Incompatibility is more fragile than entanglement

Yes, that’s in a nutshell the conclusion that we have found in the three recent studies arXiv:1504.05768, arXiv:1505.00833 and arXiv:1508.04612.

But wait, does it even make sense to compare a property of collections of measurements to a property of states of a composite systems? Isn’t it as misleading as comparing Finnish vodka to Slovak borovicka?

The basic idea in our comparison is simple: a quantum channel can be seen alternatively in the Schrödinger picture or in the Heisenberg picture. Consider, for instance, the following kind of experimental setting.

correlation experiment

Here the red thing is a preparator that produces bipartite systems, blue tubes are quantum channels and on each side there is a collection of three measurement devices. For each run, both observers choose which measurement devices they use. An experimental run goes like this:

preparointi putkeenputkestaledit

Observers are collecting measurement data, and in the end they have a correlation table which contains the information about their chosen measurements and the results they obtained. To violate a Bell inequality, the bipartite state must be entangled and the measuremens A, B, C must be incompatible. (And the same for A’, B’, C’.)

Let’s ignore the right hand side part of the picture and think about the situation from the perspective of the observer on the left. Now we come to the fact mentioned earlier: the blue channel can be seen alternatively in the Schrödinger picture or in the Heisenberg picture:

Schrödinger Heisenberg

Now, if the blue channel is very noisy, it certainly spoils our correlation experiment. Hence, we want to know how much noise our experimental setting tolerates. Looking at the channel in the Schrödinger picture, we are worried that the state becomes separable. Meanwhile, looking at the channel in the Heisenberg picture, we hope that the collection of measurements does not become compatible. (Note, however, that these are just necessary criteria which are not enough to violate Bell inequality.)

I hope you are now convinced that it can make sense to compare noise tolerances of incompatibility and entanglement. Please read more from our arxiv papers!

(Thanks to Daniel Nagaj for his comments.)





About heinosaari
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