Did Bell's Theory Contain a Mathematical Flaw?

[Gordon Watson]

Is the right side of equation 9 supposed to involve probabilities?

No, (9) is meant to represent a discrete normalized probability distribution. So LHS says what it is and RHS gives the normalized distribution.

If this were a continuous case then RHS would be written with delta-functions.

Does this last comment help?

If so I think it would make more sense to write it as P_3(\lambda_{k}, \lambda'_{k}) = (1/2)[P(\lambda_{k} = \lambda_{a}, \lambda'_{k} = \lambda'_{a}) + P(\lambda_{k} = \lambda_{ai}, \lambda'_{k} = \lambda'_{ai})]...is that OK?

As explained in earlier reply: These equalities have probability zero prior to measurement.

Do you now agree?

Even so the equation is a little unclear, does the 1/2 just mean you assume P(\lambda_{k} = \lambda_{a}, \lambda'_{k} = \lambda'_{a}) = 1/2 and P(\lambda_{k} = \lambda_{ai}, \lambda'_{k} = \lambda'_{ai}) = 1/2, i.e. these are the only two possible ways the source can emit spin vectors and both ways are equally probable?

Yes, but you have to see that we simply cannot go with those equality signs.

We have to go with the characteristic spin-related transformation symmetries that define the equiprobable etc. equivalence classes (EC).

Then the sum of the two would be 1, and you'd have to multiply that by 1/2 to get P_3(\lambda_{k}, \lambda'_{k}) for any specific value of k (k=a or k=ai). If that's the idea, I think it would be a lot less convoluted to just say that k can only take one of two values, k=a or k=ai, (which could perhaps be represented by the numerals 1 and 2, so the sum would just be \sum_{k=1}^2) and that either way P_3(\lambda_{k}, \lambda'_{k}) = 1/2.

Check what I said in earlier reply, please.

k_1 now denotes the defining condition for EC_1, k_2 the defining condition for EC_2.

Do you go with that?

In that case we could go from equation (8) to a simplified version of (10):

P''(G,G'|H,a,b') = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k)*(cos^2(b',\lambda_k)|H,b,\lambda_k)

Well I don't see how this gives any output??

If you change it to this, well OK:

P''(G,G'|H,a,b') = \sum_{k=1}^2 (1/2)*(cos^2(a,\lambda_k)|H,a,\lambda_k = \lambda_1)*(cos^2(b',\lambda_k)|H,b, \lambda_k = \lambda_1,

where lambda_1 is defined by the \uparrow business -- which is a discriminator between the ECs.

And, for completeness you include the equivalent expression for lambda_2 -- which is also is defined by the \uparrow business -- it being also a discriminator between the ECs. Though this addition = 0, on summation, it is needed for completeness.

Which is all getting messy for me, and surely for you?

Thanks, as always,

Jenni

 

[JesseM]

As I see it, we are dealing with Probability Functions. They map a subset of the Sample Space to [0, 1].

λ_k is the angle of the spin vector given to the particle when it is first sent out by the source.

OK, so you were using the horizontal arrow in P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) to represent the possibility that the initial spin vector λ_k was changed to the new spin vector λ_a as the particle passed through the polarizer, correct? If so my earlier interpretation of the equation was wrong so much of my subsequent discussion of your equations was based on wrong premises, we should step back and start from here.

If so we can say that at the moment the two particles first begin their journey from the source, their initial spin vectors should give them a predetermined answer to what direction they would end up if they met a polarizer at any given angle, right?

That is correct.

Is that what is required?

Yes, that's what Bell would assume must be true under local realism where the experimenters choose detector angles randomly and always find that same detector angle=same result...it must be true (under local realism and the given experimental conditions) that the particles had identical predetermined responses for each detector angle (so presumably the source always emits pairs of particles with hidden variables that give them identical predetermined responses).

