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As was pointed out in the comments most photons are produced individually from deexcitation of electrons from higher energy orbitals, where they had been pushed by heat ( as with heat filament lamps) to lower ones. Since I am a little new to this topic, a little background of entanglement would be appreciated as I might be wrong in my conceptualization. ...

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Just consider the gauge transformation after Fourier transforming everything. A Fourier transform turns derivatives into momenta, such that we get $$\tilde A_\mu \rightarrow \tilde A_\mu - \frac1e k_\mu \tilde\alpha \;.$$ This mean that only the component parallel to $k_\mu$ (the longitudinal one) will change, while the ...

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An experimentalists answer: If you divide the energy of the electromagnetic wave by hv you will have the number of photons that are building up the electromagnetic wave.

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You only need to rewrite $\mathbf B$ and $\mathbf E$ in terms of field $A_{\mu}$ (here $\hbar = c = 1$), $$\tag 1 \hat{\mathbf B} = [\nabla \times \hat{\mathbf A}], \quad \hat{\mathbf E} = -\frac{\partial \hat{\mathbf A}}{\partial t} - \nabla \hat{A}_{0},$$ which is written as infinite "sum" of photons: $$\tag 2 A_{\mu} = \sum_{\lambda} \int ... 1 In most quantum processes you cannot get just particles (without antiparticles) as products. That would violate some conservation laws (charge conservation mainly). So the quark-gluon plasma was a mixture of quarks and antiquarks. As a consequence, after the QCD cooled, you get both hadrons and antihadrons. These annihilated, but there was certain assymetry ... 0 First clearing up a misconception. I understand that protons and neutrons (which are baryons, which are hadrons) are made out of quarks, and quarks are held together by gluons (at a high level). So those are all the ingredients listed in the title "quark-gluon plasma". Nope. There are no hadrons in the QGP phase (regarded as a pure phase, I'm ... 5 Gluons and quark-antiquark pairs, like photons and lepton-antilepton pairs, are excitations of the vacuum. Any volume of space with enough energy density to contain a quark-gluon plasma also has, by definition, enough energy density to contain a gas of photons and electron-positron pairs. The difference is that the quark-gluon plasma is governed by the ... 2 The chronology in Wikipedia is emphasizing what is changing in each epoch. Particles that are lighter than the ambient temperature are presumed to be in thermal equilibrium, but not doing much of interest at that time. In particular, the light leptons and photons are created and annihilated all the time, so there is a sea of electrons, positrons, and ... 1 You can think of light as the carrier of the electromagnetic interaction. The particles interact with light, not directly with each other. It is an experimental fact that light does not interact with itself. Note that this is not the case with quantum chromodynamics (the theory of nuclear matter). This theory is built along the same lines as quantum ... 1 Light waves cannot be longitudnal in free space. Transverse EM waves do exert pressure, but it's too small to be perceived by humans. 0 Martin wrote: ", hence they can only interact via the weak force, which is, as the name says, weak." Makes me wonder, is "weak" a good name? Or is this interaction better described as short, and still full of surprises? I'm thinking of {Dirac, Weyl, Majorana} spinors, rates of chiral oscillation, inertial-mass... ... which makes me wonder about the ... 1 Unfortunately, in quantum mechanics "ordinary" reasoning does not get you anywhere. The photon, like any other particle, is neither a particle nor a wave; it is an entity that we can only describe mathematically. It's only when we observe it that it shows up as either particle or wave. Or senses, and hence our logic, evolved to make sense of the real ... 2 What constituent of internal energy does an electron excitation represent? You can think of electrons as just like planets orbiting the sun and get the correct answer to this question. An electron in a higher energy level has less kinetic energy, but more potential energy as it is (generally) farther from the nucleus. The net result is more energy. ... 2 The change in electronic excitation represents both a potential and a kinetic energy term in classical physics, but there is no simple correspondence to classical physics terms, when you are looking at quantum systems. All we really care about is the total energy difference between electronic states. Those energy differences correspond to the energies at ... 2 Not merely can it transfer its momentum as well as its energy when it interacts, but it must. If the target atom is in a fluid context (liquid, gas, plasma), then that energy and momentum must end up in the target or some other reaction products(s). In a solid context the Mössbauer effect can be an issue, allowing the transfer of that momentum to a much ... 3 It is not very clear to me if you are asking about energy or momentum. You should also ask about a specific interaction process as there are many, this is required especially to answer your last, quantitative, question. However, generally speaking, a \gamma photon cannot give some of its energy to anything else: it is all or nothing. Even in the Compton ... 2 This is an instructive video on the double slit experiment with photons which is experimenting with single photons. At about 2.5 minutes it explains how the experiment is done with single photons. In a nutshell, by lowering the intensity of a light source to the point of zero emitted by the source and then slowly increasing it. The detection hinges on the ... 3 Generally speaking, the emitted frequency is not necessarily the same as that absorbed but let's suppose that it is. As the white light (composed of all the different frequencies) encounters the material (the red box, below), certain frequencies are absorbed and then re-emitted but in random directions. So they're scattered. Therefore, far fewer photons of ... 