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Quantum Wave Function to Momentum Amplitude

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data
    I was hoping someone could verify that I've set up the integral correctly for this problem..

    Suppose that at t=0, the wavefunction of a free particle is
    [tex]ψ(x,0) = \sqrt{b} e^{-|x|b + ip_0 x/\hbar} [/tex]

    a) what is the momentum amplitude for this wave function?

    2. Relevant equations
    *see below

    3. The attempt at a solution

    So, I think I know how to complete this, and just want to make sure I'm setting up the integral correctly before I dive into solving it.

    a) I will use [tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_∞^∞ ψ(x,0) e^{-ipx/\hbar}\,dx [/tex] ---> that's minus ∞ to ∞

    so, for my wavefunction...
    [tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_\infty^\infty \sqrt{b} e^{-|x|b + ip_0 x/\hbar} e^{-ipx/\hbar}\,dx [/tex] ---> again, -∞ to ∞

    Here, it looks like i can't cancel [itex]e^{ip_0 x/\hbar}[/itex] and [itex]e^{-ipx/\hbar} [/itex], and I'm stuck with what looks like a very nasty integral...

    [tex]\phi(p,0)= \frac{1}{\sqrt{2\pi\hbar}}\int_\infty^\infty \sqrt{b} e^{-|x|b} e^{ix/\hbar (p_0-p)}\,dx [/tex] ---> again, -∞ to ∞
    Last edited: Sep 12, 2014
  2. jcsd
  3. Sep 12, 2014 #2
    Often the thing to do when you have an absolute value is to break whatever it is into piece-wise segments. Will that work here?
  4. Sep 12, 2014 #3
    Yeah, that's what I was planning on doing - dividing it into two integrals, one from 0 to infinity (|x|=x) and one from -infinity to 0 (|x|=-x).
  5. Sep 13, 2014 #4


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    Nasty? It's essentially integrating ##e^{-kx}## where ##k## is a constant.
  6. Sep 13, 2014 #5
    ok, so for the integral I got..

    [tex]\phi(p,0)= \frac{\sqrt{b}}{\sqrt{2\pi\hbar}} \frac{2b}{b^2 + (p_0-p)^2/\hbar^2}[/tex]

    Now, since my original wavefunction was normalized, is this [itex]\phi(p,0) [/itex] already normalized?

    The next part had me find the probability for finding the momentum between p and p+dp, so I just found what [itex]|\phi(p,0)|^2[/itex] equaled.
    After some simplification, this equaled [itex]\frac{2 b^3 \hbar^3}{\pi (b^2 h^2+p_0^2-2 p_0 p +p^2)^2}[/itex] ----> so this is the probability that I will find a momentum between a small momentum range.

    If all looks good up until this stage, I'll get started on the next parts having me find the values of the uncertainties Δx and Δp. Right now, I'm trying to find a way to do it with the things I've already solved for
    Last edited: Sep 13, 2014
  7. Sep 13, 2014 #6
    ok, so I plan on using this equation to find Δp at t=0

    [tex]<(p-p_0)^2> = \int_∞^∞ (p-p_0)^2 |\phi(p,0)|^2 dp [/tex]

    Then, [itex] Δp = \sqrt{<(p-p_0)^2>} [/itex]
    Last edited: Sep 13, 2014
  8. Sep 13, 2014 #7
    Then for Δx at t=0, I could do

    [tex]<(x-x_0)^2> = \int_∞^∞ (x-x_0)^2 |\psi(x,0)|^2 dx [/tex]

    Then, [itex] Δx = \sqrt{<(x-x_0)^2>} [/itex]

    I don't see that particular equation in my book, but it's just the root mean square, so it should work for finding Δx?

    Now, when I multiply these together, I should get [itex]\Delta x \Delta p >= \frac{\hbar}{2}[/itex]
    Last edited: Sep 13, 2014
  9. Sep 13, 2014 #8


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    Actually, the probability to find the particle with momentum between p and p+dp is ##\lvert \phi(p,0) \rvert^2 dp##. The amplitude squared alone isn't the probability.

    You might find it a little less work to use ##\Delta p^2## = ##\langle p^2 \rangle - \langle p \rangle^2## and ##\Delta x^2## = ##\langle x^2 \rangle - \langle x \rangle^2##.
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