Why can't rho^o decay into two pi^o?

[brainschen]

As title, why does the following decay is forbidden?
\rho0\rightarrow\pi0\pi0

[Vanadium 50]

Isospin. Take a look at the Clebsch-Gordon coefficient table.

[brainschen]

Isospin. Take a look at the Clebsch-Gordon coefficient table.

Thank you. You are correct. I look at the CG 1x1 table.
|10> = (1/2)|11>|1-1>-(1/2)|-11>|11>. The coefficient of |10>|10> is zero!

But why is that physically? The isospin is still conserved for |10>|10> case.

[Vanadium 50]

But why is that physically?

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.

[clem]

You can also use Bose statistics and conservation of angular momentum.
The rho has spin one. There is no pi0-pi0 state with J=1.

[memyself]

Hello,
I don't understand why pi0-pi0 can't have J=1.
I thought (probably wrong) that they could have some relative angular momentum L for example L=1 so that then pi0-pi0 would have J=1.

Maybe it's trivial but could someone please explain me why that isn't possible :)

Thank you!

[jdstokes]

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.

Isn't isospin conservation just a manifestation of nucleon SU(2) invariance of the Hamiltonian?

[Vanadium 50]

If you put it in a J=1 state the total wavefunction is antisymmetric, and it has to be symmetric because of Bose statistics.

JDStokes, yes, isospin is SU(2). Which is why you use the Clebsch-Gordon coefficients as opposed to something else.

[clem]

It is important to note that Ispin symmetry is broken, while Bose statistics and conservation of angular momentum are not.

[hamster143]

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 %.

[clem]

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 %.
1. You also need Bose statistics for the pi0s.
2. Once there is a gamma, Ispin is irrelevant.
2. The rho is u-ubar. It has C=-, but no symmetry property.

[hamster143]

1. Gamma is isospin invariant. u and d quarks couple to the gamma with different strength, because isospin is not a perfect symmetry.

2. Rho is (u\bar{u} - d\bar{d})/\sqrt{2}.

[humanino]

Rho is (u\bar{u} - d\bar{d})/\sqrt{2}.If there is a difference between (u\bar{u} - d\bar{d})/\sqrt{2} and u\bar{u}, it must break isospin ! This is why (I think) you miss clem's point.

[hamster143]

:confused:

We're located in different points on the surface of the Earth, but that does not mean that rotational symmetry of the Earth is broken. (u\bar{u} - d\bar{d})/\sqrt{2} and u\bar{u} 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.

[clem]

I know the rho is not simply u-ubar. I just gave a simple example to show that symmetry didn't enter.

[humanino]

(u\bar{u} - d\bar{d})/\sqrt{2} and u\bar{u} 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.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 photon :smile:

[clem]

I am confused by the last few posts. u-ubar is not an Ispin eigenstate.
The combination (u\bar{u} - d\bar{d})/\sqrt{2} is.