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What do you mean with "quite differently"? The answer to your first question - does this not mean that the whole universe must be a mess of entanglements would be "yes", that's what it means. But, it seems that entangled particles cannot be used to transmit information without also having a classical information channel available, which is bound by ...


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Even though you think of it as a single particle -- each of it's different properties like momentum, spin, etc (corresponding to each valid quantum number) sits in a Hilbert space of their own and the possible configurations of a particle sits in a tensor product of those Hilbert spaces. $$\mathcal{H_{particle}} = \mathcal{H_{momentum}} \otimes ...


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Treat this factoring problem just like you would factor an algebraic expression. In this case, your expression is equivalent to $AX + AY + BX + BY$, so factor it similarly: $$\begin{aligned} |\Psi\rangle_{12}&=\frac12(|+_z\rangle_1|+_z\rangle_2+|+_z\rangle_1| -_z\rangle_2+|-_z\rangle_1|+_z\rangle_2+|-_z\rangle_1|-_z\rangle_2) \\ &= \frac12 ...


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It's not exactly what you ask for, I guess, but there is a quite recent proposal by two very renowned physicists linking wormholes and entanglement: Cool horizons for entangled black holes by Maldacena and Susskind. A more accessible review of the idea can be found here: http://quantumfrontiers.com/2013/06/07/entanglement-wormholes/ Unfortunately, my ...


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Apologies to SE for the fact that I am reusing parts of an answer I wrote earlier, but to answer the core question of why you can't assume the system has a particular state before measurement I think this is the easiest example to think of. Consider two entangled quantum systems which are described by three properties, A, B and C each which can take a ...


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Lubos's answer is excellent, but a simpler answer is that after you measure the particles, they're not entangled anymore. If the first is known (after a measurement) to be in some state A and the second is known to be in some state B, then (quite regardless of whether you measured the same or different observables), the pair is in state $A\otimes B$, which ...


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Consider two entangled quantum systems which are described by three properties, A, B and C each which can take a value of up or down. The two systems can be entangled such that if you measure the same property on both systems then you will get the same result 100% of the time, and each case occurs 50% of the time. That is to say both systems give up 50% of ...


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I would just like to emphasize that the word "completely" in the question makes a big difference. If a state A is maximally entangled with a state B, then A cannot be maximally entangled with a third state C as well. This is known as the "monogamy of entanglement". In general, the more states something is entangled with, the less the entanglement will be. ...


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Once all is known we have an entropy of 0. Of course we would deal with a continuous function for waves, but am I on the right track of thought? Information entropy is not the same concept as thermodynamic entropy. Information (Shannon) entropy describes degree of lack of knowledge of some system given probabilities for its states. Thermodynamics ...


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The reduced matrix is defined as the partial trace of the density matrix. Be $A$, $B$ finite dimensional Hilbert spaces and be $T$ $\in$ $L(A \otimes B)$ (Linear operators on $A \otimes B$), then the partial trace of T is defined as $\rm{Tr}_B [T]$ in $L(A)$ is defined by \begin{equation} \langle a | \rm{Tr}_B [T]| b \rangle = \sum_n \langle a | \langle n ...


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The reduced density matrix can be found by taking the trace over the subspaces of the Hilbert space that represent systems you're not interested in. For the Bell state the density matrix of the whole system is $$\tfrac{1}{2}(|00\rangle+|11\rangle)(\langle 00|+\langle 11|)\\ = \tfrac{1}{2} (|00\rangle\langle 00|+|00\rangle\langle 11|+|11\rangle\langle ...



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