Measurement problem and computer-like functions

[jk22]

Suppose we define the measurement of an observable A by v(A) v being an 'algorithm giving out one of the eigenvalues each time it is called' (we accept the axiom of choice)

In this context we have in particular v(A)≠v(A) since when we call the left hand side and then the right handside the algorithm could give different values.

Is this not a way out from Bell's theorem since one cannot factorize the measurement results ?

But is this not at the same time the end of logical writing we are used to in maths ?

 

[bhobba]

You have defined a deterministic observable - it has nothing to do with Bell.

Thanks
Bill

 

[jk22]

Since we have v(A) different from v(A) then the expression v(A)v(B)+v(A)v(B') is not equal to v(A)(v(B)+v(B')) and Bell's theorem do not apply ?

 

[bhobba]

Since we have v(A) different from v(A) then the expression v(A)v(B)+v(A)v(B') is not equal to v(A)(v(B)+v(B')) and Bell's theorem do not apply ?

Bells theorem proves - there is no escaping it - if QM is correct you can't have both locality and reality. All you have done is define some kind of deterministic observable which has nothing to do with bell.

Here is the proof of Bells theorem:
http://www.johnboccio.com/research/quantum/notes/paper.pdf

If you think you have found an out specify exatly which step is in error.

Thanks
Bill

 

[jk22]

There is no error my aim is to understand how quantum mechanics violates the inequality. There is no escape from Bell derivation.

The problem local or global is not an issue we are used in physics to have global/local correspondances like Euler lagrange equations, Maxwell equations have local or global formulations.

Leggett showed that nonlocality and realism does not correspond to quantum neither.

Hence we conclude that it is not local or global the problem, but the realistic hypothesis.

But what does realistic mean ? I thought it was that elements of reality can exist independently of observation ?

 

[bhobba]

\But what does realistic mean ? I thought it was that elements of reality can exist independently of observation ?

Technically counter-factual definiteness as defined in my linked paper.

Thanks
Bill

 

[jk22]

Does this mean that we could have known the result without having actually performed the measurement ?

Could we say that this were equivalent to say : it is possible in principle to collect enough data to be able to predict any result, data which is coded in the variable ?

 

[bhobba]

Does this mean that we could have known the result without having actually performed the measurement ?

Of course not. QM says that's impossible.

Could we say that this were equivalent to say : it is possible in principle to collect enough data to be able to predict any result, data which is coded in the variable ?

Again QM says that's impossible.

May I suggest a bit of reading about QM:
https://www.amazon.com/dp/0465062903/?tag=pfamazon01-20

Having a look at some of your other posts your math background may be enough to follow the following which axiomatically is the essence of QM (see post 137):
https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

As you can see right at its foundations QM is probabilistic.

Thanks
Bill

 

[jk22]

What disturbs me is that Bell's theorem is also valid when we use probabilities instead of measurement results. Using probabilities would make in some way that the result is no more defined counterfactually.

 

[bhobba]

What disturbs me is that Bell's theorem is also valid when we use probabilities instead of measurement results. Using probabilities would make in some way that the result is no more defined counterfactually.

Sure. QM is two things. First an extension to probability theory. Secondly a theory about that extension applied to observations.

So?

Thanks
Bill

 

[stevendaryl]

Bells theorem proves - there is no escaping it - if QM is correct you can't have both locality and reality. All you have done is define some kind of deterministic observable which has nothing to do with bell.

Here is the proof of Bells theorem:
http://www.johnboccio.com/research/quantum/notes/paper.pdf

If you think you have found an out specify exatly which step is in error.

Thanks
Bill

I don't think that there is any error in that paper, but I think that it can be misleading to replace, as the author does, the assumption of realism with the assumption of "counterfactuality" (counterfactualness?). The usual definition of "counterfactual" is that the counterfactuals are the answers to questions of the form "If I had (counter to fact) used device setting X instead of Y, what result would I have obtained?" The assumption that such questions have definite answers is usually called (though not in this paper) the assumption of "counterfactual definiteness". For a theory to be counterfactually definite, measurement results have to be a deterministic function of the values of physical variables describing the system being measured and the measuring devices. The author tacitly makes this assumption by conflating the measurement results A, B, and C with the "hidden variables" describing the system's state prior to being measured.

