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  Does one have to account for identical particles twice

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David Griffiths: "Introduction to Elementary Particles", Second Revised Edition, formula (6.47) reads $ \frac { d \sigma } { d \Omega } = \left( \frac { \hbar c } { 8 \pi } \right) ^ 2 \frac { S | {\cal M} | ^ 2 } { ( E _ 1 + E _ 2 ) ^ 2 } \frac { | \mathbf { p } _ f | } { | \mathbf { p } _ i | } $ and expresses the differential cross section of a process $ 1 + 2 \rightarrow 3 + 4 $.

According to Griffiths, $ S $ is a statistical factor that corrects for double-counting when there are identical particles in the final state. In particlar, in Formula (6.47), $ S $ equals $ 1 $ if particles 3 and 4 are different and equals $ 1 / 2 $ if the particles are identical. Griffiths says no more about $ S $.

I have tried to Google for more information about $ S $ and exactly how to use it, but all I have found is that some web pages like https://www.nikhef.nl/~i93/Master/PP1/2017/Lectures/Lecture2017.pdf omit $ S $ and others like http://www.physics.utah.edu/~belz/phys5110/lecture23.pdf include it.

My question concerns the following thought experiment: Suppose $ A $, $ B $ and $ C $ are types of elementary particles and that $ A + B \rightarrow C + C $ is a possible process. Suppose we let a beam of $ A $-particles intersect with a beam of $ B $-particles. Suppose the luminosity of the collision (the thing to multiply $ \sigma $ by to get an event rate) is $ L $. Suppose we have a detector which detects $ C $-particles which are created by the process and travel in a particular direction within a small, spherical angle $ d \Omega $.

Now the question is: Is the expected event rate $ \frac { d \sigma } { d \Omega } L d\Omega $ or $ 2 \frac { d \sigma } { d \Omega } L d\Omega $. The argument for the latter would be that the detector could detect either of the $ C $'s in the final state. If the latter is correct, then we first have to multiply by $ S = 1 / 2 $ to account for identical particles and then multiply by $ 2 $, again to account for identical particles, and the two factors would cancel out in this particular case. Is that correct?

asked Jul 30 in Theoretical Physics by grue (5 points) [ no revision ]

it is trivial if you think about the detection count with the same device. In scattering matters, the wording is often confusing. Taking account of S in this equation is not very important.

@igael Thanks for the answer. I suppose you say the answer to "Is that correct?" is "Yes". Maybe I chose a bad example (two-body scattering) where $ S $ happens to cancel out with the number of particles. If the process was $ A + B \rightarrow C + C + C $ then one should use (6.37) in Griffiths which also includes an $ S $. In that case $ S $ would be $ 1 / 6 $ and one should multiply by 3 because the detector could detect any one of the three $ C $'s. Is it correct that in that case one shall multiply by first $ S = 1 / 6 $ and then the number $ 3 $ of particles?

counting events by length², 1/3 seems more appropriate since you detect 3 times more. Griffith was  rigorous but consider "if the particles are identical" && "and detected with the same device". If, for any reason, you are using distinct detectors for the products, the condition is not fullfilled and ideally S=1. It's why S is not very important and might be considered as a setup parameter.

@igael Thanks again. I think your answer could give a clue to me about what Griffiths means. The "and detected with the same device" is not something Griffiths wrote. That was just a thought experiment of mine. And Griffiths would indeed assign $ S = 1 / 3 ! = 1 / 6 $ to $ A + B \rightarrow C + C + C $. At least, Griffiths assigns $ S = ( 1 / 2 ! ) ( 1 / 3 ! ) = 1 / 12 $ to $ A \rightarrow B + B + C + C + C $.

But is the following what you mean: If we consider $ A + B \rightarrow C + C + C $, and if we have three detectors D1, D2 and D3 which can detect $ C $-particles, and if we set $ S $ to $ 1 $, then we get a differential cross section $ \sigma_{\mathrm{igael}} $ (events counted by $ \mathrm{length}^2 $ as you say) for all three detectors saying "ping" simultaneously. Then $ \sigma_{\mathrm{griffith}} = \sigma_{\mathrm{igael}} / 3! $ would be the cross section for D1 detecting the first $ C $, D2 the second and D3 the third, simultaneously. The latter of course does not make sense physically, but $ \sigma_{\mathrm{griffith}} $ could nevertheless be what Griffiths means by "cross section".

Of course it is unfair if I ask you what Griffiths means, but could the interpretation above sort of explain matters? In any case, it would explain to me why you say one can just set $ S $ to $ 1 $ and forget about it.

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