# Possible explanation for a Bell experiment?

#### PR0

The following assessment of a Bell experiment is based on N. David Mermin's example and is intended for persons with very little understanding of mathematics and physics (myself included).

Assumptions
(A1) A source emits a pair of particles with some opposite pieces of information.
(A2) The pieces of information from (A1) remain constant over time.
(A3) A detector can measure the pieces of information from (A1) via three methods [X, Y and Z], but only one method at a time can be used. Each method measures a specific part of the information.
(A4) There are only two values possible for the measurements from (A3). [+ and -]

Theoretical Results
A source emits pairs of particles as per the above assumptions.
One particle is sent to a detector (Alice) and the other to another detector (Bob).
There is no predefined arrangement for what type of particle pair the source emits (it can be the same type for the entire duration of the experiment or randomly different with each consecutive emission or any other combination) and for what methods Alice or Bob will use, just that each combination of measurements is used equally.
There are maximum 8 combinations of particles and 9 combinations of measurements:
Code:
    Particles for          Results [Alice|Bob]          Opposite
Alice and Bob   1   2   3   4   5   6   7   8   9   results
X Y Z   X Y Z  X|X X|Y X|Z Y|X Y|Y Y|Z Z|X Z|Y Z|Z  probability
A   + + +   - - -   O   O   O   O   O   O   O   O   O   1
B   + + -   - - +   O   O   I   O   O   I   I   I   O   5/9
C   + - +   - + -   O   I   O   I   O   I   O   I   O   5/9
D   + - -   - + +   O   I   I   I   O   O   I   O   O   5/9
E   - + +   + - -   O   I   I   I   O   O   I   O   O   5/9
F   - + -   + - +   O   I   O   I   O   I   O   I   O   5/9
G   - - +   + + -   O   O   I   O   O   I   I   I   O   5/9
H   - - -   + + +   O   O   O   O   O   O   O   O   O   1
From the above we get the following equivalent of Bell's inequality:
Probability of getting opposite ("O") results at the detectors > 5/9

Actual Results
We consider the case of two entangled electrons, for which their spin is measured at three angles: X=0°, Y=120°, Z=240°.
Probability of getting opposite results at the detectors = 1/2

Conclusions
(I) First of all I would like to state my lack of understanding on how probabilities can be meaningfully compared with actual results. Take for example the case of a dice for which we have the probability of 1/6 to get any of the faces. Does that mean that necessarily after many throws we'll get each face for about 1/6 of the number of total throws?

(II) Returning to the experiment, I think the above probability chart applies only for the case in which the types of particle pairs are measured by a complete set of measurement combinations (9). But this is something that can't be guaranteed as per how the experiment is performed, so IMO the following chart would be more appropriate:
Code:
    Particles for          Results [Alice|Bob]
Alice and Bob   1   2   3   4   5   6   7   8   9
X Y Z   X Y Z  X|X X|Y X|Z Y|X Y|Y Y|Z Z|X Z|Y Z|Z
A   + + +   - - -   O   O   O   O   O   O   O   O   O
B   + + -   - - +   O   O   I   O   O   I   I   I   O
C   + - +   - + -   O   I   O   I   O   I   O   I   O
D   + - -   - + +   O   I   I   I   O   O   I   O   O
E   - + +   + - -   O   I   I   I   O   O   I   O   O
F   - + -   + - +   O   I   O   I   O   I   O   I   O
G   - - +   + + -   O   O   I   O   O   I   I   I   O
H   - - -   + + +   O   O   O   O   O   O   O   O   O
Opposite results    1  4/8 4/8 4/8  1  4/8 4/8 4/8  1
probability
Probability of getting opposite ("O") results at the detectors > 4/8 = 1/2

(III) And finally, let's consider the following case (one of many) which AFAIK fully complies with the requirements of the experiment:
If the source emits pair B of particles when it just happens for Alice and Bob to do combinations 1, 3, 5, 6, 7, 8, 9 of measurements and pair C of particles when it just happens for Alice and Bob to do combinations 2 and 4 of measurements, we get a ratio of 1/3 for opposite results. Considering that this is a real possibility shouldn't only the following inequality be considered for a meaningful comparison with the actual results:
The ratio between the number of opposite results and total number of measurements > 3/9 = 1/3 ?

Related Quantum Physics News on Phys.org

#### Heinera

(I) First of all I would like to state my lack of understanding on how probabilities can be meaningfully compared with actual results. Take for example the case of a dice for which we have the probability of 1/6 to get any of the faces. Does that mean that necessarily after many throws we'll get each face for about 1/6 of the number of total throws?
Yes. But for any finite number of throws, the result will typically not be exactly 1/6.

