As already mentioned in this Phys.SE post, the PDG booklet says that the isospin of a photon is $0,1$. Is it referring to the strong or to the weak isospin of the photon? What would be the 3rd component of the isospin? And finally, what would be the total $I$ and $I_3$ quantum numbers on the right side of the equation in the case of the following decay:

$$\eta \rightarrow \gamma + \gamma$$

and what would be the type of interaction?

  • $\begingroup$ Thank you for your comment. So you mean that I can rewrite Gell-Mann-Nishijima formula in order to get $I_3$? Since $Q=Ba=S=0$, then I conclude that $I_3=0$. Is that right? $\endgroup$ – Jxx Jan 10 '18 at 21:09
  • $\begingroup$ Then the strong isospin $I$ is 0,1 or a combination of both? $\endgroup$ – Jxx Jan 10 '18 at 21:11
  • $\begingroup$ I'm sorry, I don't know the formula you are using for the coupling of the photon to the e.m. current. I am only trying to understand why the decay that I described in my original post can only be electromagnetic and not strong or through Z interaction? Besides, isn't parity not conserved since P=-1 for $\gamma$ (thus $P=(-1)\cdot(-1)=+1$ for the right side of the equation) and P=-1 for $\eta$? $\endgroup$ – Jxx Jan 10 '18 at 21:25
  • $\begingroup$ Ah nevermind the correct formula is $P=(-1)^l$... Sorry for that. $\endgroup$ – Jxx Jan 10 '18 at 21:28

The decays involving photons are, of course, electromagnetic. You may forget about weak intermediaries.

The e.m. current coupling to the photon in the lagrangian in terms of the lightest quarks is $$eA_μ(2/3\bar{u}γ^μu−1/3\bar{d}γ^μ d)=eA_\mu \bar{q}(\tfrac{1}{6} \mathbb {1}+I_3)\gamma^\mu q,$$ since diag(2/3,-1/3)=diag(1/6,1/6) + diag( 1/2,-1/2). Each quark couples to the photon through its (different) charge. So the photon couples to both isoscalars (I=0) and isovectors (I=1), and electromagnetism violates strong isospin. However, if you imagine this coupling term to preserve strong isospin, then you must represent the photon as a linear combination of (strong) isoscalar and isovector.

As a consequence, if you had to preserve strong isospin, as a way to track its violation, both vertices $\eta \gamma \gamma$ and $\pi^0 \gamma \gamma$ are allowed (they both have isospin zero) in this accounting, and, indeed, both particles decay to two γs as specified by the vertices.

Considerations of weak isospin, spontaneously broken anyway, are irrelevant in these decays.

| cite | improve this answer | |
  • $\begingroup$ Thank you for your answer. I must admit that I don't understand the formula you mention regarding the e.m. current coupling to the photon, it was not in my courses so far. However would you mind explaining me why the strong isospin is conserved regarding the two decays you mention? And especially what is the value of the isospin in those cases? Maybe you mean that $I$ and $I_3$ can only be retrieved from the equation you typed, then I am sorry if that is the case. $\endgroup$ – Jxx Jan 10 '18 at 21:57
  • $\begingroup$ Hope the edit made it better. Adding isospins, the ηγγ vertex is 0+0+0=0, or 0+1+1=0 , and the πγγ one, 1+1+0=0 , 1+1+1=0 , where the additions are vector additions of Is. $\endgroup$ – Cosmas Zachos Jan 10 '18 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.