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65

In short, the answer is: because gluons behave in a way that makes them useless for this purpose. To understand why, let's back up a little and look at how photons are useful, and then see how gluons behave differently. We (animals pretty broadly) evolved to see photons because they allow us to move around in and respond to our environment more efficiently....


18

The paths of the Feynman path integral are not actually taken. The phrase "takes every possible path" is a mangled statement of the mathematical instruction to take the integral of $\exp(-\mathrm{i}S)$ over all possible paths for the action $S$ to get the probability amplitude of something happening. It is a fact of quantum mechanics that this integral ...


17

Very reasonable question. I will try to answer it in an intuitive way. If you have a scattering medium, photons are reflected in random directions; but when you have a refractive medium, something else happens. The photon is not absorbed and re-emitted: instead, the photon interacts with the electrons in the medium, and since these electrons are somewhat ...


16

Even though there is a single photon in a volume of your choice the light is still a wave. An experiment was performed which proved this. In this experiment a Michelson interferometer was set up and the incident light is so weak that only one photon was in the whole setup at a time. A photographic plate was used to detect the interference pattern. Now just ...


12

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 v}{\sqrt{1-\frac{v^2}{c^2}}}\right)...


11

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 ...


11

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, ...


10

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 ...


10

The fact that the theory is not gauge invariant implies that all degrees of freedom of $A_\mu$ must have physical meaning: This is not the theory of photons where only transverse degrees of freedom make sense. This way you must tackle some non-trivial issue like the negative norm associated with temporal modes. This could be avoided by adding a mass to $A_\...


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, "...


6

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 (...


6

First of all, note that different authors disagree what should be the Coulomb potential $V$ in $d$ spatial$^1$ dimensions. Here we will assume that it satisfies Gauss's law, i.e. $$\tag{1} V(r)~\propto~\left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \\ \ln(r)&\text{for}& d~=~2. \end{array}\right.$$ Here we will only discuss the ...


6

Huygens's Wave Theory is what you call a first order scalar diffraction theory of light. So what does it describe and what does it fail to describe? First order means that electromagnetic effects like induced currents in surfaces etc. are ignored. These can be described by solving Maxwell's equations for the same system instead of working with the wave ...


6

Let me explain @ACuriousMind 's answer with some verbiage. The short, regrettably oracular, answer is that the Fabri-Picasso theorem does not hold in a finite superconductor, since translational invariance fails at its boundaries. Really, I do appreciate this is aggressively obscure: will strive to explain. First of all, if you have a chunk of warm ...


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 ...


5

I believe it would be incorrect to term this as "photon-splitting". What is really happening is that a photon is being annihilated and subsequently two new photons are created by means of a non-linear optical process. Such a process that is routinely harnessed in quantum optics laboratories is spontaneous parametric downconversion (SPDC). The reason why ...


5

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)


5

A comment about coherence in general I find the definition of coherence as some sort of "unrelated phase" problematic for a couple of reasons: This formulation somewhat implies that coherence is discrete, i.e. there is incoherent and coherent. Of course that is not true, you can have a continuum partially coherent states. But what quantity are you going ...


5

Correction, a single photon does not have a circular polarization. It has spin +1 or -1 to the direction of its motion. Qualitatively Left and right handed circular polarization, and their associate angular momenta. The way the classical wave emerges from the quantum mechanical level of photons is given in this blog entry, and it needs quantum ...


5

I would like to add a few things to ACuriousMind's answer. What Greene most certainly intends to say is every path(even faster than light ones i.e. those which are not time-like everywhere) contributes to the propagation amplitude. In fact,since every path in space-time contributes with equal weight, there are also paths which go "back in time" and "come ...


4

.Your title asks about the electric field. The content is about the electromagnetic waves, two different entities. Electromagnetic waves emerge from an innumerable number of single photons. As one cannot have water waves with just a few molecules but need of the order of $10^{23}$ (avogadros number) one cannot measure electromagnetic waves if the photons ...


4

It fails when the photon number is small. Since the electromagnetic field can never be zero because of the third law of thermodynamics this automatically couples the temperature, the effective volume and the photon number to each other. As a result it is experimentally impossible to do experiments with single photons at low frequencies because we can't ...


4

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

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 will ...


4

Consider for the shake of simplicity a free neutral scalar field $\phi$. Passing to the second quantization picture, it is a operator valued distribution $$C_0^\infty(M;\mathbb R) \ni f \mapsto \phi(f)$$ where $M$ is Minkowski spacetime and $\phi(f)$ is a densely defined symmetric operator on the Hilbert space $$F_+(\cal H) = \mathbb C \oplus \cal H \oplus(\...


4

In the quantum mechanical description of any physical system, including a quantum field or a collection of interacting quantum fields, there is always one state vector – one collection of numbers (probability amplitudes) that generalizes what is referred to as the "wave function" in quantum mechanics of particles. In quantum field theory, a better name is a ...


4

See, the thing is that spin is actually a vector — it has also a direction. When considering such vector in quantum mechanics, 2 observables describe it completely: its norm ($S$) and projection on one of the axis (usually, $S_z$). For a spin-$\frac12$ particle the norm is $S=\frac12$ and $S_z = \pm \frac12$. Then, for a system of 3 spin-$\frac12$ particles,...


4

Yes it is correct. The derivation in P&S is straightforward but I will expand on it a bit. The key observation is that \begin{equation} \int\frac{d^4k}{(2\pi)^4}e^{-ik\cdot(y-z)}\frac{i\gamma^{\mu}k_{\mu}}{k^2+i\epsilon} =-\gamma^{\mu}\partial_{\mu}\int\frac{d^4k}{(2\pi)^4}\frac{1}{k^2+i\epsilon}e^{-ik\cdot(y-z)}, \end{equation} where the integral on the ...


4

Method One: \begin{eqnarray*} & & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}+i\epsilon}e^{-ik\cdot x}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\cdot\mathbf{x}}\int dk_{0}e^{-ik_{0}x_{0}}\frac{1}{[k_{0}+(|\mathbf{k}|-i\epsilon)][k_{0}-(|\mathbf{k}|-i\epsilon)]}\\ & = & \frac{i}{(2\pi)^{4}}\int d^{3}ke^{i\mathbf{k}\...


4

Field interaction does not "give rise to particles". In fact, field interaction makes it particularly difficult to speak of particles. To understand how particle states and fields are interrelated, we must employ quantum field theory. This answer of mine roughly sketches how a particle state is defined in QFT - and the fact is that such particle states are ...



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