# Tag Info

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Coherent states, although strictly quantum, are "isomorphic" to classical states. They are also isomorphic in the same way to one-photon states. There are bijective maps between any pair of the following three sets: (i) the set of all quantum coherent states (ii) the set of all one-photon states and (iii) and the set of all solutions of Maxwell's equations. ...

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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 ...

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It is all about what meaning you put into the words "quantum" and "classical". Fock space and elements of this space are notions that belong to quantum theory of radiation and have no direct relation to states of radiation in classical electromagnetic theory, so the coherent state may be called "quantum" with good reason. However, coherent states have ...

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If coherent state are indeed the most classical states (which means that the mean value of the EM fields obeys the classical Maxwell equations), the state used in the paper you mentioned are not coherent state (at least in the arXiv paper), but cat states ! The state $|\alpha\rangle+|-\alpha\rangle$ is not a coherent state ! It is the superposition of two ...

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Back action noise and shot noise both arise from the discrete nature of quantum mechanics, but they are not the same. In a quantum optical system shot noise is the noise due to counting photons at the output of the detector while back action noise is due to the photons imparting momentum to the mirrors in the system. Both arise from the statistical nature ...

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This does not have anything to do with the bandwidth of your gain medium. Considering the one-dimensional wave equation any traveling wave is a solution, including half-cycle pulses. You can virtually plug any solution $f(t-\frac z c)$ into the one-dimensional wave equation and it will be an exact solution, it does not matter if it's a sine-wave or a wave ...

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The dipole transition matrix elements $\mathbf{d}_m$ and $\mathbf{d}_n$ are complex-valued vectors which are relatively easy to define. Their "direction" is a mathematical convenience, and it is essentially given by the vector divided by its modulus, for an appropriate interpretation of the latter. Consider first the case of a single molecule, with a ...

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Relative permittivity or Dielectric const. depends on the medium and on the frequency of the electromagnetic radiation. The bound electron has a natural frequency usually in the uv region. So the relative permittivity will be large at radio frequencies(>80 for water) but small at optical frequencies.

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The problem with the reasoning above is that it considers only the effect the measurement has on the covariance matrix and does not consider the displacement vector. While the covariance matrix of the resulting state does not depend on the particular measurement result, this is not true for the displacement vector. As a result, when doing the partial trace, ...

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It all boils down to how much information there is to, in principle, distinguish which way the photon went from the final state of the beam splitter, as encoded in the overlap between its two possible final states. The interference is destroyed because the photon gets entangled with the beam splitter, and the amount of entanglement depends on this overlap. ...

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This is actually a very interesting question, along the lines of the Bohr-Einstein debates about quantum mechanics. Answers to these take a little time to work out, though. Another way to interpret the loss of coherence is that if the splitter is light enough (no pun intended), the velocity change in the splitter from the momentum kick, effectively measuring ...

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Though it's too late to answer, I was looking for answer and saw your question. I observed the same thing through my experiments today. Zeroing the imposed magnetic field leads to vanish the dark state due to a circularly polarized light. My reasoning is related to the most basic concepts of quantum mechanics; based on my knowledge it's impossible to ...

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