A new simulation of the EPR-Bohm correlations

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Thu Jul 09, 2015 11:44 am

And let me just add a question to minkwe; what is the point that this shuffling is supposed to explain?
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Thu Jul 09, 2015 12:01 pm

Heinera wrote:And let me just add a question to minkwe; what is the point that this shuffling is supposed to explain?

Heine, like I've explained to you, address all 8 points, and you'll get the point. Anything else will be ignored.
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Thu Jul 09, 2015 12:24 pm

minkwe[ wrote:
Heinera wrote:And let me just add a question to minkwe; what is the point that this shuffling is supposed to explain?

Heine, like I've explained to you, address all 8 points, and you'll get the point. Anything else will be ignored.

I think it should suffice to comment the last sentence in that list:
The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets.

I agree. And I never made such a claim. In fact, that's why QM can predict a violation of the upper bound of 2. So what's next?
(And by the way, four sets generated by a LHV model are obviously not statistically independent).
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Thu Jul 09, 2015 1:47 pm

Jochen wrote:OK, so let's get this over with. Point 1: I agree that for the hidden variable value assignments, every row is in fact bounded by 2. Point 2: I agree that if you separate out the data into your 2xN list, any reshuffling of rows does not change the expectation value (though of course, you can't then recombine the shuffled lists without undoing the shuffling, at least in such a way that the new 4xN list is just a row-shuffled form of the original). Can we now move on?

Good that you now agree with both point (1) and (2). Now on to your response to point (3), where you say:

Jochen wrote:
minkwe wrote:3. Do you agree that for 4 independent particle pairs , the expression

is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is

For independent particle pairs, it is indeed the case that they are not bound by the value 2, but from there, it does not follow that hence, .

Let me remind you what you said with respect to point (1):

Jochen wrote:the inequality can be extended to averages because no average of a set of values can't be greater than the maximum value of that set. So you have a set of values, all below two, and immediately you know that the average value will also be below two.

Now instead of a 4xN spreadsheet, you have a 8xN spreadsheet, with columns Since the pairs are independent of each other, the maximum value of the expression is 4 and by the same logic which you stated yourself above, the average of the values of set is bounded above by the maximum value each member in the set can attain. This is completely uncontroversial and your claim that it does not follow that is clearly misguided.

But you proceeded to give a contrived example of dice

Jochen wrote:Consider two independent dice, together with the observables T=value on top of the die after a throw, and B=value on the bottom after a throw. For each throw of a single die, then B + T = 7. This also holds in the average: . But for a single throw of both, the value on top of one plus the value at the bottom of the other is not 7, but bounded above by 12. So if you keep throwing both die, and record the value on top of one, and the value at the bottom of the other, does it then follow that ? (Hint: the answer is no.)


I will show you exactly how the example fails
* In your example, for the toss of a single die, <B> + <T> = 7 is an absolute equality (absolute upper and lower bound) which can never be violated, not even by a single iota. If you toss the single die, 1 billion times, you will never violate the number 7 for any individual throw (B+T), nor will you ever violate the number 7 for the averages <B> + <T>. If anyone claims to have violated it by even 1x10^-50, then there is a mathematical error involved. This is the meaning of the sign in mathematics. And when we talk of upper-bound or lower-bound, that is what it means -- it can never be violated by any amount. Similarly then, the inequality in my point (1) is the same, the symbol also means mathematically that the inequality can never be violated.

* Now you say that if two separate independent dice are tossed many times, reading B from one and T from the other, you think that the equality <B> + <T> = 7 should still apply. But this is quite naive. It contains a hidden assumption that the two separate tosses of two independent dice have exactly the same probability distributions for the outcomes B, T. While <B> + <T> = 7 would hold for any single die tossed many times irrespective of its probability distribution of outcomes, if the two dice had different probability distributions of outcomes, the equality <B> + <T> = 7 would fail woefully. Take for example, if the first die used to obtain the T value, was biased such that it always gave T= 6, then if I toss that die alone 1 billion times and calculate <B> + <T> I will never violate <B> + <T> = 7 by even an iota. But then take another die which is biased such that it always produces B = 6. Even this die tossed by itself would not violate the equality. But toss the two of them together, always reading T from the first and B from the second, and you will end up with <B> + <T> = 12, way above 7. Note that the dice may even be identical but the tossing and reading mechanisms may lead to T producing a bias in one way, and the process of reading B produces a bias another way. Therefore your little example is too contrived, and even helps to make my point. The only sure statement you can make about the upper bound of two independent die tosses is . Similarly, as explained in my point (3), the correct upper bound for 4 independent particle pairs is and therefore, the correct upper bound for the average is also

