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There are certain so-called "No-go" theorems in physics: http://en.wikipedia.org/wiki/No-go_theorem - one classic one is Earnshaw's Theorem, showing that a collection of point charges interacting with Coulomb forces is always unstable (http://en.wikipedia.org/wiki/Earnshaw%27s_theorem). (Of course there are caveats.) However, it is important to be ...

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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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I disagree with Danu. This is a serious question being considered by theoretical physicists involved with the foundations of quantum mechanics. Counterfactual definiteness is an epistemological property that essentially allows you to ask what-if questions about experiments. For example, in a forensic analysis of a car crash, it might be important to know ...

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Consider an EPR experiment where an entangled pair of electrons is created. One of them hits a detector which finds its spin to be up. The other hits a detector at some distance. The first detector sends a light signal to the vicinity of the distant detector. That signal arrives before the other electron. Near that second detector sit two observers. ...

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It is incorrect to say that $$U(\Lambda) \left|p,\sigma\right> \propto |\Lambda p, \sigma\rangle~~~~~~ \text{WRONG!!}$$ Here is the correct logic. Consider the state $U(\Lambda) |p,\sigma\rangle$. We have just shown that (in eq. 2.5.2) that this state has a momentum eigenvalue $\Lambda p$. Now, there are a whole bunch of states with momentum \Lambda ... 2 For the sake of completeness, let's first unravel some definitions and prove the identity \begin{align} \langle x_b, t_b|x_a, t_a\rangle = \langle x_b|\hat U(t_b,t_a)|x_a\rangle. \end{align} The definition of each position eigenstate|x\rangle$is that it's a normalized eigenvector of the position operator$\hat x$corresponding to eigenvalue$x\$; ...

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In a multiverse, everything happens (at least in the many worlds view you seem to be talking about). Probability is applied to bundles of trajectories through the state space. Say you are an observaer at point P in this abstract parameter space, then if you specity a hypervolume in this space you could, in principle, get the probability of being in one of ...

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The sum of the probabilities of all possibilities being equal to 1 is necessary for probability to be well defined. In other words, I wouldn't have it any other way.

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You should specify what type of multiverse you're talking about. Anyway, I don't see the problem. The important thing is that the probability is lower than one in each universe. So probability is useful en each universe.

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