# Tag Info

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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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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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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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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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This question was cross-posted to physics forums word for word. I'll give the same basic answer I gave there. Consciousness is never part of any quantum mechanical explanation. Every experiment runs the same whether or not a person is in the room. Retrocausality is also not required here. For example, the Copenhagen interpretation explains the delayed ...

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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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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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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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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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There is no communication because the sender has no way of influencing the eigenvalue i.e. they can influence the collapse, but they cannot influence the subsequent state of "our" entangled particle, and this not the state of the particle on Alpha C

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Am I missing an important detail in my understanding of how the delayed choice quantum eraser experiment is done? Yes. I'll respond to your list of your understanding. How does one account for what takes place in the experiment without using the concept of "retrocausality" (effect before cause)? This question doesn't make sense, if ...

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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 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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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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