1
$\begingroup$

I've attached a Feynmann diagram of two photons scattering with the nuclear reaction: γ + γ --> γ + γ

enter image description here

May I ask how would one interpret this diagram?

My attempt in interpreting is that the photon in the bottom left interacts with the photon in the top left by some form of virtual particle (shown by the black arrow line, which I can't seem to wrap my head around what virtual particle can exist between two photons scattering; in fact can photons scatter?). However, for some reason the two photons disappear into some virtual particles, and then reappear back in the bottom right and top right corners.

May I also ask is it correct to interpret any particles that begin and end in a basic interaction vertex shown in a Feynmann diagram to be virtual particle (as is the case in my interpretation)?

$\endgroup$
  • $\begingroup$ Two photons (diagonal on the diagram) decay each into a virtual electron-positron pair. The pairs cross-annihilate producing two new photons. $\endgroup$ – safesphere Aug 24 '18 at 20:46
  • $\begingroup$ @safesphere Thanks for the answer, but are you saying that the photons in the bottom left and top right each decay into a virtual electron-position pair, of which the pair annihilate to produce two new photons? But I thought Feynmann diagrams are read from left to right by convention, since time increases from left to right. Therefore, we should consider the two photons in the left as the 'reactant' of the nuclear reaction? $\endgroup$ – Bøbby Leung Aug 24 '18 at 20:52
  • 1
    $\begingroup$ I think you can be interpret it either way. The way you have it drawn, the upper left photon is absorbed by the electron produced by the lower left photon. And the lower right photon is emitted by the positron produced by the lower left photon. Then the electron and positron annihilate at the upper right corner. I think this makes no difference how you interpret the process, as long as the overall energy and momentum conserve. Hopefully QFT experts will clarify with formulas. $\endgroup$ – safesphere Aug 24 '18 at 21:12
1
$\begingroup$

As far as my understanding of QED it goes like this:

All we can know are the initial and final states: 2 initial and 2 final states of definite momentum.

What happens in between is: everything. However, that is too much, so we do a perturbative expansion of the amplitude. This diagram is the leading order, where the photons scatter off of a virtual electron-position pair (or any other charged particle, but let's stick with $e^+e^-$).

An important feature of Feynman diagrams is that four-momentum is conserved at all vertices: hence the electrons is off-shell:

$$ p^{\mu}p_{\mu} \ne m_e^2 $$

That is, it's a virtual particle.

So what is the four momentum? Based on the earlier statement that "everything" can happen, it can have any four momentum, as long as it's conserved at the vertices. So, you have to integrate over $d^4p$.

Regarding the interpretation the 2 photons are absorbed by a particle, and then later readmitted: that may be misleading.

In the t-channel (scattering), the diagram is not time-ordered. The exchange particle has a space-like four-momentum and the Feynman diagram represents 2 old-school time-ordered diagrams. ("A" emits a photon that "B" then absorbs and "B" emits a photon that "A" then absorbs).

In the s-channel (annihilation) the diagram represents both cases:

1) The initial state particles annihilate into a virtual particle which then decays into the final state particles

2) the final state particles are emitted while creating a virtual particle which is then is destroyed by absorbing the initial state particles.

So I would assign an order to the operations in a Feynman diagram with some trepidation. The key is to include all crossing symmetries (the u-channel) because the final state photons are identical particles. (I think you have to permute all the indices to compute the leading order amplitude--so if you can time order the emission and absorption, why bother, it's only half, or a quarter, of the story anyway.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much for the detailed answer, but may I ask what do you mean by time-ordered in "In the t-channel, the diagram is not time-ordered"? Also, in the first case of the s-channel you proposed, you mentioned "the initial state particles annihilate into a virtual particle." But how can two photons annihilate when photons are particles and also their own antiparticles. $\endgroup$ – Bøbby Leung Aug 25 '18 at 0:48
  • $\begingroup$ before Feynman (and Schwinger, Tomonaga, and Dyson), there was old fashion perturbation theory. It took a year to compute $e^-e^+\rightarrow e^-e^+$. They had diagrams. Many of them, because they were time ordered. You had to include different diagrams for a virtual electron-positron-photon appear and the photon annihilates the exiting particles, leaving different real ones, on top of normal annihilation to a photon which becomes a pair. Both those processes are in 1 Feynman diagram. Likewise for scattering: the particle exchange doesn't have a time-direction. See: Mandelstam Variables. $\endgroup$ – JEB Aug 25 '18 at 2:28
  • $\begingroup$ @JEB I do not think they had diagrams before Feynman. I had the fortune to meet Feynman at a workshop and hear him describe how the concept of feynman diagrams came to him whole. Ihave shared it here physics.stackexchange.com/questions/14028/seeing-the-solution/… $\endgroup$ – anna v Aug 25 '18 at 5:23
  • $\begingroup$ @annav Awesome. I took his last class (QCD), and he didn't mention it. I thought OFPT diagrams conserved momentum but not energy at the vertex with different ones for particle/anti-particle, so they were not manifestly covariant, and even a tree-level was process was many diagrams with many integrals. Then Feynman does his thing, and the tree level diagram is basically short-hand for the answer, and as you said: he just blew every one's mind. Especially crossing symmetry, where, e.g., Compton scattering is just annihilation with $t \rightarrow s$. I was barley hanging on in that class, though. $\endgroup$ – JEB Aug 25 '18 at 14:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.