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0

I assume that you mean you want the general idea to be presented "Oppan Feynman-style" rather than for me to start talking about a QED Lagrangian that describes the sort of prism which would generate refraction. Photons which go into the prism will in general have the same frequency $f$, because wave-fronts and troughs cannot enter the material any faster ...


-1

As you know white light is made of many different wavelengths. Each wavelength of light takes a slightly different path through the prism. In QED we have a spinning clock which turns at different rates for different wavelengths of light. We calculate the probability of a certain path the light can take using these clocks which have associated probability ...


2

Quantum electrodynamics is part of quantum mechanics and is the mathematical method used to calculate quantities rather than hand wave explanations. What you call "quantum mechanics " is a hand waved explanation, not wrong, but no numbers can come out of it because light is composed of zillions of photons impinging on other zillions of electrons making up ...


2

You've confused two things. The business with adding vectors whose direction is determined by the stopwatch tells you what kind of interference pattern you'd see on a screen after the light diffracts through an aperture. In this case you're assuming you have an opaque screen with which to view the diffraction pattern. This is entirely independent of the ...


9

Virtual particles are not real. They come, as I've said in many answers on this site, from a naive interpretation of Feynman diagrams which should not be taken as an actual, exact description of how the physics works. The actual description of an interaction in the quantum field theory is more complicated than "photons are exchanged". In particular, ...


0

unrenormalized vertex diagram and unrenormalized self-energy diagram are as below: Now, let me summarize what Peskin and Schroeder try to express. Feynman propagators for electron $$ S_F=\frac{i}{\require{cancel}\cancel p-m} $$ end photon $$ D_F=\frac{-ig_{\mu\nu}}{q^2} $$ and propagator must hold $$ S_F(p)S_F^{-1}=\mathbb{1} $$ simply differentiate ...


0

As pointed in the comment by @CuriousOne, there is indeed a paper providing such explanation ("The photoelectric effect without photons" by Lamb and Scully). It was however later found to be faulty. A very good review and references on this can be found in these answers to a different question posted on this website, but apparently there is no consensus on ...


1

Your statement that the integral "is actually" $\propto g^{\mu\nu}\Pi$ is incorrect because you can clearly see the $q^\mu q^\nu$ in the numerator of the integrand. The correct expression that you have in your final equation is not a choice, it is the result of calculating the loop using dimensional regularization. Finally, indeed $q_\mu \Pi_2^{\mu\nu}$ is ...


0

I was just looking up your level of education in order to gauge my further replies. I bumped into this, and whilst it's an old question, but I felt moved to answer it. My understanding is that a photon propagating in vacuum has a small probability to spontaneously create a particle/antiparticle pair that will then quickly recombine to emit a photon ...


0

QED is rarely concerned with precise forces or fields, although you may calculate e.g. the Coulomb force in a non-relativistic classical limit, see my answer here. QED is a quantum field theory. As such, the electromagnetic field does not possess a definite "value" at any point, it is to be thought of as an operator-valued distribution and what you may want ...


9

Would you dig a ditch with a surgeons scalpel? Yes, quantum mechanics ultimately underlies all physical observations but the mathematical expressions for large dimensions with respect to $\hbar$ become cumbersome and are replaced by the simplest ones for the appropriate study. Thermodynamics, for the study of bulk matter, blends smoothly with quantum ...


2

As to your question, yes, the QED Lagrangian is indeed invariant under charge conjugation. You may have found differently because your transformations under charge conjugation are faulty. The prefactors are correct, however, under charge conjugation $\psi$ goes to $\bar{\psi}$ and vice versa, i.e. $$ \hat{C} \, \psi \, \hat{C} = -i(\bar\psi \gamma^0 ...



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