How does a charged field's coupling to the electromagnetic field translate to a worldline coupling?

Question

Suppose we're interested in (minimally) coupling a heavy scalar particle to the electromagnetic field. In quantum field theory, we describe the particle as an excitation of a complex scalar field $\phi$. Including the simplest possible coupling leads to the theory of scalar QED, with Lagrangian $$\mathcal{L} = - \frac14 F_{\mu\nu} F^{\mu\nu} + |D_\mu \phi|^2 - m^2 |\phi|^2.$$ By expanding out the kinetic term, we see that there are vertices where a $\phi$ emits a photon, and where a $\phi$ emits two photons. The latter is required by gauge invariance.

On the other hand, if the particle is localized, we could also describe it as a worldline supplying a classical current. For a trajectory $y^\mu(t)$, the current is $$J^\mu(x) = e \int d \tau \, \frac{dy^\mu}{d\tau} \, \delta(x - y(\tau))$$ and the action is $$S = - \int dx \, \frac14 F_{\mu\nu} F^{\mu\nu} + \int dx \, J^\mu A_\mu + m \int d \tau.$$ Everything is gauge invariant, since $J^\mu$ is.

Both of these pictures are common, but I'm confused how to translate between them. On the quantum field side, we know that $\phi$ must be a complex scalar, and that it must have a two-photon coupling to ensure gauge invariance. But on the worldline side these facts aren't manifest at all! This doesn't make sense, because in both cases it should be possible to calculate, the probability of producing two hard photons when the particle is accelerated. The results should match, but since the worldline picture is "missing" the two-photon coupling, I don't see how it can.

More specifically, it's clear that the amplitude to produce one hard photon matches between the two formalisms -- what's puzzling to me is that it seems like this implies the amplitude to produce two hard photons in the worldline formalism should match the corresponding amplitude in the field theory formalism, without the two photon vertex.

What's going on?

Answer 1

I think the best way to see where the four-point interaction disappears is to follow the derivation of the worldline formalism. Christian Schneider's PhD thesis has a good walkthrough for scalars and spinors in section 4.2; the majority of this answer paraphrases it.

Starting with the scalar QED Lagrangian you write down at the beginning of the question, you get the path integral

$$\mathcal{Z} = \int \mathcal{D} \phi \mathcal{D} \phi^* \mathcal{D}A \, \mathrm{e}^{iS[\phi, \phi^*, A]}$$

where $S$ is the action from the scalar QED lagrangian. The next step is to rewrite the lagrangian in a Gaussian form: by essentially completing the square we find

$$S \supset \phi^* (D^2 + m^2) \phi.$$

The next trick is to Wick rotate to Euclidean space (changing $D^2 \to -D^2$ and getting rid of the oscillatory integral), so we can integrate out the matter field to give

$$\mathcal{Z}_\mathrm{E} = \int \mathcal{D} A \det (-D^2 + m^2) \mathrm{e}^{- \frac{1}{4}\int F_{\mu \nu} F^{\mu \nu}}. $$

This is where the four-point interaction appears to vanish: we have integrated out over all the scalar field loops and hidden the interactions in the functional determinant. What follows is a series of integral tricks to write the functional determinant as an integral over all worldlines. We eventually get

$$\mathcal{Z}_\mathrm{E} = \mathcal{N} \int_0^\infty \frac{\mathrm{d}T}{T} \int \mathcal{D} x \int \mathrm{D} A \, \mathrm{e}^{-\frac{1}{4} \int F_{\mu \nu} F^{\mu \nu}}\,\mathrm{e}^{-\int_0^T \mathcal{L}_\mathrm{eff}[x, A] \, \mathrm{d} \tau},$$

where $\mathcal{L}_\mathrm{eff}$ is the worldline effective action, and $T$ is a parameter introduced that can be thought of as the worldline length. The effects of the four-point interaction are hidden in the $x$ and and $T$ integrals; if you sum them to all orders you'll get the same result as the standard QFT one.

As noted on p6 of this paper, the fact that the worldline method integrates over diagrams including the scalar interactions is one of its strengths; it can greatly simplify certain calculations.

Answer 2

For what it is worth, the seagull term in scalar QED

$$\begin{align} \exp&\left(i\Gamma_{\text{1-loop}}[\phi_{\rm cl}\!=\!0; A_{\rm bg}]\right)\cr ~=~&\int \!{\cal D}\phi~\exp\left[i\int\!d^dx\left\{-\left|\left(\frac{1}{i}\partial_{\mu} - qA_{{\rm bg},\mu} \right) \phi\right|^2 - m^2 |\phi|^2\right\}\right] \end{align}$$ has a formal analog in the Hamiltonian worldline (WL) formalism $$\begin{align}&\int_{\mathbb{R}_+} dT \int \!\frac{{\cal D}x{\cal D}p{\cal D}e}{\text{ Vol(Gauge)}}~\cr &\exp\left[i\int_{\tau_i}^{\tau_f}\!d\tau \left\{p_{\mu}\dot{x}^{\mu} - \frac{e}{2}\left((p - qA_{\rm bg})^2+m^2 \right) \right\}\right]\delta[\text{Gauge-fixing}]. \end{align}$$ For more information, see Refs. 1+2.

References:

1. F. Bastianelli, Constrained hamiltonian systems and relativistic particles, 2017 lecture notes; chapters 2 + 3 + 4.

2. C. Schubert, Perturbative Quantum Field Theory in the String-Inspired Formalism, arXiv:hep-th/0101036; sections 3.2 + 4.3.1.

--

Notation & Conventions: Minkowski signature $(-,+,+,+)$.