For example, we might have a situation where one particle's initial spin vector (the spin vector assigned to it by the source, prior to encountering the polarizers) is such that it if it encountered a polarizer at a=30 it would be predetermined to end up parallel to it, predetermined to come out orthogonal a polarizer at a=60, and predetermined to come out parallel to a polarizer at a=120. Is this how you see things,...

Yes.

IS THAT HOW IT SHOULD BE? Sorry for caps - hit wrong key.

Yes, according to local realism. OK, so let's focus on this issue for now. Given that each λ_k gives predetermined responses to each polarizer angle, would you agree that if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this:

L₁ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, and to pass through setting 120. We can denote this 0+, 60+, 120+

L₂ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, but not to pass through setting 120. We can denote this 0+, 60+, 120-

L₃: 0+, 60-, 120+

L₄: 0+, 60-, 120-

L₅: 0-, 60+, 120+

L₆: 0-, 60+, 120-

L₇: 0-, 60-, 120+

L₈: 0-, 60-, 120-

So no matter what direction λ_k is on any given trial, it must be associated with a single value of L_i, one of the 8 possible values above. Agreed? And to explain the fact that the experimenters always get identical results when they both choose the same polarizer angle, we must assume that on each trial, each particle has an identical value of L_i...e.g. if the particle sent to Alice has an initial spin vector λ_k that can be classified as L₅, then the particle sent to Bob must also have an initial spin vector that can be classified as L₅. Would you agree with this so far?

 

[Gordon Watson]

OK, so you were using the horizontal arrow in P(G \uparrow H , a, \lambda_{k}\hookrightarrow \lambda_{a}) to represent the possibility that the initial spin vector λ_k was changed to the new spin vector λ_a as the particle passed through the polarizer, correct? If so my earlier interpretation of the equation was wrong so much of my subsequent discussion of your equations was based on wrong premises, we should step back and start from here.


Yes, that's what Bell would assume must be true under local realism where the experimenters choose detector angles randomly and always find that same detector angle=same result...it must be true (under local realism and the given experimental conditions) that the particles had identical predetermined responses for each detector angle (so presumably the source always emits pairs of particles with hidden variables that give them identical predetermined responses).


Yes, according to local realism. OK, so let's focus on this issue for now. Given that each λ_k gives predetermined responses to each polarizer angle, would you agree that if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this:

L₁ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, and to pass through setting 120. We can denote this 0+, 60+, 120+

L₂ means the original λ_k was such that the particle would be predetermined to pass through setting 0, to pass through setting 60, but not to pass through setting 120. We can denote this 0+, 60+, 120-

L₃: 0+, 60-, 120+

L₄: 0+, 60-, 120-

L₅: 0-, 60+, 120+

L₆: 0-, 60+, 120-

L₇: 0-, 60-, 120+

L₈: 0-, 60-, 120-

So no matter what direction λ_k is on any given trial, it must be associated with a single value of L_i, one of the 8 possible values above. Agreed? And to explain the fact that the experimenters always get identical results when they both choose the same polarizer angle, we must assume that on each trial, each particle has an identical value of L_i...e.g. if the particle sent to Alice has an initial spin vector λ_k that can be classified as L₅, then the particle sent to Bob must also have an initial spin vector that can be classified as L₅. Would you agree with this so far?

OK, thank you very much for these clarifications.

I think I am saying Yes to all the above; but maybe you should explain this small piece:

if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this

Q1: Why only 8?

Q2: Would it be better to discuss GHZ?

PS: Like, have you checked your mail-box?

Thank you,

JenniT

 

[JesseM]

OK, thank you very much for these clarifications.

I think I am saying Yes to all the above; but maybe you should explain this small piece:

if the experimenters are choosing between three polarizer angles 0, 60, and 120, we can define a new variable L_i with only 8 possible values, like this

Q1: Why only 8?

