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brainschen
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As title, why does the following decay is forbidden?
[tex]\rho0\rightarrow\pi0\pi0[/tex]
[tex]\rho0\rightarrow\pi0\pi0[/tex]
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Vanadium 50 said:Isospin. Take a look at the Clebsch-Gordon coefficient table.
brainschen said:But why is that physically?
Vanadium 50 said:What kind of answer are you looking for? Why are there Clebsch-Gordon coefficients at all? Why does isospin use the same Clebsch-Gordon coefficients as angular momentum? Why is there isospin at all? I don't know how to answer you.
1. You also need Bose statistics for the pi0s.hamster143 said:It can't decay into pi0 pi0 because of conservation of angular momentum.
It can decay into pi0 pi0 gamma. In a world with unbroken isospin symmetry, that decay would also be prohibited (because rho0 is symmetric under u-d interchange, and pi0's are antisymmetric). In our world, it's merely suppressed. Relative probability of this decay is 4.5 * 10^-5 %.
If there is a difference between [itex](u\bar{u} - d\bar{d})/\sqrt{2}[/itex] and [itex]u\bar{u}[/itex], it must break isospin ! This is why (I think) you miss clem's point.hamster143 said:Rho is [tex](u\bar{u} - d\bar{d})/\sqrt{2}[/tex].
Sure, I think we agree. The states would be impossible to distinguish IF isospin were exact. Of course, the proton and the neutron have different interactions with the photonhamster143 said:[itex](u\bar{u} - d\bar{d})/\sqrt{2}[/itex] and [itex]u\bar{u}[/itex] are degenerate and there is a symmetry transformation that takes one into the other, but that does not mean that they are the same state.
Rho^o and pi^o are both subatomic particles, also known as mesons, that are made up of quarks. Rho^o is a neutral meson made up of an up quark and a down antiquark, while pi^o is a neutral meson made up of an up quark and an anti-up quark.
Rho^o cannot decay into two pi^o because of conservation of quantum numbers. Rho^o has a quantum number called isospin, which is conserved in all particle interactions. Pi^o does not have this quantum number, so it is not possible for rho^o to decay into two pi^o particles.
Yes, rho^o can decay into other particles such as a pi^+ and a pi^- or a K^+ and a K^-. These decay modes are allowed because they conserve quantum numbers like charge and strangeness.
This decay mode is important in understanding the strong nuclear force, which binds quarks together to form particles. The decay of rho^o into pi^+ and pi^- is a manifestation of this force and helps to confirm the theories of quantum chromodynamics.
There have been some rare cases where rho^o has been observed to decay into two pi^o particles. However, these decays are highly suppressed and occur through a different mechanism called the OZI rule. This rule allows for particles to decay into lighter particles that have a different quark composition, but it is a much less common process.