# Tag Info

24

Stan Rogers' answer on photography.SE seems to be claiming that QED is not just sufficient but also necessary to explain the the effect of the lens's shape. This is wrong. Ray optics suffices at ordinary magnifications, and even at high magnifications, classical wave optics suffices. Let's say you use a rectangular lens rather than a cylindrical one. First ...

13

Your intuition is correct, you don't need quantum electrodynamics to explain/model/engineer camera lenses. When considering the propagation of light, the results of geometric optics can be interpreted in terms of path integrals, as Feynman does in his QED: The Strange Theory of Light and Matter, but this is not necessary for lens design. Geometric optics ...

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Thanks for the clarification. Your question makes sense to me now. I'm not really going to be able to answer it. In general, if you start with a photon number state, and put it through linear optics, I believe the state you get looks like a big, ugly mess if you try to write it down in any reasonable basis. I don't think you'll be able to get most quantum ...

10

For (1), there is a theorem of Holevo that implies you cannot extract more than one bit of information from one qubit. You can indeed encode one bit of information, since the two inputs $| 0 \rangle$ and $| 1 \rangle$ (or any two orthogonal states) are distinguishable. If the sender and receiver share an entangled state, they can use superdense coding to ...

10

The rotating wave approximation (RWA) is well justified in a regime of a small perturbation. In this limit you can neglect the so-called Bloch-Siegert and Stark shifts. You can find an explanation in this paper. But, in order to make this explanation self-contained, I will give an idea with the following model $$H=\Delta\sigma_3+V_0\sin(\omega t)\sigma_1$$ ...

9

Squeezing of laser light generally involves a non-linear interaction, where the nature of the interaction depends on the intensity of the light that is present. An easy to understand example is frequency doubling, which takes two photons from a pump laser, and sends out one photon of twice the frequency. You can think of the input beam as a stream of ...

9

Basically: What intrinsic property causes the differences between how the varying wavelengths of light are reflected at the atomic scale? Also, how do photons factor into this? These are absorption lines in the solar spectrum Fraunhofer lines coincide with characteristic emission lines identified in the spectra of heated elements.6 It was ...

8

The vacuum state is the thermal state for $T=0K$. How to compare if a state is close enough to the vacuum state? By counting photons (for vacuum it is zero). The occupation for photons is given by Bose-Einstein distribution: $$n = \frac{1}{\exp( E/(kT)) - 1},$$ where $E$ is the photon energy ($E = \hbar \omega = h \nu$) and $k$ is the Boltzmann constant. ...

8

A. All light sources (even lasers) are subject to a diffraction limit, so any light beam will eventually diverge with an angle $\theta$ given by $$\theta \approx \frac{\lambda}{A_T}$$ where $\lambda$ is the wavelength of the light and $A_T$ is the aperture of the light beam source (and "eventually" means for distances much greater than $A_T$). Any beam ...

8

The photon model of light may be the most frequently over-applied model in physics. Lamb expresses my opinion fairly clearly here: "The photon concepts as used by a high percentage of the laser community have no scientific justification." In my experience, many physicists who answer simple questions about matter without unnecessary reference to ...

8

If you actually discuss with people working on quantum memories, you will notice (at least I did) that they share a vague definition : "a quantum memory is something which stores a quantum state" better than a classical memory could do. Beyond that, they have vastly different ideas on how to implement a quantum memory (single qubits, collective degrees of ...

8

You have in fact put your finger on the reason for the refractive index change. It is related to moving electrons in the direction of the fields. NB dispersion is a complex phenomenon, so this is necessarily going to be an arm-waving explanation - do not take it too literally! There is a discussion of the phenomenon in this article. Basically the ...

7

A single photon can only interfere with "itself". However, "itself" is ill-defined because all photons are identical in quantum mechanics. Because of their Bose-Einstein statistics, the wave function of all photons is symmetric - invariant under all permutations of the individual photons. So the states in which some photons are permuted actually do interfere ...

