Minimum distance for annihilation

1. Jun 1, 2014

How close does does a particle and anti-particle pair have to be with each other in order to achieve annihilation?

2. Jun 1, 2014

my2cts

My best guess is that the closer they are the more likely annihilation gets. In positronium the separation is typically 1 ångström. For parallel spins and the lifetime is 0.12 ns and for antiparallel 0.14 µs. In excited states where the expected distance is larger, the lifetime increases.

3. Jun 1, 2014

kurros

It depends on many factors, such as the relative momentum of the two particles, their masses, what particles they are etc. But I think roughly speaking you should just imagine the two particles as wave-packets, and if those wave-packets overlap, then there is some probability that the particles will annihilate. So typically they will need to be separated by a distance of the same order as their wavelength.

4. Jun 2, 2014

Bill_K

Cross sections and reaction rates in QED are conveniently expressed in terms of a distance:

$$r_0 = \frac{e^2}{mc^2}$$
which goes by the (most unfortunate!) name of "classical electron radius." Its value happens to be 2.82 x 10-13 cm.

Also, if a particle is placed in a box of side a, its wavefunction at the origin is order of magnitude |ψ(0)|2 = 1/a3.

These two remarks lend intuitive support to the following answer obtained from QED:
The e+e- annihilation probability per unit time is

$$\Gamma = \frac{r_0^2 \,c}{a^3}$$
where a is their average distance apart. For positronium, a is approximately the Bohr radius, 10-8 cm, and if you put these values together you'll get the positronium lifetime that my2cts quoted, 0.1 ns.

Last edited: Jun 2, 2014
5. Jun 2, 2014

Meir Achuz

Close enough for their wave functions to overlap.

6. Jun 2, 2014

Bill_K

Seems obvious, but actually not! Look at the Feynman diagram, there's two vertices x1 and x2. The electron arrives at point x1 and emits a photon, the positron arrives at point x2 and emits a photon. In between, a virtual particle. The amplitude is obtained by integrating over all x1, x2, but there's no requirement that they coincide.

7. Jun 2, 2014

Meir Achuz

Your Feynman diagram is for free particles. Include spatial bound state wave functions in the full calculation, and then "close enough for their wave functions to overlap" is relevant.
That is why P waves don't annihilate, but S waves do, with the rate proportional to |\psi(0)|^2

8. Jun 2, 2014

Bill_K

It's not "my" Feynman diagram. But I appreciate the offer!

"Relevant", I guess, but not accurate. The Feynman diagram approach still works. Just take the result for plane waves and integrate it over the momentum distribution for a bound state. Doesn't change the fact that there are two vertices that need not coincide.

P-wave states most certainly do annihilate, even though |ψ(0)|2 = 0. However the rate is suppressed by the usual factor for the centrifugal barrier, (p/mc)2L, since the particles spend more of their time farther apart.

For a p-wave, L = 1, this factor is basically v2/c2, or the ratio of the potential energy to the rest energy, 6 eV/0.5 MeV, about 10-5. Instead of nanoseconds, the lifetime for p-wave annihilation is therefore in the microsecond range. But the radiative decay to s-wave via electric dipole transition takes place in 10-8 sec. So the direct annihilation from p-wave is perfectly possible, but has too small a branching ratio to be observed.

Last edited: Jun 2, 2014
9. Jun 3, 2014

snorkack

Is it a bare prediction, or is direct p-wave annihilation something that can be and has been observed, and measured to have a branching ratio matching the predictions?