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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?

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  • $\begingroup$ In rthe wiki link you give, it is the total I, not I3. The I3 separates the charges.en.wikipedia.org/wiki/Isospin#mediaviewer/… $\endgroup$
    – anna v
    Jan 17, 2015 at 10:45
  • $\begingroup$ I think you are confusing the total isospin with just its "z-component". $\endgroup$
    – Phoenix87
    Jan 17, 2015 at 11:58

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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".

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  • $\begingroup$ How does one show that $I=1$ for $\Sigma^0$? $\endgroup$
    – Ryan Unger
    Jan 17, 2015 at 13:53
  • $\begingroup$ @0celo7: I've edited my answer to give an example of justification. $\endgroup$
    – Paganini
    Jan 17, 2015 at 16:36
  • $\begingroup$ So is the difference between the $\Sigma^0$ and $\Lambda^0$ baryons their total isospin? Is there any way to come by $I=1$ for $\Sigma^0$ theoretically or is it an experimentally justified fact because the $\Sigma^i$ have similar masses? $\endgroup$
    – Ryan Unger
    Jan 17, 2015 at 20:50
  • $\begingroup$ yes, the only difference between $\Sigma^0$ and $\Lambda^0$ in terms of quantum numbers is the total isospin. Assuming the $SU(3)$ flavor symmetry (based on the 3 quarks $u,~d,~s$), you see that a baryon made of 3 quarks should belong to the multiplets $3 \otimes 3 \otimes 3 = 10 \oplus 8 \oplus 8 \oplus 1$. It turns out that both $\Sigma^0$ and $\Lambda^0$ are at the center of the spin 1/2 octet meaning same $I_3=0$. So same spin, same quark contents (because same strangeness and $I_3$), and same parity. What's the difference? what remains is just total isospin $I$. $\endgroup$
    – Paganini
    Jan 17, 2015 at 21:18
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    $\begingroup$ and $I_3=0$ can come from $I=0$ or $I=1$ (with $u$ and $d$ carrying $\pm 1/2$. The baryon with $I=0$ has been called $\Lambda^0$, the one with $I=1$, $\Sigma^0$. $\endgroup$
    – Paganini
    Jan 17, 2015 at 21:22

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