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xiMy
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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 [itex]\pi^{+}\pi^{-}[/itex] but [itex]\eta^0[/itex] does not. Why is that?
c) [itex]\omega[/itex] decays through electromagnetic interaction into [itex]\pi^0[/itex][itex]\gamma[/itex]. C([itex]\omega[/itex])?
d) [itex]b_1(1235)[/itex] decays almost solely into [itex]\omega\pi[/itex] Reason what the isospin, parity ,C, and spin from b might be.
Homework Equations
conservation laws
[itex]P=(-1)^{L+1}[/itex]
[itex]S=(-1)^{L+S}[/itex]
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 [itex]J^{PC}[/itex] 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 [itex](-1)^l\cdot P_1\cdot P_2[/itex]
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([itex]\pi^{+}\pi^{-}[/itex])?
c)
With c([itex]\gamma[/itex])=-1 and ([itex]\pi^0[/itex])=1 it should be c([itex]\omega[/itex])=-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.