What's wrong with this local realistic counter-example to Bell's theorem?

[vanesch]

Please see my emphasis above:

Help please: I am not aware, EVER, of making any such claim.

I don't know how to interpret your post #18 differently.

So my question to you then AGAIN:
Do you agree with the number of clicks as predicted by QM in my post # 21 on Monday, Tuesday and so on ?

Now, this thread was started by your claim to produce 8 numbers P1...P8 that can REPRODUCE these predictions. You've given 8 numbers, and they DO NOT predict the numbers predicted by QM (they do so for 2 of the 3 proposed experiments, and give HALF of the actual result QM result for the third one).

So is it then your numbers which, after all, don't do what you claimed they did (namely explain the QM results) ; or is it the QM predictions (or our way of calculating them) which are, according to you, wrong ?

In other words, on Tuesday, do you think we will see (according to our way of calculating QM results) 250 000 hits, or (according to your numbers), only 125 000 hits ?

Because you can't have it both ways, right ? You can't have your numbers be different from the QM predictions as worked out here and still claim that you have provided us with a set of numbers which agree with QM.

 

[Gordon Watson]

I don't know how to interpret your post #18 differently.

So my question to you then AGAIN:
Do you agree with the number of clicks as predicted by QM in my post # 21 on Monday, Tuesday and so on ?

Now, this thread was started by your claim to produce 8 numbers P1...P8 that can REPRODUCE these predictions. You've given 8 numbers, and they DO NOT predict the numbers predicted by QM (they do so for 2 of the 3 proposed experiments, and give HALF of the actual result QM result for the third one).

So is it then your numbers which, after all, don't do what you claimed they did (namely explain the QM results) ; or is it the QM predictions (or our way of calculating them) which is, according to you, wrong ?

Because you can't have it both ways, right ? You can't have your numbers be different from the QM predictions as worked out here and still claim that you have provided us with a set of numbers which agree with QM.

Jesse and vanesch; thank you; detailed mathematical response on the way.

JenniT

 

[Gordon Watson]

DRAFTRESULTS DERIVED from, and offered as, a local realistic counter-example to Bell's theorem.

JenniT

With apologies for the formatting; this DRAFT being hurried on-line to address serious concerns above.​

1. Pab++ denotes the Probability (P) of outcome ++ across the angle ab, where ab is a clearly-defined (unambiguous) angle; etc. C and S functions are defined in the OP.

2. For such angles, all the probability functions [below; lines (A)-(C)] agree with QM.

3. Equations (X)-(Z) are NOT such functions; they are seen here as arising from (unrecognized) bi-angles when the Bell-Zakurai Table is compared with those below. Thus bi-angles have their origin, here, in this analysis of BT. They represent an attempt to identify and classify the source of Bell's Inequality.

4. Further, (X)-(Z) appear to be required to derive a Bell Inequality (BI).

5. The challenge for Bell's supporters, it seems to me, is to derive a BI without recourse to (X)-(Z); or explain why such "grotesque" functions are justifiably equated to functions from (A)-(C). I suggest that a rational BI cannot now be constructed, in that each P can now be matched to a value for any 3-particle EPRB test; and any "equating" now challenged!

6. Alternatively, show how (A)-(C) might be modified to reproduce (again) all the correct QM results, and derive a rational inequality.

7. Since these results derive from local realism, I conclude: Bell's Theorem is not an impediment to local realism and a local-realistic view of the world.

8. These results, extending beyond QM in analytic detail; being wholly based on local realistic considerations; are the reason that I am reluctant to consider specific examples -- thereby losing the beauty, and the check on each result, that generality here provides.

E & OE. ROa: Angles ab, ac. Bi-angle: bc = ab + ac, or bc = ab – bc; so 2Pbc = the average over the bi-angle.

(A1) +++ ––– P1 = Cab.Cac/2. Pab++ = P3 + P4 = Sab/2. (X1) Pbc++ = P2 + P6 = (Cab.Sac + Sab.Cac)/2.
(A2) ++– ––+ P2 = Cab.Sac/2. Pab+– = P1 + P2 = Cab/2. (X2) Pbc+– = P1 + P5 = (Cab.Cac + Sab.Sac)/2.
(A3) +–+ –+– P3 = Sab.Cac/2. Pab–+ = P7 + P8 = Cab/2. (X3) Pbc–+ = P4 + P8 = (Sab.Sac + Cab.Cac)/2.
(A4) +–– –++ P4 = Sab.Sac/2. Pab–– = P5 + P6 = Sab/2. (X4) Pbc–– = P3 + P7 = (Sab.Cac + Cab.Sac)/2.
(A5) –++ +–– P5 = Sab.Sac/2. Pac++ = P2 + P4 = Sac/2.
(A6) –+– +–+ P6 = Sab.Cac/2. Pac+– = P1 + P3 = Cac/2.
(A7) ––+ ++– P7 = Cab.Sac/2. Pac–+ = P6 + P8 = Cac/2.
(A8) ––– +++ P8 = Cab.Cac/2. Pac–– = P5 + P7 = Sac/2.ROb: Angles ab, bc. Bi-angle: ac = ab + bc, or ac = ab – bc; so 2Pac = the average over the bi-angle.

(B1) +++ ––– P1 = Cba.Cbc/2. Pab++ = P3 + P4 = Sab/2. (Y1) Pac++ = P2 + P4 = (Cab.Sbc + Sab.Cbc)/2.
(B2) ++– ––+ P2 = Cba.Sbc/2. Pab+– = P1 + P2 = Cab/2. (Y2) Pac+– = P1 + P3 = (Cab.Cbc + Sab.Sbc)/2.
(B3) +–+ –+– P3 = Sba.Sbc/2. Pab–+ = P7 + P8 = Cab/2. (Y3) Pac–+ = P6 + P8 = (Sab.Sbc + Cab.Cbc)/2.
(B4) +–– –++ P4 = Sba.Cbc/2. Pab–– = P5 + P6 = Sab/2. (Y4) Pac–– = P5 + P7 = (Sab.Cbc + Cab.Sbc)/2.
(B5) –++ +–– P5 = Sba.Cbc/2. Pbc++ = P2 + P6 = Sbc/2.
(B6) –+– +–+ P6 = Sba.Sbc/2. Pbc+– = P1 + P5 = Cbc/2.
(B7) ––+ ++– P7 = Cba.Sbc/2. Pbc–+ = P4 + P8 = Cbc/2.
(B8) ––– +++ P8 = Cba.Cbc/2. Pbc–– = P3 + P7 = Sbc/2.ROc: Angles ac, bc. Bi-angle: ab = ca + cb, or ab = ca – cb; 2Pab = the average over the bi-angle.

(C1) +++ ––– P1 = Cca.Ccb/2. Pac++ = P2 + P4 = Sac/2. (Z1) Pab++ = P3 + P4 = (Cac.Sbc + Sac.Cbc)/2.
(C2) ++– ––+ P2 = Sca.Scb/2. Pac+– = P6 + P8 = Cac/2. (Z2) Pab+– = P1 + P2 = (Cac.Cbc + Sac.Sbc)/2.
(C3) +–+ –+– P3 = Cca.Scb/2. Pac–+ = P1 + P3 = Cac/2. (Z3) Pab–+ = P7 + P8 = (Sac.Sbc + Cac.Cbc)/2.
(C4) +–– –++ P4 = Sca.Ccb/2. Pac–– = P5 + P7 = Sac/2. (Z4) Pab–– = P5 + P6 = (Sac.Cbc + Cac.Sbc)/2.
(C5) –++ +–– P5 = Sca.Ccb/2. Pbc++ = P2 + P6 = Sbc/2.
(C6) –+– +–+ P6 = Cca.Scb/2. Pbc+– = P4 + P8 = Cbc/2.
(C7) ––+ ++– P7 = Sca.Scb/2. Pbc–+ = P1 + P5 = Cbc/2.
(C8) ––– +++ P8 = Cca.Ccb/2. Pbc–– = P3 + P7 = Sbc/2.PS: I will be away from the web for a week or more, starting tomorrow.

