Sigma+ Decay: Feynman Diagrams for Proton + Photon

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Homework Help Overview

The discussion revolves around the decay process of Sigma(plus) into a proton and a photon, specifically focusing on the representation of this process through Feynman diagrams. The subject area includes particle physics and weak interactions, particularly the conservation laws involved in particle decay.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the decay process and the requirements for drawing Feynman diagrams, questioning the conservation of momentum and the representation of particles in the diagrams. There is a discussion about the transition between quark and hadron representations and the implications for the external lines of the diagrams.

Discussion Status

The discussion reflects a mix of interpretations regarding the Feynman diagram representation. Some participants express confusion about the correct approach, while others provide clarifications about the nature of internal vertices and the conditions under which certain interactions can occur. There is no explicit consensus, but guidance has been offered regarding the representation of particles in the diagrams.

Contextual Notes

Participants mention the need to adhere to physical laws such as conservation of momentum, and there is a reference to potential misconceptions about Feynman diagrams that may require further clarification from instructors.

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Homework Statement


The question is to draw possible Feynman diagrams of the following decay.
Sigma(plus) ----> proton + photon



Homework Equations


Quark content of sigma(plus) is uus
Quark content of proton is uud


The Attempt at a Solution


I know that the decay has to proceed by the weak interaction. After all strangeness is not conserved. So the the strange quark in the sigma plus must first decay into a W(minus) and an up quark. The W- will then produce a down quark plus and an anti-up quark.

With this I will now have the uud for the proton...which is good but I will have an up quark and an anti up quark left over. These can annihilate each other to form 2 photons but not one (due to conservation of momentum). I can see no way of getting rid of this "extra" photon. Do you think its a typo?
 
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No, it's not a typo.

You say they can annihilate each other to form 2 photons but not one (due to conservation of momentum). Do you know what a Feynman diagram is? It is not a step by step description of the process - "first this decays, then these pieces interact", etc. is most assuredly not what a Feynman diagram is supposed to represent, and it sounds like you think it is.

It's best to get this misconception cleared up sooner rather than later. I'd make an appointment with my professor so he or she can clear this up. It should take an hour or so.
 
I see what your saying but the reaction, sigma(plus) ----> proton + photon, is a physical process (at least according to the question) and should obey the physical laws like conservation of momentum. So the Feynman diagram should have a sigma(plus) coming in and proton and a photon going out for the external lines of the diagram. So far, I see nothing wrong with the Feynman diagram except that I get a proton and two photons coming out for the external lines. I can see no way of drawing my diagram to get a single photon coming out as an external line by drawing correct vertices.
 
<sigh>

Like I said, It's best to get this misconception cleared up sooner rather than later. I'd make an appointment with my professor so he or she can clear this up. It should take an hour or so.

If you want to keep struggling, and see how permanently you can ingrain in this misunderstanding before removing it, hey, be my guest. The job market is plenty tight as it is.

By the way, you switched from the quark to the hadron representation between the first and third messages. In the hadron representation, you have the sigma in, and a proton and photon out. In the quark representation, you have it as you described above, but with one photon, not two.
 
I did not have a misconception of what a Feynman diagram. I was simply not seeing the way to draw it correctly and it was an easy clarification when I asked my professor. It appears you did not understand my question.
 
As free on-shell particles, u+ubar->gamma can't happen. But it's a perfectly valid internal vertex in a feynman diagram where all of the lines aren't external (and the particles can be off-shell). I don't think that would have been terribly hard to explain and shouldn't take a one hour conference.
 
Exactly. That was all I needed to hear and it was a simple fix. I considered doing that vertex but at first it didn't seem right. My professor took all of five seconds to answer my question and it immediately made sense. Thanks.
 

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