Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Amplitude to go from a location to another

  1. Jul 10, 2014 #1
    In the Feynman lectures on physics Vol.III (P.3-4), Feynman gives an equation for the amplitude for a free particle of definite energy to go from r1 to r2 to be proportional to

    [itex]\dfrac{e^{ip\cdot r_{12}/\hbar}}{r_{12}}.[/itex]

    Where does this equation come from, especially the r12 in the denominator?
     
    Last edited: Jul 10, 2014
  2. jcsd
  3. Jul 10, 2014 #2
    This amplitude comes from solving the Schrodinger equation, which I think Feynman introduces some time well after stating this formula.

    The ##r_{12}## in the denominator indicates that the probability (which is the square of the amplitude) falls off like ##1/r_{12}^2##. This is because the particle has an amplitude to go in any direction. So the probability of ending up at any point on a sphere of radius ##r_{12}## is the same. So the probability of ending up at a specific point has to fall like the surface area of that sphere, which goes like ##r_{12}^2##.
     
  4. Jul 11, 2014 #3
    The [itex]\vec{r}_{12}^2[/itex] dependence in the denominator comes due to the conservation of probability on an expanding spherical wavefront.

    Mathematically, the amplitude to go from [itex]|\vec{r}_1\rangle[/itex] to [itex]|\vec{r}_2\rangle[/itex] is given by the Green's function
    [tex]
    \langle\vec{r}_2|\hat{G}|\vec{r}_1\rangle,
    [/tex]
    where, for a free particle, the Green's operator [itex]\hat{G}[/itex] is given by
    [tex]
    \hat{G} = \lim_{\epsilon \to 0^+}\left(\int_0^\infty dE'\frac{|E'\rangle\langle E'|}{E-E'+i\epsilon}\right).
    [/tex]
    The term [itex]+i\epsilon[/itex] is added to enforce outward going waves.

    Now the explicit form of Green's function becomes
    [tex]
    G(\vec{r}_1,\vec{r}_2) = \lim_{\epsilon \to 0^+}\left(\left\langle\vec{r}_1\left|\int_0^\infty dE'\frac{|E'\rangle\langle E'|}{E-E'+i\epsilon}\right|\vec{r}_2\right\rangle\right).
    [/tex]

    Using calculus of residues, one can evaluate the total amplitude to be
    [tex]
    -\frac{m}{2\pi\hbar^2}\frac{e^{i k|\vec{r}_1-\vec{r}_2|}}{|\vec{r}_1-\vec{r}_2|}.
    [/tex]

    As a test, when you insert this amplitude (wavefunction) in the time-independent Schrodinger equation, you will see that it is the eigen-function with energy [itex]\frac{\hbar^2k^2}{2m}[/itex].

    You can also evaluate the probability current density vector and show a positive divergence from [itex]\vec{r}_1[/itex].
     
    Last edited: Jul 11, 2014
  5. Jul 11, 2014 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    I'm a little confused by the phrase "the amplitude to go from [itex]\vec{r_1}[/itex] to [itex]\vec{r_2}[/itex]" in the original post.

    The time-dependent green's function

    [itex]G(\vec{r}, t, \vec{r_1}, t_1)[/itex]

    is the amplitude for going from [itex]\vec{r_1}[/itex] at time [itex]t_1[/itex] to [itex]\vec{r}[/itex] at time [itex]t[/itex]. That function is of course time-dependent. For a free particle, it is given by:

    [itex]G(R,T) = \sqrt{\dfrac{m}{2\pi i \hbar T}} e^{i m R^2/(2 \hbar T)}[/itex]

    where [itex]T = t - t_1[/itex] and [itex]R = |\vec{r} - \vec{r_1}|[/itex]

    That formula looks very different from the one people have been talking about. I think that they are related as follows (but I don't actually know how to do the math to prove it):

    Do a Fourier transform to write [itex]G(R,T)[/itex] as a superposition of states with definite energy:

    [itex]G(R,T) = \frac{1}{2 \pi} \int d\omega\ G(R,\omega)\ e^{-i \omega t}[/itex]

    Then if we substitute [itex]\frac{\hbar k^2}{2m}[/itex] for [itex]\omega[/itex] in [itex]G(R,\omega)[/itex] to get [itex]G_k(R)[/itex], we have (I conjecture):

    [itex]G_k(R) = -\dfrac{m}{2 \pi \hbar^2} \dfrac{e^{i k R}}{R}[/itex]
     
  6. Jul 11, 2014 #5
    Could we say that the r in the denominator is due to the fact that the amplitude is a spherical wave?
     
  7. Jul 11, 2014 #6
    The presence of [itex]r[/itex] in the denominator doesn't make the wave spherical. For a spherical wave, the phase must be independent of the polar and azimuthal angles ([itex]\theta[/itex] and [itex]\phi[/itex]).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook