I'm trying to understand formula from Anthony Zee qft in nutshell for beta decay: $$\langle p'|{J_5}^{\mu} |p\rangle = \bar u(p')\left[\gamma^\mu\gamma^5F(q^2) +q^\mu\gamma^5G(q^2)\right]u(p)$$

it is stated, that term with $$(p' + p)^\mu \gamma^5 A(q^2)$$ is missing because of some charge and isospin symmetry. This is not clear because weak interactions should violate isospin and at the same time if one applies charge conjunction$$C=i \gamma^0\gamma^2$$ to both $u(p)$ and $\bar u(p')$ the current stays invariant and thus $A(q)$ doesn't have to be zero. How can this be solved?

In many other books term $$\sigma^{\mu\nu} q^\nu$$ is used instead. And in this case it is clear, that this term makes the corresponding part of the current with positive G-parity, and it explains everything. Why G-parity for terms with $$(p' + p)^\mu$$ and $$(p' - p)^\mu$$ is different is completely unclear.



it seems, that one can use one of the Gordon identities:

$$0 = \bar u(p')\left[(p' + p)^\mu\gamma^5 +iq_\nu\sigma^{\mu\nu}\gamma^5\right]u(p)$$ to replace $(p' + p)^{\mu}$ with $q_\nu\sigma^{\mu\nu}\gamma^5$, which would correspond to positive G-parity term of the current, while only negative are allowed for this reaction .


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