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I have a problem computing the ratio $$\frac{P(\pi^0 P\rightarrow\Delta^+)}{P(K^- P\rightarrow\Sigma^{*0})}.$$ The problem demands reducing the $S$-matrix first but I really don't see how to get this result. I tried looking for an example of this kind of decays but nothing. Can anybody tell me how to compute just one probability and I'll figure out the rest.

My guess is that $P(\pi^0 P\rightarrow\Delta^+) = \lambda_1\cdot\alpha$, where $\alpha$ is a Clebsch-Gordan Coefficient and $\lambda_1$ is completely determined by the physics and will get cancelled in the ratio.

But still my question, how to compute the Clebsch Gordon Coefficient in this case?

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  • $\begingroup$ 1) What SU3 irreps are you talking about? 2) Consider homepages.physik.uni-muenchen.de/~vondelft/Papers/ClebschGordan for numerical answers, 3) There's also some literature for simpler cases: deepblue.lib.umich.edu/bitstream/handle/2027.42/32049/… ; journals.aps.org/rmp/pdf/10.1103/RevModPhys.36.1005 ; iopscience.iop.org/article/10.1088/1751-8121/ab4b70/pdf $\endgroup$ Dec 13, 2019 at 16:55
  • $\begingroup$ My question is, using tensor representation of SU(3) how would you compute $P(\pi^0P\rightarrow\Delta^+)$? What is the result of this probability? I can't find any example that shows you how to compute this transition probability. $\endgroup$
    – devCharaf
    Dec 13, 2019 at 17:50
  • $\begingroup$ You need to know the irreps to tensor first... By what irrep does $\pi$ and $P$ and $\Delta$ transform? $\endgroup$ Dec 13, 2019 at 18:03
  • $\begingroup$ Well, $\pi^0$ and $P$ are both 8=(1,1), i.e $v^i_j$. And $\Delta^+$ is a 10=(3,1), i.e $w^{ijk}$. So $P(\pi^0P\rightarrow\Delta^+)=\lambda\cdot\alpha$ where alpha (the Clebsch-Gordan coefficient) is a scalar made aout of $\langle W_{ijk}|S|u^a_bv^c_d\rangle$. Pease if you know how to compute the result of $P(\pi^0P\rightarrow\Delta^+)$ can you tell me with some steps. I am really stuck here. Thanks a lot $\endgroup$
    – devCharaf
    Dec 13, 2019 at 18:10

1 Answer 1

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The tensor product \begin{align} (1,1)\otimes(1,1)=(2,2)+2(1,1)+(0,0)+(3,0)+(0,3) \end{align} and in fact $(3,1)$ has dimension $24$ where $(3,0)$ has dimension 10 so it's likely $\Delta$ is in $(3,0)$, not (3,1).

You need to work out the remaining quantum numbers for your particles - let's call them $\alpha,\beta$ and $\gamma$ respectively. Once you have this you need to get the CG $C_{(11)\alpha;(11)\beta}^{(30)\gamma}$.

This CG is given in this table. You can also use Table 2 of the "canonical" deSwart paper

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  • $\begingroup$ Thank you for your response. Can I ask what's $\alpha$, $\beta$ and $\gamma$? $\endgroup$
    – devCharaf
    Dec 13, 2019 at 19:03
  • $\begingroup$ whatever other quantum numbers not the representation labels. I guess they would be the isospin number, its projection, and the other U(1) quantum number. I'm not quite sure how you label your states but whatever is not the irrep labels. $\endgroup$ Dec 13, 2019 at 19:09
  • $\begingroup$ Ok, So I know for $\Delta^+$ has Isospin 3/2, projection 1/2 and U(1) is 1. For proton, it's respectively 1/2, 1/2, 1 and finally for pion0 it's 1, 0 and 0. Does that mean that the C-G coeff is -$1/\sqrt(3)$? So I found the ratio of the two transitions to be 2. But still the exercice asks something about the reduced S-matrix. Can you explain me how can introduce it in my calculations. And thank you so much, this has been so helpful. $\endgroup$
    – devCharaf
    Dec 13, 2019 at 19:45
  • $\begingroup$ Looks like your $-1/\sqrt{3}$ is ok. Since I don't know about the reduced $S$ matrix that's all the help I can provide. $\endgroup$ Dec 13, 2019 at 19:52
  • $\begingroup$ Thank you so much, it’s been very helpful :) $\endgroup$
    – devCharaf
    Dec 13, 2019 at 20:42

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