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For now I will only give you an overview of the ideas involved and show you how you should interpret the idea of a "local realistic theory" that cannot exist at the microscopic scale. Once you've read it, and if you feel you need more mathematical rigor to be convinced, then I will draw you step by step the proof of Bell's inequality (it is not the only one ...


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Bell's argument makes very weak assumptions about the behavior of the two particles (which is why it's interesting). In effect, the particles are black boxes that take an angle as input and produce a spin direction as output. There is no restriction on how they choose the spin direction; there could be a source of true randomness in there, or a human being ...


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I think you misunderstood the significance of could for a classical theory. The text below the picture you took from Wikipedia says: "Many other possibilities exist for the classical correlation subject to these side conditions", so classicality does not imply linearity. It does, however, rule out the cosine, by the following (slightly heuristic) argument: ...


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I can answer your question regarding the eigenvalue equation of the reduced density matrix. This is basically Helmholtz integral equation of the first kind. This is an eigenvalue of equation for a continuous variable. You can do the quadrature approximation of the integral and get the matrix eigenvalue equation which is just an approximation due to the ...


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You seem to think of entanglement as a property of individual particles. That is not true: Let $\mathcal{H}$ be the Hilbert space of states of a single particle (electron, photon, whatever, doesn't matter). Then the space of states of two particles is given by the tensor product $\mathcal{H}\otimes\mathcal{H}$, and the space of states of $N$ particles is ...


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Yes, we can have entanglement between different degree of freedom of same particle or system. That is known as ''hybrid entanglement'' and that is experimentally demonstrated also. http://arxiv.org/pdf/1007.1322v1.pdf


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A state like $ \frac{1}{\sqrt{2}}(a^\dagger_+(\vec p)a^\dagger_+(-\vec p) + a^\dagger_-(\vec p)a^\dagger_-(-\vec p))|0\rangle$ would be an example. It is both entangled in spin and entangled in momenta.


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Let $f(x,y)\in L^2(\mathbb{R}^{2d})$ and $\Omega$ the vacuum of the symmetric Fock space $\Gamma_s(L^2(\mathbb{R}^d))$. Suppose there is no $f_1,f_2\in L^2(\mathbb{R}^d)$ such that $f(x,y)=f_1(x)f_2(y)$: then $f_s$ (the symmetrized of $f$) is an "entangled" two particle state of $\Gamma_s(L^2(\mathbb{R}^d))$. This is created by $$\frac{1}{\sqrt{2}}\int ...


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In the context of your paper, the time-like Bell states are : $|\psi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau v_b^{\tau'}\rangle \pm |h_a^\tau v_b^{\tau'}\rangle)$ $|\phi^\pm\rangle^{\tau, \tau'}_{ab} = \frac{1}{\sqrt{2}}(|h_a^\tau h_b^{\tau'}\rangle \pm |v_a^\tau v_b^{\tau'}\rangle)$ So, equation $(3)$ just reexpress $|S\rangle = ...


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Firstly the link you have provided for the paper is only accessible for those who have a subscription to the APS journal. So maybe instead you could just give a preview of the equation you are referring to. But I can give you a general overview on projections in quantum mechanics and Bell states. I will use Dirac's Bra-Ket notation for the rest of this ...


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A useful reference is http://arxiv.org/pdf/0906.1663.pdf, by Peschel and Eisler. A common approach is to make use of the fact that the two point function you calculated is independent of whether one uses the full density matrix or the reduced density matrix, provided one looks at operators that are local to the region that one is not tracing over. If one ...


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You've completely misunderstood the impact of adding a potential to an entangled state. Atoms a and b should roughly have the same distribution ... because each single measurement of position for each entangled pair in the ensemble should yeild $x_a=x_b$ with respect to their local axes ... because this is what it means to be entangled. (Need a check ...


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You describe a communication protocol exploiting EPR/Bell-type setups. If I understand your protocol correctly, you envisage a 'stream' of entangled ensembles, and preagreed measurements being performed on each block of the stream to yield a 'bit'. There are various ways we could do this: you use harmonic oscillator potential. Fine. We could do the ...


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There is no reason in principle, that I can think of, for this experiment to be impossible. The entanglement could be achieved by aligning the source and the slits so that the upper slit on one side, the source and the lower slit on the other side are in a line. If the particles are produced in pairs with no total momentum then they will be emitted in ...


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You have not described entanglement properly. What sort of thing do you have to explain to understand what's going on in an entanglement experiment? The problems often look somewhat like this. (1) There are observables on A and B, call them Acorr, Bcorr such that if you compare the results of the measurements after they have been completed and the ...


