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8

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 ...


7

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 ...


6

Refer to the nice complement on coherent states in the book by Cohen-Tannoudji, Diu and Laloë, volume 1. It starts off defining coherent states as neither of the ones you mention, and then derives all properties. To answer the question, if you start with definition 2, you can easily show 1, and then from 2, 3. First expand the exponential using ...


6

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 ...


6

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) ...


5

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 ...


5

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 ...


5

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 ...


5

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 ...


5

There are two reasons why it may not be necessary...firstly the intensity of light from the candle is low unlike that of the laser. Secondly, the laser encounters two mirrors while light from the candle only encounters one before reaching the detector.


5

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.


4

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. ...


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

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

This will not happen because color filters don't work like this. A red color filter does not convert blue photons to red photons. It absorbs photons that are not red (most of them) and lets red photons pass unaffected. If you use a red filter for one slit then a blue photon will not go through that slit at all, so you fill effectively have a single-slit ...


3

Presumably you have measured your spectrum as a function of wavelength, so you have $\mathscr{F}(\lambda)$, which is an power per unit wavelength. You must now convert this power per unit frequency spectrum. So we seek $\mathscr{G}(f)$ where $\mathscr{G}(f)\,|df| = \mathscr{F}(\lambda)\,|d\lambda|$; given $c = f\,\lambda$ we have: $$d\lambda = ...


3

In general this is not a problem if you are careful, but if you are not then it can certainly ruin your experiment and wash out any possible interference. The problem is not quite ensuring that the lasers be in phase, since at any given time the two sources will have some definite phase relationship with each other: the problem is making sure that they stay ...


3

To start with the double slit experiment gives interference even when the beam is composed by one photon at a time. The spot on the screen a photon/particle the statistical accumulation the interference seen as expected classically too. The joint comes because the photon as a quantum mechanical entity has a wavefunction that is the solutions of Maxwell's ...


2

I agree with @Floris that the statement doesn't make sense at face value, but since I know what he is trying to say, I might be able to translate. No signal has a single frequency. There is always a very small spread in a signal's frequency content. (Pure single frequency implies a signal that started in the infinite past and will continue into the ...


2

As you said, a coherent state is defined by the equation $$a \Psi_{\alpha} = \alpha \Psi_{\alpha} \, .$$ Therefore, to check whether a particular expression is a coherent state, just act $a$ on it and see if you get the same thing back multiplied by a complex number. Let's try it $$ \begin{align} a \Psi_{\alpha} &= a \left[ e^{-|\alpha|^2/2} ...


2

I am not sure that I get the full point of your question, but I'll answer anyway: the limiting factor is the coherence length of the incoming radiation. In the visible range this is easily obtained with lasers, and the resulting coherent interference patterns are known as holograms. In the X-ray regime, intrinsically coherent sources have only recently ...


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

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 ...


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

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

Let's suppose the light from the Sun was perfectly coherent. In that case you would get interference, however the optics used in the sort of solar tower you show are manufactured to nowhere near single wavelength of light precision. That means the phase of the light arriving at the collector would vary wildly with position across the collector, and the ...


2

Short answer: On something operating with non-coherent light on the scale of a solar power collector, there is no net effect from interference. The increase in light intensity is not additive interference at all, but is simply that the light falling on a wide area is collected and focussed on a small area. However, I'm guessing that there is some confusion ...


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 ...


1

I don't think this is a quantum optics problem. Just look up the van-Cittert-Zernike theorem . The (complex) visibility is the Fourier transform of the mutual coherence function of the source.



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