Is the Joint-State of Alice and Bob's Instruments Separable in Spacetime?

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In summary: Alice and Bob".In that thought experiment, it is stipulated that if Bob's setting would have been b' instead of b, then each of Alice and Bob would have obtained a definite outcome.Is this statement true in general, or is it only true in the instance where Bob's setting is actually b'?It is true in general.
  • #1
Eye_in_the_Sky
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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.
 
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  • #2
Of course, in the above, I assume NO retro-causality.
 
  • #3
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?
____

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
____

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
____

... Any comments?
____
____

ASIDE:

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

(I'm sure @Demystifier can answer this.)
 
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  • #4
Eye_in_the_Sky said:
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.All four of the above conditions, as well as CFD, hold true in BM ... correct?

(I'm sure @Demystifier can answer this.)
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
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.
 
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  • #6
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.

... Does anyone have anything to say about this statement?
 
  • #7
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.
_____

Shortly, I will start a new thread.
_____
_____

EDIT: The new thread is Seeing the essence of Bell.
 
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  • #8
Eye_in_the_Sky said:
All four of the above conditions, as well as CFD, hold true in BM ... correct?
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
Eye_in_the_Sky said:
I will start a new thread.

That is not necessary, nor is it desirable.

Eye_in_the_Sky said:
The new thread is https://www.physicsforums.com/threads/seeing-the-essence-of-bell.918624/.

Which I have closed and deleted. Please keep the discussion here.
 
  • #10
PeterDonis said:
That is not necessary, nor is it desirable.
... Please keep the discussion here.
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
Eye_in_the_Sky said:
The two topics are very distinct

They might seem that way to you, but to me they look close enough that they can happily coexist in the same thread.
 
  • #12
PeterDonis said:
they can happily coexist in the same thread
Let it be so.
_____
_____

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.

_____

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.
_____

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
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
morrobay said:
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

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.
 
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  • #15
morrobay said:
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
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
Reread my post. I made no such reference to a = b or a = -b
 
  • #17
morrobay said:
Reread my post. I made no such reference to a = b or a = -b
So what? These are particular examples of your more general case, not?
 
  • #18
morrobay said:
condition 2 ( locality + CFD )

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
Eye_in_the_Sky said:
Consider the following two conditions

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.
 
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  • #20
PeterDonis said:
The great virtue of Bell's papers was that he gave explicit, precise mathematical conditions instead of vague ordinary language ones.
Agreed.
You should state them using precise math. What is a "joint state"? What does "separable" mean? What does "no dependency" mean?
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.
_____

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.
_____

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.
_____

... I will stop here for any comments and to think more about the next stage of clarification.
 
  • #21
Eye_in_the_Sky said:
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.
_____

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.

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
Eye_in_the_Sky said:
This "joint-state" is said to be 'separable' with respect to the subsystems X and Y.

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

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
 
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  • #24
ueit said:
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?
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
PeterDonis said:
As far as I can tell, in any classical (i.e., non-quantum) physical theory, every joint state will be separable by this definition.
Yes, every system in classical physics is separable down to its finest grain.
 
  • #26
Eye_in_the_Sky said:
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.

OK, but in the case there is interaction between subsystem X and Y, Alice's outcome would depend upon Bob's setting. So, your entire argument refers to classical non-interacting systems, right?
 
  • #27
ueit said:
OK, but in the case there is interaction between subsystem X and Y, Alice's outcome would depend upon Bob's setting.
Can you explain that more clearly? I don't understand.
So, your entire argument refers to classical non-interacting systems, right?
I don't quite get what you are asking. So, I will make some remarks.

Here is improved wording for the first condition in the argument:

1) The joint-system composed of the measuring instruments of Alice and Bob is a separable system.

(Thus, the relationship that the two instruments share in spacetime is a relationship of classical objects.)

The second condition in the argument is:

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

(Andrei, are you asking about how "no dependency" is to be understood? I haven't explained that yet.)