Of course, in the case of coins, which the author uses to illustrate the principle, this conflation is natural: You're not going to observe that a coin is gold (for example) unless it was already gold before you looked. However, conflating the hidden variables (the author doesn't use this term, but I think his "counterfactual properties" is the same thing as what people usually mean by local hidden variables) with the measurement results (or more precisely, assuming that measurement results are a deterministic function of the hidden variables) gives the misleading impression that nondeterminism is a loophole to Bell's theorem. What's more general than assuming counterfactual definiteness is to assume that the outcome of a measurement is a random variable, with probabilities influenced by the variables describing the system and the measurement device. However, allowing this extra generality doesn't change the conclusion: No local realistic theory, counterfactual or not, can reproduce the predictions of QM for the EPR experiment.

So the author's focus on counterfactuality unnecessarily weakens the conclusions of Bell's proof.

 

[jk22]

Maybe the conclusion to this theorem should be : This local realistic theory cannot reproduce all predictions of qm ?

There is also the excess ontological baggage : the hidden variables were out of reach if it is that what is meant ? http://www.sciencedirect.com/science/article/pii/S1355219804000048

 

[DirkMan]

Here is the proof of Bells theorem:
http://www.johnboccio.com/research/quantum/notes/paper.pdf

If you think you have found an out specify exatly which step is in error.

"The terms in equation (1) refer to three measurements on the same set of coins. The terms in equation (4) refer to measurements on 3 separate disjoint sets of coins.

In probability theory, whenever you add and subtract probabilities, the expression is only meaningful if all the probabilities are from the same sample space. While you can guarantee that a set of triples of measurements A,B,C each on a pair of coins will be able to generate P(A,B), P(A,C) and P(B,C) that are from the same sample space, there is no way to guarantee that the same can be true for 3 separate measurements in which you only measure A,B on one set of coins, A,C on another set of coins and B,C on yet a different set of coins."

 

[bhobba]

That doesn't matter - the equality holds regardless. I can take a number of sample spaces and add the probabilities of events in those spaces up to get a number. It turns out to be 3/4 which violates Bell. indeed they can't be from the same sample space because equation 1 says it can be >= 1.

However your concerns have shifted from the measurement problem and computer like functions to issues with the proof of Bells theorem. I suggest you start a new thread about it. We have a number of regular posters here such as Dr Chinese that are experts in it and they are best positioned to address your concerns.

Thanks
Bill

 

[jk22]

I think the point to understand how qm violate CHSH for example is that we can't write $$A(\theta_A,\lambda)$$ see https://en.m.wikipedia.org/wiki/Bell's_theorem For the result but it should be $$A(\theta_A,\Psi)$$

Indeed in qm we don't have lambda only psi. Since the wavefunction does not in general determine the result we can see the violation arising.

 

[bhobba]

I don't quite understand what you are saying.

However what's going on is well known. QM is the simplest extension to standard probability theory that allows continuous transformations between pure states:
http://arxiv.org/pdf/quant-ph/0101012.pdf

It turns out due to that extension allowing entanglement you get a different type of correlation than classical probability theory. If you want it like classical probability theory with properties independent of observation then you need to violate locality. However if you accept nature as is and say, for example, locality isn't even a valid concept of correlated systems, then there is no issue - we have a different kind of correlation - big deal.

As I often say about QM the real issue is simply letting go of classically developed intuition about how the world is. Simply accept QM as it is. We have met the enemy and he is us - Pogo.

Anyway this has gone way off topic. What you proposed in your original post won't resolve anything. If you want to discuss general QM issues best to start other threads.

Thanks
Bill

 

[bhobba]

So the author's focus on counterfactuality unnecessarily weakens the conclusions of Bell's proof.

Yes - although I wouldn't express it that harshly.

