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Hardy's paradox kills all "realistic" models of quantum phenomena

There exists a widespread misconception among the readers of popular physics books - and sometimes even unpopular physics books - that Bell's inequalities have only falsified "local realist" models of quantum phenomena but some "nonlocal realist" models could still be true and replace the probabilistic quantum mechanics.



The nice picture is taken from an article hyping "time-like entanglement" which is nothing else than the ordinary entanglement without the apparent "spooky action at a distance"

I just encountered one or two fanatical anti-quantum bigots at the Physics Stack Exchange. (One of them is a freshly graduated student from the hockey stick's and LQG's PSU.) My God, those people are so annoyingly stupid and obnoxious. I can't stand them. And they're such cowards. They would only attack my answer even though Peter Shor writes the very same thing about QM as your humble correspondent.

There exist several "toy models" or "thought experiments" that have been designed to check the difference between the classical reasoning and the quantum reasoning. Bell's inequalities showed that the correlation measured in a system with two spinning particles has to belong to a certain interval according to all local realistic theories. Quantum mechanics predicts correlations that are often outside this interval - and these correlations may be verified experimentally. It follows that the local realistic theories are falsified.




For additional years, people thought that Bell's inequalities, relevant for pairs of spins, were the most spectacular toy model demonstrating that our world genuinely deviates from the classical intuition. However, other people continued to play with entanglement and they found some other, even more spectacular thought experiments.

All of them have been made into real experiments and quantum mechanics was shown to be correct in all cases. It also means that the classical models and the classical intuition were proved incorrect in all cases.

The GHZM state, invented by Greenberger, Horne, and Zeilinger, and streamlined by Mermin, is an arrangement of three spins for which the product of three spins (taken to be +1 or -1) has exactly the opposite value than what the classical intuition would lead you to believe.

Hardy's paradox: history

Hardy's paradox, designed by Lucien Hardy in 1992, is perhaps even more spectacular than Bell's inequalities and the GHZM state because in the experimental situation, the classical logic would allow us to deduce that some events can't occur at all. Instead, they do occur.

In the original Hardy's setup, one talks about an electron and its antiparticle, a positron. Both of them may be detected in one of two detectors. However, a particular combination of detectors could only be chosen by the pair if the two particles have previously traveled along trajectories that would guarantee that the particles annihilate. Because they annihilate, they can't get to the detectors. Except that in a significant fraction of cases, they actually do end up in this pair of detectors.

A lot of confusion has been said about this experiment. People discussed specific properties of antiparticles and annihilation. However, the surprising content of this thought experiment - and real experiment - has nothing whatsoever to do with antimatter. It is just another example of the difference between the incorrect classical logic and the correct quantum logic (including the rules of entanglement).

So it doesn't matter how we interpret the physical states and the operators; in particular, we don't need any antimatter.

Hardy's paradox: presentation

I will follow Frank Laloe's excellent explanation (page 35) and I will try to make it even more comprehensible.

Consider a system of two particles which are associated with letters "a" and "b", respectively. On each particle, we may perform two kinds of measurements, either A or A' for the article "a" and either B or B' for the particle "b". To be specific, imagine that the un-primed measurements A,B are spin measurements of two spin-1/2 particles with respect to the vertical z-axis and the primed measurements, A',B', are with respect to a different axis in the xz-plane that makes angle 2δ with respect to the vertical one.

I want to train your brain a little bit more so in this article, θ=δ everywhere; thanks, Mikael. :-)



Each of these measurements may produce the result −1 or +1. There are three qualitatively different types of measurements: either both "a" and "b" are asked about their "unprimed" up/down polarization (doubly unprimed choice), or one of them is measured by the "primed" gadget and the other by the "unprimed" gadget (mixed choice), or both of them have their "primed" quantities measured (doubly primed choice).

It's up to the two experimenters which experiments they perform with the particles.

Now, as we can show, it is possible to design the apparatus and the two particles so that the following three conditions are guaranteed. Each condition constrains the results with respect to one of the three kinds of a measurement - either the "two unprimed measurements" or "mixed" or "two primed measurements". All of the conditions say something about the "+1,+1" or "-1,-1" outcome - some of them are forbidden and some of them are required to occur.

The conditions are:

  1. After the doubly unprimed measurements, A=+1 and B=+1 sometimes occurs
  2. After the mixed measurements, A=+1 and B′=+1 never occurs, and A′=+1 and B=+1 never occurs, either
  3. After the doubly primed measurements, A′=-1 and B′=-1 never occurs.
Can you prepare the particles "a" and "b" so that regardless of the two experimenters' choice of "primed" vs "unprimed" axes, all conditions above will be universally satisfied even if the experiment is repeated many times? The answer of any realistic theory, whether it's local or not, is "no". Classical logic itself is enough to show that the number of events in which all three conditions are satisfied is zero.

The first condition guarantees that it may happen that the particles are sometimes prepared to be measured as A=B=+1. However, for these cases, the second condition guarantees that B′=−1 - because B′=+1 is impossible since we already have A=+1. And similarly, classical logic implies that A′=−1 - because A′=+1 is impossible since we already have B=+1.