* But I will use another dice example that contains another important property you have missed in your example. I have given it to you before:

If I toss a die such that it settles with the 5 exposed sides facing up(U), north(N), south(S), east(E) and west(W). I ask two experimenters (Alice and Bob) to pick a direction from the exposed sides to observe and the outcomes which we will denote as (A) and (B) respectively, one after the other. Let us suppose that the act of reading the die destroys the chosen face and makes it unavailable for the second experimenter to observe the same face. Once a direction is picked, say N, the remaining experimenter can only pick one of the remaining 4 directions U, S, E, W. Now based on the description of the experiment, we derive a certain inequality . For the purpose of this illustration, it does not matter what is. Now it is obvious that the two expectation values <A> and <B>, are not independent in a number of ways. First, the fact that the paired outcomes (A,B) are both faces of the same die carries with it a specific kind of dependence due to symmetry. But in addition, there is a dependence because whatever Alice picked is unavailable to Bob and vice versa. Therefore the relationship will include those dependencies. If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.

Your example misses the mark. My point (3) remains valid , even better I would say there is a non-zero probability that for 4 independent particle pairs.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Thu Jul 09, 2015 2:45 pm

Jochen wrote:
minkwe wrote:4. You have claimed that the CHSH expression
    ,
    where represents a single set of particle pairs, is statistically equivalent to

    where represent disjoint independent sets of particle pairs. Because according to you,
    and and

Yes I've claimed that, but you can scratch the 'according to you', since it's just according to basic statistics: given the same probability distribution , then indeed the correlation between two different experimental runs will be the same, because it's just derived from that distribution.

As shown above in my response to your response to point (3), you are mistaken. It is not always the case that and and . Simply because the nature of the experiment may not allow you to sample the same distribution. See my dice example above. But I already anticipated that you may not understand/agree to this, in my remaining points.

Jochen wrote:
5. Do you agree that the expression
    means that for every single individual pair of particles in the series which produced the outcome pair , there is an equivalent pair of particles in the series, such that as the number of particles approaches infinity, it should be possible to find a function , which rearranges the sequence of outcomes to match the sequence of , so that we have the same numbers of +1's and -1's and the same pattern of occurrences of those numbers? Yes or no?

This is just what the expectation value means: for large number of measurements n, so in order to yield the same EV, there need to be approximately the same numbers of +1 outcomes and of -1 outcomes. However, equality of these numbers only holds for , and for any two finite experiment runs i and j, both and , as well as and , will generally differ. Hence, in (almost) all real world experiments, such a function won't exist.


I'm not asking you about real world experiments. You stated that , where and represent independent sets of outcome pairs. In point (5), I'm simply asking you to acknowledge what this implies about the two sets of outcome pairs. In fact, the numbers of outcome pairs is completely irrelevant to point (5), and I already asked you to consider as N approaches infinity. But if you say , then it means for every pair in the set, there is a corresponding pair in the set and a map exists between them, whatever N is. N could be anything and that will not change the question or the validity of claim (5). So do you agree with point (5) or not?

Jochen wrote:
6. Do you agree that from point (4) above (which you believe), it follows that there must also exist other functions
    , , and . Such that after applying the functions, and , where the prime , represents the fact that the sequence of outcomes has been rearranged using the appropriate function? That is, the equivalent sets of outcomes are for all practical purposes "identical". Yes or no?

Again, only for the case, which is never attained in experiments.