The difference between realism and counter-factual definiteness has been discussed in a number of threads. It a bit subtle, but I don't think someone starting out really needs to worry about it. Understand what going on in a general sense first.

Thanks
Bill

 

[jk22]

Simply accept QM as it is. We have met the enemy and he is us - Pogo

So a global variable says for the CHSH operator S=AB-AB'+A'B+A'B' shall we write that the values are 4,2,0,-2,-4 and do we have for example the probability $$p(-4)=(\frac{1}{2}(1+\frac{1}{\sqrt{2}}))^4$$ ?

This was pbtained using quantum probabilities for the pairs and assuming them independent https://www.physicsforums.com/threads/in-bell-are-pairs-independent.827997/

Whereas qm predicts the results $$0,\pm 2\sqrt{2}$$ with $$p(-2\sqrt{2})=1$$

My problem with qm is that adding 1 and -1s does give a decimal result. How do you explain that ?

 

[bhobba]

My problem with qm is that adding 1 and -1s does give a decimal result. How do you explain that ?

Can you be a bit clearer - I can't follow what your issue is.

Also we are getting off topic here - please start a new thread about the proof of Bells theorem - we have wandered well and truly from what you posted about.

Thanks
Bill

 

[gill1109]

"The terms in equation (1) refer to three measurements on the same set of coins. The terms in equation (4) refer to measurements on 3 separate disjoint sets of coins.

In probability theory, whenever you add and subtract probabilities, the expression is only meaningful if all the probabilities are from the same sample space. While you can guarantee that a set of triples of measurements A,B,C each on a pair of coins will be able to generate P(A,B), P(A,C) and P(B,C) that are from the same sample space, there is no way to guarantee that the same can be true for 3 separate measurements in which you only measure A,B on one set of coins, A,C on another set of coins and B,C on yet a different set of coins."

The point is that Bell's inequality assume local hidden variables (or local realism). So even though you can't actually measure A, B and C on the same coins, they are all defined. Secondly, Bell also assumes a "no conspiracy" or freedom assumption or fair sampling assumption. When you actually observe A and B and look at the correlation, you get the same result (up to statistical sampling error) as you would have got from, say, the correlation between A and B when observing A and C.

 

[gill1109]

Bells theorem proves - there is no escaping it - if QM is correct you can't have both locality and reality. All you have done is define some kind of deterministic observable which has nothing to do with bell.

Here is the proof of Bells theorem:
http://www.johnboccio.com/research/quantum/notes/paper.pdf

If you think you have found an out specify exatly which step is in error.

Thanks
Bill

Perhaps worth mentioning that Lorenzo Maccone's paper can be found on arXiv http://arxiv.org/abs/1212.5214 and was published in Am. J. Phys. 81, 854 (2013).

There are similar absolutely elementary proofs of the "four variable" Bell inequality aka CHSH inequality. My favourite is on slide 7 of http://www.slideshare.net/gill1109/epidemiology-meets-quantum-statistics-causality-and-bells-theorem. And note that at long last they finally did the definitive (loophole free) experiment: http://arxiv.org/abs/1508.05949

 

[stevendaryl]

Perhaps worth mentioning that Lorenzo Maccone's paper can be found on arXiv http://arxiv.org/abs/1212.5214 and was published in Am. J. Phys. 81, 854 (2013).

There are similar absolutely elementary proofs of the "four variable" Bell inequality aka CHSH inequality. My favourite is on slide 7 of http://www.slideshare.net/gill1109/epidemiology-meets-quantum-statistics-causality-and-bells-theorem. And note that at long last they finally did the definitive (loophole free) experiment: http://arxiv.org/abs/1508.05949

I have the same complaint about those slides that I had about another paper linked to. Slide number 6 says: "Realism = existence of counterfactual outcomes". That doesn't seem right to me. A realistic model could allow for a nondeterministic relationship between "hidden variables" and measurement outcomes. For example, you could have a model along the following lines:

• Each twin pair is associated with a hidden variable [itex]\lambda[/itex], randomly produced according to a probability distribution [itex]P(\lambda)[/itex]
• Alice will measure spin-up with probability [itex]P_A(\lambda, \alpha)[/itex] where [itex]\alpha[/itex] is the setting of her measuring device.
  
