# Measurement problem and computer-like functions

1. Aug 22, 2015

### 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 ?

2. Aug 22, 2015

### Staff: Mentor

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

Thanks
Bill

3. Aug 22, 2015

### 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 ?

4. Aug 22, 2015

### Staff: Mentor

Bells theorem proves - there is no escaping it - if QM is correct you cant 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

5. Aug 22, 2015

### 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 ?

6. Aug 22, 2015

### Staff: Mentor

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

Thanks
Bill

7. Aug 22, 2015

### 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 ?

8. Aug 22, 2015

### Staff: Mentor

Of course not. QM says that's impossible.

Again QM says that's impossible.

https://www.amazon.com/Quantum-Mechanics-The-Theoretical-Minimum/dp/0465062903

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):

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

Thanks
Bill

Last edited by a moderator: May 7, 2017
9. Aug 25, 2015

### 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.

10. Aug 25, 2015

### Staff: Mentor

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

So?

Thanks
Bill

11. Aug 25, 2015

### stevendaryl

Staff Emeritus
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.

12. Aug 25, 2015

### jk22

13. Aug 25, 2015

### DirkMan

"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."

14. Aug 26, 2015

### Staff: Mentor

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 cant 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

Last edited: Aug 26, 2015
15. Aug 26, 2015

### 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.

16. Aug 26, 2015

### Staff: Mentor

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 wont resolve anything. If you want to discuss general QM issues best to start other threads.

Thanks
Bill

Last edited: Aug 26, 2015
17. Aug 26, 2015

### Staff: Mentor

Yes - although I wouldnt 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

18. Aug 26, 2015

### jk22

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 ?

Last edited: Aug 26, 2015
19. Aug 26, 2015

### Staff: Mentor

Can you be a bit clearer - I cant 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

20. Aug 26, 2015

### gill1109

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.