Again, you do not say if you agree or not. You invoke infinity despite the fact that the claim is true irrespective of the number of outcome pairs. Once you claim that , points (5) and (6) follow immediately. This should not be controversial at all. But even if you bring in real-world experiments, you may not be able to exactly match everything, but with large enough N, if you claim the two averages are almost statistically equivalent , then again you mean such a map exists such that the two sets can be almost exactly matched. In other words, the resulting rearranged sets should not be significantly mismatched. I don't expect that you will be arguing that the inequality does not apply to real-world experiments because nobody can measure an infinite number of particles, so your point about infinity is misguided. You cannot relax the requirements that the sets match, without at the same time relaxing the requirement that averages are equal. Similarly you cannot claim the averages are almost equal, without at the same time agreeing that the sets must almost match. So do you agree with point (6) or not.

Jochen wrote:
7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),
    to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get


    Which can be factorized just like in point (1) to

    since and

There's two problems here. The first one is the erroneous nature of the argument in point 1, which I've already noted.
The second is that even if the reordering functions did exist (which they won't in general), then you can't perform such reorderings and expect everything to stay the same: the correlations are generally not preserved under such a reordering.

Given that you have since changed your response to point (1) from "disagree" to "agree", you will need to amend your responses to point (7) which is obviously based on a misunderstanding, before I respond any further.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Fri Jul 10, 2015 11:37 pm

Heinera wrote:(And by the way, four sets generated by a LHV model are obviously not statistically independent).

Again, address all 8 points. Your statement above shows your head is still in the sand. If you toss a single local realistic coin many times, each time writing down the result above the table on column A and the result below the table on column B, the two columns are not independent. But if you toss two local realistic coins, each time writing down the result from above the first coin on column A, and the result from below the second coin on column B, then the two columns are independent. Yet both are local realistic.

But I won't take you seriously until you address all 8 points like Jochen is currently doing. That's the only way you will get the point.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sat Jul 11, 2015 9:47 am

There was a notation typo in the last example of my previous post, I used (B, T) instead of (A, B) in the inequalities. It should have read:

If I toss a die such that it settles with the 5 exposed sides facing up(U), north(N), south(S), east(E) and west(W). I ask two experimenters (Alice and Bob) to pick a direction from the exposed sides to observe and the outcomes which we will denote as (A) and (B) respectively, one after the other. Let us suppose that the act of reading the die destroys the chosen face and makes it unavailable for the second experimenter to observe the same face. Once a direction is picked, say N, the remaining experimenter can only pick one of the remaining 4 directions U, S, E, W. Now based on the description of the experiment, we derive a certain inequality . For the purpose of this illustration, it does not matter what is. Now it is obvious that the two expectation values <A> and <B>, are not independent in a number of ways. First, the fact that the paired outcomes (A,B) are both faces of the same die carries with it a specific kind of dependence due to symmetry. But in addition, there is a dependence because whatever Alice picked is unavailable to Bob and vice versa. Therefore the relationship will include those dependencies. If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Sun Jul 12, 2015 3:41 am

minkwe wrote:Now instead of a 4xN spreadsheet, you have a 8xN spreadsheet, with columns Since the pairs are independent of each other, the maximum value of the expression is 4 and by the same logic which you stated yourself above, the average of the values of set is bounded above by the maximum value each member in the set can attain. This is completely uncontroversial and your claim that it does not follow that is clearly misguided.

As you say, the re-ordering does not change the expectation values. So if, for the original table, , then it directly follows that for this one also . I mean, from the original table, you compute (say) , then from the re-ordered table, again , as only the sequence of pairs has been changed. Of course, the argument that any row is bounded does not hold for the new table; but it need only hold for the old, 4xN table, and from there, since the reorderings do not change the expectation values, it follows that any sum of these expectation values also does not change, and that hence, the value of the CHSH-quantity does likewise not change.


minkwe wrote:
Jochen wrote:Consider two independent dice, together with the observables T=value on top of the die after a throw, and B=value on the bottom after a throw. For each throw of a single die, then B + T = 7. This also holds in the average: . But for a single throw of both, the value on top of one plus the value at the bottom of the other is not 7, but bounded above by 12. So if you keep throwing both die, and record the value on top of one, and the value at the bottom of the other, does it then follow that ? (Hint: the answer is no.)


