# Tag Info

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As you note, spacelike-separated (what you call "FTL") events can't be ordered unambiguously in time. That means such events can't be said to 'cause' one another in any reasonable sense. One way around this is to abandon the idea of causation between events. If events don't really cause one another, but are actually just all equally caused by some sort of ...

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then, for the "A" case, detect the two beams $a$ and $a^{\prime}$, counting $D_{a}\oplus D_{a^{\prime}}$ (where $\oplus$ is a XOR gate). Its unclear what the inputs to the xor gate are supposed to be. Usually in a delayed choice experiment you get a bunch of results from each detector (fired or didn't fire) and based on the lengths of the paths and the ...

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Entanglement is not necessarily equivalent to measurement, if the entanglement is reversible. Measurement has to do with an irreversible event, like photon absorption or emission. Until an irreversible event occurs with X or Y, the entanglement of X with Y just results in the creation of an entangled, composite, XY wave function (and composite XY density ...

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A great college-level quantum mechanics textbook is John S. Townshend's Introduction to Quantum Mechanics. It starts with spin and angular momentum, and includes an entire chapter on EPR and entanglement. It sounds like that will be your best bet. A few other options include: J. J. Sakurai, Modern Quantum Mechanics. This is a famous textbook praised for ...

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A very recent paper from Anton Zeilinger's group describes an entanglement swapping experiment with two pairs of entangled photons 143 km apart, between the islands of La Palma and Tenerife (Canary Islands). They claim an expectation value for the entanglement-witness operator that is more than 6 SDs beyond the classical limit. Remarkably, this is a ...

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I am posting these notes following a request for further information regarding this question. Should not affect the OP's choice of answer. Notes added in proof: On the meaning of quantum coherence: Quantum coherence is a direct extension of the classical concept of wave coherence. Two classical waves are said to be coherent if they can produce a ...

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Alice makes the observation of her choice. Bob makes the observation of his choice. The pair of observations has an outcome with probability distribution determined by the initial joint state of the two particles. The spacetime locations of the two observations, and the states of motion of the observers (relative to each other or anything else) have ...

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To my knowledge, Monogamy means that If particle A and B are in maximal entangled state, for example one of the Bell states, then particle A or B can not have entanglement with the third particle C. This can be extend to more particle's systems. For example, In GHZ state, the entanglement between all possible bi partitions like A-BC are equal to one (by ...

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But, what if you could create a galactic network of faster-than-light transmitter/receiver hubs using Quantum Entanglement? Say you have four pairs of quantum particles and you separate them in 8 "jam jar" nodes. There are four local broadband digital devices which can be connected to two nodes at a time and work at the normal speed of light. AB CD EF GH ...

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Remembering that entanglement is a property of the global quantum state, the statement "one of the entangled objects is at some finite temperature" makes no sense without some additional information. Either the local (marginal) states of $A_1$ and $A_2$ are thermal, or the global state is thermal, but not both. If the two objects are called $A$ and $B$, ...

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You can represent $\rho_{AB}$ by $tr_B(\rho_{AB})\otimes tr_A(\rho_{AB})$ only if the two subsystems are not entangled, i.e. $\rho_{AB}=\rho_A \otimes \rho_B$.

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I tried to ask you to clarify the problem so I'll assume a strong external classical field. The Hamiltonian for a magnetic moment $\vec \mu=\gamma \vec S$ (where $\gamma$ is the gyromagnetic ratio) in an external magnetic field is $$H=(P_x^2+P_y^2+P_z^2)/2m-\vec \mu \cdot \vec B$$ Which assumes no other interactions, such as electric charge or electric ...

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This one trial might be considered one instance of an ensemble of $N$ trials whose quantum state is an entangled state as described above, with $c_{(j, k)} \ne 0$. It not about considering. If $|\psi\rangle := \sum_{j = 1}^{N_A}~\sum_{k = 1}^{N_B}~c_{(j, k)}~|\phi_j^A\rangle \otimes |\phi_k^B\rangle,$ is the state it was in and then you did a ...

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You asked three questions. My answers focus on those questions without attempting to introduce the notation of quantum states. As described below, I show in a Wikiversity article that the behavior of entangled particles can be modeled with virtually no quantum mechanics, except that the concept of the "photon" is required. Have I misunderstood a ...

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I thought entanglement involved sharing a similar (or the same?) wave function. Nothing could be farther from the truth. I could entangle the polarization of a photon with the position of an electron. Or entangle the energy of an electron with the direction of propagation of a photon. Its really just superposition. A state with an electron on the left ...

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It is important to make a distinction between the "entanglement" they are talking about, and the entanglement that occurs in quantum mechanics. The way I see it, the usual definition of entanglement is purely quantum, in virtue of explicitly referring to the Hilbert space structure of quantum mechanics. An entangled state is, by definition, a state for which ...

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The reason your argument is wrong is that one could imagine something like the following: We create an entangled pair of electrons, say in state $UU-DD=(U+D)(U-D)+(U-D)(U+D)$. You put one in your pocket (unobserved), I put one in my pocket (unobserved) and we travel to distant places, arriving at noon on January 1, 2016. At that moment, you decide to ...

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The claim "Entangled electrons share the same quantum state" is not correct. In an entangled state there is no well-defined notion of the states of the individual components, this is the very definition of an entangled state: A composite state $\chi\in\mathcal{H}_1\otimes\mathcal{H}_2$ is called entangled, if it cannot be written as $\chi=\psi\otimes\phi$ ...

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Firstly, entanglement isn't magic. And it is next to meaningless to just say something is entangled. Entangled just means not factorizable. But is the lack of factorizability from the spatial degrees of freedom? From the polarization degrees of freedom? Is it super close to factorizable? Is it maximally entangled? Just saying it is entangled isn't really a ...

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