# Scalar electrodynamics "seagull" vertex factor

By expanding the covariant derivative of the Scalar QED lagrangian one gets the following term, sometimes called "seagull" vertex. $$\mathcal{L}_{seagull} = -q^2A_\mu A ^\mu \phi^\dagger \phi$$ Most of the literature give $$2iq^2g_{\mu\nu}$$ as the Feynman rule for this vertex, saying that the factor $$2$$ is coming from the "symmetry of the two $$A$$ fields".
However, even if it actually makes sense at a first look, I can't reproduce the same result by contracting the $$S$$-matrix with appropriate initial and final states, since I get $$iq^2g_{\mu\nu}$$. In other words, I can't understand how the symmetry in the $$A^\mu A_\mu$$ term actually affects the $$\mathcal{T}$$-product on the $$S$$-matrix. How should I reproduce this result?

Here are my calculations:
I choose the following initial $$|i\rangle$$ and final $$|f\rangle$$ states, in order to evaluate $$S_{fi}$$: $$|i\rangle = |\gamma_\lambda(k_1),\gamma_\sigma(k_2)\rangle$$ $$|f\rangle = |s(p_1),\bar s(p_2)\rangle$$ Where $$\gamma_\lambda(k_1)$$ describes a photon of momentum $$k_1$$ and polarization $$\lambda$$, $$s(p_1)$$ a scalar particle of momentum $$p_1$$ and $$\bar s(p_2)$$ a scalar anti-particle of momentum $$p_2$$. Now I can write: $$S_{fi} = iq^2 g^{\mu\nu}\langle s(p_1),\bar s(p_2)|\int d^4x \, \mathcal{N}[A_\mu(x) A_\nu(x) \phi^\dagger(x) \phi(x)]\, |\gamma_\lambda(k_1),\gamma_\sigma(k_2)\rangle$$ Expanding each field in terms of its "positive energy" and "negative energy" parts, and selecting the only contributing term one gets: $$S_{fi} = iq^2 g^{\mu\nu}\langle s(p_1),\bar s(p_2)|\int d^4x \,\mathcal{N}[A_{\mu+} A_{\nu+} \phi^\dagger_- \phi_-]\, |\gamma_\lambda(k_1),\gamma_\sigma(k_2)\rangle$$ Making use of the $$\mathcal{N}$$-product, applying the fields to the states and finally integrating over $$d^4x$$ one gets the final amplitude: $$S_{fi} = (2\pi)^4 \delta^4(p_1+p_2-k_1-k_2)iq^2 g^{\mu\nu} \epsilon_{\mu,\lambda}\epsilon_{\nu,\sigma}$$ From which one can deduce the vertex factor $$iq^2g^{\mu\nu}$$. The "symmetry" in the $$A$$ fields actually reflects in the fact that I could have obtained $$\epsilon_{\nu,\lambda}\epsilon_{\mu,\sigma}$$ instead of $$\epsilon_{\mu,\lambda}\epsilon_{\nu,\sigma}$$, but since the $$\mu\nu$$ indexes are contracted through $$g^{\mu\nu}$$ I would guess that it would not make any difference. Where am I failing?

• Can you show or describe some of your calculations? The $2$ as stated in the literature is correct. Jun 18, 2019 at 15:12
• Sure, I'll edit the question
– Dave
Jun 18, 2019 at 15:20
• Are you able to get the factor 4 right for the scalar 4 point vertex? Jun 18, 2019 at 18:05
• There are two choices of how you apply your photons to your $\gamma$ states, hence the factor of 2. Jun 18, 2019 at 22:40

$$\mathcal{M}=i e^2 \epsilon_{j \nu} \epsilon^{* \nu}_j = i e^2 \eta_{\mu \nu} \epsilon^\mu_j \epsilon^{* \nu}_j = ie^2 ( \eta_{11} \epsilon^1_1 \epsilon^{* 1}_1 + \eta_{22} \epsilon^2_2 \epsilon^{* 2}_2 ) = -2i e^2$$
Here, $$\mathcal{M}$$ is the invariant scattering amplitude; $$\epsilon_j$$ is the polarization mode of the photon (which can be taken to be $$\epsilon_1^\mu = ( 0, 1, 0, 0 )$$ and $$\epsilon_2^\mu = ( 0, 0, 1, 0 )$$ for the two transverse modes of a photon traveling in the $$+\hat{z}$$-direction); $$\eta_{\mu \nu}$$ is the usual Minkowski metric; and $$e$$ is the elementary charge.
In your computation you are missing a factor of 2 coming from two different contractions of $$A(x)$$ with the two photons. In other words: \befin$$$$\langle s(p_1), \bar s (p_2)| \phi(x) \phi^\dagger (x)A_\mu(x) A_\nu(x) |\gamma(k_1), \gamma(k_2) \rangle = 2 e^{-i(k_1+k_2-p_1-p_2)x}$$$$