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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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Okay, this is getting even more into depth, which is great stuff! I heartily recommend anyone who is this dedicated take a few courses on the subject, if you haven't already. The state matrix formulation of Quantum Mechanics Here's the most basic formulation of quantum mechanics which adequately shows all of these properties, called the density-matrix or ...

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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 $\lvert c\rangle = \frac{1}{\sqrt{2}} (\lvert0\rangle + e ^{i\phi}\lvert1\... 9 The word "stimulated" means that the emission of the photon is "encouraged" by the existence of photons in the same state as the state where the new photon may be added. The "same state" is one that has the same frequency, the same polarization, and the same direction of motion. Such a one-photon state may be described by the wave vector and the polarization ... 9 The visibility of interference fringes from a double slit depends on the how correlated the fields from the source at the two slits are. The typical slit-distance for which the fringes are manifestly visible is called the transverse coherence length$d_{\rm coh}$of the source. For a spatially extended source, this quantity primarily depends on two ... 8 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 Baker-... 7 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 ... 7 I am posting these notes following a request for further information regarding this question. Should not affect the OP's choice of answer. Notes added in proof: On the meaning of quantum coherence: Quantum coherence is a direct extension of the classical concept of wave coherence. Two classical waves are said to be coherent if they can produce a well-... 6 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 ... 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 $$|E(t) +... 5 Huygen's Principle. If the slit is thin enough and you are only going to use light from one plane, then the new wavefront is dominated by contributions from only one part of the original wave and all parts of it (the wavefront) share the same phase. 5 There are two reasons why it may not be necessary...firstly the intensity of light from the candle is low unlike that of the laser. Secondly, the laser encounters two mirrors while light from the candle only encounters one before reaching the detector. 5 You are correct that every photon will interfere with itself for sure. But for the whole interference pattern to be observed, you need a large number of photons independently interfering with themselves. And, these large number of photons should be identical in every respect so that they can be represented by the same single photon wavefunction. This makes ... 5 No. Coherence means "fixed, definite phase relation", or in a polychromatic context it can mean "definite phase relation at any given frequency". What would be the phase relation of the horizontally- and vertically-polarized waves? 0^\circ? 90^\circ? There's no reason for any definite phase relation. Here's a more specific way to think about it. ... 5 Is the coherence length consistently defined as the length at which these two waves achieve a phase difference of 1 radian? No Coherence length is the maximal difference in way traveled by the to rays without losing the phase relation which allows interference. This lenghts can be some µmeters (eg for white light from a glowing body) or several ... 5 You're conflating two different views on the description of the attosecond pulse train; in particular, you're flitting back and forth between the time-domain and frequency-domain descriptions, and it's not doing you many favours. Let's look at a few things first: The total field is$$ E(t) = \cdots + E(t-2\pi/\omega)-E(t-\pi/\omega)+E(t)-E(t+\pi/\... 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 ... 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 ... 4 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\,\lambdawe have: $$d\lambda = -\frac{df}{f}... 4 A wavefront is the crest of the wave. When you go down to the beach, and see those things called waves, the front is the whole line that is at the same height. In electromagnetics, it's the same thing. It's the points that are at the same height. In your diagram, the black curves represent a unit of cycle, and the two waves through b and c can either ... 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 following: ... 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 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 ... 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 Here are a few things. Not a complete list. Coherence - This makes holography possible. Stability - Lasers are monochromatic because the light is produced by reflecting back and forth in a cavity. It is possible to select just one mode. With care, it is possible to keep that wavelength very stable over long periods of time. This is the basis of atomic ... 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 ... 3 Waves can be coherent and yet not have the same wavelength. It is sufficient that they have the same frequency - because that is sufficient to imply a constant phase difference. If you make a Michelson interferometer where you split an incoming light beam into two arms, and you send half the light through a column of water and the other half through air, ... 3 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} \sum_{n=0}... 3 The star acts as an "effective point source" since it is so far away and its angular extent is so small - in other words, the optical signal arriving at earth is "very nearly" a plane wave with the same phase over a large extent. This is what enables us to do interferometry. Think of waves, not photons. 3 Let's shift on a simpler quantum system: take a spin. Assume that your spin has been prepared aligned along x axis (eigenstate of\sigma_x$). Your spin is in a superposition of state having$\sigma_z\$ eigenvalue +1 and -1. When you actually measure the spin along z it will collapse on one of the eigenstate.It will then remain in this state (as long as you ...

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