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Currently studying Georgi's Lie Algebras in Particle Physics and problem 5.C in the isospin chapter asks to compare the probability of producing $\Delta^{++}$ in $\pi^+ P \rightarrow \Delta^{++}$ and $\Delta^{0}$ in $\pi^- P \rightarrow \Delta^{0}$.

I see the charge conservation and isospin conservation here, but I'm not sure how to use only the isospin $SU(2)$ formalism to figure out any other information regarding the probabilities. What initial state would I use in order to apply the isospin equivalent of a highest weight procedure to find Clebsh-Gordon coefficients? Given whatever state that would be, how do I construct isospin "raising" or "lowering" operators? Are they combinations of the x and y isospin generators $T_a = a^{\dagger}_{x,m} [J_a^{jx}]_{m m'} a_{x,m'}$ like they are with the spin $SU(2)$ generators?

Qmechanic
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DingleGlop
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1 Answers1

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The probability ratio is proportional to the ratio of the square of the amplitude ratio; in turn, since the spins are identical, the amp ratio is the ratio of isospin Clebsches.

Since, for the Δ isoquartet, I=3/2; for the π isotriplet I=1; and for the nucleon isodoublet I=1/2; the amp ratio is just $$ {\langle \pi^- p| \Delta ^{0} \rangle \over \langle \pi^+ p| \Delta ^{++} \rangle }={\langle 1,1/2;-1,1/2|1,1/2; 3/2,-1/2 \rangle \over \langle 1,1/2; 1,1/2|1,1/2;3/2,3/2\rangle }= 1/\sqrt{3}, $$
where the r.h.side is defined by the Clebsches $ \langle i_1=1,i_2=1/2;m_1,m_2\mid i_1=1,i_2=1/2;I=3/2,I_3\rangle $, the ms being the third components of the respective isospins, of course; evaluated, e.g., from here. If you were insecure about the irrelevant signs, use your PDG listing table.

Squaring this ratio, yields a ratio of respective production rates equal to 1/3.

Cosmas Zachos
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