# Non-locality and Counterfactual definiteness

1. Dec 10, 2015

### Daniel K

I understand that through Bell's theorem both locality and counterfactual definiteness cannot exist within physics. However shouldn't the double slit experiment give physicists a clue that losing counterfactual definiteness is actually the best way of interpreting the problem? The double slit experiment proved that the wave that guides light is in fact probabilistic. How does an advocate of particles having definite preexisting qualities account for this?

2. Dec 10, 2015

### Daniel K

Just realized that it could be explained by pilot wave theory. However is this the only way out?

3. Dec 10, 2015

### wle

You don't need to assume determinism or counterfactual definiteness in order to derive Bell inequalities.

4. Dec 10, 2015

### Daniel K

I'm aware. However if you do, then non-locality is the only way in which you can explain the correlations.
I am asking how do advocates that say particles have definite properties explain the probabilistic manner that the double slit experiment proved

Last edited: Dec 10, 2015
5. Dec 10, 2015

### zonde

The point is that FTL is the only way in which you can scientifically explain the correlations.
Definite properties or not is unrelated to the question.

Experiments do not prove anything. They can confirm or falsify some model or theory.

6. Dec 11, 2015

### Demystifier

Try http://lanl.arxiv.org/abs/1112.2034
which is a sort of hybrid between pilot wave theory and denial of counterfactual definiteness.

7. Dec 11, 2015

### Demystifier

Not necessarily. One logical possibility is superdetermisism, in which the laws of physics are objective, deterministic and local, but (there is always a "but") the initial conditions are fine tuned. Not many physicists find that idea reasonable, but among them there is one Nobel Prize laureate.

8. Dec 11, 2015

### DrChinese

Not sure why a couple of the other posters passed on this point, but we usually say:

Every interpretation of Quantum Mechanics is an attempt to provide a "way out". MWI, Time Symmetric/Retrocausal, and as you point out, the explicit FTL mechanisms such as Bohmian Mechanics (pilot wave).

9. Dec 12, 2015

### billschnieder

The Perfect ant-correlation assumption is Counterfactual Definiteness. Without it you don't have Bell's inequalities.

Bell wrote: "if measurement by Alice along axis "a" produces outcome +1, then according to quantum mechanics, measurement by Bob along axis "b" must produce outcome -1 and vice versa"

Counterfactual Definiteness means that a measurement that was not performed had a single definite result, which is the case for the perfect ant-correlation assumption.

10. Dec 12, 2015

### stevendaryl

Staff Emeritus
I think that there are two ingredients in the derivation of Bell's inequality: Perfect anti-correlation + local realism. The first is a consequence of QM (in the spin-1/2 EPR experiment), so it shouldn't be considered an assumption of Bell's theorem.
1. QM predicts perfect anti-correlations in the spin-1/2 EPR experiment.
2. Perfect anti-correlation + the assumption of local realism implies counterfactual definiteness.
3. Local realism + counterfactual definiteness implies Bell's inequality.
4. QM predicts the violation of Bell's inequalities.
So putting this altogether:
QM + Local realism is inconsistent (since it predicts both Bell's inequality and the violation of Bell's inequality)​
which is logically equivalent to:
QM implies that local realism is false​

I'm sure there must be an alternative derivation of Bell's inequalities (or some other related inequality) that doesn't assume counterfactual definiteness, but I don't know what it is.

Counterfactual definiteness comes into the derivation when Bell assumes that there are two functions:

$A(\alpha, \lambda)$
$B(\beta, \lambda)$
that return $\pm 1$ as deterministic functions of the detector settings $\alpha$ and $\beta$, and the hidden variable $\lambda$

Local realism by itself doesn't imply the existence of such functions. Instead, what it implies is the existence of two functions:
• $P_A(\alpha, O_A, \lambda)$ : the probability of Alice measuring +1, given her detector setting $\alpha$, other local conditions relevant to the detection [/itex]O_A[/itex] and hidden variable $\lambda$
• $P_B(\beta, O_B, \lambda)$ : the probability of Bob measuring +1, given his detector setting $\beta$, other local conditions relevant to the detection [/itex]O_B[/itex] and hidden variable $\lambda$
The perfect anti-correlation prediction of quantum mechanics implies that $O_A$ and $O_B$ are irrelevant, and implies that these two probabilities must in fact must be 0 or 1. In other words, the outcomes are deterministic functions of $\alpha, \beta$ and $\lambda$ (which is basically counterfactual definiteness).