Happy studying,

JenniT

 

[Gordon Watson]

Is there a simple way to format tables on PF, please ... ??

while I'm still in the editing phase .. ??

 

[JesseM]

The following may be of interim help:

1. I thought this was OK; from earlier post:

Second: To the 4 angles, termed, for convenience: 2 conventional angles and 1 bi-angle.

Consider any 3 orientations (a, b, c ), taken to be the principal axes of the 3 SGM devices in our study here. We want to define their relation by specifying the 2 angles between them as simply as possible. That is, we want to go beyond QM and its one angle between 2 orientations to a local realistic study using 2 angles between 3 orientations.

We can do this readily: Let the required angles be ab and bc. But then note: WE ARE FORCED (or so it seems) to work with 2 angles and a bi-angle. For we have: ab, bc, ac = ab + bc, ac = ab – bc. So we need to be aware of bi-angles, essentially by-products from our need for the two adjacent or over-lapping angles ab and bc.

I still have no idea why you think "we are forced" to do this, you don't explain your reasoning at all. Look, below I've drawn a crude diagram showing how you can define the angle of any given SG orientation relative to some fixed x-axis, and likewise define the angle between any two SG devices geometrically (and also using the simple mathematical definition that if one has angle x and the other has angle y, the angle between them xy is just x-y). Do you see any problem with this system? On a diagram like this, do you think it would be possible to draw in an SG orientation that we couldn't label with a unique angle, or possible to draw two SG devices with orientations such that there wouldn't be a unique angle between them? If you do think there would be any problems with non-uniqueness, any chance you could supply your own analogous diagram showing a horizontal x-axis along with one SG device which you think couldn't be assigned a unique angle, or two SG devices for which you don't think there'd be a unique angle between them?

(and once again, a page giving the basics of what a SG device is and how it works can be found here...incidentally, realistically it's not the central axis of the SG device that's moved to different angles as in my diagram, but rather the device is rotated about that central axis as depicted here, but my diagram is just to illustrate the general concept of labeling orientations with angles so it was easier to draw that way)

 

[JesseM]

Is there a simple way to format tables on PF, please ... ??

while I'm still in the editing phase .. ??

Best way would probably be to do it with LaTeX, go to yuiop's [post=2945088]post #5 on this thread[/post] and click the table he posted to see its LaTeX code, hopefully you can figure out how they work from that.

 

[JesseM]

With apologies for the formatting; this DRAFT being hurried on-line to address serious concerns above.

Since this draft appears to address none of the concerns I raised in posts #29 and #30, I would appreciate it if you would respond to those posts individually when you have the time (along with the question involving the diagram in post #35)

 

[vanesch]

ROc: Angles ac, bc. Bi-angle: ab = ca + cb, or ab = ca – cb; 2Pab = the average over the bi-angle.

(C1) +++ ––– P1 = Cca.Ccb/2. Pac++ = P2 + P4 = Sac/2. (Z1) Pab++ = P3 + P4 = (Cac.Sbc + Sac.Cbc)/2.
(C2) ++– ––+ P2 = Sca.Scb/2. Pac+– = P6 + P8 = Cac/2. (Z2) Pab+– = P1 + P2 = (Cac.Cbc + Sac.Sbc)/2.
(C3) +–+ –+– P3 = Cca.Scb/2. Pac–+ = P1 + P3 = Cac/2. (Z3) Pab–+ = P7 + P8 = (Sac.Sbc + Cac.Cbc)/2.
(C4) +–– –++ P4 = Sca.Ccb/2. Pac–– = P5 + P7 = Sac/2. (Z4) Pab–– = P5 + P6 = (Sac.Cbc + Cac.Sbc)/2.
(C5) –++ +–– P5 = Sca.Ccb/2. Pbc++ = P2 + P6 = Sbc/2.
(C6) –+– +–+ P6 = Cca.Scb/2. Pbc+– = P4 + P8 = Cbc/2.
(C7) ––+ ++– P7 = Sca.Scb/2. Pbc–+ = P1 + P5 = Cbc/2.
(C8) ––– +++ P8 = Cca.Ccb/2. Pbc–– = P3 + P7 = Sbc/2.


PS: I will be away from the web for a week or more, starting tomorrow.

Happy studying,

JenniT

I highlighted what we need: Pac++, Pab++ and Pcb++=Pbc--

Pac++ is what is measured on monday, and equals 0.073... in agreement with your numbers
Pab++ is what is measured on tuesday, and equals 0.25. Your number gives 0.125
Pcb++=Pbc-- is what is measured on thursday, and equals 0.073 in agreement with your numbers.

You will NEVER be able to get Pab++ equal to 0.25 (you actually have 0.125), simply because it can't be larger than (Pac++) + (Pbc--) which equals 0.146...
Note that indeed, your number (0.125) is, as it can't be otherwise, smaller than 0.146.
Simply because this 0.146 is made up of your 4 positive numbers P2 + P3 + P4 + P7 as you give it yourself, and Pab++ is equal to only P3+P4 (your 0.125). You ADD to your 0.125 still your P2 and P7 to obtain 0.146, so it has to be smaller (as indeed it is).

Now, QM predicts not 0.125, but rather 0.25. It is bigger. So it can't come from numbers P1...P8 in this way.

There's nothing more to say about this.

 

[Gordon Watson]

I highlighted what we need: Pac++, Pab++ and Pcb++=Pbc--

Pac++ is what is measured on monday, and equals 0.073... in agreement with your numbers
Pab++ is what is measured on tuesday, and equals 0.25. Your number gives 0.125
Pcb++=Pbc-- is what is measured on thursday, and equals 0.073 in agreement with your numbers.

You will NEVER be able to get Pab++ equal to 0.25 (you actually have 0.125), simply because it can't be larger than (Pac++) + (Pbc--) which equals 0.146...
Note that indeed, your number (0.125) is, as it can't be otherwise, smaller than 0.146.
Simply because this 0.146 is made up of your 4 positive numbers P2 + P3 + P4 + P7 as you give it yourself, and Pab++ is equal to only P3+P4 (your 0.125). You ADD to your 0.125 still your P2 and P7 to obtain 0.146, so it has to be smaller (as indeed it is).

Now, QM predicts not 0.125, but rather 0.25. It is bigger. So it can't come from numbers P1...P8 in this way.

There's nothing more to say about this.

Dear vanesch, thank you,

And apologies for where my exposition has been unclear; especially re bi-angles. [I personally think of them as BI-angles, where BI = Bell Inequalities.]

However, as I wrote: They are important, IMHO, to any local-realistic analysis of BT.

To recap: I brought my model here, responding to a post of yours, to see if it was in error. Very much appreciating the correspondence here, I believe that the model is not in error.

To support this view, I am attaching a one-page PDF. I am of the view that the model is both valid and important. I am of the view that it addresses the challenge provided by your post.

PS: If you and Jesse would be OK with this: I would like to open another thread, asking the question: Bell's theorem refuted?

I believe that such should be in order. As you will see from the PDF, there is published, peer-reviewed material on the subject. (One reviewer quite famous.)

The OP would have this PDF attached, possibly with amendments that you and Jesse might suggest.

With thanks again to you both,

please excuse me if spasmodic replies soon follow for a while -- I will soon be away from ready access to the net -- for a week or more,

JenniT

 

[JesseM]

PS: If you and Jesse would be OK with this: I would like to open another thread, asking the question: Bell's theorem refuted?

