# I State Separability & CFD

1. Jun 20, 2017

### Eye_in_the_Sky

I am thinking of the usual "Alice and Bob" type scenario.

If I choose to view the measuring instruments of Alice and Bob as "classical objects in spacetime", am I not then forced to say the following as well?

1) The joint-state of the measuring instruments is separable throughout the whole of spacetime.

2) The principle of CFD applies (validly) to the outcomes which those instruments can register.

2. Jun 20, 2017

### Eye_in_the_Sky

Of course, in the above, I assume NO retro-causality.

3. Jun 22, 2017

### Eye_in_the_Sky

I will improve upon the expression of my query, and answer it ... then add a little more.

Consider the four conditions below:

1) Alice and Bob have free choice of their instruments' settings.

2) All influences which propagate within spacetime are limited by c.

3) There are no retro-causal influences propagating within spacetime.

4) The joint-state of the measuring instruments of Alice and Bob is separable throughout the whole of spacetime.

Are these conditions sufficient to allow for CFD to apply (validly) to the outcomes which those instruments can register?
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I think the answer to the question is "No", because it is still possible to say:

Although the joint-state of the instruments is 'separable', nonetheless, the outcome that each instrument registers (for the paired events) is 'nonseparably linked' to the setting of other instrument.

I am making this statement on the basis of what @RUTA wrote not long ago:

CFD - Counterfactual Definiteness, post #105
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Moreover, I would add this also:

The above four conditions imply 'nonlocality' in the FTL sense.

I think RUTA concurs on this:

CFD - Counterfactual Definiteness, post #113
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ASIDE:

All four of the above conditions, as well as CFD, hold true in BM ... correct?

(I'm sure @Demystifier can answer this.)

Last edited: Jun 22, 2017
4. Jun 23, 2017

### ueit

As far as I can tell your condition 1 (free choice) has no justification in any theory, classical or quantum. BM is no exception. According to the theory once you specify the state of Alice and Bob some time before the experiment, their "choices" are predetermined, so you have no choices whatsoever.

Andrei

5. Jun 25, 2017

### Eye_in_the_Sky

Thanks for that, Andrei. I should not have included "free choice" in the ASIDE.

Let the ASIDE be restricted to conditions 2,3,4 only.

Last edited: Jun 25, 2017
6. Jun 25, 2017

### Eye_in_the_Sky

Consider an interpretation for which

the above four conditions hold true

but

the "principle of CFD" is invalid.

Then

according to that interpretation one must say

Alice's outcome has an intrinsically 'nonseparable' dependency upon Bob's setting, and vice versa.

7. Jun 26, 2017

### Eye_in_the_Sky

Okay, now I see what my problem is.

The required conjunction is this:

The joint-state of the measuring instruments of Alice and Bob is separable throughout the whole of spacetime;

AND

Alice's outcome has an no dependency upon Bob's setting, and vice versa.

These two conditions, when taken together, are then sufficient conditions for one to be able to derive a Bell Inequality.
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Shortly, I will start a new thread.
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EDIT: The new thread is Seeing the essence of Bell.

Last edited: Jun 26, 2017
8. Jun 26, 2017

### Denis

No. There is no CFD in BM. Or, more accurate, there is also a very restricted CFD in BM, namely the configuration is defined even if not measured. Everything else is contextual, that means, what is measured is not predefined by the system alone, but depends also on the configuration of the measurement device, thus, experiments which are not done have no result - the result would have to depend on a non-existing configuration of the non-existing measurement device.

9. Jun 26, 2017

### Staff: Mentor

That is not necessary, nor is it desirable.

Which I have closed and deleted. Please keep the discussion here.

10. Jun 26, 2017

### Eye_in_the_Sky

But that thread was not about an exploration of the relationship between the notions of "state separability" and "CFD".

It was about a particular perspective from which a Bell Inequality can be derived.

The two topics are very distinct (but, of course, related).

... Are you sure you'd rather keep both topics in one thread?

11. Jun 26, 2017

### Staff: Mentor

They might seem that way to you, but to me they look close enough that they can happily coexist in the same thread.

12. Jun 26, 2017

### Eye_in_the_Sky

Let it be so.
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By hypothesis (in the thought experiment about "Alice and Bob"), the following is true:

If Bob's setting would have been b' instead of b,

then

each of Alice and Bob would have obtained a definite outcome.

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Consider the following two conditions:

1) The joint-state of the measuring instruments of Alice and Bob is separable throughout the whole of spacetime;

2) Alice's outcome has no dependency upon Bob's setting, and vice versa.

If both conditions, 1 and 2, hold true (in a given interpretation), then (according to that interpretation) the following principle is valid:

α) It is 'admissible' to suppose that Alice's outcome for (the hypothetical setting) b' would have been the same as that for (the actual setting) b.

Also, there is an axiom:

β) It is 'admissible' to apply the theory to any 'admissible' scenario.
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The above two principles, α and β, can then be used to derive a Bell Inequality (when the theory in question is QT).

THEREFORE (for any correct interpretation of QT):

If condition 1 holds true,

then

condition 2 holds false.

