# Can the process $u\overline{u} \rightarrow s\overline{s}$ be mediated by the EM interaction?

I’m working on a homework problem which asks for the dominant contribution (e.g. EM, strong, or weak) to the process $p + \overline{p} \rightarrow \Lambda + \overline{\Lambda}$).

I know that the EM interaction cannot change flavor, but I’m not sure how this rule applies to the $u\overline{u} \rightarrow s\overline{s}$ annihilation indicated in this diagram. Is it valid to say that since there is no ‘net’ flavor on either side, the photon could be the boson involved?

(This almost doesn’t matter for my purposes—since the process can be strong, I think, that’ll be my answer to the question anyway. I was still curious about whether this could be an EM process though.)

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I think that you are right to be suspicius, given that for instance pi0 to photons is classically forbidden, but here there is a lot more of particles so surely you can arrange for all the conservation laws to work. AS for when strong interactions are dominant, look for "OZI rule". –  arivero Sep 12 '11 at 23:44

You have to look at the actual vertices involved in the Feynman diagram, not just the net flavor. Here you have an up quark changing to a strange quark, which is a change of flavor. It doesn't matter that there's an anti-up quark changing to an anti-strange quark to balance it out.

When we say "the EM interaction cannot change flavor," what that really means is that there is no elementary vertex which couples the photon (the boson that mediates EM interactions) to two fermions of different flavor. In other words, any time a particle has a fundamental interaction with a photon, it has to come out with the same flavor it had going in. That's not the case in the diagram you've shown: the elementary vertex between the photon and the third quark of the top proton has a photon, an up quark, and a strange quark. That is a flavor-changing vertex involving a photon, exactly the kind that does not exist in the Standard Model.

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This isn't quite right. There are two diagrams. The one with space-like photons have a change of flavor at each vertex and are forbidden, those with time-like photons are a annihilation/creation at each vertex and and are allowed. Compare with Drell-Yan. –  dmckee Sep 12 '11 at 0:17
I don't see what you mean, there's only one (forbidden) diagram in the question. Which other diagram are you talking about? –  David Z Sep 12 '11 at 0:32
There is only one diagram in the question, but the process includes the rotated diagram. –  dmckee Sep 12 '11 at 0:46
Right, but I was only talking about the diagram in the question, not the process. I made a point of not mentioning other diagrams. –  David Z Sep 12 '11 at 1:14
What’s the physical difference between the diagram I posted and the rotated one? –  bdesham Sep 12 '11 at 3:32

Can't include pictures in a comment, so to clarify what David and I were discussing in the comments...

At vertex level you are interested in the process $u + \bar{u} \to s + \bar{s}$, though you are of course observing it in a baryonic context. The process can (possibly!) proceed along two tree level diagrams for each allowed boson:

The diagram on the left corresponds to the diagram you exhibit in the question and is disallowed for electromagnetic processes because both vertexes are flavor changing. The diagram on the right produces the same observables, but is allowed with a photon intermediary.

Higher order diagrams also contribute but are suppressed by the extra coupling constants.

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I’m guessing the process on the right is only applicable for the process $\pi^0 \rightarrow \phi$, and wouldn’t occur if the $u$, $\overline{u}$, etc. were parts of baryons? –  bdesham Sep 13 '11 at 1:11
No, it is generally applicable. The classic Drell-Yan process is the right hand diagram except that the strange quarks are replaced by leptons. Muon Drell-Yan was the primary physics probe of NuSea (E866) at Fermi-lab. Obviously it is energy limited in a context like $\pi_0$ decay. –  dmckee Sep 13 '11 at 1:55