(III) And finally, let's consider the following case (one of many) which AFAIK fully complies with the requirements of the experiment:
If the source emits pair B of particles when it just happens for Alice and Bob to do combinations 1, 3, 5, 6, 7, 8, 9 of measurements and pair C of particles when it just happens for Alice and Bob to do combinations 2 and 4 of measurements, we get a ratio of 1/3 for opposite results. Considering that this is a real possibility shouldn't only the following inequality be considered for a meaningful comparison with the actual results:
The ratio between the number of opposite results and total number of measurements > 3/9 = 1/3 ?
The Bell inequalities apply to expectations, i.e. they hold in the limit for an infinite number of particle pairs. In an experiment with a finite number of trials (like you suggest above), then by chance the inequality can be violated even with a local hidden variable model. But if you do millions of trials, the chance that you just happen to get a significant violation of the inequality becomes vanishingly small.

#### .Scott

Homework Helper
The point of the Bell Inequality was to describe the limit, given any hidden value assumptions.
So applying your 8 hidden value sets, we get this:

Code:
    Particles for          Results [Alice|Bob]                    Bell
Alice and Bob   1   2   3   4   5   6   7   8   9          Inequality
X Y Z   X Y Z  X|X X|Y X|Z Y|X Y|Y Y|Z Z|X Z|Y Z|Z  p(X,Z) - p(X,Y) - p(Y,Z)  <= 1
A   + + +   - - -   O   O   O   O   O   O   O   O   O     -1   -   -1   -   -1    =  1
B   + + -   - - +   O   O   I   O   O   I   I   I   O      1   -   -1   -    1    =  1
C   + - +   - + -   O   I   O   I   O   I   O   I   O     -1   -    1   -    1    = -3
D   + - -   - + +   O   I   I   I   O   O   I   O   O      1   -    1   -   -1    =  1
E   - + +   + - -   O   I   I   I   O   O   I   O   O      1   -    1   -   -1    =  1
F   - + -   + - +   O   I   O   I   O   I   O   I   O     -1   -    1   -    1    = -3
G   - - +   + + -   O   O   I   O   O   I   I   I   O      1   -   -1   -    1    =  1
H   - - -   + + +   O   O   O   O   O   O   O   O   O     -1   -   -1   -   -1    =  1
Notice that A, B, D, E, G, and H put us at the limit of the Bell Inequality, so any mixture of them will keep us there. But if we mix in any C or F, we will get closer to -3.
Using hidden values, there is no way to get p(X,Z) - p(X,Y) - p(Y,Z) > 1.

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#### PR0

@Heinera For me it's not obvious that millions or infinite trials will necessarily vindicate a probability. Is there some sort of proof for this, or it must be taken as an axiom?

@.Scott Unfortunately I don't understand what that p function represents (I did search about it, but didn't find something I could understand). Also, are you implying that the inequality with 5/9 isn't an equivalent to Bell's?

Mentor

#### .Scott

Homework Helper
@Heinera@.Scott Unfortunately I don't understand what that p function represents (I did search about it, but didn't find something I could understand).
The p function is simply a correlation between the measurements of the two values specified. If you have Excel, I can show it this way:

* Put "A" through "H" in column B, rows 2 to 9, with "A" at the top (cell B2) and "H" at the bottom (B9).
* In columns D, E, and F, put the corresponding plus and minus signs showing how X, Y, and Z are measured by Alice. So, for example, D5, E5, F5, will be '+, '-, '-, Alice's readings for your "D" particles.
* In columns H, I, and J, put the corresponding plus and minus signs showing how X, Y, and Z are measured by Bob. So, for example, H7, I7, J7, will be '+, '-, '+, Bob's readings for your "F" particles.
* Column "L" will be the p(X,Z) column. Put =IF(D2=J2,0.5,-0.5)+IF(F2=H2,0.5,-0.5) in cell L2.
* Column "M" will be the p(X,Y) column. Put =IF(D2=I2,0.5,-0.5)+IF(E2=H2,0.5,-0.5) in cell M2.
* Column "N" will be the p(Y,Z) column. You can actually copy cell M2 to N2 to get =IF(E2=J2,0.5,-0.5)+IF(F2=I2,0.5,-0.5).
* Now copy L2 to N2 to L3 to N9 to fill in the rest of the correlations.

You may notice that each p() computation is adding two values together that always end up being either +0.5 + +0.5 or -0.5 + -0.5. This is because you always have Alice read the opposite value of what Bob would have read.

