Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 1:14 amby Joy Christian
This may be one of the most significant papers in the foundations of quantum mechanics I have seen in a very long time:
http://arxiv.org/abs/1509.03641:
Abstract:
The interpretation of quantum theory is one of the longest-standing debates in physics. Type-I interpretations see quantum probabilities as determined by intrinsic properties of the world. Type-II interpretations see quantum probabilities as not directly dealing with intrinsic properties of the world but with relational experiences between an observer and the world. It is usually believed that deciding between these two types cannot be made simply on purely physical grounds but it requires an act of metaphysical judgement. Here we show that, although the problem is undecidable within the framework of quantum theory, it is decidable, under some assumptions, within the framework of thermodynamics. We prove that type-I interpretations are incompatible with the following assumptions: (i) the decision of which measurement is performed on a quantum system can be made independently of the system, (ii) a quantum system has limited memory, and (iii) Landauer's principle is valid. We consider an ideal experiment in which an individual quantum system is submitted to a sequence of quantum projective measurements that leave the system in pure quantum states. We show that in any type-I interpretation satisfying (i)-(iii) the system must reset its internal state, which implies that a minimum amount of heat per measurement has to be dissipated into the system's environment. We calculate a lower bound to the heat dissipated per measurement assuming that the measurements are chosen from a set of size 2^n. Then, we show that this lower bound becomes infinite in the limit of n tending to infinity. This leads to the conclusion that either type-I interpretations are untenable or at least one of the assumptions (i)-(iii) has to be abandoned.
Re: Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 5:46 amby jreed
I agree. If correct, this could answer some long standing questions about the interpretation of quantum mechanics. Just by coincidence, there's an article in Physics Today, September 2015: "Information: From Maxwell's demon to Landauer's eraser" that explains Landauer's principle.
Re: Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 7:52 amby Ben6993
A pair of entangled particles share a pair of opposite spin states. Under a hidden variable model, the particles are not entangled but each have opposite hidden spin states. If measured in the same direction they will have opposite spins measured and Alice and Bob will feel equal thermodynamic heat. If they are not measured in the same direction then they may not necessarily have opposite spin states when measured. Anyway, Alice and Bob gain heat from measurement independently of each other.
Under QM, the system does not know the states of Alice and Bob's particles, only knowing that the states are opposite. Assuming first that the two measurements are made in the same orientation, if Alice and Bob measure at exactly the same time then they share the measurement heat equally. If Alice measures before Bob, does Alice feel all the measurement heat and Bob gets none, because as soon as Alice makes a measurement, the state of Bob's particle is known before Bob makes the measurement?
Under QM, if the measurements are made in different orientations and Alice measures before Bob, knowing the outcome of Alice does not predict the outcome for Bob as they could be -- or ++ and not necessarily -+ or +-. So is the thermodynamic heat shared equally between Alice and Bob because the measured outcomes are not the simple opposite entangled states shared under QM. Or does Alice get all the heat again because the entanglement information is resolved on Alice's measurement. But is the entanglement spin states of |+ , -> ever resolved if the actual measurements of Alice and Bob were say ++.
Re: Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 12:17 pmby Joy Christian
The concluding sentence in the paper that interests me is the following:
However, if no heat is radiated, this would support those interpretations in which quantum probabilities are not determined by intrinsic properties of the world.
It would be interesting if no heat radiation is observed in the experiment.
Re: Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 2:26 pmby FrediFizzx
Joy Christian wrote:The concluding sentence in the paper that interests me is the following:
However, if no heat is radiated, this would support those interpretations in which quantum probabilities are not determined by intrinsic properties of the world.
It would be interesting if no heat radiation is observed in the experiment.
Doesn't that statement run contrary to the second law of thermodynamics?
Re: Thermodynamical cost of quantum mysticism
Posted:
Tue Sep 15, 2015 2:53 pmby Joy Christian
FrediFizzx wrote:Joy Christian wrote:The concluding sentence in the paper that interests me is the following:
However, if no heat is radiated, this would support those interpretations in which quantum probabilities are not determined by intrinsic properties of the world.
It would be interesting if no heat radiation is observed in the experiment.
Doesn't that statement run contrary to the second law of thermodynamics?
No, it does not, if the quantum state of the observed system does not represent any intrinsic property of the observed system. Their main point is that if the quantum state merely represents the knowledge or expectations of an external observer, then his / her measurement does not cause heat emission from the observed system.