## I was wrong and Lubos Motl was right

## but quantum correlations are not like Bertlmann's socks

It finally happened. I was too careless and I

**stupidly**challenged Lubos to*"Show me a non-factorizable state in CM for spatially separated physical systems!!!"*
And he indeed presented one:

*"Bertlemann's socks? The state is described by the probability distribution that has 0%,50%,50%,0% for left-and-right-green, left-green right-red, left-red right-green, left-and-right-read. The state, P(colorLEFT, colorRIGHT) isn't factorized. "*

This is obscurely written but it is ultimately correct. Glup, glup, glup, he sank my battleship, bruised my ego, won the battle, but not the war :)

**I will attempt to show that quantum correlations are**__not__like Bertlmann's socks correlations because**quantum correlations depend in an essential way on the observer.****First let me clarify Lubos example and why he is right in his example.**
First what is this business of Bertlmann's socks? This came from a famous Bell paper:

Mr Bertlmann is a real person who uses to wear socks of different colors. The moment you see that one of his socks is pink you know that his other sock is not pink. (Lubos is using red and green) and there is a correlation between the sock colors. In general, you learn from statistic 101 that a joint probability \(P(A, B)\) factorizes: \(P(A,B)=P(A)P(B)\) if the two probabilities are independent. Therefore it was blatantly stupid on my part to ask Lubos for a non-factorizable example. But I had something else in mind for which I carelessly skipped the essential details and now I will explain it properly.

The Bell theorem factorization condition is not on independent probabilities but on

__residual__probabilities:
\(P(A,B, \lambda) = P(A, \lambda) P(B, \lambda)\)

In the case of Bertlmann's socks there are no residual probabilities! But what does this mean?

Correlations can be generated due to physical interaction between two systems or because they share a common cause. One way people explain this is by inventing silly games where two players

**agree in advance on a strategy**but they are not allowed to communicate during the actual game. For example consider this: Alice and Bob each have a coin and they flip it heads or tails. Then they both have to guess the result of each other and they win the game when successful. If they guess randomly the best odds of wining the game is 25% of the time. However there are two strategies they can agree beforehand which increases their odds of winning to 50%. (Can you guess what those strategies might be?).**Strategies, common causes and interactions, outcome filtering, all generate correlations and ruin the factorization condition.**

**But quantum mechanics is probabilistic and even if you account for all interactions the outcome is still random.**After accounting for all those factors in the form of a generic variable \(\lambda\), the remaining probability is called a

**residual probability**. Lambda can be a variable, or a set of variables of an unspecified format.

**The point is that after accounting for all common causes, if the physical systems A and B are spatially separated, they cannot communicate and in the quantum case it seems reasonable to demand**\(P(A,B, \lambda) = P(A, \lambda) P(B, \lambda)\).

But such a factorization is at odds with both quantum mechanics predictions, and experimental results.

__This is the basis on which people in foundations of quantum mechanics call nature nonlocal.__

I think I know how Lubos may attack this. I bet he will say that \(\lambda\) is basically a hidden variable and quantum mechanics does not admit hidden variables. This line of argument is faulty. Carefully read (several times) Bell's paper from the link above and you see that \(\lambda\) represents the coding of the usual way correlations are accounted by this common \(\lambda\) which is present in all 3 factors.

Let me spell out better Bell's argument following a well written paper by Bernard d'Espagnat. Bell considers the singlet state and in there you have Alice and Bob in two spatially separated labs measuring the spins on directions a, and b and obtaining the outcomes A, and B respectively, Let \(\lambda\) represent a common source of the correlation between A and B. Then one can write the standard rule of statistics: \(P(M,N) = P(M|N)P(N)\) like this:

\(P(A,B|a,b,\lambda) = P(A|a,b,B,\lambda)P(B|a,b,\lambda)\)

then

**because what happens at Alice's side does not depend on what happens on Bob side and the other way around**:
\(P(A|a,b,B,\lambda) = P(A|a, \lambda)\)

\(P(B|a,b,\lambda) = P(B|b, \lambda)\)

yields:

\(P(A,B|a,b,\lambda) = P(A|a, \lambda) P(B|b,\lambda)\)

From this the usual Bell theorem follows and disagreement with experiment is used to point out that

