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

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

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

As a test, when you insert this amplitude (wavefunction) in the time-independent Schrödinger 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
$$-\frac{m}{2\pi\hbar^2}\frac{e^{i k|\vec{r}_1-\vec{r}_2|}}{|\vec{r}_1-\vec{r}_2|}.$$

As a test, when you insert this amplitude (wavefunction) in the time-independent Schrödinger 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 $\phi$).