13. Jun 27, 2017

### morrobay

You could check the validity of condition 2 ( locality + CFD ) : Given P ++ , P-- = 1/2(sin a-b/2)2 :
Measure a stream of entangled particles with settings A = a and B = b. alternating A = a and B = b'. If outcome at Alice does not vary as B setting varies then Alice outcome is independent of setting at Bob

14. Jun 27, 2017

### ueit

Alice will obtain an apparently random sequence of spin up and spin down. Even if the individual values Alice gets are influenced by Bob's setting you cannot observe it.

15. Jun 27, 2017

### Denis

This makes no sense.

Compare B = a and B = -a. The average outcome at Alice would be the same 50% up and 50% down. But in the first case this would be perfectly anticorrelated, in the second perfectly correlated. To think that in such a situation of perfect correlation/anti-correlation there is independence makes no sense.

16. Jun 27, 2017

### morrobay

Reread my post. I made no such reference to a = b or a = -b

17. Jun 27, 2017

### Denis

So what? These are particular examples of your more general case, not?

18. Jun 27, 2017

### Eye_in_the_Sky

Condition 2 is not the same as "locality + CFD". From the latter, one can derive a Bell Inequality. Condition 2 alone is insufficient.

It is the conditions 1 and 2 together that are sufficient to establish a derivation of a Bell Inequality.

But there is an important difference between the pair 1 + 2 versus the pair "locality" + "CFD". Each of 1 and 2 can stand on its own as a well-defined concept, whereas each of "locality" and "CFD" cannot. One parsing is 'clean'. The other is 'dirty'.

Only when you think in terms of "particle properties" (as opposed to "measurement outcomes") does "CFD" become a well-defined concept.

19. Jun 27, 2017

### Staff: Mentor

These are stated in vague ordinary language. You should state them using precise math. What is a "joint state"? What does "separable" mean? What does "no dependency" mean?

The great virtue of Bell's papers was that he gave explicit, precise mathematical conditions instead of vague ordinary language ones.

20. Jun 28, 2017

### Eye_in_the_Sky

Agreed.
Let us see just how sharply those concepts can be defined, and then look back to see what will stand, what will fall, and what (if anything) will remain fuzzy.
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Begin with the notion of 'separability'.

Let X and Y be two systems with associated state spaces SX and SY.

Now, consider X and Y together as a single system and designate this (combined) system as the "joint-system" consisting of X and Y.

Then, the "joint-system" is said to be 'separable' with respect to X and Y iff

a) The state space of the joint-system is given by

SX x SY (Cartesian product);

and

b) The "state of X" is u and the "state of Y" is v

iff

the "joint-state of X and Y" is (u,v).

This "joint-state" is said to be 'separable' with respect to the subsystems X and Y.
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Let the state space of Alice's instrument be given by the Cartesian product of the set of unit vectors in R2 with the set {0, U, D}.

The unit vector represents the 'setting'.

The elements of the other set represent the 'values':

0 - ready for registration ,

U - an "up" has been registered ,

D - a "down" has been registered .

Likewise, for Bob's instrument.
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... I will stop here for any comments and to think more about the next stage of clarification.

21. Jun 28, 2017

### ueit

Would you agree that if some interaction takes place between X and Y (electromagnetic, gravitational, or some other type) their joint system (XY) cannot be separable with respect to X and Y?

22. Jun 28, 2017

### Staff: Mentor

As far as I can tell, in any classical (i.e., non-quantum) physical theory, every joint state will be separable by this definition. Quantum mechanics is the only theory I'm aware of that has joint states that are non-separable by this definition (which basically corresponds to the definition of non-entangled states in QM).

23. Jun 28, 2017

### morrobay

When detectors are aligned a = b the outcomes are a particular exception to all other cases.
For example perfect anti correlations when a = b for entangled particles created from a neutral particle.
Pre existing values, realism and CFD are the only explanations for perfect (anti) correlations ie. zero total angular momentum for the pair and together with locality are the two major premises for the derivation of Bell's inequality. *
Take for example settings a , b, c for entangled pair.
Particle 1 Settings a , b ,c = 00 , 1200 , 1600 a+ b+ c+
Particle 2 Settings a , b, c = 00 . 1200 , 1600 a - b- c-
P+- = P-+ = (cos a-b/2)2 When detectors in special case are aligned inequality cannot be violated: cos 0 + cos 0 ≥ cos 0
However in general cases , a ≠ b inequality is violated :
(a+c-) + (a+b-) ≥ (b+c-)
(.030 ) + ( .25) ≥ (..88 )

* CFD is related to the EPR paradox that Bell used in deriving inequality:
If particle 1 is measured to be spin up on x axis then particle 2 must be spin down on x axis.
If particle 2 is measured to be spin up on y axis then if particle 1 had been measured on y axis
it would have been spin down. Knowing the outcomes simultaneously of axis y and axis x for both particles is in conflict
with QT

Last edited: Jun 29, 2017
24. Jun 29, 2017

### Eye_in_the_Sky

Interaction will have nothing to do with it. Separability just means

a specification of a joint-state to the joint-system

is equivalent to

a specification of a state to each of the subsystems.

25. Jun 29, 2017

### Eye_in_the_Sky

Yes, every system in classical physics is separable down to its finest grain.