# Guessing of the vertex factor of derivative interactions

I want to ask if my "empirical" method of guessing the vertex factor is correct. As far as I understand in canonical QFT the Feynman rules are only a way to rearrange the important parts of the perturbative expansion. So, if I have an interaction lagrangian (possibly also involving derivative terms) like scalar QED: $$\mathcal{L} = -i e A_{\mu}(\psi^{\dagger} \partial^{\mu} \psi - \psi \partial^{\mu} \psi^{\dagger})$$ one has to guess what is the vertex term in order to avoid doing the explicit calculation.

In a vertex like the one above I would consider an initial state like $b^{\dagger}(p) |0 \rangle$ and a final state $a^{\dagger}(k) b^{\dagger}(q) | 0 \rangle$ (one photon and one scalar). The vertex should bring me from the initial state to the final state. Only the first part of the interaction is $-i e A_{\mu} \psi^{\dagger} \partial^{\mu} \psi$. The $\psi$ field annihilate the scalar in the initial state through $b(p) e^{-i p x}$ so the derivative brings down a factor $-ip$. The second part has $b^{\dagger} (p) e^{i p x}$ acting on the initial state that brings down a factor $ip$. So due to the minus sign the two momentum get summed and up to an overall phase the vertex is $e (p+q)^{\mu}$. As $\psi$ and $\psi^{\dagger}$ are different fields I don't have to worry about combinatorial factors.

I would apply a similar reasoning to every vertex (even if the legs are propagator) for every interacting field theory. Is it sensible? If not how can I guess every vertex without having to do long calculations? I want to avoid the use of the functional formulation of QFT. Anyone?

• That sounds pretty reasonable to me! It might not be 'rigorous' but this is how it's done most of the time in practice. – knzhou Oct 17 '17 at 9:49

The most important one is that the momentum in $b^\dagger(p)|0\rangle$ is on-shell, $p^2=m^2$, while the momentum in the Feynman rules is off-shell, $p^2\neq m^2$. This distinction is very important because Feynman intergrals are over all $p$, not over the mass-shell $\delta(p^2-m^2)$, so the $p$ in $b^\dagger(p)|0\rangle$ is very different from the $p$ in $-iep^\mu$. Even if you use the same symbol, they are fundamentally different objects. The first one is the eigenvalue of $\hat P^\mu$ and the second one is a Fourier variable.
That being said, your derivation is the best you can do in the canonical formalism. For a better justification you need to use the functional formulation of QFT, something you don't want to do. For completeness, let us mention that in the functional formalism a Feynman vertex is just a functional derivative, $$\frac{\delta}{\delta\psi}\frac{\delta}{\delta\bar\psi}\frac{\delta}{\delta A}\mathcal L(\psi,\bar\psi,A)$$ from which the factor $-iep^\mu$ readily follows. It is an immediate result, and much more easily derived than your heuristic calculation. The functional formalism is more rigorous and more convenient -- although I can see why you want to avoid it at first.