The purpose of the argument is to show:

In any correct interpretation of Quantum Theory at least one of the conditions, 1 or 2, must hold false.
 
  • #28
ueit said:
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?

Let subsystems X and Y be space like separated so there can be no classical interactions: electromagnetic, gravitational, magnetic.

Capture.jpg

The non - classical correlations (in conflict with separability) of entangled particles do not involve any faster than light signals. Nor any dependency of outcomes at Alice on settings or outcome at Bob and visa versa. But rather on the non separability of QM :The QM wave function joint probabilities do not factor into individual probabilities.
http://www.mathpages.com/home/kmath731/kmath731.htm
 
  • #29
Eye_in_the_Sky said:
Can you explain that more clearly? I don't understand.

I don't quite get what you are asking. So, I will make some remarks.

Here is improved wording for the first condition in the argument:

1) The joint-system composed of the measuring instruments of Alice and Bob is a separable system.

(Thus, the relationship that the two instruments share in spacetime is a relationship of classical objects.)

The second condition in the argument is:

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

(Andrei, are you asking about how "no dependency" is to be understood? I haven't explained that yet.)

The purpose of the argument is to show:

In any correct interpretation of Quantum Theory at least one of the conditions, 1 or 2, must hold false.

As you said, in any classical theory condition 1 holds true. On the other hand if the subsystems X and Y interact with each other (and virtually all matter interacts electromagnetically and gravitationaly) condition 2 seems false. So, I guess you are right, I don't understand how this "non-dependency" is going to be enforced in any theory except for the most naive ones.
 
  • #30
morrobay said:
Let subsystems X and Y be space like separated so there can be no classical interactions: electromagnetic, gravitational, magnetic.

OK, let the two subsystems be an electron (X) and a proton (Y). How would you make them space like separated? As far as I can tell space like separation refers to events, not to physical systems that are more or less eternal.
 
  • #31
ueit said:
OK, let the two subsystems be an electron (X) and a proton (Y). How would you make them space like separated? As far as I can tell space like separation refers to events, not to physical systems that are more or less eternal.
Bob's fixing of setting is spacelike separated from Alice's registration of outcome, and vice versa.
_____
ueit said:
As you said, in any classical theory condition 1 holds true. On the other hand if the subsystems X and Y interact with each other (and virtually all matter interacts electromagnetically and gravitationaly) condition 2 seems false.
But Bob's (free) choice of setting is spacelike separated from the registration of Alice's outcome, so just what kind of 'interaction' could possibly establish such a dependency?
 
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  • #32
Eye_in_the_Sky said:
1) The joint-system composed of the measuring instruments of Alice and Bob is a separable system.

(Thus, the relationship that the two instruments share in spacetime is a relationship of classical objects.)
Each of the instruments (in and of itself) can, for the sake of making the argument, be regarded as a classical object altogether.

(This is because the argument does not involve, in any way, any of their subsystems).

But the scope of the argument remains in the broader sense.
 
  • #33
Eye_in_the_Sky said:
Bob's fixing of setting is spacelike separated from Alice's registration of outcome, and vice versa.
_____

But Bob's (free) choice of setting is spacelike separated from the registration of Alice's outcome, so just what kind of 'interaction' could possibly establish such a dependency?

In your post #5 you said:

I should not have included "free choice" in the ASIDE.

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

I understand that you have changed your mind. In this case, what is your take on my observation:

"As far as I can tell your condition 1 (free choice) has no justification in any theory, classical or quantum."

I am especially interested in what mechanism do you think is responsable for the property of free choice so that we can asses if it is compatible with the classical theory that you intend to test.

Andrei

Andrei
 
  • #34
ueit said:
what is your take on my observation:

"As far as I can tell your condition 1 (free choice) has no justification in any theory, classical or quantum."
I agree.
I am especially interested in what mechanism do you think is responsable for the property of free choice so that we can asses if it is compatible with the classical theory that you intend to test.
I do not think it is a 'mechanism'. I think it is a 'soul' and that the 'soul' has a character which is out of the scope of any of the known "physics" of today. I would like to believe that there is a subject matter which, in a very deep way, is somehow like 'physics' and is also somehow about the 'soul', but I have no idea what that could be.
 
