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The ray theory of light is equivalent to the Eikonal Equation, which in turn is essentially a slowly varying envelope approximation to Maxwell's equations. If we write the electric and magnetic field vectors as $\mathbf{E}\left(\mathbf{r}\right) = \mathbf{e}\left(\mathbf{r}\right) e^{i\,\varphi\left(\mathbf{r}\right)}$, $\mathbf{H}\left(\mathbf{r}\right) = ... 0 From your description, I think you are talking about the diffraction pattern from a pinhole, and that is an Airy Disk. https://en.wikipedia.org/wiki/Airy_disk It shows a central maximum with surrounding rings which decrease in intensity as you get further from the center of the pattern. -1 I am not a specialist of quantum optics but I am familiar with laser and classical optics. I would say that coherent decay is stimulated emission and population decay rate is simply spontaneous emission. When atoms/electrons are interacting with light they have a certain probability of absorbing photons to get into higher energy level states. Once they are ... 0 According to the diagram you provided,$|a\rangle$is injected into a beam-splitter, resulting in the out-put state, $$|\psi_{o}\rangle =B|a\rangle= e^{\frac{i\pi}{2}}|b\rangle +|c\rangle.$$ Here the beam-splitter unitary is denoted by$B$. I have expressed$i$as$e^{\frac{i\pi}{2}}$, to highlight the fact that$|b\rangle$picks ... 1 There isn't really an answer to this. If your initial conditions are spherically symmetric then the system remain spherically symmetric and the emitted light will be in a superposition of all directions. The superposition will collapse, and the symmetry be broken, only when something interacts with your system e.g. a CCD detector. On the other hand, if the ... 0 When the mean number of photons is huge, the Heisenberg uncertainty becomes negligible and "disappears" (formally it looks like$\hbar\to 0$). Thus, such a coherent state becomes quite classical one. 0 I think that it in this case is a bit of semantics. In an ideal beam-splitter there is really no "true" reflection. Rather the beam is split into two output directions. Which one of the two directions you call transmitted and deflected is a matter of taste. Naively I would say the example with transition$t$on the diagonal looks more natural. In this case ... 0 I believe your professor's trying to tell you that the general solution will be a sum of three terms, and he could have just as easily used$f_{ij}$,$g_{ij}$and$h_{ij}$for the three superscript terms. Edit: In other words, you're Matrix$\rho$can be expressed as a sum of matrices$\rho^{(1)}$,$\rho^{(2)}$, which is proportional to$e^{i\Omega t}\$ and ...

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Photons indeed have phase (or phases, in the plural) as folloows: one can turn Hossein's answer on its head in the following way. A "lone photon" is simply a particular pure state - a so called Fock State - of the quantum electromagnetic field (see the "Quantization of the Electromagnetic Field" Wikipedia article) where the number observable is certain to ...

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Please recall how the concept of a photon came to life. 1.You have Maxwell's equations, 2. you look for wavy solutions (all classically),3. you get a classical Lagrangian for EM fields and the classical action accordingly. You quantize this classical action. Frequency in this case is an energy (Planck constant is unity in selected system of units). The ...

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