What's the isospin of photons? Why PDG book says $0, 1(1^{--})$ for its $I(J^{PC})$? What does $0, 1$ mean here? Is it that the isospins aren't determined and can be 0 or 1?

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    $\begingroup$ Isospin is a strong interactions thing, while the weak isospin is an analog for weak interactions. I wish they didn't have the same name. $\endgroup$
    – Ron Maimon
    Dec 20, 2011 at 10:20

1 Answer 1


It's a (known) linear combination of 0 and 1. The electromagnetic field operator can be written as something like

$$A_\mu = B_\mu\cos\theta_W + W_\mu^3\sin\theta_W$$

where $B_\mu$ is the $U(1)$ hypercharge operator and $W_\mu^3$ is the measured component of the weak isospin operator. $B_\mu$ is not associated with any weak isospin, so that's where the 0 comes from, but $W_\mu^3$ is, so that's the 1.

By the way, it's worth mentioning (as Ron pointed out) that weak isospin is not the same thing as just isospin. The latter corresponds to a particular $SU(2)$ subgroup of the $SU(6)$ flavor "symmetry," but weak isospin corresponds to its own completely separate $SU(2)$ symmetry group. Basically they are completely different physical properties that just happen to share the same group structure. Some physicists have developed an unfortunate habit of calling any property that follows this group structure "isospin" (unless it is actual spin).

Since the photon does not have any up/down flavor, it exists in the trivial representation of non-weak isopsin $SU(2)$, and thus its actual isospin is zero.


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