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In the general case, "no" as there is a angular momentum transfer involved (meaning there are preferred directions relative the original and/or final angular momentum of the atom). That said, for most matter at room temperature the atoms have random orientation so you can treat the answer as "yes" for experimental purposes. Now, I see that you are ...

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It is not a good idea to see a Feynman diagram as some sort of collision process really happening. The diagram is just a term in the perturbative expansion of a quantum mechanical transition amplitude (in other words, a nice "graphical" way to represent a bunch of integrals). The only actual observed objects are two incoming photons with a certain energy, ...

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No - assuming they don't hit anything they don't decay. The distance dependant "decay" is the drop in the number of photons per volume as the volume gets bigger - it's not a decay of individual photons. It's the same as a crowd dispersing as it leaves a subway exit - nobody is disappearing. Photons can lose energy as they collide with gas or dust in space ...

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It depends on the mechanism by which the photon is emitted. Stimulated Emission: Yes, the emitted photon will inherit the characteristics of the photon that stimulated it, including its propagation direction. That's how lasers get coherent light. Spontaneous Emission: No, this should be random orientation. Another way to think about spontaneous emission is ...

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A photon is a unit ("quantum") of excitation of the quantum electromagnetic field. Thinking roughly of the quantum field as a vast collection of quantum harmonic oscillators, each oscillator corresponding to a mode of vibration of the field, we specify the quantum field's state by stating how many quantums above the QHO ground state each mode oscillator is ...

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So how is it possible to have a quantized outcome from a symmetric continuous event? Easily. So easily that I'll describe the easiest example to me. Which is to describe what happens when a Stern-Gerlach device interacts with a spin 1/2 particle. You could have a particle with any spin whatsoever, but no matter what single particle state you pick it ...

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I should like to capture the excellent comment by user CuriousOne (emphasis mine): There is no such size. A smaller aperture will merely reduce the transmission probability, but there is no known cutoff. Indeed, there is currently a lot of interest in deep sub-wavelength imaging techniques. A photon, by the way, is not an object. It's a quantum, i.e. the ...

