Does annihilation differs from a pair production from two interacting particles?

Question

By reading this: "Can two photons annihilate?" and answers to that question, I have came to strange question - Does annihilation in principle is different than a pair production process ? My intuition says that at least energy boundary conditions needed for pair-annihilation and pair-production should be different. That's why we need LHC collider to form more interesting particle pairs. But reverse process - annihilation - should go much easier at lower energy levels, isn't it ?

But, if we for example take a Feynman diagram of two photon decoupling into fermion-antifermion pair :

we can find a somewhat similar process, only involving just 1 photon in quantum vacuum fluctuations:

Given that vacuum fluctuations are spontaneous process, then 2-photon breaking into fermion-antifermion pair doesn't seem to have a high energy requirements too. This may show that not necessary for a pair production we need high energies. And if low energy is enough sometimes, then How to differentiate a pair-production from two interacting particles from a pair-annihilation ? (If it's different at all). That's the main question, second is that I still haven't understood - Do a pair of photons annihilate or not ? Thanks.

Answer 1

How to differentiate a pair-production from two interacting particles from a pair-annihilation ? (If it's different at all). That's the main question,

The beauty of Feynman diagrams is that they can be read with different definitions of the time and space axes. It all depends on what one defines as input four vectors , and output four vectors.

Your diagram above can be read as a photon scattering on an electron, and a photon and electron leaving ( compton scattering) . Or for time going from right to left, it is electron positron annihilation. In general the term annihilation is used in the case where the incoming particle quantum numbers add to zero so that any type of new pairs could come out if the energy is enough.

In this sense , if your read the diagram from left to right, you can say that the photons are annihilating. The photons should have enough energy to be able to create real fermions. In general photon photons scattering is highly improbable that is why light beams may superimpose but not scatter.

Your second diagram with the fermion antifermion loop has the fermions off mass shell, so it is a useful diagram for calculations of two particle interactions, but the loop cannot be measured: the gamma has mass zero, the two fermions if not virtual would have an invariant mass.

second is that I still haven't understood - Do a pair of photons annihilate or not

Not if they do not have enough energy to create a new pair. That is the reason the term annihilation is not used for photons. For low energy photons two photon interactions due to the 1/137 of the electromagnetic coupling have very low probability, because they have to go through a fermion loop.

Answer 2

"Pair creation" is usually something that happens in curved backgrounds (in gravity, cf Hawking radiation) or in electro-magnetic fields (Schwinger effect), because the background has enough energy so that the process of pair-creation, which is otherwise just a way to depict the vacuum of a quantum theory, can actually give rise to two observable particles.

Two photons cannot desintegrate into a single massive particle, for kinematic reasons ; try to do a simple conservation of energy-momentum and you'll see the problem. You need at least two other particles, as in your Compton-type diagram #1. If you have enough energy (you're above the mass threshold), a single photon can give rise to a pair electron/positron, which is essentially the pair creation process of describe above. The energy you need is the same as the one you obtain when an electron and positron meet to disintegrate. They could annihilate into a single graviton, but the cross section is very small (suppressed by the Planck mass). They could not disintegrate into a gluon because of color charge conservation constraints.

The reason why the LHC needs to go to very high energy is to be able to create very massive particles as intermediate states, and for this to happen you need high collision energy. The result of the disintegration has the same energy as the incoming particles.

If a Higgs boson was seen to wander around in your lab, it would disintegrate as something very energetic, as energetic as the process which would have created it. Physics is invariant by CPT conjugation (you reverse charge, time and parity, and the interactions are the same).