# Tag Info

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The word "stimulated" means that the emission of the photon is "encouraged" by the existence of photons in the same state as the state where the new photon may be added. The "same state" is one that has the same frequency, the same polarization, and the same direction of motion. Such a one-photon state may be described by the wave vector and the polarization ...

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Let's recall the easy way to make a diffraction pattern from a double slit. You just shine a laser on it. Why does this create a diffraction pattern? Well we know that the intensity of the light is determined by the electric field, so to figure out why the intensity of the light is the way it is, we just need to figure out why the electric field is the way ...

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Is the coherence length consistently defined as the length at which these two waves achieve a phase difference of 1 radian? No Coherence length is the maximal difference in way traveled by the to rays without losing the phase relation which allows interference. This lenghts can be some µmeters (eg for white light from a glowing body) or several ...

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There is a phenomenon called decoherence in quantum mechanics which is largely responsible for this. Basically (the following is a simplification), all the strange behavior that occurs in QM tends to happen when the wavefunctions of different particles are in phase. Decoherence occurs when the phases are randomized, so there's no special correlation between ...

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The coherence length is just the coherence time multiplied by the propagation speed. To understand the coherence time, say you have a wave described, in complex notation, by $$E(t) = A(t) e^{i \omega t}$$ where $A(t)$ is a slowly varying complex amplitude. You make this wave interfere with a delayed version of itself and collect the intensity  |E(t) ...

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This is just the comments formalised into a partial answer. First, it's important to realise that entanglement is a type of quantum correlation between two distinct systems. So, it's not useful to consider a single two-level system, and there is no such thing as entanglement between states. As Lubos Motl points out, you need to consider a system that has a ...

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You are correct that every photon will interfere with itself for sure. But for the whole interference pattern to be observed, you need a large number of photons independently interfering with themselves. And, these large number of photons should be identical in every respect so that they can be represented by the same single photon wavefunction. This makes ...

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Huygen's Principle. If the slit is thin enough and you are only going to use light from one plane, then the new wavefront is dominated by contributions from only one part of the original wave and all parts of it (the wavefront) share the same phase.

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Coherent (or pure) state Consider 2 basic states $\lvert0\rangle$ and $\lvert1\rangle$. (If you never heard about states, treat them as ordinary complex vectors.) Here we suppose that $\lvert0\rangle$ and $\lvert1\rangle$ are orthogonal ($\langle 0\lvert1\rangle =0$). Now, consider $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ... 3 What is coherence and quantum entanglement? Does it mean that two particles are the same? No, coherence means a mathematical relationship that remains invariant . Entanglement is related to coherence, as it is a description of particles that have an invariant relationship in time and space. An every day description of entanglement would be the ... 3 I think it's deeply related to the fact that photons are bosons, ergo they follow the Bose-Einstein Statistics, or in this case they make a Bose-Einstein Condensate. If you are not familiar with this exciting concept, I suggest you make a look at this Wikipedia article or any other statistical mechanics textbook you have around. Anyway, the two photons ... 3 At the time of Young there were no lasers to provide coherent light. Incandescent light is incoherent because it comes from many sources and the same is true for sunlight. By passing the light through the one slit he created a single coherent wave front . It is worth reading his description "on the nature of light and colors" in the link above. Edit: I ... 3 No. Coherence means "fixed, definite phase relation", or in a polychromatic context it can mean "definite phase relation at any given frequency". What would be the phase relation of the horizontally- and vertically-polarized waves?$0^\circ$?$90^\circ$? There's no reason for any definite phase relation. Here's a more specific way to think about it. ... 2 Environmentally induced decoherence makes wave function collapse unnecessary. As far as I know, that's still a conjecture. I might be wrong about that, though. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) It's the rest frame of the system + reservoir, ... 2 The beam of white light is refracted by the prism, but the different wavelengths in the light are refracted by different angles. That means when the light falls on the screen the different wavelengths are spread out over a region of the screen. This is what is meant by dispersion. The dispersion is linear if the different wavelengths of light are spread ... 2 Yes, coherent radiation means that the phases of two ( or more ) waves representing the radiation differ by a known constant. Incoherence means that the phase differences are unknown/random. Laser radiation is coherent because stimulated emission assures phase differences are constant . Radiation from an incandescent lamp is incoherent because the ... 2 I believe, that the derivation is wrong... If you assume a translationally invariant state, such that$G^1(r, r') = G^1(r - r')$then you can get the result. Rewrite the exponential as$p r - p' r' = p( r- r') + r'(p - p')$. Since, in this case, the left-hand-side of Eq. (2.27) can only depend on$r - r'$it must be such that$p = p'$from the second term. ... 2 They are two different but closely related concepts Given two sine waves of equal frequency once can ask what is their relative phase, and are they in-phase or out of phase. So$\sin(\omega t)$and$sin(\omega t+\pi)$have a relative phase$\pi$or 180 degrees and so we would say they are out of phase. The question of coherence is: how stable is the ... 1 I am assuming for simplicity that either the waveguides are one-moded or, if not, the input to the system is in one mode alone and the system design is such that coupling between modes is negligible. Usually with resonant ring systems like this one used as interferometers, we are using them to sense changes in the ring's optical phase delay. Actually, if ... 1 I think a more basic answer would be helpful. When we think of particles and wavefunctions we think of each particle having its own wavefunction$\psi(x)$- the tendency for a particle to be found at some position$x$. When we have two particles, it is natural to think that we have two wavefunctions,$\psi_1(x)$and$\psi_2(x)\$, each describing its own ...

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For two waves to be 'perfectly' coherent (I'm assuming electromagnetic, transverse waves) they have to have the same wavelength, polarization and phase. To go a bit deeper, the coherence can never be perfect. Only an infinitely long wave has a wavelength that is 'fully defined', ie. has no uncertainty (or width) associated with it. One therefore defines a ...

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If I understand your question correctly, you wish to use quantum interference to observe phenomena that otherwise have a low probability of occurring? If so, exploiting this is well known and well established in quantum optics. For example, take phenomena that occur with coherently prepared atomic systems, such as Lasing without inversion, ...

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Jones Matrices cannot express partial polarization. For that you need Stokes Vectors and the related Mueller Matrices (for transforming one Stokes vector into another). Interestingly a Stokes Vector can be trivially separated into a fully polarized vector and a fully unpolarized vector.

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Brandon, the simple truth is that you have just asked one of the hardest and least understood questions in all of physics. So, don't feel bad if you don't understand it very well, because, er... no one else really does either? It's not that we can't model this stuff mathematically. Shoot, Richard Feynman's version of something called Quantum Electrodynamics ...

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