7

The supspaces $V_n = Span \{ (a_1^{\dagger})^{n_1}, . . . (a_d^{\dagger})^{n_d} |0>\}$, $n_i \ge 0$, $n_1 + . . . n_d = n$, constitute of invariant subspaces of the operator $S S^{\dagger}$ action. The dimension of $V_n$ is $\frac{(d+n-1)!}{(d-1)! n!}$. Thus the operator can be represented on each of these subspaces as a square matrix of size $... 7 There is only one electromagnetic field in the Universe – it's the function that assigns each point in${\mathbb R}^4$with two vectors$\vec E,\vec B$. When we say that we quantize the electromagnetic field, it doesn't mean that we quantize a particular configuration of the electric and magnetic vectors. It means that we quantize the whole function, namely ... 7 As Claudius suggests, vacuum does not absorb. But that is not a material. You can have light that travels through a material without absorption; that happens in nonlinear optics with self-induced transparency. The full theory behind that is rather involved and you need really high intensities for that. The basic picture is that the front of the light pulse ... 7 Yes. Consider quantizing electromagnetic fields in a box. This corresponds to photons being trapped inside of said box since photons are just the mode quanta of the EM fields. The Hilbert space (called Fock space in this case) of the quantized radiation is found to be spanned by states $$|\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ... 7 Starting with:$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$If t_i=0:$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$Using the identity: \sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$Since the exponential of an operator is (by ... 7 I'd like to add to Ruslan's, Gregsan's and Oscar Lazo's answers, particularly Oscar's. All these answers are perfectly valid. The multiple bounces in a laser raise the probability of a given photon's stimulating another in a stimulated emission event AND shape the output spectrum. But why is there a spectrum to shape? And how does the cavity shape the ... 6 A single photon can easily be detected by a photomultiplier. The basic idea is that a photon hitting a metal plate in the tube ejects an electron from the metal plate by the photoelectric effect. An electric field inside the photomultiplier then accelerates the electron until it slams into another metal plate, releasing a bunch of electrons. These are ... 6 Just a few random thoughts. There is something important in your observation that the Born-Infeld model is essentially a free-space model. It is known to Boillat and Plebanski (separately in 1970) that the Born-Infeld model is the only model of electromagnetism (as a connection on a U(1) vector bundle) that satisfies the following conditions Covariance ... 6 To keep things simple, let's talk about two-qubit states. A single qubit could have an orthonormal basis \{|0\rangle, |1\rangle\}. But it could also have a different orthonormal basis \{|+\rangle,|-\rangle\}, where$$|+\rangle = \large(\normalsize|0\rangle \small+\normalsize |1\rangle\large)\normalsize / \sqrt{2}|-\rangle = ... 6 In a normal conductor the electrons sit in energy bands, so you can change the energy of an electron by an arbitrarily small amount. By contrast, in a superconductor there is an energy gap between the ground state energy and the first excited state energy of the electron pairs. This means you cannot raise the energy of an electron in the ground state by an ... 6 If such a material exists and it absorbs no light at any frequency, then it must have absolutely no optical activity. This is a consequence of the Kramers-Kronig relations, which are very, very basic constraints on how absorption and dispersion in a material can be related to each other, and represent mathematically the physical principle of causality. (That ... 6 You have to use the eigenstates$|n\rangle $of the operator$\hat{n} = a^\dagger a$. You have, then, that$a \sqrt{\hat{n}} ~|n\rangle = a \sqrt{n} ~|n\rangle = \sqrt{n} ~ a |n\rangle = \sqrt{\hat{n}+1} ~ a |n\rangle ,$where the last equality is because$a |n\rangle \sim |n-1\rangle$. So,$\left[a, \sqrt{\hat{n}}\right]~ |n\rangle = ...

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Yes, the velocity of light can exceed $c$, but this is a somewhat technical situation and does not represent a violation of relativity. As you say, the change in the speed of light in some material is due to an interaction of light with the electrons in that material. When you're well away from any electronic transitions the interaction is relatively small, ...

6

The quadrature is a process – any process – of turning something into a "square". "Quadro" in Latin is "make square", "quadrus" is a "square". It comes from "quattors", four, because that's the number of vertices of a square. So integration of a function is also known as "quadrature" because we are calculating the area i.e. looking for a well-known area ...

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A qubit is a unit of information, so the information included in one qubit is exactly one qubit. For most purposes, the information may be identified with the information of one bit. We call it a "quantum bit" because the two possibilities may be combined into arbitrary complex combinations such as $a|0\rangle + b|0\rangle$. However, the complex amplitudes ...

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