I will show you exactly how the example fails
* In your example, for the toss of a single die, <B> + <T> = 7 is an absolute equality (absolute upper and lower bound) which can never be violated, not even by a single iota. If you toss the single die, 1 billion times, you will never violate the number 7 for any individual throw (B+T), nor will you ever violate the number 7 for the averages <B> + <T>. If anyone claims to have violated it by even 1x10^-50, then there is a mathematical error involved. This is the meaning of the sign in mathematics.

Yes, this is because I essentially forced the two sides to be perfectly correlated. But the argument would work equally well with two separate dice that are only very highly correlated, i.e. which show the same face up almost every time.

minkwe wrote:* Now you say that if two separate independent dice are tossed many times, reading B from one and T from the other, you think that the equality <B> + <T> = 7 should still apply. But this is quite naive.

No, as long as we're talking about expectation values, the equality applies exactly; however, in every real experiment with a finite number of runs, it will of course apply only approximately. The confusion between the two is the same problem you also have in connecting Bell's theoretical work with actual experiment.

minkwe wrote:It contains a hidden assumption that the two separate tosses of two independent dice have exactly the same probability distributions for the outcomes B, T.

Yes, this is the same assumption as the one underlying Bell inequalities, that the two systems are sampled from the same probability distribution.

minkwe wrote:If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.

But this is disanalogous to the Bell case: there, the experimenter each only have access to one of the particle pairs, i.e. in your example, one set of directions; thus, they can never carry out measurements on the other particle.

minkwe wrote:Your example misses the mark. My point (3) remains valid , even better I would say there is a non-zero probability that for 4 independent particle pairs.

This simply isn't true. Any local hidden variable model must admit to write down a table of possessed values, from which the measured one are sampled. In fact, that such a table exists is equivalent to the requirements of value definiteness and nondisturbance: if these hold, then I can write down a set of values every observable possesses immediately before the measurment, and that table will be independent of what measurements are carried out at the distant particle, i.e. no observable can change its value based on what happens at spatial separation. And the existence of such a table implies the existence of the joint probability distribution, which implies that Bell inequalities hold.

Or, to demonstrate it exemplarily, the measurment table I gave above might for instance have been obtained from the following table of values (where square brackets denote the measurements actually performed, and hence, the values obtained):
Code: Select all
 
   Alice   |    Bob   
  A  |  C  |  B  |  D 
 [+1]| +1  | -1  | [-1]
 [-1]| -1  |[-1] |  +1 
  +1 |[-1] | +1  | [+1]
.
.
.
  -1 |[-1] |[-1] |  -1 
 [+1]| +1  |[+1] |  +1 
 [+1]| -1  | -1  | [+1]


Now, as you can see, every row here is bound by 2, and hence, so is the CHSH quantity. If you now perform the reordering as you proposed, then every row won't be bounded by two anymore; but this does not matter, as every correlator computed from the re-ordered table still will have the same value as computed from the above table, and hence, so does the CHSH quantity. What you have achieved in performing the reordering is simply that values in the same row no longer pertain to the same run of the experiment, and hence, don't respect the correlations as present in the table above. But this does not change anything globally. There is no way to start with a table like the above and obtain a value greater than 2 for the CHSH quantity (well, or rather, all the ways in which one could do that simply mean that one has done something wrong, that doesn't respect the correlations between different observable), and every local hidden variable model leads to such a table (and hence, a joint PD, which is just given by (or experimentally approximated by) the relative frequency of all possible value assignments in the table). Hence, for any local hidden variable model, the CHSH inequality holds.


minkwe wrote:As shown above in my response to your response to point (3), you are mistaken. It is not always the case that and and . Simply because the nature of the experiment may not allow you to sample the same distribution.

It is always the case if you have a joint PD, which you have if you can write down a table such as the above, which you can always do for LHV models.

minkwe wrote:Given that you have since changed your response to point (1) from "disagree" to "agree", you will need to amend your responses to point (7) which is obviously based on a misunderstanding, before I respond any further.