A diagram showing four (out of eight) possible outcomes of the two experimenters' measurement and 4 pairs of outcomes that may, must, must, or mustn't occur for the same event according to the 3 conditions.

We have just derived that in the situations in which the particles are prepared to be observed with A=B=+1, we have A′=B′=−1 which violates the third condition. So the conditions can't be simultaneously satisfied. It sounds like an impenetrable logic. Whenever you assume that right before the measurements, the pair of particles already objectively possesses some properties prepared for any experiments that the experimenters may choose to perform (and we may even allow the particles to superluminally communicate before the measurements), it follows that the three rules above are incompatible.

I want to emphasize that even if there's some nonlocal mechanism - a gadget that correlates the particles in any classical way and tells them how to behave - the classical logic will still hold and it will lead to the wrong conclusion that the conditions are incompatible. Locality isn't the problem; the classical realism is the problem.

Quantum mechanics: conditions are compatible

However, in quantum mechanics, they're totally compatible, and the compatibility has been demonstrated experimentally, too. With respect to the primed bases, the right two-particle state that satisfies all the three conditions may be written as
|Ψ⟩ = −cosθ (|A′=+1,B′=−1⟩+|A′=−1,B′=+1⟩) + sinθ |A′=+1,B′=+1⟩
where the angle between the "primed" and "unprimed" axes was chosen to be 2δ (one axis is "z" and both belong to the "xz" plane) and we consider the spin equal to 1/2. The state vector above is not normalized, so you should normalize it if you need it.

Note that one of the four basis vectors, the vector with A′=B′=−1, is omitted, so we automatically satisfy the third condition. By a proper transformation to the rotated, unprimed basis for one of the two particles, either "a" or "b", we may also check that the second condition (a pair of similar conditions, in fact) is satisfied. That's why the coefficients were chosen to be cosines and sines of θ.

However, the first condition is also satisfied because the coefficient of the A=B=+1 can be seen to be nonzero - after the simple rotation of both bases. The coefficient of the A'=B'=+1 state is sinθ in the formula above and the similar coefficient stays nonzero even after the rotation. We may choose the angle so that the probability of the classically forbidden A=B=+1 is as high as 9.0169944%, a figure that may be enhanced further if additional particles are added.

If you care, the probability of A=B=1 is M(1-M)^2/(2-M) where M is sin(θ)^2.

So quantum mechanics implies that two particles may be arranged in a state such that all three conditions - for all three types of paired measurements that the two experimenters may decide to perform - will be satisfied. Some events will follow the condition 1 despite the seemingly contradictory conditions 2 and 3. This rules out all "realistic" interpretations of reality, whether they're local or not.

Where do the quantum novelties hide?

What is different about the quantum logic that doesn't allow us to derive the classical conclusion? Well, the second condition says that the state |Ψ⟩ is annihilated by two projection operators,
P(A=+1)P(B′=+1)|Ψ⟩ = 0,
P(A′=+1)P(B=+1)|Ψ⟩ = 0.
It's also annihilated by the following operator, as dictated by the third condition,
P(A′=−1)P(B′=−1)|Ψ⟩ = 0.
Classically, we could derive that the state must also be annihilated by
P(A=+1)P(B=+1)|Ψ⟩ = 0 ... (no in QM!)
which is a projection operator relevant in the first condition. The classical logic was showed above. We could try to translate the classical logic into the quantum mechanical formalism by realizing that
P(Q=+1) = 1 − P(Q=−1)  for Q=A,B,A',B'.
Let us do it in some detail. Using the "complementarity" of the answers +1 and −1 above, the two parts of the second condition may be rewritten as
P(A=+1)P(B′=-1)|Ψ⟩ = P(A=+1)|Ψ⟩,
P(A′=-1)P(B=+1)|Ψ⟩ = P(B=+1)|Ψ⟩.
Classically, the state vector may be "cancelled" because the projection operators have "classical values" 0 or 1. So by substituting the newest two equations, we get the following for the probability of A=B=+1:
P(A=+1)P(B=+1)|Ψ⟩ = (!!!)
= P(A=+1)P(B′=-1) P(A′=-1)P(B=+1) |Ψ⟩
= 0
The sloppy classical manipulations would allow us to say that the projection operator for A=B=+1 annihilates the state because we rewrote it in a form that contained P(B′=-1)P(A′=-1)|Ψ⟩ which vanished because of the third condition.

However, quantum mechanically, the state doesn't have to be annihilated by the A=B=+1 projection operator because the step highlighted by (!!!) is incorrect. We haven't proved any identity for the projection operators themselves; we have only assumed and proved the identities of the PP=P form when they act on the state vector |Ψ⟩.

