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Matter wave / quantum stuff

  1. Sep 22, 2004 #1
    G'day guys,
    Just looking for a bit of help.... I'm not sure that I fully understand the question here either.... but here goes:

    a) Using the Fourier integral,
    [tex] \Psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}a(k)e^{ikx}dk[/tex]
    Show that a matter wave having a wave-vector distribution given in the diagram below (see attached), has the form:
    [tex]\Psi(x)=\sqrt{\frac{\Delta k}{2\pi}} \frac{\sin{\frac{\Delta kx}{2}}}{\frac{\Delta kx}{2}} e^{ik_0 x}[/tex]

    (note that [itex]\Delta k[/itex] is a constant in the above diagram (attached))

    b) Calculate the probability of finding a particle given by the above wavefunction in the region [itex] -\infty<x<+\infty[/itex]

    For part a) I'm just assuming that I am supposed to work out the Fourier integral for the given function a(k), and that should work out to what's above right??.... Whenever I do that, however I can't get it! Is this the right way to approach this? Or is there simply a mistake somewhere in my workings? (My maths is a bit scratchy at the moment) Here it is..... (I'm new to this LaTex game too by the way, so please be gentle)

    [tex] \Psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}a(k)e^{ikx}dk[/tex]

    [tex]\Psi(x)=\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{k_0 -\frac{\Delta k}{2}}a(k)e^{ikx}dk + \int_{k_0 -\frac{\Delta k}{2}}^{k_0 +\frac{\Delta k}{2}}a(k)e^{ikx}dk + \int_{k_0 +\frac{\Delta k}{2}}^{+\infty}a(k)e^{ikx}dk\right][/tex]

    [tex]\Psi(x)=\frac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{k_0 -\frac{\Delta k}{2}}(0)e^{ikx}dk + \int_{k_0 -\frac{\Delta k}{2}}^{k_0 +\frac{\Delta k}{2}}(1)e^{ikx}dk + \int_{k_0 +\frac{\Delta k}{2}}^{+\infty}(0)e^{ikx}dk\right][/tex]

    [tex]\Psi(x)=\frac{1}{\sqrt{2\pi}} \int_{k_0 -\frac{\Delta k}{2}}^{k_0 +\frac{\Delta k}{2}}e^{ikx}dk [/tex]

    [tex]\Psi(x)=\frac{1}{\sqrt{2\pi}}\left[\frac{1}{ix}e^{ikx}\right]_{k_0 -\frac{\Delta k}{2}}^{k_0 +\frac{\Delta k}{2}}[/tex]

    Am I making the mistake here? ^^^ Is the integral correct?

    [tex]\Psi(x)=\frac{1}{\sqrt{2\pi}}\left[\frac{1}{ix}\left(e^{i(k_0 +\frac{\Delta k}{2})x}-e^{i(k_0 -\frac{\Delta k}{2})x}\right)\right][/tex]

    [tex]\Psi(x)=\frac{1}{ix\sqrt{2\pi}}\left(e^{ik_0 x +i\frac{\Delta k x}{2}}-e^{ik_0 x -i\frac{\Delta k x}{2}}\right)[/tex]

    [tex]\Psi(x)=\frac{1}{ix\sqrt{2\pi}}\left(e^{ik_0 x}e^{i\frac{\Delta k x}{2}}}-e^{ik_0 x}e^{-i\frac{\Delta k x}{2}}}\right)[/tex]

    [tex]\Psi(x)=\frac{1}{ix\sqrt{2\pi}}\left[e^{ik_0 x}\left(e^{i\frac{\Delta k x}{2}}}-e^{-i\frac{\Delta k x}{2}}}\right)\right][/tex]

    [tex]\Psi(x)=\frac{1}{ix\sqrt{2\pi}}\left[e^{ik_0 x}\left(\sin{\frac{\Delta k x}{2}}\right)\right][/tex]

    Where do I go from here? This is obviously not correct, but I just don't know what is going on....... I would love some guidance!