• Bob will measure spin-up with probability [itex]P_B(\lambda, \beta)[/itex] where [itex]\beta[/itex] is the setting of his measuring device.
• For fixed lambda, the probabilities are independent; the probability of both Alice and Bob getting spin-up is given by: [itex]P_{A\&B}(\lambda, \alpha, \beta) = P_A(\lambda, \alpha) \cdot P_B(\lambda, \beta)[/itex]

I would call such a model "realistic"; it's just not deterministic, and the values of the "hidden variables" are not directly measurable. But in such a model, there are no counterfactual outcomes (there is no definite answer to a question of the form "What measurement result would Alice have gotten if she had chosen setting [itex]\alpha'[/itex] instead of [itex]\alpha[/itex]). So I don't think it's correct to equate realism with counterfactual outcomes.

On the other hand, for the purposes of Bell's proof as applied to EPR, it doesn't actually matter, because the perfect correlations (or anti-correlations) between Alice's and Bob's results when their settings are equal implies that any local hidden-variables explanation must, in fact, be counterfactually definite. But the counterfactuality is a conclusion, rather than an assumption.

 

[bhobba]

I have the same complaint about those slides that I had about another paper linked to. Slide number 6 says: "Realism = existence of counterfactual outcomes". That doesn't seem right to me.

Its not right. But I think going into it would confuse the beginner more than illuminate. Once they grasp the essentials of it fine points like that can be gone into.

Thanks
Bill

 

[gill1109]

I have the same complaint about those slides that I had about another paper linked to. Slide number 6 says: "Realism = existence of counterfactual outcomes". That doesn't seem right to me. A realistic model could allow for a nondeterministic relationship between "hidden variables" and measurement outcomes. For example, you could have a model along the following lines:

• Each twin pair is associated with a hidden variable [itex]\lambda[/itex], randomly produced according to a probability distribution [itex]P(\lambda)[/itex]
• Alice will measure spin-up with probability [itex]P_A(\lambda, \alpha)[/itex] where [itex]\alpha[/itex] is the setting of her measuring device.
  
• Bob will measure spin-up with probability [itex]P_B(\lambda, \beta)[/itex] where [itex]\beta[/itex] is the setting of his measuring device.
• For fixed lambda, the probabilities are independent; the probability of both Alice and Bob getting spin-up is given by: [itex]P_{A\&B}(\lambda, \alpha, \beta) = P_A(\lambda, \alpha) \cdot P_B(\lambda, \beta)[/itex]

I would call such a model "realistic"; it's just not deterministic, and the values of the "hidden variables" are not directly measurable. But in such a model, there are no counterfactual outcomes (there is no definite answer to a question of the form "What measurement result would Alice have gotten if she had chosen setting [itex]\alpha'[/itex] instead of [itex]\alpha[/itex]). So I don't think it's correct to equate realism with counterfactual outcomes.

On the other hand, for the purposes of Bell's proof as applied to EPR, it doesn't actually matter, because the perfect correlations (or anti-correlations) between Alice's and Bob's results when their settings are equal implies that any local hidden-variables explanation must, in fact, be counterfactually definite. But the counterfactuality is a conclusion, rather than an assumption.

I agree that my definition seems to exclude stochastic hidden variable models. But every probabilistic model can be recast as a deterministic model simply by adding another layer of hidden variables. Remember: in Kolmogorovian probability theory, every random variable is "just" a deterministic function of a hidden variable omega and it is omega which is picked by chance. I claim that mathematically, every local realistic model *can* be mathematically represented as a deterministic model with counterfactual outcomes. Of course the mathematical representation need not be physically very interesting. Those counterfactual outcomes exist in a theoretical mathematical description of our experiment. I am not claiming that they exist in reality (whatever that might mean).

