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5

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed ...


9

A photon is an elementary particle. As much elementary and as much particle as the electron . A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a ...


0

In the semi-classical approach you treat the electromagnetic field of the incoming light as a (sinusoidal, with angular frequency $\omega$) perturbation $H^{\prime}$ in the Hamiltonian of the atom. The wavefunction of the atom can be expressed in terms of a linear combination of its eigenstates, each with a multiplying coefficient. If we start on the ground ...


3

There are two factors at play here. The Lorentz force which causes the paths to bend with a radius proportional to the particles velocity and with a sense that dependent on both the particles charge and the direction of the particles velocity. In high energy (compared to $m_e$ events) such as the one pictured, the particles are nearly co-linear at the ...


-4

The electron has three well known properties, its electric charge, its magnetic dipole moment and its intrinsic spin. All three are constant quantities. And to prevent contradition about the reality of this intrinsic spin, it was shown in the Einstein-de-Haas experiment, that this spin really has to do with a rotation of the electron. It has to be stated, ...


9

But what we never seem to see is why the electron and positron move the way that they do. Saying "they move like they do because of the force on them" doesn't explain anything at all. It's a non-answer. The equation of motion for charge particle (electron,positron) in magnetic field is $$ m\frac{d}{dt}\left(\frac{\mathbf ...


5

Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated. This is ...


2

The violation of gauge invariance by this term is the "only" reason why it's never written down – as long as we define the word "only" to include all other reasons that may be shown to be "physically equivalent" to gauge symmetry. Gauge symmetry is extremely important and its violation would make a similar theory inconsistent, especially at the quantum ...


0

As with most of the questions in this forum, the answer is right in the same section of the book where you read about the problem. Why don't you go back and find it? First you talk about a photon hitting an electron, and then you suddenly change to a photon hitting a photon, a completely different problem (photons don't interact with other photons, only ...


1

Photons and electrons are elementary particles and their behavior is predicted by quantum mechanical equations. Quantum mechanics predicts probabilities for a reaction to happen. Photon photon interactions have very very small probabilities of happening. Thus photons pass through each other for all measurable purposes at low energies ( light and below ...


1

If you had a laser you wouldn't see it unless it was aimed at your eye (ouch). Or if there is dust or such around for it to scatter off of. And scattering is the key. If you want to see something then it either has to get to your eye or it needs to deflect something towards your eye. If you have a beam of electrons you could try to get something to ...


1

If you want to think of a free on-shell particle as a mode that propagates like a soliton far from interactions then you can think of it as a thing that interacts with the vacuum so as to make more of itself. But really you should stick to asking questions the formalism is designed to answer. The vacuum is not a thing like a house that provides space for ...


-1

The oscillating electromagnetic wave, why is the wave oscillating between electric and magnetic waves? I'll focus my answer on electromagnetic radiation. Light - a small part of electromagnetic spectrum - could be produced by accelerating electrons. The easiest way is to take a thin current carrying wire. The electrons inside the wire will bounce on ...


0

Quantum mechanics is the reality of things,not classical mechanics(including EM waves). Having this in mind, the photon is not a different thing from the EM wave. They are the same entities. The craziness of quantum mechanics(well one of them) is the wave-particle duality. It's not that the photon travels with the electromagnetic wave, it is the EM wave(as ...


1

But I want to know, if an incoming photon is unable to excite an electron, then why not all the photons pass through glass? i.e. photons should not reflect off glass, all the photons should pass through glass. While I don't know the exact details of reflection off glass (it is related to solid state physics from what I recall, with plasmon and such ...


2

In classical electrodynamics, the process of how much light refracts, passing through the glass, and how much light reflects, is determined by the Huygens-Fresnel principle. This principle, named after Christiaan Huygens and Augustin-Jean Fresnel, is a method of analyzing the wave propagation patterns of light, especially in diffraction and refraction. It ...


0

Instead of thinking about it deterministically think about it terms of the probability that a photon will be reflected. These probabilities can be calculated using Quantum Electrodynamics. Check out this lecture series by Richard Feynman for a detailed explanation. https://www.youtube.com/watch?v=eLQ2atfqk2c


1

However, I'm having trouble understanding this formulation when applying it to a free field. A free field is just a cavity with infinitely large boundaries, right? That would seem to imply that for each frequency, there is only one harmonic oscillator. A free field is a field that obeys homogeneous Maxwell equations (with charge and current density ...


3

The propagator as defined by the time-ordered 2-point VEV ⟨0|T[ψ^(x)ψ^(x+y)]|0⟩ is clearly not invariant under the local gauge transformation $$A_{\mu}(x)\to A_{\mu}(x)+\partial_\mu\Lambda(x)\ \ \ ; \ \ \ \psi(x) \to \exp(ie\Lambda(x)\psi(x)$$ since the VEV acquires an extra phase factor $\exp(ie\Lambda(y))$ under the transformation. Schwinger, among many ...


1

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 ...


0

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 ...



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