I would not be OK with this, I think you are using this forum as a platform for advertising a crackpot theory that you are already totally confident is right, rather than just exploring the issue in an open-minded way that acknowledges the strong likelihood that it is you who have made an error somewhere in your analysis, and making a sincere effort to listen to critiques in order to identify likely errors. You have not given any coherent explanation of what you intend the probabilities P1-P8 to actually represent, despite my asking this question repeatedly; it is clear they do not have the standard meaning that they do on the Sakurai's Bell inequality page if you both reject the idea of predetermined values for each of the three angles, and also reject the idea that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7. So I would recommend not starting any new threads until you have provided clear answers to these questions, along with the others I brought up in posts #29, 30, and 35.

 

[Gordon Watson]

Since this draft appears to address none of the concerns I raised in posts #29 and #30, I would appreciate it if you would respond to those posts individually when you have the time (along with the question involving the diagram in post #35)

Dear Jesse,

AS ALWAYS, your contributions are, for me, fantastic. And, here, above, I so very much appreciate that offer of time. Whew! I've been in a time-bind for some time -- soon to get worse -- so time is what I need.

Time-poorness is also a small part of some missed answers to you; for which I apologize.

(BUT I also found your questions quite complex -- with so many on a page! If you break questions into a one-at-a-time sort of thing, that will be much easier for me; and easier for you to point to answers I might accidently miss providing. I AM committed to answer any and every question on the model and its development; perhaps too often hoping that answers given on one matter may help the questioner see the answer on another matter.)

However, responding to vanesch above, you will see that I have attached a one-page PDF file. I am hoping it will go a long way to providing all the answers that you seek. I am also hoping it will help us develop a common 'lingo' -- me being fairly jargon-based and not over-familiar with QM.

Have a look at it, please. Then point me to questions that you would like answered; in the model's context, please, if possible.

Thank you, again ... and did you ever reply to my question re acknowledgment? (Do so privately, if you wish.)

JenniT

 

[Gordon Watson]

I would not be OK with this, I think you are using this forum as a platform for advertising a crackpot theory that you are already totally confident is right, rather than just exploring the issue in an open-minded way that acknowledges the strong likelihood that it is you who have made an error somewhere in your analysis, and making a sincere effort to listen to critiques in order to identify likely errors. You have not given any coherent explanation of what you intend the probabilities P1-P8 to actually represent, despite my asking this question repeatedly; it is clear they do not have the standard meaning that they do on the Sakurai's Bell inequality page if you both reject the idea of predetermined values for each of the three angles, and also reject the idea that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7. So I would recommend not starting any new threads until you have provided clear answers to these questions, along with the others I brought up in posts #29, 30, and 35.

OK; thank you; your post came in as I was posting my last.

Q: Do you not see that all the sums that you could possibly require are included in -- or easily derived from -- the base-data in the PDF.

Q: I left many more sums for the reader to complete. Do you want to see any of them?

NB: I ABSOLUTELY REJECT:

P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7

IF IT IMPLIES: P3 + P4 = P3 + P4 + (POSITIVE NUMBER)! (XXX)

Q: Seriously: In case I misinterpret: Is equation (XXX) what you want me to accept?

Please: Is this not the very SUM that I specifically included as the worked example in the PDF?

Q: You suggest crackpot theory? OK. But would my question be ruled out by a fair interpretation of PF rules? Just as i started this thread with a question; a sincere one; with helpful answers.

PS: I believe in my theory; and QM; and I will answer every question.

Perhaps some of your confusion arises because I did not understand a question, or I answered wrongly or inconsistently.

You mentioned a game, which had me confused. I had hoped that the PDF showed my response to ANY game played with EPRB?

 

[Gordon Watson]

Since this draft appears to address none of the concerns I raised in posts #29 and #30, I would appreciate it if you would respond to those posts individually when you have the time (along with the question involving the diagram in post #35)

Would you see, please, if the PDF shows that I finally get to an interpretation of, and use of, angles correctly.

As an engineer, I'm confident that I do,

I have probably made a big mistake in talking about BI-angles. That is what I used personally, to get my head around BT -- me believing BT was wrong the instant I read a Mermin account of it.

It probably reflects my approach to engineering, now I think about: Explaining my thinking, believing the results speak for themselves.

Considering the PDF, the result, may I take it that post #35 is satisfactorily answered now?

And that the answer is not crackpot?

If not, give me another crack at it.

Thanks.

 

[Gordon Watson]

I don't know what you mean by "Bell type probability" since you have said that you aren't calculating the probabilities that the source will emit particles with different combinations of predetermined results for each angle; or did I misunderstand you there? You certainly haven't provided a "counterexample" to Bell in the form of a local realistic physical model, i.e. one where you can give us some local hidden variables associate with the particle and rules for how the variables together with the polarizer angle determine (in a probabilistic or deterministic way) the outcome of each measurement, with the rules obeying locality (so that all values of variables and other events can only be causally influenced by values/events in their past light cone). If you had an actual local realist physical model you would be able to use it to meet the challenge I offered earlier:

But you said you wouldn't be able to win at this challenge. So please don't continue to assert you have a local realist model or a counterexample to Bell if you don't even understand the notion of "local realism" well enough to see what this would actually entail. As you know this forum is not meant to be a platform for people who think they have made some brilliant discovery which destroys some mainstream result, when you asked if I thought it would be appropriate to start a thread like this I offered the opinion that it would be OK if you were here in a spirit of learning and being willing to listen to explanations as to why your argument doesn't falsify Bell's theorem, if you aren't willing to do that and just want to confidently assert that you have done so, then I don't think the discussion should continue on this forum.

As I said in my previous post #25, I would like you to use the standard type of notation for angles, where individual angles are defined relative to some fixed coordinate angles and differences between two angles are defined in some fixed way, like ab=a-b. If you think the terminology of "bi-angles" still makes sense in this context, then please explain clearly what you mean, hopefully using a numerical example where we have definite angles for a,b,c and can thus calculate any angles like ab and ac. If you're just saying that the probabilities of certain observable results may be identical for some combinations of angles, then certainly I agree, for example if we pick a=240, b=120 and c=0 then the probability of ++ for any combination of these angles will be 3/8 according to equation (7) on the Bell inequality[/url] page. But if you're saying something more than that, please explain more clearly.

Can you be more specific? What precise outcome does P1 give the probability of? And likewise for P2, P3, P4, P5, P6, P7, and P8? And do you understand that this is not what P1-P8 are defined to mean on the Sakurai's Bell inequality page, that there they are supposed to be probabilities for different combinations of predetermined results determined by hidden variables, not probabilities of observable outcomes?

Also, if P1-P8 are simply probabilities of observable outcomes, then why is it that you still seem to calculate P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7? These equations make perfect sense if P1-P8 are predetermined results as on the wiki page, but I have no idea why an observable outcome like P(a+,c+) would be the sum of the probabilities of two other observable outcomes.

Of course, that's why I keep asking for clarification on your terms, and asking for numerical examples, and saying things like "this thread would be a lot more productive if you would respond to my requests for clarification and numerical examples instead of just repeating the same impenetrable terminology which seems to make sense only to you". You can't just refer me back to the OP because your explanations there were no use to me!

Not in any way that makes sense to me, just with some incomprehensible jargon that seems to be your own private language, not any standard mathematical terminology I'm familiar with. For example, you say "This follows from the topological fact re spatial relations here: ab may be constructed in two ways"...what is a "topological fact re spatial relations", and what does it mean to "construct" ab? Isn't ab just the angular difference between two polarizers at angles a and b, i.e. isn't it just a-b? That's what an angular difference is normally defined to mean by everyone I've ever seen talking about the "angle between" two things!

