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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?

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  • $\begingroup$ Can you show or describe some of your calculations? The $2$ as stated in the literature is correct. $\endgroup$ – knzhou Jun 18 '19 at 15:12
  • $\begingroup$ Sure, I'll edit the question $\endgroup$ – Dave Jun 18 '19 at 15:20
  • $\begingroup$ Are you able to get the factor 4 right for the scalar 4 point vertex? $\endgroup$ – RenatoRenatoRenato Jun 18 '19 at 18:05
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    $\begingroup$ There are two choices of how you apply your photons to your $\gamma$ states, hence the factor of 2. $\endgroup$ – Aaron Jun 18 '19 at 22:40
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Here's a short derivation, copied verbatim from my professor's lecture notes.

$\mathcal{M}=i e^2 \epsilon_{j \nu} \epsilon^{* \nu}_j = i e^2 \eta_{\mu \nu} \epsilon^\mu_j \epsilon^{* \nu}_j = ie^2 ( \eta_{11} \epsilon^1_1 \epsilon^{* 1}_1 + \eta_{22} \epsilon^2_2 \epsilon^{* 2}_2 ) = -2i e^2$

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.

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    $\begingroup$ Note: bounty posted since I'm also curious to see the full derivation. $\endgroup$ – wyphan Dec 4 '20 at 20:33
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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\begin{equation} \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} \end{equation}

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