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In the process of pair annihilation an electron and a positron annihilate each other to produce a pair of photons, conserving momentum and energy. As the oppositely charged particles approach each other, the field around them decreases. We may consider them as a dipole with a decreasing dipole moment and radiating energy. As the field decreases the value of E changes. Consequently, B should change: $$ \nabla \times \mathbf{B}=\epsilon_0\mu_0\frac{\partial \mathbf{E}}{\partial t}$$ and $$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}.$$ So we have a Poynting vector that drives energy in the outward direction (If it was in the inward direction the field would grow stronger with the approach of the particles which is not the case). By the time the electron and the positron are close enough to clash the classical self energy (electromagnetic self energy) has moved out to an infinite distance.

Quantum Mechanical Picture:Virtual photons emitted by the electron (or the positron) hit back the particle increasing its KE.This extra energy gets included into the mass of the particle leading to the dressed mass of the particle. In pair annihilation this energy is included in $E=mc^2$.And we get it after the collision of the particles. But the classical electromagnetic self energy, as we have seen in the previous paragraph, flows out to infinity before the collision of the particles.

So the classical self energy of the electron and its QM self energy must be physically different. Is this true?

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The classical annihilation process described in the original post is a falling-at-the-center problem with releasing the potential energy difference as radiation due to time-dependent dipole moment (Anamitra, you missed the current $\vec{j}$ in the first equation). So the total radiated energy is essentially determined with the radius at which the falling stops. If we use the classical electron radius $r_0=\frac{e^2}{mc^2}$, the released energy will be $E=\frac{e^2}{2r_0}=\frac{mc^2}{2}$. If we take into account the magnetic field contribution (an electron is a magnet), we may derive another "classical electron radius" $r_1>r_0$, so the released energy will be even less. What I wanted to underline is that it is the transformation of the interaction potential energy into radiation energy, which is not a self-energy transformation. As soon as the electron and positron remain slow (non relativistic, due to loss of energy into the radiation), the final state in CED is some sort of a still dipole "molecule" of some small size and new total mass (if they do not merge). If they merge, we can (theoretically) get a neutral system with zero rest mass and zero rest energy (the mass defect being $2mc^2$).

The notion of self-energy is rather strange. It is a theoretical difficulty of our describing interactions via fields that is resolved by a brute force - with discarding the self-energy contributions (they are not necessary, to say the least). The more important thing is the total electron energy that includes the famous $mc^2$ in the relativistic mechanics with its new conservation laws.

In QED, when a non-relativistic positronium annihilates, we use the energy conservation law that gives us the total radiated energy providing the electron and positron disappear. The difference between CED and QED in an essentially "discontinuous" spectrum of the radiation (transitions from one level to another) and in admitting the possibility for particles to disappear, the energy conservation law still holds. The interaction manifests itself here in attracting the charges in the initial state and in disappearing the charges in the final state.

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The soft photons of annihilation should lead the hard photons to always add up to less than 2mc^2 by a certain small fraction of order $\alpha$. Do you know what these corrections are experimentally or theoretically? Just curious. –  Ron Maimon Dec 30 '11 at 9:50
    
@RonMaimon: No, I do not know exactly, but I suspect that the only certain thing is an average value. One can get two less hard photons with more energetic soft contributions within the same initial energy of the electron-positron pair. –  Vladimir Kalitvianski Dec 30 '11 at 12:20
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The quantum electrodynamics process includes the distant field of an electron only in a crude way, by additional photon emissions from the electron line. The mass, however, includes the self-field contribution, in a physical renormalization, because the mass of the electron is fixed at its experimental value. The tree-level QED description is not of the far-region outside the compton wavelength of the electron, but of the behavior in the near region, close to and inside the Compton wavelength.

The field of the electron and positron is described by adding photon lines to the simple tree level annihilation diagram, and making these lines have low momentum, soft photons. The soft photon amplitude is always divergent, and this divergence is only made sensible by doing a summation of infinitely many soft photon emissions, with the classical field as a guide for the summation. This type of thing is very involved, but since the physical situation is well understood classically, along the lines you suggest, people generally do not worry about it. This line of research is generally classified as infraparticle analysis, which tries to take into account the infrared behavior of QED.