Why? What physical or geometric considerations lead you to think P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, but that P3 + P4 is only "one-half the average"? Again since you seem to be defining P1-P8 in a totally different way than on the wikipedia page, it would really be helpful if you would tell us what observable outcomes P1-P8 are each supposed to give the probability of.

This average could also be derived from the QM average, calculated over each angle value; i.e., directly from QM's one-angle calculations, done twice, one for each value of the 2-valued ab.


Your notion of "focusing" on 2 angles or "reference angles" are similarly incomprehensible to me, I'm just talking about angles in the standard way that physicists always talk about angles, defining them relative to some fixed coordinate system, see post #25. As I requested there, I would like you to start using this sort of standard definition of angles as well, if your argument really revolves around saying there is something fundamentally flawed about defining angles relative to a fixed coordinate system and that we must use your incomprehensible alternative definitions, then your argument really is hopelessly crackpot and I am not interested in continuing.

Really? What angles do you think QM can't handle? If you have trouble with the standard notation for defining angles, could you draw a picture of polarizers at different orientations and show graphically what angles in the picture you think QM can't describe?

Again it's pretty clear you don't understand what "local and realistic" even mean, if you admit you can't answer my simulation challenge above but still think you have a local and realistic "model". Again if you are here in the spirit of learning rather than just here to proclaim your glorious victory over Bell, please stop making such claims and admit there may be some fault in your understanding of the notion of local realism. For example, did you completely understand the definition of local realism I gave in my [post=3154224]post #20 to Avodyne[/post]? If not we might use that as a starting point.

No, I have no idea what either of these terms mean, that's why I keep asking for clarifications. Please explain using either a numerical or diagrammatic example, not just strange abstract verbal discussions and equations where you haven't explained the meaning of the terms (like your using equations with P1 or ab without explaining what these mean physically) or where the equations come from (like why you think it should be true that ab = ac + bc and ac = ac - bc).

If the experiment is with a Stern-Gerlach apparatus this is pretty straightforward since the apparatus creates a magnetic field and "up" would just mean in the direction of magnetic North while "down" would mean in the direction of magnetic South. With photons, I believe + or - tells you whether the photon passed through the polarizer and was detected by the detector behind it, or whether it was reflected by the polarizer and detected by the other detector (see the diagram of the "two-channel polarizer" on the CHSH inequality page)


If you use a standard notation for labeling orientations with angles, then each orientation has a unique angle. If your argument is that it is somehow impossible or forbidden to use this type of standard notation where there's a one-to-one relation between orientations and angles, then it seems to me you're just confused about basic geometry and at the very least I want to see a diagram of an orientation that you think cannot be assigned a unique angle using the standard procedure (make sure to draw both the orientation and the x-axis of the fixed coordinate system).

Um, how could you possible get that from my comment above? I said that QM only fails to give the Ps in the table if those Ps are understood to have the standard meaning of predetermined results for each possible angle[/b], but you have said this is not what the Ps in your model mean at all, with comments like (from post #10) "I am providing the Ps for different outcomes under the defined tripartite setup; introducing the third orientation to be consistent with Bell and Sakurai; recognizing that we can only test for one angle (two orientations) in anyone physical test." As I keep telling you, the Ps on the wiki page aren't supposed to represent measured outcomes at all, rather they represent predetermined facts about what the results will be for each of the 3 possible angles, these predetermined facts based on hidden variables which human experimenters have no way of knowing. What's more, in a more recent post I asked you:

And in the post of yours I am responding to right now (post #24) you responded:

Your "next post" did not actually explain your disagreement...but look, if you disagree with point #1, then you disagree that there are predetermined answers for all three angles, so you cannot possibly be defining your Ps in the same way as the wikipedia page! On the wikipedia page, P1-P8 specifically refer to the probabilities of different combinations of hidden predetermined results for all three angles! How many times do I have to repeat this, and how many times do I have to ask you what you mean when you write P1-P8 before you will give me a straight answer? If they are supposed to represent probabilities of some observable results, then your comment "So QM does not give the Ps in the Table, and the model does" is clearly wrong, since as I already stated very clearly in the quote you were responding to there, "QM certainly delivers probabilities for all measurable outcomes."

Huh? Again, if you disagree with my point #1 above, then you disagree that there are hidden "predetermined results" for each possible angle, so how can you possibly think you can "identify the Ps in the BI" when these are supposed to be the probabilities of different predetermined results for each possible angle?

No, it certainly doesn't. You have denied that you think the hidden variables associated with each particle after emission give it predetermined results for each possible angle, and if you changed your mind and accepted this, it would be very trivial to show it must be true that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

Please stop behaving like a crackpot and confidently proclaiming you have overturned Bell while ignoring all possibility that all the criticisms (such as the criticism that you don't seem to even understand what 'local realism' means) have any validity. Again, if you're here to learn and try to understand why the argument might be flawed that's fine, if you're just here to announce your earth-shattering discovery please use another forum.

What does "each ab" even mean, geometrically or physically? I know of only one ab, the angle between the polarizer (or Stern-Gerlach device) which is at angle a relative to some fixed axis, and the polarizer (or Stern-Gerlach device) which is at angle b relative to the same axis.

Jesse, I hoped that my later "tabulated" example, and the PDF (for sure), would address everyone of these points.

Would you cut and paste into separate posts questions remaining after you've studied the PDF, please?

I can see that my jargon has not helped. EG: By "Bell-type probability" I meant any that relate to EPRB and Bell. So I sure meant the examples in the Sakurai-wiki. I was just not wanting to go beyond that scenario, here.

 

[Gordon Watson]

(reply to post #24, continued)


If you're saying you disagree with #1, are you changing your mind from our earlier exchange in post #9 and 10? There I said:

To which you responded:

Have you changed your mind, or did you misunderstand my question, or do you not actually disagree with point #1?

OK, suppose the "large number of trials" in #3 is 900 trillion. And if we are omniscient beings who know all the hidden variables associated with Alice's particle on each trial, and therefore know what predetermined results each particle sent towards Alice had before it actually reached the polarizer, then we can know exactly how many trials in total there were where the particle sent towards Alice had some set of predetermined results, like (a+,b+,c-). Suppose this number is 90 trillion, which would mean P2 is 90 trillion out of 900 trillion, or 1/10. What 3 is saying is that if the hidden variables sent out by the source on each trial have no correlation with the choice of angles Alice and Bob made on that trial, then this same P2 should also give us the fraction that had predetermined results (a+,b+,c-) in the subset of trials where Alice picked angle a and Bob picked angle c. If Alice and Bob each select their angle randomly, then we'd expect a 1/3*1/3=1/9 chance of any given combination of polarizer angles on each trial, so out of 900 trillion total trials, we'd expect that there were about 100 trillion trials where Alice chose angle a and Bob chose angle c. So 3 is saying that within this subset P2 still gives the fraction of trials where Alice's predetermined results were (a+,b+,c-), so the total number of trials where Alice chose a and Bob chose c and Alice's predetermined results were (a+,b+,c-) should be (1/10)*100 trillion = 10 trillion.

To sum up, what 3 is saying is just that the probability the source will send out particles with a given combination of hidden results on a given trial is uncorrelated with what polarizer angles Alice and Bob happened to choose on that trial. If you agreed with #1 and #2 (which apparently you don't), then I don't see how you'd be able to disagree with #3 unless you thought the source had some sort of precognitive knowledge of what settings Alice and Bob were going to choose on that trial, after the two particles had already been emitted.

I believe the PDF should now provide the best answers that I can give to all these questions.

I believe that because I think the PDF moves me closer to QM terms, and hence closer to your field. The maths should be at one with the Sakurai table -- and against the Sakurai-Bell inequality in the Sakurai wiki.

Please, let me know if this answer is not now satisfactory.

 

[JesseM]

OK; thank you; your post came in as I was posting my last.

