, you'd find it's equal to sqrt(alpha). Also, you're talking about the maximum of the function, not its width.

    Anyway, I take it you haven't actually plotted the function yet.
     
  12. Feb 27, 2010 #11
    I've plotted a graph of [tex]|\phi(k)|^2}[/tex] against [tex]\alpha[/tex], but not for N against [tex]|\phi(k)|^2}[/tex]
    .

    .. sorry to be stupid, but how to I find N? I don't really get your last post.
     
  13. Feb 27, 2010 #12

    vela

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    You can't plot a graph of "[itex]|\phi(k)|^2}[/itex] against [itex]\alpha[/itex]." For each value of [itex]\alpha[/itex], you get a different graph of [itex]|\phi(k)|^2}[/itex] vs. k. The idea is to see how the shape changes as you change [itex]\alpha[/itex].

    You find it by normalizing the wavefunction, but it doesn't really matter here. Just use N=1 for now.
     
  14. Feb 28, 2010 #13
    Why am I finding the maximum not it's width?! Surely the maximum value of the function is at k=0. I thought I have to find something like the full width at half maximum.

    I'm confused now as to what I'm meant to be trying to do.
     
  15. Feb 28, 2010 #14

    vela

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    I never would have suggested you plot this function if I knew how difficult this was going to be. Let's just forget plotting the function and take a different tack.

    When k is "small," the function looks like

    [tex]\phi(k) \approx \frac{2N}{\sqrt{2\pi}}\left( \frac{\alpha}{\alpha^{2}} \right)[/tex]

    When k is "big," it looks like

    [tex]\phi(k) \approx \frac{2N}{\sqrt{2\pi}}\left( \frac{\alpha}{ k^{2}} \right) \rightarrow 0[/tex]

    What's the criterion for determining when k is 'small' and k is 'big.'
     
  16. Feb 28, 2010 #15

    vela

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    I've attached plots of |phi(k)|2 for three different values of alpha. I normalized them so they have the same maximum to make them easier to compare. The red, green, and blue curves correspond, respectively, to values of alpha equal to 2, 1, and 1/2.
     

    Attached Files:

  17. Feb 28, 2010 #16
    .. I'm not sure :|

    Well, doesn't [tex]|\phi(k)|^2}[/tex] give the 'spread' of the momentum of the particle, and the momentum is just given by: [tex]p=\hbar k[/tex].
     
  18. Feb 28, 2010 #17

    vela

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    No, [itex]|\phi(k)|^2 dk[/itex] is the probability that the particle has a momentum between k and k+dk.
     
  19. Feb 28, 2010 #18
    Thanks for the graph, it is very useful, I now get that varying the value of [tex]\alpha[/tex] changes the shape of the [tex]|\phi(k)|^2}[/tex] curve, or rather more specifically changes the width of the peak.

    So going back to the question, which is to find a characteristic width, which is the distance between either side of [tex]k=0[/tex] where the curve stops 'rapidly decreasing'.

    Since the function/curve is symmetric, then just need to consider one side and determine the value at which the function stops 'rapidly decreasing'? It is finding this that I do not understand how to do. I thought maybe when [tex]\alpha = k[/tex], or something like that. I really don't get what to do.
     
  20. Feb 28, 2010 #19

    vela

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    Don't get too hung up on the "rapidly decreasing" detail. The problem just wants you to estimate how wide the peak is. That's why it says to "think about" it rather than calculate an explicit expression for the width.

    When you're at the peak, the curve's derivative is 0, so the curve is flat there. A small change in k therefore results in a small change in phi. As k gets larger, however, the curve falls off more quickly, but the curve again flattens out when k is large. So there's a small range of values of k over which the curve rapidly decreases. Looking at the graph, you can see that this range essentially defines how wide the peak is.
    OK, so you've seen that the value of [itex]\alpha[/itex] affects the width of the peak. Now how would you say it does that? For example, does the width vary linearly with [itex]\alpha[/itex]? Is it 10 times [itex]\alpha[/itex]? You're not looking for an exact answer, but just a good enough description of how the width depends on [itex]\alpha[/itex].

    Take a look at the graphs again. Let's somewhat arbitrarily though reasonably say the peak is defined between the points where the curve falls to 1/4 its maximum value. What are the widths you get for the three curves and how are they related to the [itex]\alpha[/itex] for each curve?
     
  21. Feb 28, 2010 #20
    Considering the graph..

    For the blue line, the value of 1/4 of the curve height is at 1/2, and this blue line corresponds to alpha = 1/2.

    For the green line, the value of 1/4 of the curve height is at 1, and this blue line corresponds to alpha = 1.

    For the red line, the value of 1/4 of the curve height is at 2, and this blue line corresponds to alpha = 2.

    .. so the width are directly related to alpha?
     
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