 

[georgir]

I've always had trouble understanding what else other than non-locality could lead to Bell violations, in other words, what local but non-realistic model means.
As gill explains, any mumbo-jumbo about something not existing until it gets measured and such, is still the same as it being determined (deterministically, lol) from a hidden variable. The only question is if the variables that determine it are local or not, and I see no way to cause violations if they are local.

 

[jk22]

I think to see a violation one should consider that in Chsh (because it is used in most experiments) at least when it is written AB+AB' that the operator A is measured twice hence we can have two different results for the A operator appearing.

On the other hand if you consider that these two A have the same measurement result, like a register, then you can have no violation.

 

[stevendaryl]

I've always had trouble understanding what else other than non-locality could lead to Bell violations, in other words, what local but non-realistic model means

I haven't spent much effort trying to work out the details, but it seems to me that some kind of local world-splitting could give a local, non-realistic toy model to explain EPR-type correlations. As I said, I'm fuzzy about the details, but such a model might look like this:

1. You produce a twin pair of anti-correlated spin 1/2 particles at some location intermediate between Alice and Bob.
2. When Alice measures the spin of one particle, she splits into two noninertacting copies, one of which sees spin-up, and the other of which sees spin-down.
3. Same for Bob.
4. If Alice and Bob compare notes by sending a message telling the other what result he/she got, the message is nondeterministically delivered to one copy or the other. If Alice saw spin-up at angle [itex]\alpha[/itex], and Bob also chose angle [itex]\alpha[/itex], then the message from Alice would be delivered to the copy of Bob that saw spin-down. If they chose different angles, then there would be some nonzero chance of the messages being sent to either copy.
5. After one copy of Alice interacts with a copy of Bob, the other copies become inaccessible for future communication.

It seems that such a toy model, which would only apply to EPR and not to QM more generally, could explain the EPR correlations. I'm not claiming this as a serious possibility, but just as an example of a local, non-realistic model. It's non-realistic in the sense that there is no "hidden variable" corresponding to the true, unknown spin of a particle.

 

[gill1109]

I've always had trouble understanding what else other than non-locality could lead to Bell violations, in other words, what local but non-realistic model means.
As gill explains, any mumbo-jumbo about something not existing until it gets measured and such, is still the same as it being determined (deterministically, lol) from a hidden variable. The only question is if the variables that determine it are local or not, and I see no way to cause violations if they are local.

I think that "non realistic" means that there are no variables that determine the outcome! It is irreducibly random. It is not pre-determined.

 

[jk22]

Yes if you have $$A(a,hv)=rand()$$ where rand() is a computer like function that returns 1 or -1 each time it is called then it violates the inequality but you cannot predict with certainty in any case.

I think that the local approach is confronted to a dilemma : either you violate xor you can predict with certainty but quantum does both at least theoretically.

It is then an experimental thing to see if both arise. Violation seems okay but i have seen experimental curves that were only at 87% of prediction power for the same angles.

 

[stevendaryl]

Yes if you have $$A(a,hv)=rand()$$ where rand() is a computer like function that returns 1 or -1 then it violates the inequality but you cannot predict with certainty in any case.

I don't think that's true. Adding randomness does not allow you to violate the Bell inequality unless you add randomness nonlocally. That is, you can assume that Alice's result is random, and that Bob's result is random, but as long as Alice's random choice is independent of Bob's random choice, you're not going to violate Bell's inequality. As Richard says, localized randomness is indistinguishable from a deterministic model with an extra hidden variable.

 

[stevendaryl]

I think that "non realistic" means that there are no variables that determine the outcome! It is irreducibly random. It is not pre-determined.

To me, realism versus non-realism is about whether the theory describes the objective, observer-independent state of the system under consideration, as opposed to describing something subjective (our knowledge, for example). A classical probability theory is non-realistic in the sense that the probabilities don't reflect anything about the world (except in the case of probabilities 0 and 1), they only reflect our knowledge, or lack of knowledge about the world.