    I haven't attempted part b) yet, but I'm assuming the solution will come from


    and should equal 1 ??

    Any help would be great, thanks,


    Attached Files:

    Last edited: Sep 22, 2004
  2. jcsd
  3. Sep 22, 2004 #2
    SInce most of the problms you see in this forum are based on elementary physics, i suggest try posting in the quantum forum you may be more successful with getting a response.
  4. Sep 22, 2004 #3


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    Science Advisor

    To stunner5000p: No, if the is homework, then it should be here, not in "Quantum Physics". (In fact, if he posted it there, it would probably get moved here.)

    To Tyco05: [itex]sin(a)= \frac{e^{ix}- e^{-ix}}{2i}[/itex]. You should not have that "i" in the denominator of your final answer. Other than that, you answer is exactly the same as
    [tex]\Psi(x)=\sqrt{\frac{\Delta k}{2\pi}} \frac{\sin{\frac{\Delta kx}{2}}}{\frac{\Delta kx}{2}} e^{ik_0 x}[/tex]
    except for that
    [tex]\sqrt{\Delta k}[/tex]
    You have only [itex]\Delta k[/itex] without the square root. Check the answer again.
  5. Sep 22, 2004 #4
    Thanks HallofIvy,

    I just made that sin(x) identity incorrectly...... oh well, some things get past us.... but the [itex] \sqrt{\Delta k} [/itex] is definitely in the question I was given. Maybe it is a typo, or perhaps a genuine mistake? When I worked it through I also got what you said, just [itex]\Delta k[/itex] .
    I'll leave it as is unless you have any other suggestions, and just see what happens!

  6. Sep 22, 2004 #5
    Hey again,

    Just checked another source (Serway, Moses and Moyer. Modern Physics.) and it is definitely a typo or mistake.

    Instead of what was written on the assignment:

    [tex]\Psi(x)=\sqrt{\frac{\Delta k}{2\pi}} \frac{\sin{\frac{\Delta kx}{2}}}{\frac{\Delta kx}{2}} e^{ik_0 x}[/tex]

    It should have been :

    [tex]\Psi(x)=\frac{\Delta k}{\sqrt{2\pi}} \frac{\sin{\frac{\Delta kx}{2}}}{\frac{\Delta kx}{2}} e^{ik_0 x}[/tex]

    Thanks again.
  7. Sep 22, 2004 #6
    OK, this question is getting on my nerves now.
    I tried part b), only to have to integrate by parts over and over and over and over again. It won't stop. What's going on??......



    [tex]\Psi(x)=\frac{\Delta k}{\sqrt{2\pi}} \frac{\sin{\frac{\Delta kx}{2}}}{\frac{\Delta kx}{2}} e^{ik_0 x}[/tex]

    Now, I found that :

    [tex]\mid\Psi(x)\mid^2 = \frac{2}{\pi x^2} \sin^2{\frac{\Delta k x}{2}}[/tex]


    [tex]P(x)=\int_{-\infty}^{+\infty}\frac{2}{\pi x^2} \sin^2{\frac{\Delta k x}{2}}dx[/tex]

    [tex]P(x)=\frac{2}{\pi}\int_{-\infty}^{+\infty}\frac{1}{x^2}\sin^2{\frac{\Delta k x}{2}}dx[/tex]

    [tex]P(x)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{1}{x^2}(1-\cos{\Delta k x})dx[/tex]

    [tex]P(x)=\frac{1}{\pi}\left[\int_{-\infty}^{+\infty}\frac{1}{x^2}dx-\int_{-\infty}^{+\infty}\frac{1}{x^2}\cos{\Delta k x}dx\right][/tex]

    Is there something I should realise about these integrals?.... The one on the right is the one that just goes on and on and on....

    The one on the left seems easy enough, but what about the limits.... ? Can it be evaluated ? I'm confused....
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