Classical electron radius vs. Compton wavelength

The classical electron radius is defined, up to small factors of order unity, as the radius of a sphere of total charge e whose self-energy equals the mass of the electron. This distance is about 2.5 fermis. If the electron were a classical sphere of this size, nearly all its mass would be self-mass, and the annihilation process would proceed by the process you describe, gradual classical dipole moment relaxation, and outward Poynting flow.

But the Compton wavelength of the electron is about a thousand times larger, or 2500 fermis (the ratio of the two is the perturbtion parameter $\alpha\over 2\pi$). The annihilation process in quantum electrodynamics is not causal within the Compton wavelength, the intermediate electron line between the two photons is executing a mostly space-like motion.

So there is a separation of scales--- for scales significantly larger than the Compton wavelength, the classical dipole field picture applies, as you describe, and leads to a shower of coherent soft photons during the annihilation process whose total energy is approximately equal to the fraction of the mass-energy of the electron contained in the field at a distance larger than the Compton wavelength. This is about a tenth of a percent of the mass of the electron.

The rest of the process is the emission of two (or less often three) hard photons described by a QED tree diagram, and this happens when the electron and positron are within the compton wavelength. The hard process is not related to the classical field relaxation, it is a pure quantum effect, and it is the thing calculated in field theory books.

Classical vs. Quantum self energy

The behavior of the self-energy in quantum electrodynamics is completely different from the behavior in classical electrodynamics. The self-energy diagram for the electron is only log divergent, while the classical sphere of radius r has a self-energy which diverges as 1/r. Both calculations are well known, the classical one is just from the field of a spherical source, while the quantum one is from the Feynman diagram where an electron emits and reabsorbs a photon along its worldline, where the integrand is (Wick rotated, with factors absorbed in dk, schematic $\gamma$ structure, and ignoring the external wavefunction parts)

$$ \int {\gamma^\mu (\gamma\cdot k + m ) \gamma_\mu\over (k+p)^2(k^2 + m^2)} dk $$

Which is log divergent because the linear-in-k divergence cancels by symmetry after combining denominators.

This cancellation of the classical linear divergence is nearly self-evident in the modern Feynman formalism, which is why people often neglect to mention that this was originally a non-trivial result, due to Victor Weisskopf. Weisskopf's calculation uses old fasioned perturbation theory and is unreadable today. Old fasioned perturbation theory consists of putting all the intermediate lines on shell and summing over the intermediate momentum, which reproduces the Feynman propagator, but in a way that separates time and space. This separates out positron and electron contributions on intermediate states. The classical linear in k divergence is still present in each part alone, but cancels (up to a relatively negligible log divergence) at large k between the two parts.

There is a simple heuristic to understand the cancellation. The electron is propagating back and forth in time by a particle path integral, and when it is going backwards, it is a positron. The positron segments have positive charge, and the electron segments have negative charge. When you slice this at any time, you always have one more electron line than positron on each slice (because overall the electron is moving from past to future), but if you look close to one time slice, each individual path zig-zags a diverging number of times, making the effective instantaneous distribution a smeared out mess of positive and negative charges, which is approximately as wide as the Compton wavelength (which is the scale at which the electron path stops being strictly future oriented). This makes an effective fractal smeared out charge distribution which adds up to e, and the self-energy of a smeared out smooth charge distribution would be finite, except if irregular, it can still be divergent.

The cancellation of the positron and electron classical divergences would be suspect if you didn't have a covariant regulator, which is why the Feynman style regulator is so important. In the Pauli-Villars picture, you add additional heavy electrons with wrong sign loop contributions to cancel out the high-k divergences in the diagram. In general, this requires two regulator fields. The regulator fields allow the same formal manipulations on their divergent integrals as the physical fields, so that the end result just subtracts out the same diagram at a higher mass. The subtraction process gives a finite integral, which justifies the intermediate manipulations, and shows that, in a covariant regulator, the electron-positron cancellation is really happening.

If you just cut off positron and electron at high k, you need to make sure that the cutoff respects the cancellation. If you include more high-k electrons than positrons, you get the classical linear in the cutoff divergence again.

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+1 : Very interesting "bringing together" of pieces of the picture of the annihilation process. –  twistor59 Dec 29 '11 at 9:11
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