Isospin should be I=1, I think.

[xiMy]

Homework Statement

a) η decays into two photons, but not three. Through which interaction does the decay occur. What is C(η)?

b) ρ0 decays into $\pi^{+}\pi^{-}$ but $\eta^0$ does not. Why is that?

c) $\omega$ decays through electromagnetic interaction into $\pi^0$$\gamma$. C($\omega$)?

d) $b_1(1235)$ decays almost solely into $\omega\pi$ Reason what the isospin, parity ,C, and spin from b might be.

Homework Equations

conservation laws
$P=(-1)^{L+1}$
$S=(-1)^{L+S}$
spin either 0 or 1 for mesons - is that correct ? I know that there are technically only to constituents but the flavour wave functions have weird structures.

The Attempt at a Solution

Would someone be so kind and help me through this? I still struggle to get my head into this.

a) is the only one I am able to solve:
the c parity has to be multiplied. the photon has a c parity of -1 and therefore C is 1 here.
electromagnetic.
b) I am not sure whether I am allowed to look up the $J^{PC}$ value. I guess so, otherwise it's not possibly to solve ?
The roh has 1++ and the eta has 0-+
Using the formulas above I get l=1 for eta and l=0 for rho.
The Parity is calculated via $(-1)^l\cdot P_1\cdot P_2$
from the data booklet I get 0- for the pion thus parity(pion=-1) (is there a way to prove the parity by the way?)
How do I get to L($\pi^{+}\pi^{-}$)?

c)
With c($\gamma$)=-1 and ($\pi^0$)=1 it should be c($\omega$)=-1*1=-1

d)
Omega has 1-- and Pion 0-+. The spin follows therefore 1 and 0.
c(b)=c(omega)*c(pi)= -1.
Again I have no idea how to calculate the angular momentum between the final products.

 

[mark726]

a) The decay of η into two photons occurs through the electromagnetic interaction. The C(η) value is 1, since the two photons have a C-parity of -1 and -1, respectively, and multiplying them gives a C-parity of 1.
b) The ρ0 decays into π+π- through the strong interaction, while the η0 does not because it has a different quark composition. The η0 has a quark composition of (uubar + ddbar)/√2, while the ρ0 has a quark composition of (uubar - ddbar)/√2. Since the strong interaction conserves isospin, the ρ0 can decay into two pions, while the η0 cannot.
c) The ω decays through the electromagnetic interaction into π0γ. The C(ω) value is -1, since the π0 has a C-parity of 1 and the photon has a C-parity of -1, and multiplying them gives a C-parity of -1.
d) The b1(1235) decays primarily into ωπ through the strong interaction. Since the ω has a quark composition of (uubar + ddbar)/√2, it can decay into a ω and a pion, while the b1(1235) has a quark composition of (uubar + ddbar + ssbar)/√3, which cannot decay into a ω and a pion. The isospin of the b1(1235) is 1, since it can decay into an ω and a pion, which have an isospin of 1. The parity of the b1(1235) is (-1)^(L+1), where L is the orbital angular momentum. The spin of the b1(1235) is 1, since it can decay into an ω and a pion, which have a spin of 1.