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I would just mention here that every probabilistic system, even classical, may exhibit a kind of "entanglement" or "spooky action at a distance" features. For instance, imagine that you have $2$ boxes, and one bowl in each box. The bowl could have only a white or black color, and the two bowls have the same color. The boxes are closed, then one box stay on ...


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The Transactional Interpretation of QM suggests that Maxwell's equations work backwards in time, carrying the information of one test back to the point of entanglement, where it can affect the entangled particle. This explanation bypasses any issue of violation of SR.


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There is what appears to be a very good lecture that deals with many of the issues. It is a Google Techtalk from Ron Garret entitled "The Quantum Conspiracy:What the Popularizers of QM don't want you to know". A rather tongue-in-cheek title that should not mislead you into thinking it is not a serious and worthwhile talk. www.youtube.com/watch?v=HQIJgheuYNU ...


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"I know that much of quantum mechanics is not "intuitive". If anyone can explain why particles don't have a definite property, even before we measure them, I would be grateful. Thanks." I think the short answer is, "Because, that's the way it is." We do the experiment and are then stuck with the result. I always go back to the two slit experiment. How ...


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At that instant, the other particle "assumes" the other spin. This is where you go wrong, because things are actually weirder than that. I'll try to keep this simple, especially because if I don't I'd have to make extremely rigorous arguments given that this is a matter where different physicists may see things a bit differently. A historical note : ...


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The assumption (if you cry it or not) "but the particle does have a definite spin, we just don't KNOW what it is, until it is measured! Duh!" is called realism, or in mathier speak, a theory of hidden variables. Bell's inequalities now say that no theory that fulfills local realism (equivalently that has local hidden variables) can ever predict the correct ...


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Your thought experiment does have a major flaw. According to quantum mechanics in any measurement of two spatially separated atoms a and b what happens to b has absolutely no effect at all on the probabilities of measurement outcomes on b. I'm not going to work out exactly what the flaw is in your proposed experiment, but just indicate why quantum theory ...


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user26143 gives a correct argument about the effects of entanglement in two-atom processes, but specifically to your question, I have to reiterate the answer 1) You are not talking about entanglement 2) Yes, this effect gets effectively blocked out in the cases you consider 3) QFT does not really bring any new insight to this problem In theory, all ...


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I will try to present the case at hand as clear as possible, but I cannot avoid it using a bit of math. To clarify notation, I will also recap what entanglement and quantum states in their simplest cases are: Let us only consider spin as the only degree of freedom our quantum mechanical 1-particle spin-$\frac{1}{2}$ state has. Then, a single particle has ...


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$\newcommand{\spinup}{\vert{\uparrow\rangle}} \newcommand{\spindown}{\vert{\downarrow\rangle}}$ The "spin property" you mention is false. Since spin up and spin down by definition have different eigenvalues of the spin operator, they must be orthogonal states. Therefore for a general state, there is no relation between $\psi(r,s)$ and $\psi(r,-s)$ other ...


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My understanding is that the measurement at 45° matches the measurements at 0° and 90° more than it should (assuming local hidden variables) given how often 0° and 90° match. Think of two detectors that move between 0°, 45°, and 90°, so that you get the 90° measurement when one is at 0° and the other at 90° and the 45° measurement when one is at 45° and the ...


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As Trimok says, every photon - and for that matter, any physical two-level system - is immediately a qubit. In other words, qubits are abundant in nature and not very interesting by themselves. Also, a qubit alone is not entangled. The interesting part is, what you can do with your qubit: Can you create a qubit in always the same state? Can you entangle a ...


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Any photon (pure) state may be described by a q-bit formalism: $$|photon\rangle = \alpha |0\rangle + \beta|1\rangle$$ where $|0\rangle$ and $|1\rangle$ represent the two possible polarizations of the photon. So, any photon "is" a q-bit. You don't have to "create" q-bits. Just prepare photons is some state. An entangled state of $2$ photons may be ...


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(1) Spacetime isn't all in the mind. Spacetime is just the gravitational field and it has degrees of freedom independent of your brain, so it's not all in your mind. You can't make an even happen on Tuesday just by thinking it happened on Tuesday. (2) As commented by others, you can't ride a photon. Also, the way to figure out whether something is real is ...


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Give a physically distinguishable definition of "out there" vs. "in the mind" and we can try to discuss this further. As jinawee comments, there are no frames of reference that move with the speed of light, since the photon we "ride on" would have no speed at all by definition of a comoving reference frame, and that contradicts the constancy of the speed of ...



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