Q: Do you not see that all the sums that you could possibly require are included in -- or easily derived from -- the base-data in the PDF.

But the sums are physically meaningless if you don't explain basic definitional questions like what you mean the probabilities P1-P8 to be the probabilities of. You obviously don't mean them to be probabilities of different predetermined states according to the chart on the wiki page, because then you wouldn't say:

NB: I ABSOLUTELY REJECT:

P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

If you accepted the standard meaning of P1-P8 you would have to accept this; for example, P3 and P4 are the probabilities of the only two predetermined states where Alice is predetermined to get + if she chooses angle "a" and Bob is predetermined to get + if he chooses angle "b".

IF IT IMPLIES: P3 + P4 = P3 + P4 + (POSITIVE NUMBER)![/B] (XXX)

Not if you use the standard definition of P1-P8. I have no idea what you mean the probabilities P1-P8 to be the probabilities of, and thus no idea why you think the previous line would "imply" this silly equation.

Forget about particle experiments for a second and consider this simple analogy. Suppose we have a machine that periodically prints out pairs of scratch lotto cards, and sends one card to Alice and one to Bob. Each card contains three squares labeled "a", "b", and "c", which are covered by a silvery substance you have to scratch to see what's underneath, and anytime a square is scratched it will reveal either a + or - underneath. Suppose Alice and Bob choose randomly which square to scratch each time they receive a card, and after many trials we observer that on every trial where they choose to scratch the same square, they get opposite results (+- or -+). Wouldn't the natural local realist conclusion be that each time the machine prints a pair of cards and sends them, then before Alice and Bob choose which box to scratch there is already either a + or - under each square, with the machine always making sure to make each member of the pair the opposite of the other, so for example the hidden symbols on Alice's card are a+,b-,c+ then the hidden symbols under Bob's must be a-,b+,c-? In this case, over many trials the source will print each type of card with some well-defined frequency, so we can define the following probabilities:

P1 = Probability that source sent card (a+,b+,c+) to Alice and card (a-,b-,c-) to Bob
P2 = Probability that source sent card (a+,b+,c-) to Alice and card (a-,b-,c+) to Bob
P3 = Probability that source sent card (a+,b-,c+) to Alice and card (a-,b+,c-) to Bob
P4 = Probability that source sent card (a+,b-,c-) to Alice and card (a-,b+,c+) to Bob
P5 = Probability that source sent card (a-,b+,c+) to Alice and card (a+,b-,c-) to Bob
P6 = Probability that source sent card (a-,b+,c-) to Alice and card (a+,b-,c+) to Bob
P7 = Probability that source sent card (a-,b-,c+) to Alice and card (a+,b+,c-) to Bob
P8 = Probability that source sent card (a-,b-,c-) to Alice and card (a+,b+,c+) to Bob

Can you see that in this example, it must be true that if Alice scratches square a and Bob scratches square b, P(a+,b+) must be P3 + P4? Likewise if Alice scratches square a and Bob scratches square c, P(a+, c+) = P2 + P4, and if Alice scratches square c and Bob scratches square b, P(c+, b+) = P3 + P7. But in no way does this "imply" your strange equation saying that P3 + P4 = P3 + P4 + (POSITIVE NUMBER), here P3 + P4 = P3 + P4 and nothing more. Do you agree that all this is true of this specific example involving scratch lotto cards, ignoring for the moment your weird ideas about particle experiments and angles? Please give a clear yes/no answer here.

If you agree with this, then understand that the probabilities in the Sakurai's Bell inequality page are meant to be exactly analogous, they simply represent predetermined results (determined by the hidden variables associated with the particle, which might be arbitrarily complex) for each of the three allowed orientations of the polarizers or Stern-Gerlach devices. So if you mean the probabilities to represent something different, and this difference leads you to odd conclusions like the "implication" P3 + P4 = P3 + P4 + (POSITIVE NUMBER), you need to explain what you mean the probabilities P1-P8 to represent physically, if they are not intended to have the same meaning as on the Sakurai Bell inequality page. If you don't explain what the physical meaning of P1-P8 are supposed to be, then the "reasoning" behind your statements is going to be totally incomprehensible to anyone but yourself, it's as if you're speaking your own private language that no one else understands. But I suspect that the problem may be deeper, that even you don't really have any clear notion of what P1-P8 are supposed to represent, and that you've just invented some rather arbitrary mathematical rules for yourself without bothering to think about what the physical meaning of the symbols is actually supposed to be.

You mentioned a game, which had me confused. I had hoped that the PDF showed my response to ANY game played with EPRB?

Why did it have you confused? The rules were fairly simple, you play the role of the source, then you can load any information onto the two flash drives on each trial (representing the hidden variables associated with the two particles created by the source on a given trial), then the two flash drives are sent to Alice and Bob. They then have a choice of whether to press A, B, or C (representing three orientations of simulated polarizers or SG devices), and their computers will simulate the "laws of physics" (these laws can be anything you want them to be, you programmed the rules in advance), taking into account both the variables associated with the particle on the flash drive and the simulated orientation selected by the experimenter, to produce either + or - as output. To "win" this game, you just have to produce statistics analogous to those in QM, meaning that on every trial where Alice and Bob chose the same button to press they always got opposite results, but on trials where they chose different buttons, the frequency they both got + is the same as in QM (and can thus violate the Bell inequality). Is there any aspect of this game you find unclear? If not, do you think that the PDF shows a strategy that would allow you to "win" at this game?

 

[Gordon Watson]

I highlighted what we need: Pac++, Pab++ and Pcb++=Pbc--

Pac++ is what is measured on monday, and equals 0.073... in agreement with your numbers
Pab++ is what is measured on tuesday, and equals 0.25. Your number gives 0.125
Pcb++=Pbc-- is what is measured on thursday, and equals 0.073 in agreement with your numbers.

You will NEVER be able to get Pab++ equal to 0.25 (you actually have 0.125), simply because it can't be larger than (Pac++) + (Pbc--) which equals 0.146...
Note that indeed, your number (0.125) is, as it can't be otherwise, smaller than 0.146.
Simply because this 0.146 is made up of your 4 positive numbers P2 + P3 + P4 + P7 as you give it yourself, and Pab++ is equal to only P3+P4 (your 0.125). You ADD to your 0.125 still your P2 and P7 to obtain 0.146, so it has to be smaller (as indeed it is).

Now, QM predicts not 0.125, but rather 0.25. It is bigger. So it can't come from numbers P1...P8 in this way.

There's nothing more to say about this.

Just to be clear on my earlier reply: You are adding over a limited sample. The sample where 2 angles only (in my terms) are clearly defined.

To check any SUM that you (rightly) want to test you MUST sum over all three examples. Then take the average by dividing by 3.

That is what the PDF does. If you think it does not, please say so.

Please see what I say about P1-P8 in the PDF. It CONFIRMS THE QM PREDICTIONS for any outcome that you could wish for; though I have only completed one example: ALL the results for Pab.

Do you disagree with any of the 4 Pab results, please?

 

[JesseM]

Would you see, please, if the PDF shows that I finally get to an interpretation of, and use of, angles correctly.

The PDF doesn't say anything about how you define angles, nor does it say what P1-P8 are supposed to represent physically. If they represent the probabilities of different predetermined results for a given particle like (a+,b-,c+), then logically this would imply that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7...see my previous post #46 with the analogy of the scratch lotto cards if you still aren't clear on this. If you use any different formula for calculating P(a+,b+), P(a+,c+) or P(c+,b+), then it's clear P1-P7 must represent something totally different than what they do on the Sakurai Bell inequality page.

