# Flavor changing neutral current with photon

I'm trying to understand why the penguin diagram for photons takes its form.

$$I$$ here is the integral and some CKM elements.

The author (Ch1.6.1) says that for $$b\rightarrow s\gamma$$, we get a factor in the amplitude like $$F_{\mu\nu}\bar{s}\sigma^{\mu\nu}b$$ but not $$A_\mu\bar{b}\gamma^\mu s$$ (which we get for a normal loop of this form). This is because the second term is forbidden by EM gauge invariance.

1). Why is this term forbidden? Would this still be forbidden if we had $$Z_\mu$$ instead of $$A_\mu$$?

Secondly the author says that we have to flip the chirality of $$b_R$$ to $$b_L$$ so it can interact with the W, and thus adding $$m_b$$ to the expression.

2). But why could we not have started with $$b_L$$? So instead of $$b_R\rightarrow s_L\gamma$$ we would have $$b_L\rightarrow s_L\gamma$$. Is this not allowed?

• Can you put the RHS of the equals sign of your first equation and using LaTeX/MathJax? Commented Jul 20, 2023 at 18:39

1). These are terms in the effective lagrangian, that must have the right symmetries, including EM gauge invariance. Now note $$F_{\mu\nu}\bar{s}\sigma^{\mu\nu}b$$ is EM gauge invariant, as both factors are individually gauge invariant; but $$A_\mu\bar{b}\gamma^\mu s$$ is not, since $$\bar{b}\gamma^\mu s$$ is not a conserved current, (FCEMC!) which you know the photon must couple to for gauge invariance! Indeed a term $$Z_\mu\bar{b}\gamma^\mu s$$ would also be not gauge invariant, for the same reason, absence of FCNC in the tree lagrangian. You might argue that other clever terms could conceivably provide the non vanishing gauge variations to cancel this one, hence his hedge footnote, but you can quickly see this is impossible....
This would suggest getting an amplitude proportional to $$\overline{ u(p_s)}\left(\frac{1+\gamma_5}{2}\right)\sigma^{\mu\nu}{\textstyle \left(\frac{1-\gamma_5}{2}\right)}u(p_b)$$ which, of course, vanishes.
So, he reminds you $$b_L\rightarrow s_L\gamma ~$$ is disallowed with extreme prejudice, since a left projector $$P_L$$ under the overbar becomes a R projector one outside it, as exhibited, and slips through the $$\sigma^{\mu\nu}$$ unchanged, so annihilates the left chiral projector of the rightmost spinor (by contrast, R would flip to L if it had to pass through a mere $$\gamma^\mu$$.) This is the heart of all such arguments, and you must get in control of it before reading on!
So there must be an R in the initial or final quark. He explains that it is more likely for the heaviest one, the initial b, to flip chirality, as such flips are proportional to the mass of the quark. The amp he gives is $$m_b/m_s$$ larger than the one with an R s quark!