# Ambiguity in ordering of isospin states for Clebsch-Gordan coefficients

In studying isospin for nuclear physics, I am confused a bit by an ambiguity I found.

If a process that goes from $K^- + p \rightarrow \Sigma^0+ \pi^0$, I can write the isospin for the left hand side as $|K^- + p \rangle = | \frac{1}{2} \frac{-1}{2} \rangle | \frac{1}{2} \frac{1}{2} \rangle$ but decomposing this with Clebsch-Gordan coefficients gives us $| \frac{1}{2} \frac{-1}{2} \rangle | \frac{1}{2} \frac{1}{2} \rangle=\sqrt{\frac{1}{2}}| 1 0 \rangle-\sqrt{\frac{1}{2}}| 0 0 \rangle$.

However, if I change the ordering, which seems completely arbitrary to me, we get $|p + K^- \rangle = | \frac{1}{2} \frac{1}{2} \rangle| \frac{1}{2} \frac{-1}{2} \rangle$ but decomposing this with Clebsch-Gordan coefficients gives us $| \frac{1}{2} \frac{1}{2} \rangle| \frac{1}{2} \frac{-1}{2} \rangle=\sqrt{\frac{1}{2}}| 1 0 \rangle+\sqrt{\frac{1}{2}}| 0 0 \rangle$.

Maybe this ends up not mattering, but if for instance we want to calculate a scattering amplitude, where $|\Sigma^0+ \pi^0 \rangle = \sqrt{\frac{2}{3}}| 2 0 \rangle-\sqrt{\frac{1}{3}}| 0 0 \rangle$, $M(K^- + p \rightarrow \Sigma^0+ \pi^0)=-\sqrt{\frac{1}{6}}\langle 0 0 | M | 0 0 \rangle$, but $M(p + K^- \rightarrow \Sigma^0+ \pi^0)=\sqrt{\frac{1}{6}}\langle 0 0 | M | 0 0 \rangle$.

Obviously this still leaves the cross section invariant, but I can imagine processes where we add something else to get something like $M= A-B$ and $M=A+B$ when I flip the ordering of two initial state particles, so then my issue is that since $\sigma \sim |M|^2$, it's not necessarily true that $\sigma=|A-B|^2$ and $\sigma= |A+B|^2$ are equal, whereas intituitely they should be.

Is there a general rule for ordering of particles? I feel that the choice in ordering of particles is arbitrary, but can give different results.

This is an old subject so I must be missing something fundamentally.

The notation $\lvert \phi \rangle \lvert \psi \rangle$ is shorthand for $\lvert \phi \rangle \otimes \lvert \psi \rangle$.
What you are doing is flipping a tensor product around. Though, in general, $A \otimes B = B \otimes A$, it is not the case that $a \otimes b = b \otimes a$ for $a \in A, b \in B$ (since the left and right hand side don't even live in the same space.
You have to label the spaces and stick with it. In this case, there is the isospin space $\mathcal{H}_K$ for the $K$ and the isospin space $\mathcal{H}_p$ for the $p$. Though they are isomorphic, they are not the same. In one case, you are Clebsch-Gordan decomposing $\mathcal{H}_K \otimes \mathcal{H}_p$, in the other, it is $\mathcal{H}_p \otimes \mathcal{H}_K$.
If you stay consistent with this labeling, and write all states in the same way, i.e. you have to choose whether $\mathcal{H}_K \otimes \mathcal{H}_p$ or $\mathcal{H}_p \otimes \mathcal{H}_K$ is the isospin space for the initial states (and similarily for the final ones), the abstract isomorphy of the tensor products guarantees that you will make no error with relative signs as you fear. If you do not keep track of that, it is indeed possible to incur sign errors, since you basically use two different spaces of states for the same thing.