Isospin of the neutral sigma baryon I was typing up another answer on P.S.E. and I wanted to use the fact that the decay $$\Sigma^0\longrightarrow\Lambda^0+\gamma$$
does not occur strongly as an example of isospin conservation. But then I got to thinking about it, and realized that both are $uds$ baryons and therefore should have $I_3=0$. So I pulled out my trusty baryon octet, and sure enough, they are both at the origin. So of course I investigated further. For some reason, both Wikipedia and this MIT article (top of page 4) state that the sigmas have $I_3=1$. What gives? Have I been reading particle diagrams wrong all along? Is the horizontal axis not isospin? Is 
$$Q=I_3+\tfrac{1}{2}(S+B)$$
wrong?
Also, now that I really examine the numbers, is that even an example of isospin conservation? The mean lifetime for this decay is about $10^{-20}$s, which is pretty close to the $10^{-23}$s for strong interactions. Shouldn't it be closer to $10^{-16}$s? Or is this just a manifestation of the approximate nature of isospin symmetry?
 A: The $\Lambda^0$ and $\Sigma^0$ have respectively I=0 and I=1. The source of your confusion may come from the fact that the electromagnetism doesn't respect the isospin symmetry. In E.M. processes, only $I_3$, the third component, is conserved. The total isospin $I$ is not. A trivial example is the neutral pion decay $\pi^0 \to \gamma \gamma$. The pion is a member of an isospin triplet $(\pi^+,\pi^0,\pi^-)$ (thus $I=1$ and $I_3=0$) while photons cannot carry any isospin numbers (since not made of $u$ or $d$ quarks).
Edit (to answer a comment): how do we know that $\Sigma^0$ has $I=1$?
There are several experimental facts. The $\Sigma^+,~\Sigma^0,~\Sigma^-$ have almost the same mass (about 1190 MeV). So historically, they were considered in the same multiplet of isospin, reasonably a triplet as for $\pi^+,~\pi^0,~\pi^-$. In addition, the reaction $K^- + p \to \Sigma^0$ is seen (strong interaction). Knowing that $p$ has $I=1/2,~I_3=+1/2$ and $K^-$ has $I=1/2,~I_3=-1/2$, the initial state $K^-+p$ can have $I=0$ or $I=1$ (sum of 2 Isospin 1/2). If you admit that $\Sigma^0$ is a member of a multiplet (previous argument), you're forced to conclude that $I=1$. I'm sure that many other reactions can be found in the literature justifying $I=1$.  
In the modern quark languages, $\Sigma^+ \equiv uud,~\Sigma^0\equiv uds,~\Sigma^-\equiv dds$. With the group theory, assuming the (approximate) SU(3) flavor symmetry, $I=1$ comes "naturally". 
