# Conservation of isospin

Our particles course has two contradicting descriptions of how isospin must be conserved in strong interactions (assuming a perfect sitution with only the strong interaction so isospin symmetry is exact).

1. In $$n + p \rightarrow d + \pi^0$$. It says isospin magnitude $$I$$ is conserved as for the left hand side we have isospin states $$|\frac{1}{2}\rangle|-\frac{1}{2}\rangle = \frac{1}{\sqrt2}(|0,0\rangle + |1,0\rangle)$$ (in $$|I,I_3\rangle$$ representation) while the right hand side just has isospin with the $$\pi^0$$ which is the $$|1,0\rangle$$ state. The argument is that the isospin of either the $$|0,0\rangle$$ or $$|1,0\rangle$$ must be conserved (like in a measurement).

2. In showing that $$J/\psi$$ (with quark composition $$c \bar c$$) cannot decay into two $$\pi^+$$ and $$\pi^-$$ it uses the argument that $$J/\psi$$ has $$I=0,I_3=0$$ (i.e. it is in the singlet state) which decomposed into 2 particle states $$|I=0, I_3=0 \rangle = \sqrt{1/3} |1\rangle|-1\rangle -\sqrt{1/3} |0\rangle|0 \rangle + \sqrt{1/3} |-1\rangle|1\rangle$$. It then argues that this includes a $$|0\rangle|0\rangle$$ state which corresponds to a $$\pi^0 \pi^0$$ in the final state (while the others are $$\pi^+ \pi^-$$ final states), and since this violates charge conjugation, a decay into 2 pions is impossible and so it must decay into 3 pions instead as isospin cannot be completely conserved with a 2 pion final state.

The first argument seems to imply that a interaction is like a non-linear 'measurement' where the resulting isospin is not completely conserved but one can only measure one of the original isospin states in the superposition. The second argument seems to imply that the interaction Hamiltonian commutes with $$I$$ and $$I_3$$ so they must both be conserved as the wavefunction evolves linearly time progresses in the interaction.

I would be inclined to believe the second argument more seeing as we do scattering theory with an interaction Hamiltonian, but then how should I interpret isospin conservation in example 1?

From further investigation, it looks like the explanation 2 is more likely to be correct and possibly our lecture notes had an incorrect explanation for 1. Perhaps the superposition state $$|\frac{1}{2}\rangle|-\frac{1}{2}\rangle = \frac{1}{\sqrt2}(|0,0\rangle + |1,0\rangle)$$ does not correctly describe the $$n + p$$ initial state (which is believalbe since it does not have well defined exchange symmetry between the $$n$$ and $$p$$ which are treated as identical by the strong force).
Perhaps instead the lecturer should have used a mixed state of $$|0,0\rangle$$ and $$|1,0\rangle$$ states. Therefore only $$|1,0\rangle$$ states can decay (whilst conserving isospin) into theproduct state $$d + \pi^0$$, whilst $$|0,0\rangle$$ states of the intial particles are unable to decay in this way. However in the case where $$|1,0\rangle$$ states do decay, isospin is completely conserved as expected.