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New answers tagged quantum-electrodynamics

-1

Considering that an electron is a quantized excitation of the Dirac field, why are there still discussions regarding the "size" of an electron? Isn't the "size" of an electron simply defined as the expanse of the Dirac wave function? You could say that. I quite like saying the electron's field is what it is. The electron isn't some billiard-ball thing ...

2

in what physical pictures is the "size" of the electron, independent of the expanse of the wave function, useful or meaningful? Wave functions describing particles are in the framework of first quantization. This is useful for specific problems and boundary conditions , to get spectra of atoms for example. It is not useful for the study of elementary ...

0

I'll quote the Nobel Lecture of Hans Dehmelt: With the rise of Dirac’s theory of the electron in the late twenties their size shrunk to mathematically zero. Everybody “knew” then that electron and proton were indivisible Dirac point particles with radius R = 0 and gyromagnetic ratio g = 2.00. The first hint of cuttability or at least ...

2

There is no "physical" interpretation of internal gauge symmetries. Gauge symmetries are reflections of an overcounting of degrees of freedom1, or, equivalently, of the presence of constraints. While the gauge principle is a powerful theoretical tool, the symmetry itself is not really "physical". General relativity is a special case - and not a "proper" ...

2

Yes, there is. When you adjust the gauge you change the phase. Thus the difference in phase between two points can basically be anything you want, just adjust the phase and compensate the gauge, or adjust the gauge and compensate the phase.

-6

In quantum (or classical) electrodynamics we are free to make gauge transformations, which change the form of terms in the Feynmann diagrams (or the potentials) without affecting any physical observable. This is sometimes viewed as a flaw in the theory. I've not heard anybody else say that. Can you give a reference? A similar freedom exists in ...

3

Think I got it: One uses the anti-commutation relationship: $$\gamma^\nu \epsilon_\nu \gamma^\mu k'_\mu = 2\epsilon \cdot p - \gamma^\mu k'_\mu \gamma^\nu \epsilon_\nu$$ And then uses the fact that the spinors satisfy the Dirac equation, i.e: $$\bar{u} \gamma^\mu k'_\mu \approx 0$$ So that we are only left with $2\epsilon \cdot k'$ in the numerator. ...

2

First of all, these diagrams are no tadpole diagrams. A tadpole diagram is a diagram with exactly one external leg. Nevertheless, the QED diagram exists, of course. When you calculate it, you need to "connect" the electron propagator $S_F = \frac{i (\gamma \cdot p + m)}{p^2 - m^2}$ corresponding to the loop from both sides with the $\gamma^\mu$ from the ...

1

The problem with all the answers posted so far is that they are inconsistent with the historical narrative. There were three "nails in the coffin" of classical electromagnetism: namely, The black body spectrum The photo-electric effect The Compton effect. All of the critical experiments measuring these phenomena were done at the macroscopic level, with ...

0

I assume that in the quadrupole transition there is a 2 photon interaction, since photons only have spin 1. Hence why it is a less probable transition.

0

It is called Quantum Mechanics and to understand the processes a lot of graduate studies are necessary. A photon is a quantum mechanical entity/particle. The classical magnetic field , electric field and electromagnetic radiation emerge from the underlying quantum mechanical level by the very large number of the elementary underlying processes involved. ...

1

I think Andrey Grozin's http://arxiv.org/pdf/hep-ph/0508242.pdf works quite well enough if you are looking for a general strategy to calculate the anomalous dimension of an operator. You need to somehow define $Z$, i.e. you need to develop a scheme. Now let's say you have defined your scheme or you have simply tried one of the conventional ones. The rest is ...

5

Feynman diagrams are most definitely not a representation of what's going on between the particles. Feynman diagrams are simply a tool to help you remember formulas: if you want to calculate the probability that two electrons will scatter off each other in so-and-so angle, you draw all possible diagrams with two incoming electrons and two outgoing electrons ...

2

The angle of the particle lines is irrelevant and it's just a convention. You could as well draw them as straight vertical lines. It doesn't affect the calculation of its contribution to the probability amplitude of scattering. For the same reason, Feynman diagrams do not intent to show attraction nor repulsion. They are just a bookkeeping graphical tool for ...

1

If your photon has not enough energy to excite the electron then it will just not be absorbed and will pass by, and if you have an electron with an excess energy, it will be absorbed and a photon with the excess energy will be automaticaly emitted and the electron will jump to an excited state. So yeah in your case, you might have a photon with \$0.1 \space ...

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