# Why is this particle interaction forbidden?

Why is this interaction forbidden? $$\nu_e+\bar{\nu}_e\rightarrow K^+ + K^-$$ Lepton number is conserved, charge is conserved, baryon number is irrelevant since these are mesons. Energy is conserved since we could collide these two neutrinos (and anti) at any energy required to produce the two mesons. Neutrinos only interact via the weak force so if it were possible it would be a weak decay. Parity needn't be conserved since this would be a weak interaction. The only thing that is bugging me is that quark flavour (and strangeness) is not conserved but I thought this was allowed via a weak interaction. Maybe it is momentum?

EDIT

I think this process is allowed and here is my attempt at the lowest order Feynman diagram, does this look correct?

EDIT

• ? What makes you suspect quark flavor, including strangeness, is not conserved?? – Cosmas Zachos May 13 '16 at 0:43
• How did you conclude the decay is forbidden? Could e+ e- produce two pseudoscalar mesons weakly through a Z, in principle? – Cosmas Zachos May 13 '16 at 0:58
• I thought that that having quarks in the products but not in the reactants meant quark flavour not conserved but I guess thats silly. Also my mistake strangeness is conserved. I am just not sure how i would draw the feynman diagram. Could you have a look at the original post, I have added an attempt at the lowest order diagram, could you tell me if it looks ok? – NormalsNotFar May 13 '16 at 7:51
• @NormalsNotFar I'm confused about why you include the diagram in your question. Are you asking about the general process $\nu_e \bar\nu_e \to K^+ K^-$, or are you asking about the specific interaction shown in the diagram? – David Z May 13 '16 at 7:59
• Hi David Z, I added the diagram in response to the above comments. What I really wanted to know, was whether this reaction was allowed or forbidden. My initial thoughts was that it wasn't possible (just a gut feeling) but I couldn't put my finger on why. Then Cosmas's comment made me rethink my decision. I now think that this process is allowed, since i can't see any conservation laws being violated. Now I have tried to draw the diagram for the process, so someone could tell me if it is correct or not. – NormalsNotFar May 13 '16 at 9:01

Your first diagram is wrong, since there is no vertex in a Lorentz invariant theory where three fermions and a vector meet.

However, I don't see why you say the interaction is forbidden. It would surely be insanely suppressed since amplitudes are extremely weak, but I don't see a problem with the diagram (for instance):

Notice that quarks mix, so the vertex $Wsu$ is there, albeit suppressed by an off-diagonal element of the Cabibbo-Kobayashi-Maskawa quark mixing matrix. The interaction is allowed.

(I just saw the second diagram you drew, which is another valid -- and even much more relevant in magnitude -- contribution to the full amplitude).

EDIT: A nice reference for the Standard Model which IMHO may clarify beginner course confusion is Giunti & Kim's "Fundamentals of Neutrino Physics and Astrophysics" (2007, Oxford University Press), Chapter 3.

• this is even more improbable than the Z0 one as it has four weak vertices – anna v May 13 '16 at 14:48
• Sure, in fact, I pointed this out in my answer. The question was if it was allowed, and I drew this one to contest a previous (wrong, now deleted) answer which claimed you cannot do it with W's and electron lines. – J-T May 13 '16 at 15:10

I am very bad at drawingin "paint", but the process can go as

antineutrino +neutrino to Z0 , Z0 to s antis (or up antiup), a gluon vertex from the quark to an up antiup (or strange antistrange quark) the parenthesis are the alternate diagram. ,

So it is not forbidden, it has two weak vertices and so very small cross section.

The problem with this process is, that there is no term in the standard model which allows a coupling of a fermion a W-boson and a quark-antiquark pair. Even with the neutrino having no charge, you have to take this arrow you draw on the fermion line seriously, so the fermion flow is violated at this interaction points. Furthermore you can only couple neutrinos with the presence of an electron or positron respectively (same for higher lepton families)

• The problem I have with this is that this course is very beginner level, just to serve as an introduction. Since this was part of a question on an exam, I would expect that they would have taught us why this process is or isn't allowed. You're answer seems beyond the scope of our course, unless I am misunderstanding. – NormalsNotFar May 13 '16 at 9:05
• If they teach you anything about Feynman diagrams, "you have to follow the rules when drawing diagrams" should be within the scope of the course. – fqq May 13 '16 at 11:34
• Yes but obviously the rules can get quite complex, and some are certainly not trivial. – NormalsNotFar May 13 '16 at 12:54

## protected by Qmechanic♦May 13 '16 at 12:44

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