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

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Discussion Overview

The discussion centers on the decay process of the rho meson (\(\rho^0\)) into two neutral pions (\(\pi^0\)), exploring the underlying principles of isospin, angular momentum conservation, and Bose statistics. Participants examine why this decay is forbidden and the implications of isospin symmetry in particle interactions.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants suggest that the decay \(\rho^0 \rightarrow \pi^0 \pi^0\) is forbidden due to conservation of angular momentum.
  • Others mention the role of isospin and refer to Clebsch-Gordon coefficients to explain the constraints on the decay process.
  • One participant questions why the \(\pi^0\) pair cannot have a total angular momentum \(J=1\), proposing that relative angular momentum \(L\) could allow for this state.
  • Another participant argues that if the \(\pi^0\) pair were in a \(J=1\) state, the total wavefunction would need to be symmetric due to Bose statistics, which contradicts the requirement for antisymmetry in the case of identical bosons.
  • Some participants note that while isospin symmetry is broken, Bose statistics and angular momentum conservation remain intact in this context.
  • It is mentioned that the decay into \(\pi^0 \pi^0 \gamma\) is allowed, but in a world with unbroken isospin symmetry, this decay would also be prohibited.
  • There is a discussion about the nature of the rho meson and its relation to quark content, with some participants emphasizing the importance of isospin invariance and the differences in interactions with photons.
  • Confusion arises regarding the distinction between different quark combinations and their symmetry properties, with participants attempting to clarify the implications of isospin eigenstates.

Areas of Agreement / Disagreement

Participants express differing views on the reasons behind the decay prohibition, with some focusing on angular momentum conservation and others on isospin considerations. The discussion remains unresolved, with multiple competing perspectives on the underlying physics.

Contextual Notes

Participants reference Clebsch-Gordon coefficients and the implications of Bose statistics, but the discussion does not reach a consensus on the physical interpretation of these concepts in relation to the decay process.

brainschen
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As title, why does the following decay is forbidden?
\rho<sup>0</sup>\rightarrow\pi<sup>0</sup>\pi<sup>0</sup>
 
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Isospin. Take a look at the Clebsch-Gordon coefficient table.
 
Vanadium 50 said:
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.
 
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brainschen said:
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.
 
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.
 
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!
 
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.

Isn't isospin conservation just a manifestation of nucleon SU(2) invariance of the Hamiltonian?
 
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.
 
It is important to note that Ispin symmetry is broken, while Bose statistics and conservation of angular momentum are not.
 
  • #10
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 %.
 
  • #11
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 %.
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.
 
  • #12
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}.
 
  • #13
hamster143 said:
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.
 
  • #14
: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.
 
  • #15
I know the rho is not simply u-ubar. I just gave a simple example to show that symmetry didn't enter.
 
  • #16
hamster143 said:
(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:
 
  • #17
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.
 
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