This is adressed above: the CHSH quantity will always be below two for any model for which you could write down a table of values such as above, i.e. every hidden variable model; if you 'violate' it by using some re-ordering, then the reordering simply didn't respect the correlations between the observables. In other words, any valid reordering must ultimately be equivalent to a simple row permutation of the table I've given, which obviously doesn't change the fact that .
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sun Jul 12, 2015 4:18 am

Good, you now agree with points (1), (2), (4), (5) and (6) despite claims of disagreement earlier. You still refuse to agree with point (3)

minkwe wrote:3. Do you agree that for 4 independent particle pairs , the expression

is false, and the correct upper bound should be , and by the same logic of the argument in point (1) above, the correct upper bound for averages is


and your current responses to (7) and (8) are still contradictory since they are based on earlier responses you've since recanted.

Jochen wrote:
Minkwe's wrote:7. Do you agree that if points (5) and (6) are true, then you can apply the same argument from point (1),
    to measurements performed on 4 disjoint sets of particle pairs , precisely because after rearranging, you will get


    Which can be factorized just like in point (1) to

    since and

There's two problems here. The first one is the erroneous nature of the argument in point 1, which I've already noted. The second is that even if the reordering functions did exist (which they won't in general), then you can't perform such reorderings and expect everything to stay the same: the correlations are generally not preserved under such a reordering.
minkwe wrote:8. Do you see now that in order for
    to be true, it must be the case that and . But both and have already been rearranged independently of each other, and since any rearrangement will shuffle both outcomes in the set of pairs any new rearrangement to make agree with will undo the previous rearrangements; the same for and . Therefore for measurements on 4 independent disjoint sets to obey the the inequality, you need several different independent sorting functions to be dependent. In other words, the assumption that 4 independent sets of particle pairs should obey the inequality is a contradiction and your claim in point (4) above fails. The upper bound of 2, depends on the fact that the terms are dependent. It is a contradiction to claim that the same upper bound applies to independent sets

The only rearrangements you can validly perform on the measurement tables are row permutations;...

As explained multiple times in detail, the only rearrangements present in my argument are row permutations. So your responses to 7 and 8 are clearly misguided. Besides, you now agree with points (5) and (6) where those rearrangements are introduced into the argument.

I suspect that you did not fully understand how the 8 points fit together before voicing your disagreements. Perhaps not realising that, because I understood your argument, and your error, I've built-in your argument into my rebuttal.

I'm waiting for an amended response to points 7 and 8 before responding further.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Sun Jul 12, 2015 5:47 am

Anything I'm going to say is contained in my previous response, where I've shown that only row permutations can't change the value of the CHSH quantity.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sun Jul 12, 2015 9:12 am

Jochen wrote:Anything I'm going to say is contained in my previous response, where I've shown that only row permutations can't change the value of the CHSH quantity.

So you won't even attempt to correct the obvious errors in your responses to (7) and (8)? In your response to (7) you claimed point (1) was "erroneous", but under pressure, you've since retracted that and now agree with the argument in point (1). You claim in your responses to (8) that my rearrangements are not permitted because only row permutations are allowed. But by agreeing to points (5) and (6), you must now recognize that all rearrangements in my argument are row permutations.

So again, unless you want to concede my argument, address points (7) and (8) specifically, then I will respond to the remaining issues you have raised and show you how my original argument already anticipated and disposed of all your claims.

Take for example:
Jochen wrote:
minkwe wrote:* Now you say that if two separate independent dice are tossed many times, reading B from one and T from the other, you think that the equality <B> + <T> = 7 should still apply. But this is quite naive.

No, as long as we're talking about expectation values, the equality applies exactly; however, in every real experiment with a finite number of runs, it will of course apply only approximately. The confusion between the two is the same problem you also have in connecting Bell's theoretical work with actual experiment.

You hastily proclaim "NO", and then proceed to agree with the statement, even while dreaming up some "confusion" you imagine my argument is based on. Which is completely false. Perhaps, all along you've been arguing against imagined argument you think is my argument.