With all your confusing comments on bi-angles, I am also unclear on what symbols like ab and bc mean in your pdf...see post #35, do you agree that we can assign a unique angle to each orientation by using a fixed coordinate system, and that the angle between two orientations can simply be calculated by subtracting the angle of one from the angle of the other, so ab=a-b, ac=a-c, bc=b-c, cb=c-b etc.? If you don't agree please address post #35 directly.

 

[Gordon Watson]

But the sums are physically meaningless if you don't explain basic definitional questions like what you mean the probabilities P1-P8 to be the probabilities of. You obviously don't mean them to be probabilities of different predetermined states according to the chart on the wiki page, because then you wouldn't say:

If you accepted the standard meaning of P1-P8 you would have to accept this; for example, P3 and P4 are the probabilities of the only two predetermined states where Alice is predetermined to get + if she chooses angle "a" and Bob is predetermined to get + if he chooses angle "b".

Not if you use the standard definition of P1-P8. I have no idea what you mean the probabilities P1-P8 to be the probabilities of, and thus no idea why you think the previous line would "imply" this silly equation.

Forget about particle experiments for a second and consider this simple analogy. Suppose we have a machine that periodically prints out pairs of scratch lotto cards, and sends one card to Alice and one to Bob. Each card contains three squares labeled "a", "b", and "c", which are covered by a silvery substance you have to scratch to see what's underneath, and anytime a square is scratched it will reveal either a + or - underneath. Suppose Alice and Bob choose randomly which square to scratch each time they receive a card, and after many trials we observer that on every trial where they choose to scratch the same square, they get opposite results (+- or -+). Wouldn't the natural local realist conclusion be that each time the machine prints a pair of cards and sends them, then before Alice and Bob choose which box to scratch there is already either a + or - under each square, with the machine always making sure to make each member of the pair the opposite of the other, so for example the hidden symbols on Alice's card are a+,b-,c+ then the hidden symbols under Bob's must be a-,b+,c-? In this case, over many trials the source will print each type of card with some well-defined frequency, so we can define the following probabilities:

P1 = Probability that source sent card (a+,b+,c+) to Alice and card (a-,b-,c-) to Bob
P2 = Probability that source sent card (a+,b+,c-) to Alice and card (a-,b-,c+) to Bob
P3 = Probability that source sent card (a+,b-,c+) to Alice and card (a-,b+,c-) to Bob
P4 = Probability that source sent card (a+,b-,c-) to Alice and card (a-,b+,c+) to Bob
P5 = Probability that source sent card (a-,b+,c+) to Alice and card (a+,b-,c-) to Bob
P6 = Probability that source sent card (a-,b+,c-) to Alice and card (a+,b-,c+) to Bob
P7 = Probability that source sent card (a-,b-,c+) to Alice and card (a+,b+,c-) to Bob
P8 = Probability that source sent card (a-,b-,c-) to Alice and card (a+,b+,c+) to Bob

Can you see that in this example, it must be true that if Alice scratches square a and Bob scratches square b, P(a+,b+) must be P3 + P4? Likewise if Alice scratches square a and Bob scratches square c, P(a+, c+) = P2 + P4, and if Alice scratches square c and Bob scratches square b, P(c+, b+) = P3 + P7. But in no way does this "imply" your strange equation saying that P3 + P4 = P3 + P4 + (POSITIVE NUMBER), here P3 + P4 = P3 + P4 and nothing more. Do you agree that all this is true of this specific example involving scratch lotto cards, ignoring for the moment your weird ideas about particle experiments and angles? Please give a clear yes/no answer here.

If you agree with this, then understand that the probabilities in the Sakurai's Bell inequality page are meant to be exactly analogous, they simply represent predetermined results (determined by the hidden variables associated with the particle, which might be arbitrarily complex) for each of the three allowed orientations of the polarizers or Stern-Gerlach devices. So if you mean the probabilities to represent something different, and this difference leads you to odd conclusions like the "implication" P3 + P4 = P3 + P4 + (POSITIVE NUMBER), you need to explain what you mean the probabilities P1-P8 to represent physically, if they are not intended to have the same meaning as on the Sakurai Bell inequality page. If you don't explain what the physical meaning of P1-P8 are supposed to be, then the "reasoning" behind your statements is going to be totally incomprehensible to anyone but yourself, it's as if you're speaking your own private language that no one else understands. But I suspect that the problem may be deeper, that even you don't really have any clear notion of what P1-P8 are supposed to represent, and that you've just invented some rather arbitrary mathematical rules for yourself without bothering to think about what the physical meaning of the symbols is actually supposed to be.

Why did it have you confused? The rules were fairly simple, you play the role of the source, then you can load any information onto the two flash drives on each trial (representing the hidden variables associated with the two particles created by the source on a given trial), then the two flash drives are sent to Alice and Bob. They then have a choice of whether to press A, B, or C (representing three orientations of simulated polarizers or SG devices), and their computers will simulate the "laws of physics" (these laws can be anything you want them to be, you programmed the rules in advance), taking into account both the variables associated with the particle on the flash drive and the simulated orientation selected by the experimenter, to produce either + or - as output. To "win" this game, you just have to produce statistics analogous to those in QM, meaning that on every trial where Alice and Bob chose the same button to press they always got opposite results, but on trials where they chose different buttons, the frequency they both got + is the same as in QM (and can thus violate the Bell inequality). Is there any aspect of this game you find unclear? If not, do you think that the PDF shows a strategy that would allow you to "win" at this game?

Taking just the first part of your post:

I am confused.

In the PDF, the P1-P8 are the probabilities that attach to the outcomes that are in the two Bell-table columns to their left.

In equation (1), directly below the table, I directly derive Pab by summing over P3 and P4 (as you suggest).

I get the QM correct result, Sab/2.

In the same way: Any correct QM result may be obtained from the model.

I seem to be missing something; me thinking the table (and the consequent results), was an OK result.

Re games: I do not expect any classical game (eg, cards, balls) to deliver the correct results that are to be associated with EPRB.

I thought (was I wrong) that the so-called silly equation was mentioned somewhere in relation to this thread? I thought someone related it to BI?

Suggestion: Could we please take one key, QM-EPRB based (not classical card games etc) question at a time? One initiating post at a time?

Is there another Pxy sum that might help? I'm concerned/confused that it is not?

Is the above-quoted sum, (1) in the PDF, wrong?

 

[Gordon Watson]

The PDF doesn't say anything about how you define angles, nor does it say what P1-P8 are supposed to represent physically. If they represent the probabilities of different predetermined results for a given particle like (a+,b-,c+), then logically this would imply that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7...see my previous post #46 with the analogy of the scratch lotto cards if you still aren't clear on this. If you use any different formula for calculating P(a+,b+), P(a+,c+) or P(c+,b+), then it's clear P1-P7 must represent something totally different than what they do on the Sakurai Bell inequality page.

With all your confusing comments on bi-angles, I am also unclear on what symbols like ab and bc mean in your pdf...see post #35, do you agree that we can assign a unique angle to each orientation by using a fixed coordinate system, and that the angle between two orientations can simply be calculated by subtracting the angle of one from the angle of the other, so ab=a-b, ac=a-c, bc=b-c, cb=c-b etc.? If you don't agree please address post #35 directly.

YOU say "logically this would imply that

P(a+, b+) = P3 + P4 and

P(a+, c+) = P2 + P4 and

P(c+, b+) = P3 + P7..."

From the PDF; and in its terminology (which is consistent with yours):

Pab++ = P3 + P4. See PDF eqn (1).

Pac++ = P2 + P4; from direct observation in the PDF!

Pcb++ = P3 + P7; from direct observation in the PDF!

As DrC often says: Where's the beef?

Your logical implication agrees with the PDF.

Did you check and think that it did not?

Does this rare agreement between us cause you to reconsider your questions and their focus?

Did you not see in the OP under ANGLES: ab = the angle between orientations a and b?