**what happens at Alice's side does depend on what happens on Bob side**. In other words, nonlocality.
I disagree (with good arguments) with several points of view:

- I disagree with Lubos that nature does not follow the logic of projectors and follows the Boolean logic instead. (measurements project on a Hilbert subspace and quantum OR and quantum NOT are different than their classical counterparts)
- I disagree with Tim Maudlin who best defends the point that Bell proved nonlocality based on the argument above (
**Quantum mechanics is contextual and because of that:**\(P(A|a,b,B,\lambda) \ne P(A|a, \lambda)\)**.**As such**Bell's factorization condition is not justified.**)__Only if you think in terms of classical Boolean logic avoiding contextuality the nonlocality conclusion is inescapable.__ - I disagree with Lubos that quantum correlations are like Bertlmann's socks. To eliminate thinking about lambda as a hidden variable, picture it as fixing all conceivable sources of correlations (Bertlmann's socks type or any other type). Now add locality independence, Boolean logic, and you get correlations at odds with experiments. Pick your poison: give up locality or give up Boolean logic. The one to give up is Boolean logic.
**Unlike Bertlmann's socks correlations, quantum correlations depend in an essential way on the observer.** - I disagree with giving up on locality and that the Bohmian position represents a valid description of nature (what happens with he quantum potential of a particle after the particle encounters its antiparticle? Both vanish or not vanish result in predictions incompatible with observations)

Bell himself provided 4 possible explanations of quantum mechanics' correlations:

-quantum mechanics is wrong sometimes

-superdeterminism (lack of free will)

-faster than light causal influences

-non-realism

My take on this is that quantum mechanics is the complete and correct description of nature, there is free will, there are no faster than light causal influences, realism is incorrect, and what people call nonlocality is actually a manifestation of contextuality because the observer (but not consciousness) does play an active role in generating the experimental outcome. This active role happens even when parts of the composed system are out of causal reach because quantum mechanics is blind to space-time separations.

Dear Florin, thanks for the admission. The whole situation "classical Bertlmann's socks" and "two entangled spins" are obviously not the same because quantum mechanics and classical physics aren't the same.

ReplyDeleteBut all the difference between the two situations boils down to the nonzero commutators in QM. The nonzero commutators in QM are what makes the observer matter - and irremovable. If the observer measures L, then he collapses the system to an eigenstate of L, and that's why all the measurements of M which doesn't commute with L will be predicted with different probabilities than if he hadn't previously measured L.

The socks may in practice only be measured as classical bits - a counterpart of J_z. But the spins may also be measured when it comes to their J_x and J_y and other combinations and functions. What's new is that there are special new - non-diagonal - operators, those that don't commute with the "normal bit".

In the entangled state, nothing is changed between "socks" and "spins" when it comes to the moment when the correlation/entanglement is created, and what its consequences are given one particular set of final measurements.

Dear Lubos, I am not after gaining your approval, but when I am wrong basic scientific integrity demands that I accept I am wrong.

DeleteThank you for saying "classical Bertlmann's socks" and "two entangled spins" are obviously not the same because quantum mechanics and classical physics aren't the same. I guess we are in agreement then.

I do not deny noncomutativity is at the root of the difference between QM and CM, all I am saying is that the story is much more subtler. From category theory there are three kinds of theories possible: QM, QM, and a nonphysical (for a different reason) hyperbolic QM. Both QM and hyperbolic QM are noncomutative in the very same way. Hyperbolic QM lacks positivity and when positivity enters in the mathematical formalism it generates all kinds of highly nontrivial consequences.

Otherwise what I find remarkable is that you still don't understand that the world isn't realist. The probabilities of the type P(A,B,lambda) that your whole blog post is all about simply don't describe the most general probabilities according to quantum mechanics.

ReplyDeleteFirst, there are no lambdas. Second, it is not enough to have the probabilities for the "basic points of the phase space" because there are no points in the phase space according to quantum mechanics. Instead, for every observable A, i.e. every Hermitian matrix on the Hilbert space, there has to be a P(A). But these values of P(A) aren't independent - they can't be chosen separately for every value of the matrix A. Instead, in QM, only P(A) that may be written in the form Tr(rho*A) are allowed.