It is becoming increasingly obvious that you make claims without thinking or reading carefully only to later have to retract them. Did you even read points (7) and (8), I doubt it very much.
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Sun Jul 12, 2015 9:33 am

minkwe wrote:There was a notation typo in the last example of my previous post, I used (B, T) instead of (A, B) in the inequalities. It should have read:

If I toss a die such that it settles with the 5 exposed sides facing up(U), north(N), south(S), east(E) and west(W). I ask two experimenters (Alice and Bob) to pick a direction from the exposed sides to observe and the outcomes which we will denote as (A) and (B) respectively, one after the other. Let us suppose that the act of reading the die destroys the chosen face and makes it unavailable for the second experimenter to observe the same face. Once a direction is picked, say N, the remaining experimenter can only pick one of the remaining 4 directions U, S, E, W. Now based on the description of the experiment, we derive a certain inequality . For the purpose of this illustration, it does not matter what is. Now it is obvious that the two expectation values <A> and <B>, are not independent in a number of ways. First, the fact that the paired outcomes (A,B) are both faces of the same die carries with it a specific kind of dependence due to symmetry. But in addition, there is a dependence because whatever Alice picked is unavailable to Bob and vice versa. Therefore the relationship will include those dependencies. If instead of doing those measurements on the same die, they decide to toss 2 separate independent dice, it will be foolish to claim that the relationship continues to hold simply because the two dice are of the same type! Obviously, even if the dice are identical, the fact that you have 2 separate independent dice affords more degrees of freedom for the experimenters and therefore the expectation values calculated from a single die will generally not agree with those from 2 independent dice.


I agree with this. But the premise

"Let us suppose that the act of reading the die destroys the chosen face and makes it unavailable for the second experimenter to observe the same face. Once a direction is picked, say N, the remaining experimenter can only pick one of the remaining 4 directions U, S, E, W."

does not have any parallel in the models we are discussing. Even in a LHV model there is nothing that prevents Bob from measuring "up" on his particle, even if Alice observes "up" on her particle. No faces are destroyed. So the second experimenter can surely observe the same face.

It seems that you think that a counterfactual outcome must somehow have a different value then the actual outcome. This is not correct. Bot the actual and the counterfactual outcome may take on the same values in a LHV model.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sun Jul 12, 2015 9:44 am

Heinera wrote:I agree with this. But the premise

You don't know what models we are discussing, not having read or understood points (1) to (8). The example you are now responding to, was addressing a very specific error Jochen made. You've missed the point completely because instead of addressing the points one by one, you have jumped in the middle and misunderstood the purpose of the example.

In case you forgot, this started when I asked
minkwe wrote:Could you please tell us what Quantum Mechanics predicts for the 4 expectation values ?


A question you claimed was meaningless and had no answer. A question you haven't answered ever since. Jochen proceeded to claim that the QM predictions were , , and , respectively, the same as the QM predictions for because he believes and and . Again note that this was a question about QM, not a question about local realistic theories to begin with. He was providing a justification why he gave the exact same Quantum Mechanical predictions for both, a justification which permits him to demonstrate violation of CHSH using those same QM predictions. In other words, if the QM predictions for those two scenarios are different, the he could not reasonably use the QM predictions from one scenario to claim violation of the other.


So please, once again, I ask you to address them. There is no point discussing with you otherwise. If you want to understand the argument, you have to engage it point by point.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Sun Jul 12, 2015 9:58 am

minkwe wrote:So again, unless you want to concede my argument, address points (7) and (8) specifically, then I will respond to the remaining issues you have raised and show you how my original argument already anticipated and disposed of all your claims.

Point 7: if all you do is row rearrangings, I'm OK with it, provided I understand everything correctly. Point 8: if all you do is row rearrangings, I'm OK with that, provided I understand everything correctly. Can we now please quit this silliness and return to the actual argument?

Really, though, everything should become clear to you if you write down just some simple examples of value assignment tables as I've provided above, and convince yourself that no matter what violence you do to the data, it'll never yield a value for the CHSH-quantity greater than 2 (provided everything you do is equivalent to a simple row-rearrangement). That's in the end all there's to it.
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Re: A new simulation of the EPR-Bohm correlations

Postby Heinera » Sun Jul 12, 2015 10:01 am

minkwe wrote:
Heinera wrote:I agree with this. But the premise

You don't know what models we are discussing, not having read or understood points (1) to (8). The example you are now responding to, was addressing a very specific error Jochen made. You've missed the point completely because instead of addressing the points one by one, you have jumped in the middle and misunderstood the purpose of the example.