11. Dec 12, 2015

### wle

Two sentences later, in Bell's summary of the EPR argument from the same paper:

Since we can predict in advance the result of measuring any chosen component of $\vec{\sigma}_{2}$, by previously measuring the same component of $\vec{\sigma}_{1}$, it follows that the result of any such measurement must actually be predetermined.​

So as far as Bell was concerned, nondeterministic local hidden variable models had no chance of working, so there was no point in further considering them.

But this is moot, since exactly the same Bell inequalities hold for nondeterministic LHVs as for deterministic LHVs.

Last edited: Dec 12, 2015
12. Dec 12, 2015

### morrobay

With this table for perfect anti correlations I see two cases where Alice produces outcome +1 along axis "a" and Bob produces outcome +1 along axis "b" in disagreement with above. So it seems this conflict with the table is part of inequality violations explanation.

Alice.......................Bob
a b c.......................a b c
+++........................ ---
++-..........................--+
+-+.........................-+-
+--
......................... -++
-++.........................+--
-+-..........................+-+
--+..........................++-
---...........................+++

Last edited: Dec 12, 2015
13. Dec 13, 2015

### billschnieder

if measurement by Alice along axis "a" produces outcome +1, then according to quantum mechanics, measurement by Bob along axis "a" must produce outcome -1 and vice versa.

Even if Bob had measured along "b". According to QM, he would still have obtained the definite value -1, had he measured along "a".

If perfect anti-correlation is a consequence of QM, and perfect anti-correlation is CFD, there is no way for Bell's theorem to reject CFD without rejecting QM. BTW, CFD and realism are equivalent.

In quantum mechanics, counterfactual definiteness (CFD) is the ability to speak meaningfully of the definiteness of the results of measurements that have not been performed (i.e. the ability to assume the existence of objects, and properties of objects, even when they have not been measured). -- Wikipedia

Last edited: Dec 13, 2015
14. Dec 13, 2015

### stevendaryl

Staff Emeritus
I see it as:
1. QM implies AC (perfect anti-correlations)
2. AC + LR (local realism) implies CFD.
3. LR + CFD implies BI (Bell's inequalities)
4. QM contradicts BI
5. Therefore, QM contradicts LR
I think they should be distinguished. In the case of a measurement with two possible results, $\pm 1$, I would say that the model is locally realistic if the probability of getting $+1$ depends only on conditions local to the measurement. CFD is the special case in which the result, $\pm 1$, is a deterministic function of those conditions. So you could have a purely local stochastic model, which would not obey CFD. But such a model couldn't produce perfect anti-correlations among twin pairs unless it also obeyed CFD.

Last edited: Dec 13, 2015
15. Dec 13, 2015

### DrChinese

I actually agree with you on this. Although I know there are those who want to define Realism as different than CFD, they are used interchangeably in the literature and there is no significant difference between them. Essentially, any example of realism is also an example of CFD, and vice versa.

QM does NOT require CFD due to perfect correlations. That is an extra assumption.

16. Dec 13, 2015

### stevendaryl

Staff Emeritus
I would say that they are used indistinguishable in discussions of QM, because QM predicts perfect correlations in certain situations, and perfect correlations plus realism implies CFD.

17. Dec 13, 2015

### RUTA

Most of my foundations colleagues equate CFD with realism, but there are realistic interpretations without CFD, e.g., http://www.ijqf.org/wps/wp-content/uploads/2015/06/IJQF2015v1n3p2.pdf. In general, a realistic retrocausal interpretation is indifferent to CFD, e.g., https://www.physicsforums.com/insights/retrocausality/.

18. Dec 13, 2015

### billschnieder

I think this is not possible, I would be interested to see their definition of "realistic" and "CFD" and what the difference is.

Is it true in RBW, that if Alice measured along "a" and obtained +1, Bob would have obtained "-1" if he measured along "a", even if he actually ends up measuring along "b" instead? But this is exactly the meaning of CFD.

I do not agree with the statement that "perfect correlations plus realism implies CFD". Perfect correlations is CFD, and is Realism already.

19. Dec 13, 2015

### stevendaryl

Staff Emeritus
"Realism" is too fuzzy a concept to reason about, but Bell had a pretty precise notion of what a "local realistic theory" was. It was described in his essay "Theory of Local Beables". The fact of perfect correlations does not imply that there is a local realistic theory underlying those observations.

20. Dec 13, 2015

### billschnieder

The moon is there when nobody is looking. Particles have properties even when they are not being measured. Not fuzzy at all, very precise statement.