# 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?

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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