# Tag Info

27

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 ...

14

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 ...

14

Nobody is answering this question, so I'll take a stab at it. Consider the mirror. Suppose you started your experiment by (somehow) putting it in a nearly-exact momentum state, meaning there is a large uncertainty in its position. Now, when you send a photon at it, the photon either bounces off or passes through. If the photon bounces off the mirror, it ...

13

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 ...

13

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 ...

13

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 ...

13

Nothing is happening. At least, nothing except that a new generalized quantity suggestively called "angular momentum" was defined and subsequently measured. But nothing we know about the usual angular momentum of photons is changed by this in any way. Standard total angular momentum is $J = L + S$, where $L$ is the orbital and $S$ the spin angular momentum. ...

12

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 ...

12

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 $$H=\Delta\sigma_3+V_0\sin(\omega t)\sigma_1$$ ...

12

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 ...

11

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 ...

11

Optical pumping will at most only achieve equal population of a two-level system. This is because the probabilities for raising an electron to the upper level and inducing the decay of an electron to the lower level (simulated emission) are exactly the same! In other words, when both levels are equally populated, the numbers of electrons "going up" and ...

10

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 ...

10

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 ...

9

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 ...

9

the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of $E$ and $B$ vectors at each point. The probability density that a photon is found at a particular point is ...

9

Coherent states are quantum states, but they have properties that mirror classical states in a sense that can be made precise. To be concrete, let's consider coherent states in the context of the simple harmonic quantum oscillator which have precisely the expression you wrote in the question. One can demonstrate the following two facts (which I highly ...

9

Perhaps the question is, how do you measure an arbitrary number of photons without destroying the system? It's certainly the case that somehow measuring the number of photons in the cavity would collapse the state into a Fock state, but it's not obvious how to do that without actively destroying the system in some way; there has been some work done in ...

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

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 ...

8

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 ...

8

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 $$|\mathbf k_1, \mu_1; \dots, ; \mathbf k_N, \mu_N\rangle, \qquad ... 7 To keep things simple, let's talk about two-qubit states. A single qubit could have an orthonormal basis \{|0\rangle, |1\rangle\}. But it could also have a different orthonormal basis \{|+\rangle,|-\rangle\}, where$$|+\rangle = \large(\normalsize|0\rangle \small+\normalsize |1\rangle\large)\normalsize / \sqrt{2}|-\rangle = ...

7

This is true. The simple explanation is this: For calculating the decay rate of an excited state, you use Fermi's Golden Rule, which involves the matrix element $$|\langle f | V | i \rangle|^2$$ where $f$ and $i$ denote the final and initial state, respectively. Since the final state contains the electron in its groundstate together with a photon created ...

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 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 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 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 = ...

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 ...

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