Feynman rule for scalar QED vertex

Question

A popular problem in QFT textbooks and courses is to derive the Feynman rules for scalar QED. Usually, this theory is presented via the following Lagrangian density:

$$\mathcal{L} = (D_\mu\phi)^\dagger D^\mu\phi - m^2\phi^\dagger\phi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$

where the covariant derivative is given by

$$D_\mu\phi = (\partial_\mu + ieA_\mu)\phi,$$

and the $U(1)$ gauge field strength tensor is

$$F_{\mu\nu} = (\partial_\mu A_\nu - \partial_\nu A_\mu).$$

I find it easier to attempt to find the Feynman rule for the vertex after plugging in the expression for the covariant derivative. The Lagrangian density then assumes the form

$$\mathcal{L} = (\partial_\mu\phi)^\dagger(\partial^\mu\phi) - m^2\phi^\dagger\phi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - ieA^\mu[\phi^\dagger(\partial^\mu\phi) - (\partial^\mu\phi)^\dagger\phi] + e^2 A_\mu A^\mu \phi^\dagger\phi.$$

We see that there are two interaction terms at order $e$ which will result in an interaction between a scalar field and its conjugate and the gauge field. Hence, we should compute a $3$-point function and expand each interaction exponential and then apply the LSZ formula to get the rules.

In position space, the $3$-point function should involve time-ordered products of the form

$$A_\mu(x_1)\phi^\dagger(x_2)\phi(x_3)A_\nu(y)\phi^\dagger(y)\partial^\nu\phi(y).$$

Wick's theorem then tells us that this time-ordered product is just the sum of all possible contractions and we know that only full contractions survive in the vaccuum expectation value. The answer to the vertex should be $-ie(p - p')_\mu$.

However, it is not clear to me how to proceed with the contraction between $\phi^\dagger$ and $\partial^\nu\phi$ because of the derivative. I have taken a look at these SE questions and also at Schwartz's book, section $9.2$. In the latter, he vaguely states that the derivative becomes a factor of momentum after going to momentum space.

Well, for me it doesn't seem to be that straightforward.

Following the reasoning of Schwartz, which seems the origin of the explanation in the aforementioned SE questions, we start from the expansion of the fields in Fourier modes:

$$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p e^{-ip\cdot x} + b^\dagger_p e^{ip\cdot x}),$$

$$\phi^*(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a^\dagger_p e^{ip\cdot x} + b_p e^{-ip\cdot x}).$$

If we act with the derivative in the first expression, assuming that we can interchange the integral with the derivative operator, we will indeed pull down a factor of momentum, but which is integrated and we will also get a relative sign between the modes. Concretely:

$$\partial_{\mu}\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(-ip_\mu)(a_p e^{-ip\cdot x} - b^\dagger_p e^{ip\cdot x}).$$

Schwartz then just proceeds to state the Feynman rule of the vertex with the factors of momentum. It seems to me that he is arguing that one can do the contractions coming from Wick's theorem by considering the time-ordered product of the form

$$(-ip^\nu)A_\mu(x_1)\phi^\dagger(x_2)\phi(x_3)A_\nu(y)\phi^\dagger(y)\phi(y).$$

However, acting with the derivative, as I have showed above, induces a relative sign change and a factor of momentum which is also being integrated. I am not sure if I understand Schwartz's argument or the subsequent forum discussions which are clearly inspired by the discussion in the book. I seek enlightenment.

Answer 1

It may be easier for you to expand the interaction action term in Fourier space: $$ \begin{align} iS_I &= e\int A^\mu(x)(\phi^*(x)\partial_\mu\phi(x)-\phi(x)\partial_\mu\phi^*(x))d^4x \\ &= ie\int A^\mu(k)(q_\mu-p_\mu)\phi(p)\phi^*(q)(2\pi)^4\delta(k+p-q)\frac{d^4k}{(2\pi)^4}\frac{d^4p}{(2\pi)^4}\frac{d^4q}{(2\pi)^4} \end{align} $$ You therefore directly read out the interaction vertex with the imposed momentum conservation.