  • #35
Eye_in_the_Sky said:
I agree.

I do not think it is a 'mechanism'. I think it is a 'soul' and that the 'soul' has a character which is out of the scope of any of the known "physics" of today. I would like to believe that there is a subject matter which, in a very deep way, is somehow like 'physics' and is also somehow about the 'soul', but I have no idea what that could be.

What you are saying is that the existence of free-choice represents a falsification of all current physical theories. Are you aware of any experiment backing that up?
 
<h2>1. What is state separability in CFD?</h2><p>State separability in CFD (Computational Fluid Dynamics) refers to the ability to divide a complex fluid flow problem into smaller, more manageable sub-problems. This allows for the use of simpler and more efficient numerical methods to solve the individual sub-problems, which can then be combined to obtain a solution for the entire problem.</p><h2>2. Why is state separability important in CFD?</h2><p>State separability is important in CFD because it allows for more accurate and efficient simulations of complex fluid flow problems. By breaking down the problem into smaller sub-problems, it is easier to apply appropriate numerical methods and accurately capture the behavior of the fluid flow. This can lead to faster and more reliable results.</p><h2>3. How is state separability achieved in CFD?</h2><p>State separability is achieved in CFD through the use of numerical methods such as finite difference, finite volume, and finite element methods. These methods discretize the fluid flow domain into smaller elements or control volumes, which can then be solved individually and combined to obtain a solution for the entire problem.</p><h2>4. What are the benefits of state separability in CFD?</h2><p>State separability offers several benefits in CFD, including improved accuracy, faster computation times, and the ability to handle more complex fluid flow problems. It also allows for the use of parallel computing techniques, which can further increase the efficiency of CFD simulations.</p><h2>5. Are there any limitations to state separability in CFD?</h2><p>While state separability is a powerful tool in CFD, it does have some limitations. In some cases, the sub-problems may interact with each other, making it difficult to accurately combine the results. Additionally, the accuracy of the overall solution may be limited by the accuracy of the individual sub-problems. Careful consideration and validation of the results is necessary to ensure the reliability of the simulation.</p>

Related to Is the Joint-State of Alice and Bob's Instruments Separable in Spacetime?

1. What is state separability in CFD?

State separability in CFD (Computational Fluid Dynamics) refers to the ability to divide a complex fluid flow problem into smaller, more manageable sub-problems. This allows for the use of simpler and more efficient numerical methods to solve the individual sub-problems, which can then be combined to obtain a solution for the entire problem.

2. Why is state separability important in CFD?

State separability is important in CFD because it allows for more accurate and efficient simulations of complex fluid flow problems. By breaking down the problem into smaller sub-problems, it is easier to apply appropriate numerical methods and accurately capture the behavior of the fluid flow. This can lead to faster and more reliable results.

3. How is state separability achieved in CFD?

State separability is achieved in CFD through the use of numerical methods such as finite difference, finite volume, and finite element methods. These methods discretize the fluid flow domain into smaller elements or control volumes, which can then be solved individually and combined to obtain a solution for the entire problem.

4. What are the benefits of state separability in CFD?

State separability offers several benefits in CFD, including improved accuracy, faster computation times, and the ability to handle more complex fluid flow problems. It also allows for the use of parallel computing techniques, which can further increase the efficiency of CFD simulations.

5. Are there any limitations to state separability in CFD?

While state separability is a powerful tool in CFD, it does have some limitations. In some cases, the sub-problems may interact with each other, making it difficult to accurately combine the results. Additionally, the accuracy of the overall solution may be limited by the accuracy of the individual sub-problems. Careful consideration and validation of the results is necessary to ensure the reliability of the simulation.

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