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

1. Dec 9, 2008

### brainschen

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

Last edited: Dec 9, 2008
2. Dec 9, 2008

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

3. Dec 9, 2008

### brainschen

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.

Last edited: Dec 10, 2008
4. Dec 10, 2008

Staff Emeritus
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.

5. Dec 10, 2008

### 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.

6. Dec 10, 2008

### 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!

7. Dec 10, 2008

### jdstokes

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

8. Dec 10, 2008

Staff Emeritus
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.

9. Dec 11, 2008

### clem

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

10. Dec 11, 2008

### 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 %.

11. Dec 11, 2008

### clem

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. Dec 11, 2008

### 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 $$(u\bar{u} - d\bar{d})/\sqrt{2}$$.

13. Dec 11, 2008

### humanino

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. Dec 12, 2008

### hamster143

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. Dec 13, 2008

### clem

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

16. Dec 13, 2008

### humanino

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

17. Dec 13, 2008

### 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.

Last edited: Dec 13, 2008