# The form of QED Vertex Correction

In chapter 6 of Peskin-Schroeder's text Introduction to Quantum Field Theory it is argued that the form of vertex correction for QED can only have the following form (since we have only the constants, $$\gamma^\mu$$ and momenta + the transformation properties it must follow) :

$$\Gamma^\mu = A \gamma^\mu + B (p+p')^\mu + C (p-p')^\mu$$ with A,B and C being scalars.

Question :

If we go by the transformation property similarity requirement of $$\Gamma^\mu$$ and $$\gamma^\mu$$ then the following terms can also be included : $$X^{\mu\nu}[D (p+p')_\nu + E (p-p')_\nu]$$ and $$F\ Y^{\mu\nu\rho} p_\nu p'_\rho$$ with D,E,F being scalars and Y required to be symmetric. Why don't we include them?

Discussion & Thoughts :

One could ask what are $$X^{\mu\nu}$$ and $$Y^{\mu\nu\rho}$$. They could be made up of product of different $$\gamma^\mu$$ etc and then it would appear that they are the same as the terms included above (more on this in a bit). But there is nothing that stops $$X^{\mu\nu}$$ and $$Y^{\mu\nu\rho}$$ to be some completely random tensors with constant entries which cannot be determined analytically and must only be determined experimentally (exactly in the manner that the first two terms from the normally used $$\Gamma^\mu$$ are converted to form factors which can only be determined experimentally). Now, who's to say that they aren't already being determined from the experimental form factors included in the commonly used expression. The only way to argue otherwise would be to ensure that the $$X^{\mu\nu}$$ and $$Y^{\mu\nu\rho}$$ are independent terms. This is where I invoke the previous "in a bit" comment. Usually what we do is that we use the Gordon Identity to exchange $$(p+p')^\mu$$ for a single $$\gamma^\mu$$ and a $$\sigma^{\mu\nu}$$ which is like two gamma matrices (commutator of them) so the only way to have independent terms in $$\Gamma^\mu$$ is to have product of 3 different gamma matrices or 4 different gamma matrices, any more can be reduced to lesser number of gamma matrices. The 4 different gamma matrices term would be proportional to $$\gamma^5$$ and since it violates parity we cannot include it in a theory for QED. But what about the term with three gamma matrices? So, in effect my questions boil down to the following :

i) Why don't we include $$X^{\mu\nu}[D (p+p')_\nu + E (p-p')_\nu]$$ and $$F\ Y^{\mu\nu\rho} p_\nu p'_\rho$$ terms if $$X^{\mu\nu}$$ and $$Y^{\mu\nu\rho}$$ are taken to be random tensors? ; or

ii) If we take them to be essentially involving $$\gamma^\mu$$ then why don't we consider the term involving product of three different gamma matrices?

Edit 1 : After a comment I would like to clarify that in point ii) above I want to include three gamma matrices term directly into the $$\Gamma^\mu$$ without including the extra tensors that I am introducing.

Edit 2 : After further thought, the three gamma term problem is resolved. It is already a part of first term and since we don't know exactly how the gammas are contracting and how the terms are depending on constants it gives rise to the fact that the form factors must be determined experimentally.

Edit 3 : I am not entirely sure that we can discard the three-gamma term because an identical argument could be applied to the two-gamma term but still it exists in the commonly used expression. Somehow, the Gordon Identity exists which allows us to write down the two-gamma term independently. I don't know what I should verify/ensure to include or exclude the three-gamma term beforehand.

• If you have three gamma matrices then two of them need to be contracted with the momentum vectors and can be absorbed into the coefficient of a single gamma matrix. A similar argument holds for two gamma matrices. Apr 14, 2020 at 17:34
• @SounakSinha Sorry, I meant the three gamma term to be there without the presence of extra tensors that I am introducing. Apr 14, 2020 at 17:50
• You can't have three gamma terms alone, they need to be contracted since $\Gamma$ is a vector. Apr 14, 2020 at 17:55
• @SounakSinha Yes, I just realised that and modified the question accordingly. Also, just to be crystal clear, $\Gamma^\mu$ is not really a vector. Apr 14, 2020 at 18:01
• Have you tried reasoning through how your random tensors will satisfy the ward identity and gauge invariance? Apr 15, 2020 at 11:06