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The number of photons may indeed be finite because the energy of the photon in ${\rm J}$ is $$E = hf$$ where $h=6.626\times 10^{-34}{\rm J}\cdot {\rm s}$ is Planck's constant and $f$ is the frequency in ${\rm Hz}$. For monochromatic light, the number of photons may be determined from the energy in this simple way because $f$ is a fixed constant: $$N_{\rm ... 3 I know that photons are quantized, they are not continuous. Photons are not quantised, nor are they continuous. They are the charge carriers of the electromagnetic field as arising in quantum field theory. An accelerated charge generates an electromagnetic field whose carriers are, in turn, the photons whose energy might be quantised. So how is it ... 3 Given you've read only QED, this is a highly astute question. Conservation laws in the quantum world work a little differently from classical conservation grounded on Noether's theorem (there is a kind of quantum analogue in the Ward-Takahashi identity). If a quantum entity has state |\psi\rangle, then conserved quantities are measurement means defined ... 3 E = m \times \text{acceleration} \times \text{displacement} is not the whole story even for normal matter once you get into special relativity. The full formula is$$E^2 = p^2 c^2 + m^2 c^4 \tag{1}$$where m is the rest mass, p is the momentum, and c is the speed of light. Photons have m=0 but p \neq 0. I'm not sure why this matters. In general ... 2 We live in the age of measurements and observations and specific mathematical theories that fit measurements and observations beyond any doubt. Photons are elementary particles. . They have zero mass, and other characteristics which separate them from other elementary particles . Can photons lose a small amount of energy over time when traveling large ... 2 Yes, photons do lose energy because of the expansion of space. Their wavelengths are increased by a factor of (1+z) between when they were emitted at redshift z and when we detect them now. Their energies are therefore decreased by a similar amount. The "tired light" interpretation of this effect has been discussed, debated and disproved. In my opinion, ... 2 Kallen-Lhemann representation is just a way to expand in the momentum basis the two point correlation function of a local operator \hat{O}(x), it holds true for massive and massless theories alike except for non abelian gauge theory in which the situation is a bit more complicated. Let's demonstrate the K-L formula with a more general proof: let's start ... 2 Quantum spin of any particle is an intrinsic property of that particle. It has nothing to do with the rotation in spatial coordinates. It is called "spin" only because it follows the same angular momentum algebra followed by the orbital angular momentum. You don't measure the spin quantum number directly but the interaction of the spin with something. This ... 2 Why does the wavelength determine a photon's energy? In the 19th century, it was thought that the energy of light was determined only by its intensity. Then, experiments, particularly the photoelectric effect, showed that this was not so: a low-intensity short-wavelength light can cause reactions that intense light of a longer wavelength cannot. Thus, ... 2 Well, actually it doesn't. Knowing the wavelength allows you to calculate the energy, but it does not "determine" it in a causal way. Energy (E), wavelength (\lambda) and frequency (\nu) are related by$$E = h\nu =\frac{hc}{\lambda}$$so if you know the wavelength or the frequency you can determine the energy. I think his use of "determine" confused ... 2 The photon is an elementary point "particle". particle between quotation marks because it is not a classical point particle , it is a quantum mechanical entity. Quantum mechanical entities depending on the boundary conditions display sometime classical point-like elementary particle behavior and sometimes have a probability density for their location ... 1 How can photons/particles/objects/things be massless? Photons aren't massless the way people think. A photon has a non-zero "inertial mass" and a non-zero "active gravitational mass". But it doesn't have a "rest mass" because it's never at rest. You can't slow down a photon like you can slow down an electron. Or speed it up by pushing it. Rest mass does ... 1 The edge provides a boundary condition that the EM field must satisfy. The total EM field is "aware" of the boundary. "Photons", being quantized excitations of the EM field, are created (emitted) and destroyed (hit a screen) only where the EM field exists. If you are trying to think of photons as particles, forget it. You'll end up with all sorts of ... 1 One thing that you can be sure of is, for a large enough LED, you will get poisson statistics to a very good approximation. Neither bunching nor anti-bunching. The reason is simple: One photon comes from a certain part of the LED, the next photon is likely to come from a totally different part of the LED and head in a totally different direction. There's no ... 1 You've mixed up which time dilates for which observer, and written yourself into a paradoxical corner: if going closer to the speed of light slows time for the object going that speed, and if time slowing down means going slower, then the conclusion is that "the faster you go, the slower you go." Which obviously doesn't make sense as you've pointed out. ... 1 My question is whether individual photons also carry orbital angular momentum? Yes. See https://en.m.wikipedia.org/wiki/Orbital_angular_momentum_of_light If yes, what are the values of orbital angular momentum in one-particle states? To quote the wikipedia page In particular, in a quantum theory, individual photons may have the following ... 1 Yes, single photons can have orbital angular momentum. However, unlike spin, they are not required to have any. Just like in the classical case, the orbital momentum of single photons is determined by the shape of their EM mode- roughly speaking, the wavefront must have a helical aspect to it. In particular, this means that the eigenmodes of light in a 3D ... 1 The connection of frequency to energy comes when one considers the covariant formulation of the electromagnetic wave propagation. In Panofski and Philips "classical electricity and magnetism" second edition chapter 21. This quote in particular. This associates a zero mass particle with a fourvector, i.e. energy and momentum . Text goes on to explain ... 1 No, because we can think of the classical wave as being made up of a large number of photons. If we have a low-frequency wave with the same energy as a high-frequency wave, it simply means that there are a larger number of low-energy photons. 1 You are looking at two different things assuming they should be the same. In classical electrodynamics the Poynting vector is defined as \mathbf{S} = \mathbf{E}\times \mathbf{H} and enters the variation of energy density as$$ \frac{\partial u}{\partial t} = -\textrm{div}\,\mathbf{S} - \mathbf{j}\cdot\mathbf{E};  according to the functional form of ...

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A photon is a quantum mechanical elementary particle and follows quantum mechanical formulae, not classical ones. In quantum mechanics the only way an elementary particle can change direction is through an interaction with another elementary particle or field. The interaction is shown with feynman diagrams which give the integrals that have to be calculated ...

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The CMB radiation is the "temperature of the universe" Approximately 379 000 years after the Big Bang the ionized hydrogen (free protons and electrons) had cooled down to about 3000 K, and thus became "transparent" hydrogen due to ionization ceased. For each time point later the hydrogen cooled down "exactly" as much as the universe expanded. In other ...

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[Since none of the experts have so far posted an answer from a pedagogical standpoint, I'm going to have a try. I'll make it a community wiki, so please consider improving it if you can. Even better, post your own answer!] As Yuggib correctly points out, you shouldn't attempt to view a Feynman diagram as a description of a collision process. (You can do ...

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