Is a bipartite system necessary for the proof of the PBR theorem?

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SUMMARY

The discussion centers on the necessity of a bipartite system for proving the PBR theorem, particularly in relation to the example provided by @Demystifier. Participants express confusion over whether the requirement for two systems is a technicality or a fundamental aspect of the theorem. While Hardy's alternative proof does not rely on a bipartite system, it introduces an unclear technical assumption. The conversation highlights the distinction between entangled states, as seen in Bell's theorem, and the product states referenced in the PBR theorem.

PREREQUISITES
  • Understanding of the PBR theorem and its implications in quantum mechanics
  • Familiarity with bipartite systems and their role in quantum entanglement
  • Knowledge of Bell's theorem and its significance in no-go theorems
  • Basic grasp of quantum measurement theory, particularly the Bell basis
NEXT STEPS
  • Research the implications of bipartite systems in quantum mechanics
  • Study Hardy's proof and the additional assumptions it entails
  • Explore the relationship between entangled states and product states in quantum systems
  • Investigate the measurement techniques used in the Bell basis and their relevance to quantum contradictions
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Quantum physicists, researchers in quantum mechanics, and students studying the foundations of quantum theory will benefit from this discussion.

greypilgrim
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Hi.

I'm trying to grasp what the PBR theorem is about. I'm not tackling the full version, but rather the simple example in @Demystifier's summary.

While I think I understand the mathematical steps, my question is why you need two systems to prove it. Is this only technical or more fundamental?

I mean it's not that surprising for a no-go theorem to make use of a bipartite system, but the crucial thing about that, such as in Bell's theorem, usually is that those systems are entangled. Here they just seem to be in a product state.
 
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greypilgrim said:
my question is why you need two systems to prove it. Is this only technical or more fundamental?
It's not clear. Hardy found a different proof without assuming two systems, but he used an additional technical assumption the physical meaning of which is not entirely clear.
 
greypilgrim said:
I mean it's not that surprising for a no-go theorem to make use of a bipartite system, but the crucial thing about that, such as in Bell's theorem, usually is that those systems are entangled. Here they just seem to be in a product state
The measurements used for the contradiction are in the Bell basis though.
 

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