Physics Asked by Vonofre on December 9, 2020
In this thesis, section "1.1.4 Quantum Entanglement", page 19. It is mentioned that "for mixed states, entanglement is necessary but not sufficient to ensure the violation of the Bell inequality". I’m finding it hard to understand the meaning of this statement. What I understand is that only the states that violate the Bell inequality are entangled. How can a mixed state be entangled without violating the Bell inequality?
In the thesis, there is an example of this: The Werner state $rho = p |psiranglelangle psi| + (1-p) I/4$, $pin [0,1]$ is entangled for $frac{1}{3} < p leq 1$ but violates Bell inequality only when $frac{1}{sqrt{2}} < p leq 1$.
In the case $frac{1}{3} < p leq 1$ the only quantum correlation that the system presents is entanglement. In the case $frac{1}{sqrt{2}} < p leq 1$ there is entanglement and another type of quantum correlation (quantum discord, for example). This means that entanglement will always be present in a system that has some type of quantum correlation. Is this statement correct?
I have been reading more and found the hierarchy of entanglement and quantum correlation very confusing. "Entanglement is necessary but not sufficient to ensure the violation of the Bell inequality", this means that for the violation of Bell inequality in mixed states you need quantum correlations. Is not possible to have a system with quantum correlation but without entanglement?
"for mixed states, entanglement is necessary but not sufficient to ensure the violation of the Bell inequality". I'm finding it hard to understand the meaning of this statement.
It means what it says: there are mixed states which are entangled but which do not violate the CHSH inequality. The presentation of the Werner state, as a counterexample, is all the proof that's required to show this.
What I understand is that only the states that violate the Bell inequality are entangled.
That is correct: entanglement is a necessary condition for Bell-inequality violations (i.e. the state needs to be entangled to break the inequality) but that does not mean that it is a sufficient condition.
In case the problem is that you're mixing up "necessary" and "sufficient", it helps to think about the properties "being an octopus" and "having eight legs":
How can a mixed state be entangled without violating the Bell inequality?
That's too vague of a question to give a real answer, but in general, entanglement for mixed states is substantially more complicated than it is for pure states.
Anyways, moving on:
In the case $frac{1}{sqrt{2}} < p leq 1$ there is entanglement and another type of quantum correlation (quantum discord, for example). This means that entanglement will always be present in a system that has some type of quantum correlation. Is this statement correct?
No, this is incorrect. There are mixed states which show "quantum correlations" (particularly, nonzero quantum discord) without being entangled. For a start on the details, see the Wikipedia page for quantum discord and its references.
Two notes:
More generally, the term "quantum correlations" is an extremely broad umbrella term, which covers a wide array of properties, including (i) entanglement, (ii) quantum discord, (iii) violation of individual Bell inequalities, as individual examples from a wider class. These properties are linked by a complex web of logical implications, and they're all different, so the relationship between any two aspects of that class needs to be looked at and understood separately.
Correct answer by Emilio Pisanty on December 9, 2020
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