Because the point of L_i is just to specify which of two possible binary measurement results (+ or -) would be predetermined to occur for each of the three polarizer settings. Eight is the total number of possible combinations of +'s and -'s for the three angles:

0+, 60+, 120+
0+, 60+, 120-
0+, 60-, 120+
0+, 60-, 120-
0-, 60+, 120+
0-, 60+, 120-
0-, 60-, 120+
0-, 60-, 120-

Q2: Would it be better to discuss GHZ?

The point I'm getting at is just about whether cos^2(a-b) could be possible for the types of hidden variables theory you're assuming. I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

PS: Like, have you checked your mail-box?

Yup, sorry for the delay on that!

 

[Gordon Watson]

Because the point of L_i is just to specify which of two possible binary measurement results (+ or -) would be predetermined to occur for each of the three polarizer settings. Eight is the total number of possible combinations of +'s and -'s for the three angles:

0+, 60+, 120+
0+, 60+, 120-
0+, 60-, 120+
0+, 60-, 120-
0-, 60+, 120+
0-, 60+, 120-
0-, 60-, 120+
0-, 60-, 120-

The point I'm getting at is just about whether cos^2(a-b) could be possible for the types of hidden variables theory you're assuming. I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

Yup, sorry for the delay on that!

I'm also sorry for the delay too.

But note that lambda has to deliver results at any combination of any angular settings; so OK; please go ahead and demonstrate the point. If it fails at 8, I guess you'd agree it fails at a countable infinity of angular settings also. Right?

PS: On a related point, could you briefly explain what I should understand by spin? For example, photons are spin 1; electrons spin 1/2. What does this difference mean? What are its consequences, please? If someone said that electrons were spin 1, photons spin 2, would the world crumble? Why would they be wrong? Could an EPR-Bohm test shoot them down? Or some other test?

You guessed: I'm not very much into QM; mainly maths. Hope you can help; thanks.

Jennifer

 

[Gordon Watson]

I want to show that, given the assumption that every λ_k falls into one of the above 8 categories of predetermined responses above, it's impossible to get cos^2(a-b) for the probability they both get the same response when they choose different polarizer angles (note that for the polarizer angles 0, 60, and 120, if Alice and Bob choose different angles, cos^2(a-b) will always be equal to 0.25)

OK, good. So that I understand what is going on: Please be sure to show the QM result for however you arrive at your conclusion with the above assumptions, please.

Must run, J

 

[JesseM]

I'm also sorry for the delay too.

But note that lambda has to deliver results at any combination of any angular settings; so OK; please go ahead and demonstrate the point. If it fails at 8, I guess you'd agree it fails at a countable infinity of angular settings also. Right?

Right, although any given Bell test will involve only a small number of possible angular settings for the experimenters to choose from, like the three possible settings 0, 60 and 120 that I assumed are available to Alice and Bob. Of course, it's arbitrary what set of angles we choose to make available, I just chose 0, 60 and 120 for convenience.

So, assume that on each trial Alice and Bob choose their settings randomly, and independently. Then in the limit as the number of trials goes to infinity, each possible combination of settings will occur equally frequently--for example (Alice:60, Bob:120) occurs on the same fraction of trials as (Alice:120, Bob:0).

On each trial, both particles have the same value for Li (for example on any trial where Alice's particle has a λ_k corresponding to L₅, Bob's particle also has a λ_k corresponding to L₅). This means that in the subset of trials where Alice and Bob both chose the same detector setting, they always get the same measurement result with probability 1. But what if we look at the subset of trials where they chose different settings, what is the probability they got the same measurement result in that case? Here we want the conditional probability that they get the same result given that they chose different settings, which we can write as P(same measurement result | Alice&Bob chose different measurement settings).

Now, under the type of hidden variables theory you are proposing, would you agree we can break this up into the following sum?