So your - and Bell's - analysis is completely worthless for studying the world around us because it assumes quantities P(A,lambda)... that simply aren't well-defined according to the actual, quantum i.e. non-realist, laws of physics. That's also why none of your claims about nonlocality can be trusted and why almost all of them are blatantly wrong.

"I find remarkable is that you still don't understand that the world isn't realist." Huh?

Delete"The probabilities of the type P(A,B,lambda) that your whole blog post is all about simply don't describe the most general probabilities according to quantum mechanics."

I predicted this reaction: " I bet he will say that λ is basically a hidden variable and quantum mechanics does not admit hidden variables."

The way to think about it is in the form of residual probabilities after fixing all factors which can cause correlations. The next section is unclear so I'll skip commenting on it.

"So your - and Bell's - analysis is completely worthless for studying the world around us because it assumes quantities P(A,lambda)... that simply aren't well-defined according to the actual, quantum i.e. non-realist, laws of physics. "

This is incorrect. Forget about Bell and anything else. Think like an experimentalist: you have experimental results and you want to attach a probability to them after accounting for all factors which you can think of. That probability is a number which I can call whatever I want including P(A, lambda). The problem is not that I cannot define such quantities-they are perfectly clear in the mind of the experimentalist (think Bayesian and the experimentalist's degree of belief in the outcome A given the entire experimental setup), the problem is on the validity of demanding the factorization condition. It is clear that QM does not satisfy Bell factorization. The nonlocality claim comes from assuming the validity of the factorization + its experimental refutation. Bell contends the factorization is acceptable, I say it is not because of contextuality. CM is noncontextual and for it Bell factorization is perfectly OK. Nature and QM are contextual and there is no basis for demanding the factorization condition. Hence the nonlocality conclusion is not valid.

So basically my point is the following: Bell demands the factorization and experimental violations prove that nature is not factorizable as required by Bell. From this people imply nonlocality. But I say the factorization condition is not justified in the case of QM and so nonlocality does not follow as a consequence.

You may define probabilities for any observation A, P(A), but their total number is higher than in classical physics, they are predicted by a totally different procedure than in classical physics, and their changes don't mean the same thing as in classical physics.

DeleteTheir number is much higher because for every projection operator (matrix) Q, one has to have a P(Q). Also, when a measurement of some L is made, not only P(L) changes but P(M) changes for all M that have a nonzero commutator for L. Nevertheless, this change may be shown to imply no action at a distance - no influence on observables N that do commute with L. The predictions for N separately do change but their change follows from the pre-existing correlation between L,N.

I forgot to mention perhaps the most important novelty that arises if you try to use the P(A,B,lambda) labels even in quantum mechanics.

DeleteYou may use them but the fun is that the corresponding values of the probability distribution are allowed to be negative - only in small regions relatively to the "unit cell of the phase space", like in Wigner's distributions. Because of this "relaxation" of the inequality "probability density is greater than zero", one cannot prove Bell's inequalities etc.

On the other hand, the pointwise negativity of the quasi probability distribution doesn't imply any real negative probabilities because for any real experiment, the probabilities are being averaged at least over one cell of the phase space, and in those regions, the averages are already non-negative. Again, it's the uncertainty principle that allows for the deviation from classical physics to remain consistent.

Lubos, you are fishing for information here. There are no Wigner's distributions in Hilbert space. They correspond to the phase space formulation of QM. In there you encounter quasi-probabilities which can be in some parts negative. In this case there is no such thing as a Bayesian interpretation.

DeleteIn phase space once you establish the C* algebra condition (which does require a bit of effort to do, but it is not terribly hard) then you can apply the GNS construction and arrive at the standard QM formulation in Hilbert space.

I haven't said anything about "distributions on the Hilbert space" and you know it.

DeleteOtherwise it's nonsense to say that I am "fishing". This is just about foundations of QM. I have perfectly understood them before I was 17. It's a closed subject, settled knowledge since the mid or late 1920s. It's only retarded people who believe to "research" it when they are adults.

ReplyDeleteIt is good to be enlightened that all this quantum nonlocality work and the experiments that support them are nonsense. The wunderkind who knew quantum foundations at 17 and also upends the science of climatology conquers the world again!

DeleteThe reason you like Donald Trump is that while you are clearly a lot more intelligent than him, your personality and emotional intentional stance is much the same.