Amplitude to go from a location to another

[damosuz]

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

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

Where does this equation come from, especially the r₁₂ in the denominator?

 

[The_Duck]

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 $r_{12}^2$.

 

[Ravi Mohan]

The \vec{r}_{12}^2 dependence in the denominator comes due to the conservation of probability on an expanding spherical wavefront.

Mathematically, the amplitude to go from |\vec{r}_1\rangle to |\vec{r}_2\rangle is given by the Green's function
<br /> \langle\vec{r}_2|\hat{G}|\vec{r}_1\rangle,<br />
where, for a free particle, the Green's operator \hat{G} is given by
<br /> \hat{G} = \lim_{\epsilon \to 0^+}\left(\int_0^\infty dE&#039;\frac{|E&#039;\rangle\langle E&#039;|}{E-E&#039;+i\epsilon}\right).<br />
The term +i\epsilon is added to enforce outward going waves.

Now the explicit form of Green's function becomes
<br /> G(\vec{r}_1,\vec{r}_2) = \lim_{\epsilon \to 0^+}\left(\left\langle\vec{r}_1\left|\int_0^\infty dE&#039;\frac{|E&#039;\rangle\langle E&#039;|}{E-E&#039;+i\epsilon}\right|\vec{r}_2\right\rangle\right).<br />

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

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 \frac{\hbar^2k^2}{2m}.

You can also evaluate the probability current density vector and show a positive divergence from \vec{r}_1.

 

[stevendaryl]

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

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 \frac{\hbar^2k^2}{2m}.

You can also evaluate the probability current density vector and show a positive divergence from \vec{r}_1.

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

The time-dependent green's function

G(\vec{r}, t, \vec{r_1}, t_1)

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

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

where T = t - t_1 and R = |\vec{r} - \vec{r_1}|

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 G(R,T) as a superposition of states with definite energy:

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

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

G_k(R) = -\dfrac{m}{2 \pi \hbar^2} \dfrac{e^{i k R}}{R}

 

[damosuz]

Could we say that the r in the denominator is due to the fact that the amplitude is a spherical wave?

 

[Ravi Mohan]

Could we say that the r in the denominator is due to the fact that the amplitude is a spherical wave?

The presence of r in the denominator doesn't make the wave spherical. For a spherical wave, the phase must be independent of the polar and azimuthal angles (\theta and \phi).