# $\Sigma^0$ baryon decay

I’ve seen it stated that the sigma baryon $$\Sigma^0$$ only decays to $$\Lambda^0 \gamma$$, and then $$\Lambda^0$$ decays to $$p\pi^{-}$$ or $$n\pi^0$$.

I understand that the weak interaction conserves weak isospin, making $$\Sigma^0 \rightarrow p\pi^{-}$$ or $$n\pi^0$$ impossible. But why does emitting a photon change the isospin of $$\Sigma^0$$ from 1 to 0, i.e. from $$\Sigma^0$$ to $$\Lambda^0$$?

Also, I assume this is just because there are no good options, but why is the decay $$\Sigma^0 \rightarrow \Lambda^0 \gamma$$ 100% (I’ve seen it stated that it is 100%, but I want to assume it is really $$\sim$$ 100%)?

I’ve seen it stated that the sigma baryon, $$\Sigma^0$$, only decays to $$\Lambda^0 \gamma$$, and then $$\Lambda^0$$ decays to $$p\pi^{-}$$ or $$n\pi^0$$.
But why does emitting a photon change the isospin of $$\Sigma^0$$ from 1 to 0, i.e. from $$\Sigma^0$$ to $$\Lambda^0$$? Also, I assume this is just because there are no good options, but why is the decay $$\Sigma^0 \rightarrow \Lambda^0 \gamma$$ 100% (I’ve seen it stated that it is 100%, but I want to assume it is really $$\sim$$ 100%)?
The (strong) isospin of $$\Sigma^0$$ is $$I=1, I_3=0$$, of $$\Lambda$$ is $$I=0$$, and the photon has both isoscalar and isovector pieces, since electromagnetism violates isospin. The BR is almost 100%, as the other channels, 2γ or $$e^+e^-$$ , are very suppressed, as they must be. There is no need for weak channels.
The weak decays $$\Lambda^0 \rightarrow p\pi^{-}$$ or $$n\pi^0$$ violate (strong) isospin by 1/2, the celebrated dominant $$\Delta I=1/2$$ piece of the weak hamiltonian. There is no paradox, if you imagined one.
At the quark level, the weak Hamiltonian breaks weak isospin, but there might be "accidental/incidental" preservation of $$T_3$$, which, however, as you must appreciate, cannot really be well-defined for hadrons which also contain right-handed quarks, weak isospin singlets. You might be tempted to pull limited technical stunts, but these should be avoided unless you are in full control.