So please, once again, I ask you to address them. There is no point discussing with you otherwise. If you want to understand the argument, you have to engage it point by point.

There is no point in me addressing all eight points, since I basically agree with all of them. So what? They simply show that you don't understand Bell's proof of his theorem, since your chain of reasoning in no way invalidates that proof. It would be much better if you succinctly explained exactly where Bell (or Richard Gill) made a mathematical mistake in their proofs.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Sun Jul 12, 2015 10:04 am

minkwe wrote:A question you claimed was meaningless and had no answer. A question you haven't answered ever since. Jochen proceeded to claim that the QM predictions were , , and , respectively, the same as the QM predictions for because he believes and and .

My answer to you was:
Jochen wrote:As such, the question is meaningless. You could ask 'What does Quantum Mechanics predict for the 4 expectation values if the measurements are carried out on a singlet state, and the measurements are those in the standard CHSH setting?', in which case the prediction would be , , and , respectively; but in different settings, wildly different outcomes are possible. Does that help?

On its own, the question makes indeed no sense and has no answer; only once the state is specified can an answer be given. And of course the answer is the same in different runs---that's just saying that the particles in those runs were in the same state.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sun Jul 12, 2015 10:06 am

Jochen wrote:Point 7: if all you do is row rearrangings, I'm OK with it, provided I understand everything correctly. Point 8: if all you do is row rearrangings, I'm OK with that, provided I understand everything correctly. Can we now please quit this silliness and return to the actual argument?


This is not silliness, unless you mean your actions. You claim point (7) fails because of the errorneous nature of point (1). Then you later recant and agree with point (1), and yet have the audacity to accuse me of silliness because I give you the opportunity to fix your response to point (7). It is you who is being silly, given what you've argued here so far, you will not agree to point (8) if you've read it carefully and understood it. Because it hammers the final nail into your argument. So again I give you another opportunity to read it carefully and make sure you understand it before you admit that is correct, which effectively ends the discussion.

Like I suspected, you've not understood the argument at all, yet you were proclaiming disagreements left and right.
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Re: A new simulation of the EPR-Bohm correlations

Postby minkwe » Sun Jul 12, 2015 10:10 am

Heinera wrote:There is no point in me addressing all eight points, since I basically agree with all of them. So what?

Good! You agree with all the 8 points! That wasn't so hard that it needed almost a week to extract.

Point (8) proves that point (4) is false. In other words, the QM predictions for are not the same as the QM predictions for. In other words, you guys (including Bell) have been using the wrong QM predictions to claim violation.
Last edited by minkwe on Sun Jul 12, 2015 10:19 am, edited 3 times in total.
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Re: A new simulation of the EPR-Bohm correlations

Postby Jochen » Sun Jul 12, 2015 10:12 am

minkwe wrote:This is not silliness, unless you mean your actions. You claim point (7) fails because of the errorneous nature of point (1). Then you later recant and agree with point (1), and yet have the audacity to accuse me of silliness because I give you the opportunity to fix your response to point (7). It is you who is being silly, given what you've argued here so far, you will not agree to point (8) if you've read it carefully and understood it. Because it hammers the final nail into your argument. So again I give you another opportunity to read it carefully and make sure you understand it before you admit that is correct, which effectively ends the discussion.

OK, how exactly do you think that point 8 'hammers the final nail' into my argument? Just use the table of values like I provided to illustrate what you think you're doing in that point, because to be honest, it appears about as clear as mud to me.

And did you get why the existence of such a table implies ?
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Re: A new simulation of the EPR-Bohm correlations

Postby FrediFizzx » Sun Jul 12, 2015 10:16 am

minkwe wrote:
Heinera wrote:There is no point in me addressing all eight points, since I basically agree with all of them. So what?

Good! You agree with all the 8 points! That wasn't so hard that it needed almost a week to extract.

Point (8) proves that point (4) is false. In other words, the QM predictions for are not the same as the QM predictions for. In other words, you guys (including Bell) have been using the wrong QM predictions to claim violation.

For lurkers, the 8 points by minkwe are here,

viewtopic.php?f=6&t=168&start=230#p4731
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