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

If you agree this sum is valid, consider an individual term like P(same measurement result | Alice&Bob chose different measurement settings AND L₅). L₅ was 0-, 60+, 120+. Would you agree that since two predetermined results are + and one is -, the only way they can both get the same measurement result is if they both chose settings with predetermined result +? And would you agree that if we know Alice chose her setting randomly, there should be a 2/3 chance she chose a setting with predetermined result + (i.e. either 60 or 120), and if we know Bob also chose randomly and we are dealing with a trial where his setting was different from Alice's, there is a 1/2 chance he chose the other setting with predetermined result +? In other words, do you agree that P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = (2/3)*(1/2) = 1/3? Another way of seeing this is just by considering all possible ways they can choose different settings, and what their results will be given particles in state L₅:
(Alice:0, Bob:60): different measurement results (Alice-, Bob+)
(Alice:0, Bob:120): different measurement results (Alice-, Bob+)
(Alice:60, Bob:0): different measurement results (Alice+, Bob-)
(Alice:60, Bob:120): same measurement results (Alice+, Bob+)
(Alice:120, Bob:0) different measurement results (Alice+, Bob-)
(Alice:120, Bob:60) same measurement results (Alice+, Bob+)

So, in 2 out of the 6 possible cases where they choose different settings, they get the same result, meaning P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = 2/6 = 1/3.

If you agree with this, it's not hard to see why P(same measurement result | Alice&Bob chose different measurement settings AND L₂) would also equal 1/3 (since here you also have two + and one - for the predetermined results), and likewise for L₃, L₄, L₆ and L₇, which we might call the "inhomogenous states" since they each involve two predetermined results of one type and one of the other. On the other hand, for the "homogenous" states L₁ and L₈ where all three predetermined results are the same, then even if they choose different settings they are guaranteed to get the same result, so P(same measurement result | Alice&Bob chose different measurement settings AND L₁) = P(same measurement result | Alice&Bob chose different measurement settings AND L₈) = 1.

Let me know whether you agree with all of this so far, and if so I'll continue, if not I'll try to elaborate on whatever you're unclear about in the above.

 

[Gordon Watson]

Right, although any given Bell test will involve only a small number of possible angular settings for the experimenters to choose from, like the three possible settings 0, 60 and 120 that I assumed are available to Alice and Bob. Of course, it's arbitrary what set of angles we choose to make available, I just chose 0, 60 and 120 for convenience.

So, assume that on each trial Alice and Bob choose their settings randomly, and independently. Then in the limit as the number of trials goes to infinity, each possible combination of settings will occur equally frequently--for example (Alice:60, Bob:120) occurs on the same fraction of trials as (Alice:120, Bob:0).

On each trial, both particles have the same value for Li (for example on any trial where Alice's particle has a λ_k corresponding to L₅, Bob's particle also has a λ_k corresponding to L₅). This means that in the subset of trials where Alice and Bob both chose the same detector setting, they always get the same measurement result with probability 1. But what if we look at the subset of trials where they chose different settings, what is the probability they got the same measurement result in that case? Here we want the conditional probability that they get the same result given that they chose different settings, which we can write as P(same measurement result | Alice&Bob chose different measurement settings).

Now, under the type of hidden variables theory you are proposing, would you agree we can break this up into the following sum?

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

If you agree this sum is valid, consider an individual term like P(same measurement result | Alice&Bob chose different measurement settings AND L₅). L₅ was 0-, 60+, 120+. Would you agree that since two predetermined results are + and one is -, the only way they can both get the same measurement result is if they both chose settings with predetermined result +? And would you agree that if we know Alice chose her setting randomly, there should be a 2/3 chance she chose a setting with predetermined result + (i.e. either 60 or 120), and if we know Bob also chose randomly and we are dealing with a trial where his setting was different from Alice's, there is a 1/2 chance he chose the other setting with predetermined result +? In other words, do you agree that P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = (2/3)*(1/2) = 1/3? Another way of seeing this is just by considering all possible ways they can choose different settings, and what their results will be given particles in state L₅:
(Alice:0, Bob:60): different measurement results (Alice-, Bob+)
(Alice:0, Bob:120): different measurement results (Alice-, Bob+)
(Alice:60, Bob:0): different measurement results (Alice+, Bob-)
(Alice:60, Bob:120): same measurement results (Alice+, Bob+)
(Alice:120, Bob:0) different measurement results (Alice+, Bob-)
(Alice:120, Bob:60) same measurement results (Alice+, Bob+)

So, in 2 out of the 6 possible cases where they choose different settings, they get the same result, meaning P(same measurement result | Alice&Bob chose different measurement settings AND L₅) = 2/6 = 1/3.

