I have a homework problem asking whether the decay $\Sigma_0 \to n + \gamma$ is allowed, and if so, we are asked to draw a Feynman diagram for the decay (in the quark level). Clearly the strangeness is violated so the decay must involve weak interaction. It is also very fishy that the strange quark turns into a down quark but I cannot really show, why this decay should not be allowed. However I also cannot draw a Feynman diagram. The single photon seems to be the problem.
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$\begingroup$ The photon can emanate from any charged quark, and carry away excess energy and momentum. You might well be barking up the wrong tree. $\endgroup$– Cosmas ZachosJul 20, 2017 at 13:59
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2$\begingroup$ Note that 0.1% of the time, $\Sigma^+ \to p \gamma$. Now apply to this W-exchange diagram an isospin-lowering step. $\endgroup$– Cosmas ZachosJul 20, 2017 at 18:51
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2$\begingroup$ @gonenc It's just a penguin diagram :) $\endgroup$– dukwonJul 20, 2017 at 20:44
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$\begingroup$ I'm going to delete my answer because the above two comments have pointed out other possibilities than the two-$W$ case that I am considering... $\endgroup$– CR DrostJul 20, 2017 at 22:16
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1 Answer
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When in doubt, look at the particle data group listings .It says it goes 100% to lamda gamma, and gives limits for lamda gamma gamma and lamda e+e-. The reason is that it can decay electromagnetically, because Lamda and Sigma zero have the same quark content, (instead of weakly as Sigma + and Sigma -), and the difference in the couplings is a factor of 10000.
Note also that the s quark decays into an up and a W-. It cannot turn into a down.
The Sigma + decay, to accomodate charges need two hadrons. As seen in the comments a gamma can always be produced, but can one get away from needing two hadrons for charge conservation?
please note that in homework questions a direct answer is frowned upon on this site.
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$\begingroup$ Thank you for your answer. I discuss this problem also with my friends. Why would the this (sorry for the messy diagram) diagram not allowed? The problem asks, if it would theoretically be possible and says nothing about it being extremely unlikely. $\endgroup$– GonencJul 21, 2017 at 6:17
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$\begingroup$ @annav yeah right W- makes $s \to u$ and $u \to d$ and after the transition $\gamma$ "takes the excess energy away" similar to compton/bremsstrahlung. $\endgroup$– GonencJul 21, 2017 at 10:07
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$\begingroup$ I found this , as suggested by Zachos. researchgate.net/publication/… . It has the photon before the decay of the s. One has to be clear on the direction of the arrow of theW exchange to get the charges correct. $\endgroup$– anna vJul 21, 2017 at 12:40
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$\begingroup$ strange quark can not turn into down quark. This is impossible. Check flavor changing neutral currents. physics.stackexchange.com/questions/549494/… $\endgroup$ May 18, 2020 at 1:39