What is the physical difference between maximally mixed and maximally entangled bipartite states?

Asked Jan 19 '21 at 18:09

Active Jan 24 '21 at 16:24

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If we have two bipartite quantum states, one maximally mixed (2) and the other maximally entangled(1):

$ |\psi_{1}\rangle =1/\sqrt{2} ( |00\rangle+|11\rangle) \rightarrow \rho_{1} = |\psi_{1}\rangle \langle \psi_{1} |$
$ \rho_{2} = 1/2 (|00\rangle \langle 00 |+|11\rangle \langle 11 |)$

In both cases, if we measure one of the qubits, it seems like that the other one would be determined too. In the first case, entanglement does the work for us, and in the second case, we have prepared the ensemble in such a way that this happens. The question I have is that what is the physical difference between these two systems? they seem to behave in the same way in the experiment.

edited Jan 24 '21 at 16:24

Qmechanic

201,751

asked Jan 19 '21 at 18:09

Sobhan

1

Try measuring both spins in another basis! – Norbert Schuch Jan 19 '21 at 18:11
@NorbertSchuch so you say if I measure in another basis, I would always get the same results for both qubits for the maximally entangled state but not the same for the mixed one? – Sobhan Jan 19 '21 at 18:16
Not for this maximally entangled state, but for a singlet you would always get the opposite outcome. – Norbert Schuch Jan 19 '21 at 18:52
related on qc.SE: What is the difference between superpositions and mixed states?, https://quantumcomputing.stackexchange.com/q/1365/55, and links therein. – glS Jan 24 '21 at 14:19

What is the physical difference between maximally mixed and maximally entangled bipartite states?

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