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Fourier Transform of wavefunction - momentum space

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    2q9g8lh.png

    Find possible momentum, and their probabilities. Find possible energies, and their probabilities.

    2. Relevant equations



    3. The attempt at a solution

    First, we need to Fourier transform it into momentum space:

    [tex]\psi_k = \frac{1}{\sqrt{2\pi}} \int \psi_x e^{-ikx} dx [/tex]
    [tex] = \frac{1}{\sqrt{2\pi \hbar}} \int \psi_x exp(-i\frac{p}{\hbar}x) dx[/tex]

    (Why is there an extra factor of ##\sqrt \hbar## at the denominator?)

    [tex] = \frac{A}{\sqrt{2\pi \hbar}} \int \left[ cos (\frac{p}{\hbar}x) + cos(\frac{2p}{\hbar}x)\right] exp (-i\frac{p}{\hbar}x) dx [/tex]

    Now the fourier transform of ##cos (qx)## is ##\pi(\delta_{(k-q)} + \delta_{(k+q)})##:

    [tex]\frac{A\pi}{\sqrt{2\pi \hbar}}\left[ \delta_0 + \delta_{(2\frac{p}{\hbar})} + \delta_{(\frac{p}{\hbar})} + \delta_{(3\frac{p}{\hbar})}\right] [/tex]

    [tex] \psi_{(p)} = A \sqrt{\frac{\pi}{2\hbar}}[/tex]

    It's strange that I got a constant..

    I tried a different approach to see if it will work:

    The wavefunction is a combination of state 'k' and '2k':

    Probabilities of getting each state is simply ##A^2##.
    The two possible momenta are ##\hbar k## and ##2\hbar k##.

    Likewise for energy, probabilities are simply ##A^2##.
    The two possible energies are ##\frac{\hbar k^2}{2m}## and ##\frac{\hbar (2k)^2}{m}##.
     
  2. jcsd
  3. May 7, 2014 #2

    strangerep

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    Science Advisor

    You seem to be using "k" inconsistently. It was a constant in the question, but here you're using it as a variable.

    And now you switch to a "p" variable without explanation. You need to write down the definitions of Fourier transform (and inverse) that you're working from. (It should have been in the "relevant equations" section, which you left empty.) Doing that might also help answer your question about ##\sqrt \hbar## at the denominator.

    Also, the original question didn't specify the domain of x. Is it supposed to be the whole real line, or just a finite region?

    That looks wrong. Why has the original "k" in ##\psi(x)## disappeared?

    Well, that's because you got it wrong (afaict). Attend to some of the things I mentioned above, and then re-do the FT, showing the detail of your attempt here (i.e., don't just quote the result you get).

    Another approach is this: what is the representation of the momentum operator? What are its eigenfunctions? Can you express the original wave function easily in terms of such eigenfunctions?
    (Of course, that's what an FT is doing, but maybe there's a shortcut in this case...)
     
  4. May 9, 2014 #3
    I think I got it.

    The eigenfunctions of momentum operator are obtained by solving:

    [tex]-i\hbar \frac{\partial}{\partial x} u_p = p u_p[/tex]

    [tex]u_p = C exp(-i\frac{p}{\hbar}x)[/tex]



    To normalize, we say ##\langle p|p\rangle = 1##.

    img1174.png

    Therefore ##u_p = \frac{1}{\sqrt{2\pi \hbar}}exp(i\frac{p}{\hbar}x) = \frac{1}{\sqrt{2\pi \hbar}} cos (\frac{p}{\hbar}x)##. Simply by comparing we deduce ##k = \frac{p}{\hbar}##.

    Thus the wavefunction given is in a state that's the superposition of ##\hbar k## and ##2\hbar k## states.

    This implies that the possible momenta are ##\hbar k## and ##2\hbar k##.

    Equal probabilities of getting either.

    Energy is found by using ##E_1 = \frac{p^2}{2m}## or ##E_2 = \frac{4p^2}{2m}##
     
  5. May 9, 2014 #4
    You can't jump from the exponential to the cosine willy-nilly like that. What happened to the imaginary part? Do you think it's OK to through it away without an explanation?
     
  6. May 9, 2014 #5
    Observables corresponding to hermitian operators are real. So we take the real part.
     
  7. May 9, 2014 #6

    strangerep

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    Science Advisor

    Wrong terminology. Observables are hermitian operators (well, actually self-adjoint operators, but I won't get into that distinction now). There is a theorem that eigenvalues of hermitian operators on a Hilbert space are real. Perhaps that's what you were mis-remembering?

    Think more carefully about what dauto said. You seem to have missed the possibility that p could be either +ve or -ve. You need a linear combination of your basic ##u_p## eigenstates to get each cos function in ##\psi(x)##.
     
  8. May 10, 2014 #7
    You meant to say that the eigenvalue of an observable is real.
    The wave function is nether an observable nor an eigenvalue. You can't just drop the imaginary part. Try expressing the cosine as a combination of exponential functions.
     
  9. May 10, 2014 #8
    The cosine is ##\frac{1}{2}\left(u_{p} + u_{-p}\right)##. Combination of forward and backward momentum
     
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