Anyway, this should show you how the inequality is computed for each of your examples.

@Heinera@.ScottAlso, are you implying that the inequality with 5/9 isn't an equivalent to Bell's?
That is correct, it is not. If I could follow your logic, I would point out where you went wrong, but I don't understand how you got where you are.

#### PR0

@bhobba If I understand correctly those proofs rely on the existence of some special random variables (?). If that's the case, is it possible that in practice those conditions aren't always fulfilled? For e.g. in the OP I wrote that any combination for how the source emits the pairs of particles is possible. Isn't this a problem?

@.Scott Thank you for your explanation on how p is calculated. I even made the excel as suggested. But I'm struggling to understand what it means? For starters why 0.5 for equal and -0.5 for different? If I get this maybe I'll finally understand that inequality.

Regarding the 5/9 probability. It's related (at least I think it is) to N. David Mermin's example to which I gave a link in the OP (he claims that's an equivalent to Bell's). It's the probability for a single pair of particles (for each row in my chart, except two), though I don't think it's correct for the experiment, and should be calculated as I suggested in the OP (for each column, instead of row).

#### bhobba

Mentor
@bhobba If I understand correctly those proofs rely on the existence of some special random variables (?).
No.

It follows directly from the Kolmogerov axioms.

However here is not the place to discuss it - the probability subforum is.

Thanks
Bill

#### PR0

@bhobba But still, for our particular case if the source is indeed *biased* to some pattern/s of the type of particle pairs it emits, doesn't it make the probability calculations violable?

#### bhobba

Mentor
@bhobba But still, for our particular case if the source is indeed *biased* to some pattern/s of the type of particle pairs it emits, doesn't it make the probability calculations violable?
The theorem I mentioned makes no assumption on the probability distribution.

Its a well known and fundamental result of probability theory.

You need to take your concerns to the proper subforum - here is not the place to discuss it. Physics simply assumes well known and fundamental results like the law of large numbers.

Thanks
Bill

#### PR0

@bhobba I won't pursue this any further. I'm not questioning the theory per se, just that in practice it may be applied incorrectly. For instance we can calculate the probability for a dice but if it turns out the dice was heavier on one side...

#### bhobba

Mentor
@bhobba I won't pursue this any further. I'm not questioning the theory per se, just that in practice it may be applied incorrectly. For instance we can calculate the probability for a dice but if it turns out the dice was heavier on one side...
So? You throw the dice a large number of times and the law of large numbers guarantees it will give you the probability.

At the elementary level our own Dr Chinese details what going on with Bell:
http://www.drchinese.com/Bells_Theorem.htm

Thanks
Bill

Homework Helper

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#### DrChinese

Gold Member
@bhobba But still, for our particular case if the source is indeed *biased* to some pattern/s of the type of particle pairs it emits, doesn't it make the probability calculations violable?
Don't you think someone would notice that? And how would that cause the Bell Inequalities to be violated anyway? You can hand pick the outcomes and it won't be violated IF you don't also choose the angle settings for Alice and Bob.

#### Nugatory

Mentor
@bhobba But still, for our particular case if the source is indeed *biased* to some pattern/s of the type of particle pairs it emits, doesn't it make the probability calculations violable?
You may have to do the calculations for yourself before you are convinced, but if you try them you will find that a biased source cannot lead to a violation of the inequality. However, an intuitive way of seeing this is to consider that you cannot even hand-pick a set of particle attributes that will lead to a violation for all detector settings. Careful hand-picking to produce a desired result is the most extreme bias possible, and if it's not enough to do the job no lesser bias will be.

#### PR0

@bhobba So no matter how the dice may be rigged (or for that matter the environments in which the throws are performed) the probabilities still remain 1/6?

@.Scott I'll try to understand how those correlations work for this case.

@DrChinese @Nugatory Yes I agree, having just the source *biased* to some patterns is not enough, also Alice and Bob's choices would have to be *biased* in a similar way, and the combination of those patterns should give another mean/average result for all possibilities of synchronization. And yes, it is very unlikely, maybe at the same level as having the results sync'ed by chance, almost zero one might argue.

Can anyone confirm whether the inequality with the 5/9 probability can be an equivalent to Bell's as suggested by David Mermin? Or this probability doesn't apply to my example the same way as it does to Mermin's?

#### bhobba

Mentor
@bhobba So no matter how the dice may be rigged (or for that matter the environments in which the throws are performed) the probabilities still remain 1/6?.
Of course not.

I am saying it doesn't matter how its rigged we can determine its probabilities.

