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Equation calculating the amplitude of a particle

  1. May 11, 2008 #1
    If I am right, it's supposed to be this:

    When a particle moves from [tex]r_{1}[/tex] to [tex]r_{2}[/tex], the amplitude of the move can be written as:

    [tex]\langle r_{2} | r_{1} \rangle = \frac{e^{ipr_{12}/\hbar}}{r_{12}}[/tex]

    where [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are vectors of particle's position (I think?).
    and [tex]r_{12} = r_{2}-r_{1}[/tex] and is the scalar( I think again?)

    So my question is.. how can you substitute vectors to get a numerical value for the equation above?
  2. jcsd
  3. May 12, 2008 #2
    Anyone? =[
  4. May 12, 2008 #3


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    I'm not familiar with this equation, but it looks like you'd have to know:
    The particle's momentum, p
    The distance between the two points, r_12

    Then just plug the numbers in.

    Edit added:
    In the exponent, that looks like the dot product of 2 vectors, so you need not only the distance r_12, you actually need the displacement vector r2-r1. Likewise, you need to know the direction of the momentum.

    I think it's a scalar, |r_12|, in the denominator of your expression.
    Last edited: May 12, 2008
  5. May 12, 2008 #4
    I didn't actually recognize the expression up to now. It seems the Green's function for the free Schrodinger equation. See:

    http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Lippmann-Schwinger_equation [Broken]

    In this case the Green's function is actually isotropic, and so the product in the exponent is the product of a scalar and another scalar.
    Last edited by a moderator: May 3, 2017
  6. May 12, 2008 #5
    Yes, I just realized that, it was hard to see the bold in that equation (it's very small)
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