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Quantum electrodynamics (QED) is the quantum field theory believed to describe electromagnetic interaction. It is the simplest example of a quantum gauge theory, where the gauge group is abelian, U(1).

1
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T. Welton gave an estimation in 1948. It was of good order of magnitude but of the opposite sign, apparently, because his estimations were non-relativistic and did not take into account virtual pairs. …
answered Nov 14 '11 by Vladimir Kalitvianski
3
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Because $\left(i\gamma^\mu\frac{\partial}{\partial x^\mu}+m\right)_{\xi\xi'}i\Delta(x-x')$ is a symbolic expression for a given analytical right-hand side. It is written so for convenience (not yet ca …
answered May 2 '11 by Vladimir Kalitvianski
4
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5answers
Let us take the standard QED ($e^-, e^+, \gamma$) as a model of QFT and ask what is its "short-distance" physics? They say the UV infinities appear because we do not know the real physics of short d …
asked Feb 20 '11 by Vladimir Kalitvianski
-1
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There is no strict proof, but a qualitative reasoning. Dysons' argument is based on an invented (mental) situation with imaginary charge values. It is still not clear whether such a pathological situa …
answered Jan 17 '13 by Vladimir Kalitvianski
1
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I would like to complement rather than to answer: one can see the pair creation as a relaxation mechanism. For example, in a usual capacitor the charges are artificially separated and there is a poten …
answered Mar 22 '11 by Vladimir Kalitvianski
1
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Do not worry, it is a constant. There is a sloppiness and misunderstanding in physics when the energy dependence of the cross sections is wrongly attributed to the "fundamental constant" whereas it …
answered Jun 15 '11 by Vladimir Kalitvianski
4
votes
Yes, in QED there are bound states. Nobody can forbid us to take the Coulomb potential into account exactly and the rest - by the perturbation theory. In this way they obtain the Lamb shift, for examp …
answered Jul 20 '11 by Vladimir Kalitvianski
1
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If the photon energy $E$ is sufficient to produce a pair ($E>2mc^2$), then a pair can be produced in a collision. It is due to equivalence of mass and energy. The same energy can be "represented" diff …
answered Mar 11 '13 by Vladimir Kalitvianski
0
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Presence of something else next to an excited atom influences the lifetime of the excited state. Any such presence is described with some additional interaction energy. In case of a cavity QED, you …
answered Apr 16 '11 by Vladimir Kalitvianski
1
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As Marek said, it is as unstable as any other radioactive atom. It is a weak decay of muon or antimuon that destroys the muonium.
answered Jul 21 '11 by Vladimir Kalitvianski
0
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Charge is a property of waves/particles to interact with each other in a certain manner. This understanding is the only one which is correct. Imagining charge existing in an empty space is useless. It …
answered Feb 25 '13 by Vladimir Kalitvianski
1
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As soon as your source is X, there is no wave propagation from Y to Z, whatever wave front looks like (spherical, plane, etc.). About "virtuality": A wave is real if it is of sufficient intensity, i …
answered Apr 5 '11 by Vladimir Kalitvianski
3
votes
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 (An …
answered Dec 29 '11 by Vladimir Kalitvianski
0
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1answer
I have a conceptual difficulty in understanding the electron spin. On the one hand, it is an experimental, observable feature of electrons. The problem is in understanding to what it belongs - to a ba …
asked Apr 29 '11 by Vladimir Kalitvianski
3
votes
An electron is never alone. It is always coupled to the quantum field oscillators. Preparing an atom in an excited state means the entire excitation energy is concentrated in the electron motion. Due …
answered Jun 15 '11 by Vladimir Kalitvianski

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