It was in response to:
@Heinera For me it's not obvious that millions or infinite trials will necessarily vindicate a probability. Is there some sort of proof for this, or it must be taken as an axiom?
You seemed to be doubting that for a large number of trials while it most likely will not be exactly as per the probabilities, it will likely get closer and closer to it as the trials get larger.

Thanks
Bill

#### Nugatory

Mentor
also Alice and Bob's choices would have to be *biased* in a similar way,
Yes, introducing bias at the detectors can lead to the violations of the inequality. There are two possibilities here:
- If the detectors can miss detecting some particles altogether so some pairs aren't counted, and there is some bias in which pairs are dropped, then it is possible for a local hidden-variable theory to violate the inequalities. This is the so-called "fair sampling" loophole, and it has been closed by the most recent experiment.
- If the detector settings themselves can be biased by something that is also biasing the source, then you can produce just about any imaginable result. Google for "superdeterminism", and it cannot even in principle be experimentally tested.

#### DrChinese

Gold Member
@DrChinese @Nugatory Yes I agree, having just the source *biased* to some patterns is not enough, also Alice and Bob's choices would have to be *biased* in a similar way, and the combination of those patterns should give another mean/average result for all possibilities of synchronization. And yes, it is very unlikely, maybe at the same level as having the results sync'ed by chance, almost zero one might argue.

Can anyone confirm whether the inequality with the 5/9 probability can be an equivalent to Bell's as suggested by David Mermin? Or this probability doesn't apply to my example the same way as it does to Mermin's?
Yes, since the chance of the outcome of Bell tests being due to a fluke diminishes to zero, it is not science to explain it that way. You can just as easily say that the speed of light varies and we just happen to measure it as c every time we check it. This issue is not unique to Bell tests.

I compare the 5/9 classical minimum (per your "Bell/Mermin" example) to the 1/2 prediction of QM. The experimental value will be below 5/9 and approach 1/2 as efficiency improves.

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#### morrobay

Gold Member
[QUOTE=".

Code:
    Particles for          Results [Alice|Bob]
Alice and Bob   1   2   3   4   5   6   7   8   9
X Y Z   X Y Z  X|X X|Y X|Z Y|X Y|Y Y|Z Z|X Z|Y Z|Z  p(X,Z) - p(X,Y) - p(Y,Z)  <= 1
A   + + +   - - -   O   O   O   O   O   O   O   O   O     -1   -   -1   -   -1    =  1
B   + + -   - - +   O   O   I   O   O   I   I   I   O      1   -   -1   -    1    =  1
C   + - +   - + -   O   I   O   I   O   I   O   I   O     -1   -    1   -    1    = -3
D   + - -   - + +   O   I   I   I   O   O   I   O   O      1   -    1   -   -1    =  1
E   - + +   + - -   O   I   I   I   O   O   I   O   O      1   -    1   -   -1    =  1
F   - + -   + - +   O   I   O   I   O   I   O   I   O     -1   -    1   -    1    = -3
G   - - +   + + -   O   O   I   O   O   I   I   I   O      1   -   -1   -    1    =  1
H   - - -   + + +   O   O   O   O   O   O   O   O   O     -1   -   -1   -   -1    =  1
[/QUOTE]
With this data set there is another simple inequality : yz + xz > xy where you can cherry pick violations/non violations for 00,1200.2400
With opposites for all 3 pairs ( at both detectors ) then cos2 θ/2 you get .25 + .25 > .25,
For all same sin2 θ/2 = .75 + .75 > .75
However for yz + xz > xy with opp + opp > same
Then there is violation: .25 + .25 > .75

#### PR0

@bhobba You're right. I did deviate from the initial objection. It was because of my confusion. But now I think I get it (at least I hope so). If the probabilities are correct it's already implied that the more trials are made the closer we get to those probabilities by averaging the results. It doesn't even need a proof (?). My problem then is about what are the sufficient conditions to correctly find/assign probabilities in practical cases.
But for now, I'm just going to assume that the (classical?) probabilities are indeed correct for the example in the OP.

@Nugatory I knew that there are loopholes for the Bell experiments but I'm in no position to either deny or affirm that all loopholes were indeed closed in the Delft (?) experiment.

@DrChinese Can you please give your input about my suggestion that the correct way to find the inequality is vertically not horizontally in the chart? It's in the OP, Conclusions (II).

PS It's still on my TO DO list to understand how the correlations work.

#### PR0

I think I found my answer, if in my OP we consider the probabilities vertically, classical minimum is 1/2 but quantum is 1/4 (?). So there's violation, though I still think the vertical analysis is the correct one for the proposed experiment.

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