Are we now further agreed about angles, especially as they appear in the PDF?

 

[JesseM]

Taking just the first part of your post:

I am confused.

In the PDF, the P1-P8 are the probabilities that attach to the outcomes that are in the two Bell-table columns to their left.

But those cannot possibly be observable outcomes because they specify + or - for three different angles, whereas on each trial there are only two experimenters who can each get a result at only one angle. Do you understand that on the Sakurai's Bell inequality page, notation like + - + in the "Alice" column is intended to mean that on a given trial, the source sent a particle to Alice whose hidden variables predetermined (even before she chose what angle to set her polarizer to) that if she chose angle "a" on that trial she was guaranteed to get +, if she chose angle "b" on that trial she was guaranteed to get -, and if she chose angle "c" on that trial she was guaranteed to get +? And do you understand that there is no way we can actually know what predetermined results were associated with Alice's particle on each trial, that + - + represents a hidden set of predetermined results, with P1-P8 representing the hidden probabilities that the source sends out each possible set of predetermined results? Please tell me, yes or no, if you understand and agree that this is the meaning of P1-P8 on the Sakurai's Bell inequality page.

If you do understand and agree with this, then are P1-P8 on your table intended to have exactly the same meaning? If not, what do they mean? For example, suppose on one trial Alice chooses angle a and got result +, while Bob chose angle c and got +. How are we supposed to know whether this trial corresponded to the "outcome" [Alice: +--, Bob: -++] or if it corresponded to the "outcome" [Alice:++-, Bob: -++]? Both of these have a + in the "a" slot for Alice, and a + in the "c" slot for Bob. Is the "outcome" of each trial supposed to be something determined by experiment, or is it supposed to be a "hidden" fact which we imagine could be known by an omniscient observer even if it's not known by us, as on the Sakurai's Bell inequality page?

In equation (1), directly below the table, I directly derive Pab by summing over P3 and P4 (as you suggest).

Then why did you say that you "absolutely reject" the claim that P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7? Which one of these do you reject? I won't be able to understand why you reject it unless you give me a straight answer to what P1-P8 are supposed to be the probabilities of, certainly all three of these equations must be true if P1-P8 have the meaning on the wiki page which I discussed above, do you disagree?

Re games: I do not expect any classical game (eg, cards, balls) to deliver the correct results that are to be associated with EPRB.

Do you include the type of computer-simulation game I mentioned as a "classical game"? If so, I'd say that shows you don't understand what "local realism" means, since the laws of a local realist universe should be possible to simulate (at least approximately, to any desired degree of accuracy) on a collection of computers where each computer is calculating what should happen in each local region of spacetime, with each computer only having access to the output of other computers which simulated local regions in the past light cone of that one. If you don't see why this should be true and want to discuss it, read over the definition of local realism I gave in [post=3154224]post #20 to Avodyne[/post] and tell me if you don't understand it or don't see how it would imply the idea that a local realist universe should be possible to simulate on a network of local computers each figuring out what happens in a small unit of space in each small increment of time (like a cellular automaton)

I thought (was I wrong) that the so-called silly equation was mentioned somewhere in relation to this thread? I thought someone related it to BI?

I was referring to this equation: P3 + P4 = P3 + P4 + (POSITIVE NUMBER). I'm sure no one seriously suggested a sum of two probabilities could fail to be equal to itself.

Is the above-quoted sum, (1) in the PDF, wrong?

It gives the correct answer but without an explanation of what the symbols P3 and P4 are supposed to mean, it is physically meaningless.

 

[JesseM]

YOU say "logically this would imply that

P(a+, b+) = P3 + P4 and

P(a+, c+) = P2 + P4 and

P(c+, b+) = P3 + P7..."

From the PDF; and in its terminology (which is consistent with yours):

Pab++ = P3 + P4. See PDF eqn (1).

Pac++ = P2 + P4; from direct observation in the PDF!

Pcb++ = P3 + P7; from direct observation in the PDF!

As DrC often says: Where's the beef?

Your logical implication agrees with the PDF.

The PDF does not give any formulas for Pac++ and Pcb++, so I don't know what you mean by "from direct observation". And if that's the case, why did you say in post #42 (after you had posted the PDF) the following?

NB: I ABSOLUTELY REJECT:

P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7.

I was just taking you at your word that you "absolutely reject" that. In any case, if you do agree with these equations, then would you not also agree that since P2, P3, P4, and P7 are all non-negative, it must be true according to your formulas that P3 + P4 ≤ P2 + P4 + P3 + P7, and thus Pab++ ≤ (Pac++) + (Pcb++) for all possible values of a,b,c?

 

[Gordon Watson]

But the sums are physically meaningless if you don't explain basic definitional questions like what you mean the probabilities P1-P8 to be the probabilities of. You obviously don't mean them to be probabilities of different predetermined states according to the chart on the wiki page, because then you wouldn't say:

If you accepted the standard meaning of P1-P8 you would have to accept this; for example, P3 and P4 are the probabilities of the only two predetermined states where Alice is predetermined to get + if she chooses angle "a" and Bob is predetermined to get + if he chooses angle "b".

Not if you use the standard definition of P1-P8. I have no idea what you mean the probabilities P1-P8 to be the probabilities of, and thus no idea why you think the previous line would "imply" this silly equation.

Forget about particle experiments for a second and consider this simple analogy. Suppose we have a machine that periodically prints out pairs of scratch lotto cards, and sends one card to Alice and one to Bob. Each card contains three squares labeled "a", "b", and "c", which are covered by a silvery substance you have to scratch to see what's underneath, and anytime a square is scratched it will reveal either a + or - underneath. Suppose Alice and Bob choose randomly which square to scratch each time they receive a card, and after many trials we observer that on every trial where they choose to scratch the same square, they get opposite results (+- or -+). Wouldn't the natural local realist conclusion be that each time the machine prints a pair of cards and sends them, then before Alice and Bob choose which box to scratch there is already either a + or - under each square, with the machine always making sure to make each member of the pair the opposite of the other, so for example the hidden symbols on Alice's card are a+,b-,c+ then the hidden symbols under Bob's must be a-,b+,c-? In this case, over many trials the source will print each type of card with some well-defined frequency, so we can define the following probabilities:

P1 = Probability that source sent card (a+,b+,c+) to Alice and card (a-,b-,c-) to Bob
P2 = Probability that source sent card (a+,b+,c-) to Alice and card (a-,b-,c+) to Bob
P3 = Probability that source sent card (a+,b-,c+) to Alice and card (a-,b+,c-) to Bob
P4 = Probability that source sent card (a+,b-,c-) to Alice and card (a-,b+,c+) to Bob
P5 = Probability that source sent card (a-,b+,c+) to Alice and card (a+,b-,c-) to Bob
P6 = Probability that source sent card (a-,b+,c-) to Alice and card (a+,b-,c+) to Bob
P7 = Probability that source sent card (a-,b-,c+) to Alice and card (a+,b+,c-) to Bob
P8 = Probability that source sent card (a-,b-,c-) to Alice and card (a+,b+,c+) to Bob

Can you see that in this example, it must be true that if Alice scratches square a and Bob scratches square b, P(a+,b+) must be P3 + P4? Likewise if Alice scratches square a and Bob scratches square c, P(a+, c+) = P2 + P4, and if Alice scratches square c and Bob scratches square b, P(c+, b+) = P3 + P7. But in no way does this "imply" your strange equation saying that P3 + P4 = P3 + P4 + (POSITIVE NUMBER), here P3 + P4 = P3 + P4 and nothing more. Do you agree that all this is true of this specific example involving scratch lotto cards, ignoring for the moment your weird ideas about particle experiments and angles? Please give a clear yes/no answer here.