 

[jk22]

I don't think that's true. Adding randomness does not allow you to violate the Bell inequality unless you add randomness nonlocally..

If i take the Chsh inequality

ab+ab'+a'b-a'b'<=2

Take a a' and b' b equals to rand() and you can obtain 4 obviously.this you can check pn a computer but it is straight forward.

This said the way out to the local dilemma is that each pair ab shares the same variable but each term receives a different variable.

 

[stevendaryl]

If i take the Chsh inequality

ab+ab'+a'b-a'b'<=2

Take a a' and b' b equals to rand() and you can obtain 4 obviously.this you can check pn a computer but it is straight forward..

I think you've misunderstood what the a, a', etc., mean in the CHSH inequality. The inequality is about correlations, not values. The real statement of the inequality is:

[itex]\rho(a,b) + \rho(a, b') + \rho(a',b) - \rho(a', b') \leq 2[/itex]

[itex]\rho(\alpha, \beta)[/itex] is computed this way: (for the spin-1/2 case)

• Generate many twin pairs, and have Alice and Bob measure the spins of their respective particles at a variety of detector orientations.
• Let [itex]\alpha_n[/itex] be Alice's setting on trial number [itex]n[/itex]
• Let [itex]\beta_n[/itex] be Bob's setting on trial number [itex]n[/itex]
  
• Define [itex]A_n[/itex] to be +1 if Alice measures spin-up on trial number [itex]n[/itex], and -1 if Alice measures spin-down.
• Define [itex]B_n[/itex] to be [itex]\pm 1[/itex] depending on Bob's result for trial [itex]n[/itex]

Define [itex]\rho(\alpha, \beta)[/itex] to be the average of [itex]A_n \cdot B_n[/itex], averaged over those trials for which [itex]\alpha_n = \alpha[/itex] and [itex]\beta_n= \beta[/itex]. Then if [itex]a[/itex] and [itex]a'[/itex] are two different values for [itex]\alpha[/itex], and [itex]b[/itex] and [itex]b'[/itex] are two different values for [itex]\beta[/itex], then the CHSH inequality says that [itex]\rho(a,b) + \rho(a',b) + \rho(a,b') - \rho(a',b') \leq 2[/itex].

It doesn't make any sense to say [itex]\rho(\alpha, \beta) = rnd()[/itex]. [itex]\rho[/itex] is not a random number, it's an average of products of random variables.

You can certainly propose a model where [itex]A_n = rnd()[/itex] for each [itex]n[/itex], and similarly for [itex]B_n[/itex]. But that would not make [itex]\rho(\alpha, \beta) = rnd()[/itex].

 

[jk22]

The demonstration uses factorization of measurement results if the measurement result of this sum is smaller than 2 then of course the average is.

The fact is that qm says we cannot factorize

$$ A_n B_n+A_nB'_n\neq A_n(B_n+B'_n)$$

Because A_n are different since measured two times.

To make sense we have $$\rho_i$$ with $$A_{i,n},B_{i,n}$$

The $$A_n$$ for the first rho are not the same as for the second.

 

[stevendaryl]

The demonstration uses factorization of measurement results if the measurement result of this sum is smaller than 2 then of course the average is.

The fact is that qm says we cannot factorize

$$ A_n B_n+A_nB'_n\neq A_n(B_n+B'_n)$$

Because A_n are different since measured two times.

Well, in the CHSH inequality, you compute [itex]\rho(a,b)[/itex] by averaging over all trials for which Alice's setting is [itex]a[/itex] and Bob's setting is [itex]b[/itex]. Then you independently compute [itex]\rho(a,b')[/itex], [itex]\rho(a',b)[/itex] and [itex]\rho(a',b')[/itex]. You never compute a quantity such as [itex]A_n B_n+A_nB'_n[/itex] (because you can't, for a single value of [itex]n[/itex], measure both [itex]B[/itex] and [itex]B'[/itex].

Anyway, my point is that the use of random number generators doesn't affect the conclusion of Bell's theorem, or at least I don't think it does.