1 The "slowing" of light in a medium can be entirely explained using a classical wave-based approach. An incoming EM wave wiggles the electron clouds around the atoms in the material. These electrons clouds re-emit a much weaker EM wave having a very small amplitude. This re-emitted wave is 90-degree phase shifted from the original wave but superposes with ... 0 To make one thing perfectly clear: photons do interact with each other. That interaction is simply very weak for low energy photons, but non-trivial at very high energies. Photon-photon interaction does, of course, not play any measurable role in optical experiments with visible light. As for the Hanbury and Twiss effect... IMHO it's trivial and classic. ... 1 There's no contradiction. Without considering high-energy physics, photons don't interact in free space in the sense that they pass through each other intact. Here's a very recent review article: http://www.nature.com/nphoton/journal/v8/n9/full/nphoton.2014.192.html Free space photons can of course lead to interference patterns and the HBT effect is still ... 0 We know four fundamental forces, three of them being included in the Standard Model. At low energies (compared to the mass of W^\pm) the forces have strengths as follows: strong force > EM force > weak force > gravity Also, photon and neutrino are really quite different. Photon is a force carrier, while neutrino is a matter particle. Now, matter ... 1 The problem with the picture (and probably with your understanding of the physical process) is that it assumes photons as classical particles on well-defined trajectories. If this were a true picture of reality, your objection would be justified. This, however, is not so. In order to describe the process properly, one has to acknowledge the quantum nature ... 1 The photons do not have a well defined trajectory. The diagram shows them as if they were little balls travelling along a well defined path, however the photons are delocalised and don't have a specific position or direction of motion. The photon is basically a fuzzy sphere expanding away from the source and overlapping both slits. That's why it goes through ... 10 The photon does couple directly to charged stuff, e.g. via Compton scattering. This is indirectly related to the spin, as direct interactions between fermions are hard to construct. The neutrino on the other hand does not couple immediately to any other matter particle. It requires a force-carrier. Now as it turns out the only force carriers that care about ... 7 Neutrinos having no charge means they don't participate in electromagnetic interactions, which are the strongest (at least long range). Them being leptons means they don't interact with the strong force (which is, as the name says, strong), hence they can only interact via the weak force, which is, as the name says, weak. Photons on the other hand do not ... 0 It's a matter of definition. For example, by photon, do you mean first or second quantization (the latter being the canonical treatment)? Although there's no ambiguity in theoretical physics, the nature of photon has generated much debate in the community of applied physics. See, for example, these articles: http://arxiv.org/pdf/quant-ph/0605102.pdf ... 0 You have really figured out the answer yourself. On one hand, you have a natural phenomenon (light) and on the other you have our models (wave description, photon description). When speaking of the reflection of rarefaction of photons, the author implicitly assumes the reader to know about the wave-particle duality (see links in the comments). I think you ... 2 Vacuum magnetic birefringence basically involves the same loop diagram as light-light elastic scattering except that two of the four photons come from a magnet. Detecting this effect is the aim of the PVLAS experiment in Ferrara, Italy. See arXiv:1406.6518 and references within. The experiment is running at the moment but the sensitivity is not good enough ... 1 What you read is correct. I am not sure if those were the exact words of your teacher but according to the general theory of relativity, sun doesn't "attract" the photon (or any other body for that matter). In fact gravity is not even a real force. Let me briefly state what the theory of relativity has to say about gravity without going into the complicated ... 6 To properly understand what is going on you need to understand general relativity. Massless particles, like photons, travel on null geodesics and mass bends spacetime so the null geodesics are not straight lines. The problem is that neither you nor your teacher understand general relativity so this isn't a very convincing argument. But here is an argument to ... 2 Briefly, the formula E=mc^2 applies only particles at rest in an inertial frame of reference. Since there is no rest frame for a photon, no inertial reference frame in which a photon is at rest, one cannot apply the formula E=mc^2 to a photon. In more detail, the four-momentum of a particle has components (\frac{E}{c}, \vec p) The four-momentum for ... 1 The correct general equation is E^2=m^2c^4+p^2c^2, since m=0, E=pc. A photon's momentum is p=\hbar k, where k is the wavenumber (k=\frac{2\pi}{\lambda}). This would be consistent with E=h\nu where \nu is the frequency. 0 Yes, all radio waves have inthe nearfield the same sequence of the E and the B field. It's the right hand grip rule (conventional direction of current) because there is no principal difference to a straight wire. 0 When an electron gains enough energy from a photon of light, it can leave the surface of the metal - leaving the metal with a positive charge. But this positively charged piece of metal will attract an electron to become neutral again. Often, if 'left to itself' the photo electron will just fall back to the metal surface it came from. If there is a ... 2 You are asking about the number of photons fired from a device such as a laser at a unit time. We cannot say what the precise number of photons would be at any interval due to a version of the Heisenberg uncertainty principle:$$\Delta E \Delta t \geq \frac{\hbar}{2} It basically states we cannot suppress our uncertainty about energy fluctuations in time, ...