If you agree with this, it's not hard to see why P(same measurement result | Alice&Bob chose different measurement settings AND L₂) would also equal 1/3 (since here you also have two + and one - for the predetermined results), and likewise for L₃, L₄, L₆ and L₇, which we might call the "inhomogenous states" since they each involve two predetermined results of one type and one of the other. On the other hand, for the "homogenous" states L₁ and L₈ where all three predetermined results are the same, then even if they choose different settings they are guaranteed to get the same result, so P(same measurement result | Alice&Bob chose different measurement settings AND L₁) = P(same measurement result | Alice&Bob chose different measurement settings AND L₈) = 1.

Let me know whether you agree with all of this so far, and if so I'll continue, if not I'll try to elaborate on whatever you're unclear about in the above.

Pray, good sir; please continue -- this is getting interesting -- must rush; J xxoo

 

[JesseM]

OK, if you agreed that P(same measurement result | Alice&Bob chose different measurement settings AND L_i) would be equal to 1/3 if L_i was L₂, L₃, L₄, L₅, L₆ or L₇, and you also agreed that it would be equal to 1 if L_i was L₁ or L₈, then this equation:

P(same measurement result | Alice&Bob chose different measurement settings) =
P(same measurement result | Alice&Bob chose different measurement settings AND L₁)*P(L₁) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₂)*P(L₂) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₃)*P(L₃) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₄)*P(L₄) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₅)*P(L₅) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₆)*P(L₆) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₇)*P(L₇) +
P(same measurement result | Alice&Bob chose different measurement settings AND L₈)*P(L₈)

...would reduce to this:

P(same measurement result | Alice&Bob chose different measurement settings) = 1/3*[P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] + 1*[P(L₁) + P(L₈)]

Now, we don't know the value of each P(L_i), they represent the frequencies that the source sends out pairs of particles with different values of L_i, which would depend on how the source works in the context of our hidden-variables theory. But we do know that the sum of all 8 probabilities must be 1 since those are the only possible values for L_i, so [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] + [P(L₁ + P(L₈)] = 1.

If [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] = 1 while [P(L₁ + P(L₈)] = 0, then that would mean P(same measurement result | Alice&Bob chose different measurement settings) = 1/3. If [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] = 0 while [P(L₁ + P(L₈)] = 1, that would mean P(same measurement result | Alice&Bob chose different measurement settings) = 1. If both [P(L₂) + P(L₃) + P(L₄) + P(L₅) + P(L₆) + P(L₇)] and [P(L₁ + P(L₈)] have values between 0 and 1 (which together add up to one), then that would mean P(same measurement result | Alice&Bob chose different measurement settings) is somewhere between 1/3 and 1. In any case, according to this type of hidden variables theory it must be true that P(same measurement result | Alice&Bob chose different measurement settings) is larger than or equal to 1/3. Do you agree?

If so, it becomes easy to see why this type of hidden variables theory can't reproduce the cos^2 relationship predicted by QM. If the three angles are 0, 60, and 120, then on any trial where Alice and Bob pick different detector angles, the probability of getting the same result according to QM will always be 0.25, since cos^2(120 - 0) = cos^2(120 - 60) = cos^2(60 - 0) = cos^2(60 - 120) = cos^2(0 - 60) = cos^2(0 - 120) = 0.25, which is smaller than 1/3.