You could try to invent another quantum proof by proving that the expectation value
⟨Ψ|P(A=+1)P(B=+1)|Ψ⟩ = 0 ... (???)
vanishes. That would actually be enough to prove that P(A=+1)P(B=+1)|Ψ⟩=0 because the PP product is a projection operator that squares to itself, so the square's expectation value i.e. the norm of the state P(A=+1)P(B=+1)|Ψ⟩ would have to vanish, implying that the state itself vanishes. In this strategy to calculate the matrix element in equation (???), P(B=+1)|Ψ⟩ would be replaced by the
P(B=+1)|Ψ⟩ ⇒ P(A'=-1)P(B=+1)|Ψ⟩
product above, while ⟨Ψ|P(A=+1) would be replaced by the Hermitian conjugation of the formula above, i.e. by this
⟨Ψ|P(A=+1) ⇒ ⟨Ψ|P(A=+1)P(B'=-1)
object. By now, we have rewritten our matrix element as:
⟨Ψ|P(A=+1)P(B=+1)|Ψ⟩ =
= ⟨Ψ|P(A=+1)P(B'=-1)P(A'=-1)P(B=+1)|Ψ⟩
Now, you're "almost" finished because the matrix element includes P(B'=-1)P(A'=-1) = P(A'=-1)P(B'=-1) which "almost" vanishes due to the third condition. Well, it doesn't: we only know that it vanishes when it acts on |Ψ⟩, but to make it act on |Ψ⟩, you have to commute the operator P(A'=-1)P(B'=-1) through P(B=+1) which blocks the approach to |Ψ⟩.

So you would also fail because you would need to permute the projection operators and the projection operators P(B'=-1) and P(B=+1) don't commute with one another, and similarly the operators that measure the A,A' spins don't commute with each other, either.

So as a more explicit quantum calculation can show very easily, and as experiments confirm, the vanishing of the A=B=+1 outcome that would contradict the condition 1 doesn't follow from the conditions 2,3 in the real (quantum) world.

The characteristic fact of quantum mechanics I mentioned a moment ago - that observables don't commute with each other, not even the projection operators describing Yes/No properties of systems - guarantees that we can't derive the third displayed equation above from the previous two lines. And the reason is not our lack of skills; the reason is that the identity that someone would like to derive is actually demonstrably invalid.

The very Yes/No properties of a particle don't commute with each other. What are the implications of this statement that is self-evident in quantum mechanics but deep and confusing away from quantum mechanics? The statement implies that we can't ever imagine that a particle is ready to react to different kinds of measurements at the same moment. In some sense, it's just another form of the uncertainty principle, optimized for the projection operators and binary properties in this case.

Different properties of a physical system can't simultaneously "exist" because of the refusal of the operators to commute with each other. And because none of the observables is privileged - experimenters may measure the spin with respect to any axis - it follows that none of the observables can "exist" prior to the measurement.

Appendix: the positron setup

The original setup looked like this:



The electron "e-" and positron "e+" arrive from the bottom. The electron always moves on the right half of the picture; the positron always moves on the left half of the picture. They may only intersect - and annihilate - at the center of the picture.

The BS gadgets are 50%-50% beam splitters (with a relative "i" phase) that partly reflect and partly transmit the incoming particle, separating the wave function into pieces. The first condition, requiring that "A=B=+1" occurs sometimes, is represented by the fact that the positron and the electron are sometimes simultaneously detected in the detectors d+ and d-, respectively. So the measurements of unprimed quantities A=+1, B=+1, A=-1, B=-1 correspond to clicks in the detectors d-, d+, c-, c+.

The primed quantities A',B' are represented by whether the electron or positron took the green "v" or the blue "w" path from the bottom beam splitters. Clearly, there is no adjustable "theta" angle here: it's been set to a particular value. The equalities A'=+1, B'=+1, A'=-1, B'=-1 correspond to the electron (-) or positron (+) going through the blue or green arcs v-, v+, w-, w+. We don't measure v/w (i.e. A', B') here but we could.

Using the dictionary, the condition 2 says that v- is incompatible with d+ and v+ is incompatible with d-. It holds because the interferometer for the electron is adjusted so that the particle from e- has to end in c-, in the absence of any other particle in the w/u half-circles, and similarly for the positron e+, c+. So classically, the particles can only appear in d+, d-, instead of the ordinary c+, c-, if both of them are disturbed (phases etc.) by the other particle in the central w/u region. However, if both of them go through the central region, they have to annihilate. So classically, they can never be detected in d+, d-.

Experiments confirm the quantum predictions that in 1/16 of cases, we get a hit in d+ and d- at the same moment.

As we can see, the classical reasoning leading to the conclusion that the absence of the electrons in the red u+, u- arcs implies that the particles cannot be detected in d+, d- is incorrect. We can't imagine that the primed, "w or u arc", questions about the electron and positron can be simultaneously answered with the unprimed question "which detector, c or d". If we ultimately measure the particles via detectors, "c or d", the value of the "u or v" arc just isn't well-defined before the measure. The projection operators refuse to commute.

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reader Gustav said...

No, Hardy's paradox does not kill "all realistic models of quantum mechanics". See Physics Letters A 375 (2011), pp. 2606-2616
DOI: 10.101/j.physleta.2011.05.042