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Seeing your last question, why would you even consider this scenario with entangled photons? Seems like the same question can be perfectly asked with one single photon. But anyway, ignoring the possible optical issues in the scenario, like the exact color of the film you use, or what Nanite has commented about, either way you will not see a superposition of ...


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Mathematically you have a superposition of two Eigenstates and you use it because you don't know all parameters of the source. And yes, you use a probabilistic source, otherwise you will not do any experiment with it. Using your special source you get always the next result: Measuring one of the photons you know immediately the Eigenstates of the twisted ...


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There exists a precise way of calculating the entanglement entropy in a conformal field theory via the Ryu-Takayanagi (RT) prescription in the context of the AdS/CFT correspondence. The RT prescription says that the entanglement entropy of a sub-system $A$ in the CFT$_{d+1}$ that lives on the boundary of AdS$_{d+2}$ is given by the minimal area surface ...


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All different QKD protocols are well covered in most Quantum Information textbooks, e.g. Quantum Information by Jaeger. Trying to give a complete, understandable coverage of the matter is impossible in a post like this, so I'll just give you an overview: In the E91 scheme, entangled photon pairs are used between Alice & Bob, and unlike the single-photon ...


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The answer is definitely yes. The ground states (and low-lying eigenstates) of many-body systems are generically entangled. Examples include the ground states of local quantum field theories (which describe the fundamental particles and forces of nature) and ground states of fermionic lattice models (which describe much of the solid matter we see around us). ...


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In short, entanglement is perfectly normal. I am sure that entanglement is ubiquitous in, say, atoms with more than one incomplete subshell, as well as in some kind of organic molecules, but I am not a quantum chemistry expert and can’t provide an easy-to-realize example. Generally, any decay process produce particle states that are entangled in some way ...


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The morally correct answer is that the measurement of one spin in the EPR-entangled pair eliminates the entanglement as it picks a particular factorized basis vector for the measured spin, and the total wave function therefore has to factorize to $\psi_\text{just measured}\otimes \psi_{\rm something}$. If you parameterize the multi-fermion states in terms ...


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Our visualization horizon is limited to our 3 dynamic dimensions. We can only observe the other dimensions effects like watching TV program which detects passing non visual waves by the antenna. The same is with quantum entanglements event. We see only the result. Therefore, if results are modulated well with good sensors, we will see from other dimensions ...


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You may be thinking of the particles as idealized coins that are in identical initial states, and measuring along a certain angle as flipping a coin in a certain way. As long as the world is deterministic, they will produce identical results if flipped in the same way, and probably similar results if flipped in similar ways. That's fine, but there's nothing ...


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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 second 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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