# Why is 3 pion decay of eta (η) strong forbidden?

## Main Question or Discussion Point

I have read that 3 pion decay of the eta is not allowed as a strong reaction, but proceeds as an electromagnetic interaction. I do not see why it is strong forbidden.

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Meir Achuz
Homework Helper
Gold Member
That is easiest to see using G parity, which is $$C e^{i\pi I_y}$$.
One or three pions have negative G, while the eta has positive G.

How do you find the result of the I2 rotation operation?

Meir Achuz
Homework Helper
Gold Member

From G-parity - Wikipedia

In general, P(G) = P(C)*(-1)^I
For fermion-antifermion systems, P(G) = (-1)^(L+S+I)
For boson-antiboson systems, P(G) = (-1)^(L+I)

To calculate exp(-i*pi*I2), use Wigner D-matrix - Wikipedia for rotation matrices between quantum-mechanical angular-momentum states. One can carry angular-momentum features over to isospin without much trouble.

For angular momentum j and state m to state m', find D(j,m,m',0,-pi,0) = d(j,m,m',-pi)

It is only nonzero if m' = - m: d(j,m,-m,-pi) = (-1)^(j+m)

One can do the calculation more directly, by using isospin conservation and pion spin-statistics. Pions have spin 0, making them bosons, with their combined wavefunction always being symmetric. It also simplifies the treatment of their spins, since their combined spin is always 0, with a symmetric combined wave function. Orbital angular momentum is usually handled by setting it to 0 (s-wave), giving the same simplification and symmetry.

Pions have isospin 1, while eta and eta' mesons have isospin 0. To find the total isospin of 3 pions, let's start with finding it for 2 pions. One gets these isospins and wavefunction symmetries:

Symmetric: 0, 2
Antisymmetric: 1

One can prove this alternation of symmetry more generally, I think.

Since their combined wavefunction must be symmetric, 2 s-wave pions must have isospin 0 or 2. Combining the isospin of the third pion gives possibilities 3, 1, and 1.

Using my SemisimpleLieAlgebras package, I find:
Symmetric: 3, 1
Mixed: 2, 1
Antisymmetric: 0

Thus, 3 s-wave pions cannot have zero isospin, and thus, an eta cannot decay into them without violating isospin. That can be done with the electromagnetic interaction, however.

Let's depart from the s-wave hypothesis, while using the smallest possible orbital angular momenta. The sum of these values must always be even from parity conservation, and we have two possibilities: 1 d-wave and 2 s-wave, and 1 s-wave and 2 p-wave.

The first one cannot have zero isospin, while the second one can, if the 2 p-wave pions have total angular momentum 1. That makes antisymmetry, and that can be compensated for with isospin antisymmetry and total isospin 1. The third pion's isospin can then combine with it to make isospin 0.

Thus, an eta can decay into 3 pions without violating isospin if 2 of them have p-wave wavefunctions.