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'The mirror is given a momentum twice that of the incoming photon. As a mirror is typically quit heavy, lets say one gram. Its kinetic energy due to momentum it received will be extremely small. However, the photon will actually change its energy by the same amount, thus its wavelength changes, but not much.

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Since the photon reflects, its momentum changes: $p_{ph}'=-p_{ph}$. But total momentum of the system is conserved: $p_m+p_{ph}=p_m'+p_{ph}'$. Thus, the mirror will change its momentum. But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass $m_2$, the energy of ...

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Tadeas Bilkas answer let me think about the sence of all and all time citing the quantum mechanics. I write his answer in terms of common mechanics and get the same result: You have an emitter of balls which radiates just one single ball but in a spherical area. You place a lot of baskets some meters apart (with same distance) from the emitter. Mathematics ...

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Absorption isn't an instant event. At the level of simple quantum mechanics, this system can be described as follows. Evolution of electron in crystal is governed by Schrödinger's equation. External electromagnetic field, namely the light which we shine on the crystal, is a periodic addition to the Hamiltonian. When you start shining light at the crystal, ...

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I would say in the same way a thick beam in a structure "knows" that you are trying to tear it off: it is temporarily displaced somewhat due to your effort, but then returns to its stable state. So the electrons are not totally immune to the photon (for example, the crystal acquires some minuscule momentum due to light pressure, if the photon is reflected ...

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According to the QM, a photon leaves an atom and reaches after some time another atom. The QM makes no statement how the photon bridges the gap and therefore QM can not explain Doppler effect and interference. A variant of QM, the "De Broglie-Bohm theory", tried to describe the path of a photon, but is considered to be refuted.

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Thats nonsense. In Spontaneous parametric down-conversion, one photon may produce two photons with half the energy. No one refers to them as components. The term "magnitude" is unusual in the context of photons. Photons have energy and spin.

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In case you "run out of photons", you must switch to probabilistic description of quantum mechanics. Let's consider an extreme case: You have an emitter of spherical waves which radiates just one single photon. You place a lot of detectors some meters apart (with same distance) from the emitter. QM says that the photon propagates as a probabilistic wave to ...

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If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called "photon" or "wave packet". Linking the appropriate formulas from QM and E&M waves, you get the diameter of the wave packet (about λ/2) but not the length. The radius and the direction of propagation do not change as long as the wave packet is not disturbed. ...

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(1) If an atom emits energy hf, it emits also an angular momentum (spin). That combination is called photon or wave packet. If you link the appropriate formulas from QM and E&M waves, the result is the diameter of the wave packet (about lambda/2). The radius and the direction of propagation do not change. If this wave packet hits your detector, you can ...

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The classical electromagnetic field given mathematically by Maxwell's equations can be proven to emerge from a confluence of individual photons, which photons are described by the Quantum Mechanical form of Maxwell's equations. Thus the classical wave is made up by zillions of photons with energy $h\nu$, where $\nu$ is the frequency of the classical wave. ...

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At very low energy, the difference between photons and E&M waves becomes clearer: After every spin-flip in a hydrogen atom, "something" with E=6 µeV and spin=h/(2*pi) is released. Is that "something" a photon? Nobody can detect a photon with such a low energy. But if you have a radio receiver at 1420 MHz, you get a clear signal. The electric component of ...

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"Running out" of photons simply means that your wavefront is absorbed or scattered in a different direction or something like that. Either way, the original wave is "consumed", so you loose intensity or photons, depending on which picture you like better. For the case of a single photon source: One photon can only interact with one electron. However, there ...

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I think you have invented the zone plate, a kind of specialized flat circular diffraction grating that acts as a lens. It consists of a set of concentric transparent zones of decreasing widths. The width you have derived is the diameter of the first zone. Subsequent zones interfere constructively by allowing paths differing by integer multiples of the ...

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