Can we violate Bell inequalities by giving up CFD?

[Derek Potter]

If I interact with paper using a pencil, the resulting picture will not be the measurement of some property of the paper, and nobody would suspect that this picture existing as part of the paper before I started my interaction.

‘In every block of marble I see a statue as plain as though it stood before me, shaped and perfect in attitude and action. I have only to hew away the rough walls that imprison the lovely apparition to reveal it to the other eyes as mine see it.’—Michelangelo

 

[Derek Potter]

Why do you think that BI is not violated in this special case?

This is a bit elementary:
- 2 <= E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) <= +2 (CHSH)
a = b and a '= b' (the special case)
Therefore E(a, b') = E(a', b); the expression evaluates to 2 (or -2) and BI is not violated.

 

[Ilja]

?
- 2 <= E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′) <= +2
a = b and a '= b'; therefore E(a, b') = E(a', b) and BI is not violated.

And why do you restrict this to a=b or a'=b'?

You obviously should distinguish the reasoning which proves that CFD holds for this type of experiment for above parts and all a, b from the measurements.

The EPR criterion requires only an ability. It is "If ... we can ..." not "if we actually do". After applying it, we have concluded - as a general fact about the particular Einstein-causal theory, not about a particular experiment - that for all directions of possible spin measurements by Alice or Bob CFD holds. Always, that means, even if we measure E(a,b) for a =/= b.

 

[zonde]

Let's say they can get the EPR result of 100% correlation at certain angles.

But at other angles the prediction of QM is violated, so there is no Bell inequality violation.

Perhaps we could still model that using QM - maybe we just got the Hamiltonian is wrong - rather amazingly at such a low energy.

This is of course hypothetical but I would consider more likely that new interpretation (or rather theory in this case) of QM can be constructed that does not violate BI if prediction about 100% correlation is taken as false. That's because CFD assumption in this case surely can be relaxed and without CFD fair sampling assumption is unjustified.

 

[Derek Potter]

And why do you restrict this to a=b or a'=b'?

because that is what you asked about.

You obviously should distinguish the reasoning which proves that CFD holds for this type of experiment for above parts and all a, b from the measurements.
The EPR criterion requires only an ability. It is "If ... we can ..." not "if we actually do". After applying it, we have concluded - as a general fact about the particular Einstein-causal theory, not about a particular experiment - that for all directions of possible spin measurements by Alice or Bob CFD holds. Always, that means, even if we measure E(a,b) for a =/= b.

We have concluded no such thing for the simple reason that the argument relies on 100% correlation. However we can certainly construct a different argument which replaces the 100% correlation with a cos-squared law. The EPRC then refers to the predictability of this correlation, which means that the correlation is a property of the system. On which bombshell I would ask you to define CFD in such a way as to tell us unambiguously whether this property means CFD holds or not.

 

[stevendaryl]

We have concluded no such thing for the simple reason that the argument relies on 100% correlation. However we can certainly construct a different argument which replaces the 100% correlation with a cos-squared law. The EPRC then refers to the predictability of this correlation, which means that the correlation is a property of the system. On which bombshell I would ask you to define CFD in such a way as to tell us unambiguously whether this property means CFD holds or not.

If Bob's experiment has adjustable setting $\Theta$, then let's say that Bob's experiment satisfies CFD for setting $\theta_1$ if the question: "What would Bob's result have been if he had chosen setting $\theta_1$?" has a definite answer, even in the case where Bob didn't choose angle $\theta_1$. This is a property of theories; it's not just a philosophical question. The theory either does or does not imply CFD.

In this case, the theory is a combination of QM, plus the assumption that definite outcomes occur (no many-worlds), plus the assumption of free-will (that is, Alice and Bob's detector settings are freely chosen parameters, and are not forced by the experimental set-up--no superdeterminism), plus the assumption of Einstein causality (nothing Alice does can instantly change the physical situation for Bob, who is far away).

The argument for CFD given these assumptions is something like:

1. Suppose that in Alice's coordinate system, her measurement takes place before Bob's.
2. Suppose that Alice chooses detector setting $\theta_1$ and gets result $A_1$. (For simplicity, let's assume that the result is binary--she either detects a particle at that filter angle, in which case $A_1 = 1$ or doesn't, in which case $A_1 = 0$)
   
3. Immediately after Alice's measurement , she knows something definite about Bob's future measurement result: Namely, "if he chooses detector setting $\theta_1$, he will get result $B_1$". (Depending on the details of the twin-pair setup, either $B_1 = A_1$ or $B_1 = \neg A_1$)
   
4. So she concludes that the implication $\theta_1 \rightarrow B_1$ is a physical property of Bob's experimental situation.
5. If she also assumes that her measurement has no effect on Bob's situation (since it is far away), then she concludes that the implication $\theta_1 \rightarrow B_1$ was a physical property of Bob's experimental situation even before her measurement.
6. So even if Bob doesn't choose detector setting $\theta_1$, Bob's situation satisfies CFD for that angle: If he chose that angle, his result would definitely be $B_1$

That's what CFD means; regardless of what setting Bob actually chooses, there is a definite answer to the question: "What would Bob's result have been if he had chosen setting $\theta_1$?"

So, after Alice's measurement, she knows that Bob's situation satisfies CFD for angle $\theta_1$. Now, she can do hypothetical reasoning on her own choice, as follows:

1. For any angle $\theta$, if Alice chooses detector setting $\theta$, then she knows that Bob's situation satisfies CFD for angle $\theta$.
2. If her choices have no effect on Bob, and after her choice, Bob has CFD for angle $\theta$, then he must have had CFD for angle $\theta$ before she made her choice.
3. Since $\theta$ is arbitrary, then Bob must have CFD for every angle $\theta$

 

[atyy]

I usually avoid using the term CFD, because I don't really understand what it means, so as you can see in my answers in this thread, I've always redefined the term.

But what is the actual definition of CFD and who defined it? The only one I know is Peres's famous definition of the negation of CFD being that "unperformed experiments have no results" - but I can't imagine he was serious, just one his jokes.

 

[zonde]

I usually avoid using the term CFD, because I don't really understand what it means, so as you can see in my answers in this thread, I've always redefined the term.

But what is the actual definition of CFD and who defined it? The only one I know is Peres's famous definition of the negation of CFD being that "unperformed experiments have no results" - but I can't imagine he was serious, just one his jokes.

I would like to turn your question on it's head. We need to give a name for this idea:

regardless of what setting Bob actually chooses, there is a definite answer to the question: "What would Bob's result have been if he had chosen setting $\theta_1$?"

Is it ok to name it CFD? If it's not the best choice what would be your choice?

 

[zonde]

I would like to say that I consider my question in OP answered thanks to Ilja, stevendaryl's nice summary of the argument and those who maintained skeptical opposition.

 

[Derek Potter]

[QUOTE="stevendaryl, post: 5155634, member: 372855
That's what CFD means; regardless of what setting Bob actually chooses, there is a definite answer to the question: "What would Bob's result have been if he had chosen setting $\theta_1$?"
[/QUOTE]
Thanks, Steve, that's pretty clear. So, given Einstein causality, CFD (as you define it and this seems to satisfy zonde) is unavoidable. i.e. A viable theory that gives up CFD must necessarily give up Einstein causality.

 

[zonde]

zonde - just call it physical realism :)

You are too radical. Physical realism allows probabilistic results just as well but we speak about CFD only in case of definite results.

 

[Derek Potter]

You are too radical. Physical realism allows probabilistic results just as well but we speak about CFD only in case of definite results.

No problem, I'd already deleted the comment when you replied :)