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Classical electromagnetism says an accelerated charge would radiate. Is this understood in QED?

Is there QED derivation of Larmor's formula?

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Yes! There are lots of ways to do it, but here is one awesome way. I'm only going to sketch it because it's a little complex. The general methods are described in many places, but one place is section 2 of the Les Houches Lectures on Effective Field Theories and Gravitational Radiation by Goldberger http://arxiv.org/abs/hep-ph/0701129.

We start with the action for a photon coupled to an external source. The idea is to look at the radiation from a big, classical lump of charge that undergoes an accelearation. You could also ask about the amplitude for a single electron to emit two photons (this is a quantum mechanical process), $^1$ but the Larmor formula as derived in classical electrodynamics is really about a classical lump of charge.

[1] The italicized text is wrong, whoops, a single electron can't emit two photons obviously because of energy conservation. Instead you could compute the amplitude for two electrons to exchange a photon, and set things up so that the photon is on shell or nearly on shell. There's subtleties about infrared divergences and so on since the photon is massless. This is pretty similar to what I'm describing below, but I still think that the simplest version of your question is about the radiation of a classical lump of charge.

So we write the classical action

\begin{equation} S[A,J]=\int d^4x-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+qA_\mu J^\mu \end{equation} where $q$ is the charge of the source and the source must be conserved $\partial_\mu J^\mu=0$.

We will input a source distribution of a point charge moving in a circle

\begin{equation} J_\mu = \delta_\mu^0 \delta^3(\vec{x}-\vec{x_q}(t)) \end{equation} where \begin{equation} \vec{x}_q(t) = r\left(\hat{x}\cos \omega t + \hat{y} \sin \omega t\right) \end{equation} is the trajectory of the charge.

We can write down an effective action that comes from integrating out $A_\mu$

\begin{equation} e^{i S_{eff}[J]} = \int D A_\mu e^{i S[A,J]+ i S_{gf}[A]} \end{equation} where $S_{gf}[A]$ is a gauge fixing term.

This action will be nonlocal because it describes the forces directly between two classical sources (since we have integrated out the photon).

The path integral is gaussian and can be done to yield

\begin{equation} S_{eff}[J] = \int d^4 x \int d^4 y J^\mu(x) G_{F,\mu\nu}(x,y) J^\nu(y) \end{equation}

where $G_{F,\mu\nu}$ is the Feynman propagator of the photon. (The gauge of the propagator is fixed by the gauge fixing term in the path integral).

The action $S_{eff}$ will have an imaginary part. Mathematically it comes from the fact that the Feynman propagator is complex. Physically it accounts for the fact that energy can be radiated into photons (but since we have integrated out $A_\mu$ we describe that process as energy of the source not being conserved).

The path integral gives the vacuum->vacuum transition amplitude in the presence of the source. The radiation amounts to a 'decay of the vacuum' (aka creating photon particles). So we can compute the probability of the vacuum decaying by computing

\begin{equation} e^{\Gamma t} = |e^{i S_{eff}}|^2 = e^{-2 {\rm Im}S_{eff}} \end{equation}

$\Gamma$ would be 0 if there were no radiation. However there is an imaginary part of $S_{eff}$ arising from integrating out the photon, and this describes the radiative process.

That leads to the actual calculation part, which is actually not hard. If you work out the imaginary part of $S_{eff}$ above [you will need the identity ${\rm Im}(1/(x+i\epsilon))=1/x+i\pi \delta(x)$ to take the imaginary part of the feynman propagator, and you also will need to do a multipole expansion of the propagator and you only want the dipole part], you will find that the rate of energy loss (aka the power radiated) is consistent with the Larmor formula.

The formalism is a little complex, but it's actually a very similar calculation to what you would do in classical electromagnetism at the end of the day, and it shows you a way to think about radiation quantum mechanically.

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  • $\begingroup$ I really like your answer, and I have a question. Because the functional integral is gaussian, it seems to me that the answer you get from the "fully quantum" calculation as you did here is equivalent to the classical (saddle point) calculation. Or does the determinant coming from the integration play a role in the quantum case ? That seems improbable because it is independent of the source... $\endgroup$ – Adam Oct 9 '13 at 23:31
  • $\begingroup$ Exactly right. Quantum mechanically, a U(1) gauge field not coupled to any matter is very boring, the only actual quantum effect is the zero point energy. So indeed, the path integral is gaussian and this is ultimately a quantum-looking way of doing a classical calculation. (But it shows some of the power of the path integral formalism, it easily includes classical effects). Even if there were interactions though we could still compute the tree level contribution--that is basically the point of the paper I linked to, they do this calculation with gravity to compute classical radiation. $\endgroup$ – Andrew Oct 10 '13 at 21:49
  • $\begingroup$ The path integral for gravity is not gaussian, there are an infinite tower of interactions. The paper computes the tree level contribution to the power but accounting for the interactions, essentially doing what would normally be called a PPN expansion in a very elegant way. They definitely aren't doing a quantum mechanical loop expansion, quantum gravity isn't relevant for pulsars :). BTW the determinant you are talking about are also called 'fadeev popov ghosts' and indeed they are irrelevant for U(1) gauge fields because the path integral is exactly gaussian. $\endgroup$ – Andrew Oct 10 '13 at 21:51
  • $\begingroup$ I was not talking about the ghost part, but the determinant at $J=0$ (zero point energy). So I guess that to have a more interesting (beyond classical) Larmor formula, one would have to include some fermions, at least at the level of the Euler–Heisenberg Lagrangian. Do you know if anyone had a look at that ? $\endgroup$ – Adam Oct 11 '13 at 3:36

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