If you agree with this, then understand that the probabilities in the Sakurai's Bell inequality page are meant to be exactly analogous, they simply represent predetermined results (determined by the hidden variables associated with the particle, which might be arbitrarily complex) for each of the three allowed orientations of the polarizers or Stern-Gerlach devices. So if you mean the probabilities to represent something different, and this difference leads you to odd conclusions like the "implication" P3 + P4 = P3 + P4 + (POSITIVE NUMBER), you need to explain what you mean the probabilities P1-P8 to represent physically, if they are not intended to have the same meaning as on the Sakurai Bell inequality page. If you don't explain what the physical meaning of P1-P8 are supposed to be, then the "reasoning" behind your statements is going to be totally incomprehensible to anyone but yourself, it's as if you're speaking your own private language that no one else understands. But I suspect that the problem may be deeper, that even you don't really have any clear notion of what P1-P8 are supposed to represent, and that you've just invented some rather arbitrary mathematical rules for yourself without bothering to think about what the physical meaning of the symbols is actually supposed to be.

Why did it have you confused? The rules were fairly simple, you play the role of the source, then you can load any information onto the two flash drives on each trial (representing the hidden variables associated with the two particles created by the source on a given trial), then the two flash drives are sent to Alice and Bob. They then have a choice of whether to press A, B, or C (representing three orientations of simulated polarizers or SG devices), and their computers will simulate the "laws of physics" (these laws can be anything you want them to be, you programmed the rules in advance), taking into account both the variables associated with the particle on the flash drive and the simulated orientation selected by the experimenter, to produce either + or - as output. To "win" this game, you just have to produce statistics analogous to those in QM, meaning that on every trial where Alice and Bob chose the same button to press they always got opposite results, but on trials where they chose different buttons, the frequency they both got + is the same as in QM (and can thus violate the Bell inequality). Is there any aspect of this game you find unclear? If not, do you think that the PDF shows a strategy that would allow you to "win" at this game?

Jesse,

1: You appear to have been mislead by an incorrectly placed full stop; possibly from a cut-and-paste. I've fixed this. But am unsure what you think my qualifying clause attached to. And why it is not mentioned in your critique?

May I take it that you do not require me to deliver one of your terms being equal to the sum of two others?

Am concerned there's a subtlety here that I do not want to gloss or miss. Do you require any manipulation of the 3 terms?

2. Re games: Excuse my being guarded here until we are clear about the PDF. Flash-drives etc are classical devices; I doubt they could encode entanglement. It might be that you just want to encode the PDF? Is it the God's-eye distribution that some look for?

PS: Could I have one question per post for a while, please.

Here's mine: Do you require any manipulation of the 3 terms that you keep citing, please?

 

[Gordon Watson]

A: The PDF does not give any formulas for Pac++ and Pcb++, so I don't know what you mean by "from direct observation".

I meant "by looking at the table".

B: And if that's the case, why did you say in post #42 (after you had posted the PDF) the following?

Please see the qualifying sentence immediately beneath

C. I was just taking you at your word that you "absolutely reject" that. In any case, if you do agree with these equations, then would you not also agree that since P2, P3, P4, and P7 are all non-negative, it must be true according to your formulas that P3 + P4 ≤ P2 + P4 + P3 + P7, and thus Pab++ ≤ (Pac++) + (Pcb++) for all possible values of a,b,c?

Help please. Could you explain the physical significance of this SUM?

Is it the case that the inequality holds (or is relevant) for some logical or physical reason?

I think explanation here might make it clearer to me where I have confused you. Or where I have erred.

Thanks.

 

[JesseM]

Jesse,

1: You appear to have been mislead by an incorrectly placed full stop; possibly from a cut-and-paste. I've fixed this. But am unsure what you think my qualifying clause attached to. And why it is not mentioned in your critique?

I thought "if it implies" was just a rhetorical phrase, like "I reject Obama's healthcare plan if it implies people are mandated to buy insurance", where the person saying this knows that Obama's plan does include such a mandate. Are you saying you are genuinely unsure as to whether the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7 hold, or whether P3 + P4 = P3 + P4 + (POSITIVE NUMBER) would be an implication of this? I don't know what possible argument would "imply" that the second would be a consequence of the first, seems like a total non sequitur to me. And in any case, if you aren't even sure how to calculate P(a+, b+) and P(a+, c+) and P(c+, b+) then you obviously can't claim that you have a well-defined "model"!

May I take it that you do not require me to deliver one of your terms being equal to the sum of two others?

If by this you mean P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, then this is an automatic consequence of the definitions of P1-P8 on the Bell inequality page, if you don't think all three of these equations hold then you need to explain what your physical definitions of P1-P8 are (as I keep asking over and over) and how you calculate P(a+, b+) and P(a+, c+) and P(c+, b+).

Am concerned there's a subtlety here that I do not want to gloss or miss. Do you require any manipulation of the 3 terms?

Don't know what you mean by "manipulation".

2. Re games: Excuse my being guarded here until we are clear about the PDF. Flash-drives etc are classical devices; I doubt they could encode entanglement.

A "local realist" model of QM would be a classical model in the sense that Maxwell's laws are classical, it would explain everything without recourse to superpositions or spooky-action-at-a-distance. Again it seems doubtful to me that you really understand the meaning of "local realism" if you don't see that local realist models should be possible to simulate to arbitrary accuracy on classical computers.

It might be that you just want to encode the PDF? Is it the God's-eye distribution that some look for?

If you want to pick your strategy in the game by "encoding the PDF" be my guest. For example, on each trial you could load the flash drives with a set of predetermined results like a+,b-,c+, with the probabilities of different predetermined results on each trial being determined by P1-P8 in your PDF. Then each computer's algorithm for generating a "result" on each trial would just be to take the button pressed by the user (Alice or Bob), then look at the corresponding slot on the predetermined results on the flash drive, and give the symbol in that slot as output (for example, if Alice chooses button "b" and her flash drive contains the predetermined results a+,b-,c+, then the output would be -). But this strategy would not be a "win" for you, because the statistics over many trials would not match those of QM.

PS: Could I have one question per post for a while, please.

Here's mine: Do you require any manipulation of the 3 terms that you keep citing, please?

Again I don't know what you mean by "manipulation", perhaps you could give an example? Anyway, my main question right now would just be how you would propose to calculate P(a+, b+) and P(a+, c+) and P(c+, b+), whether you would use the formulas P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, with P2,P3,P4,P7 all defined as in the PDF, or whether you would use some different formulas.

 

[JesseM]

I meant "by looking at the table".

Your table does not give any formulas for calculating Pac++ and Pcb++.

In any case, if you do agree with these equations, then would you not also agree that since P2, P3, P4, and P7 are all non-negative, it must be true according to your formulas that P3 + P4 ≤ P2 + P4 + P3 + P7, and thus Pab++ ≤ (Pac++) + (Pcb++) for all possible values of a,b,c?

Help please. Could you explain the physical significance of this SUM?

Is it the case that the inequality holds (or is relevant) for some logical or physical reason?

It would hold in a local realist theory where each particle pair has hidden variables that predetermine what result it would give to any of the possible polarizer angles, with P1-P8 giving the probabilities of different sets of predetermined results as shown in the wiki page's table. But Pab++ ≤ (Pac++) + (Pcb++) does not hold in the theory of quantum mechanics for certain values of a,b,c. Therefore the theory of QM cannot be explained by a local realist theory.

 

[Gordon Watson]

I thought "if it implies" was just a rhetorical phrase, like "I reject Obama's healthcare plan if it implies people are mandated to buy insurance", where the person saying this knows that Obama's plan does include such a mandate. Are you saying you are genuinely unsure as to whether the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7 hold, or whether P3 + P4 = P3 + P4 + (POSITIVE NUMBER) would be an implication of this? I don't know what possible argument would "imply" that the second would be a consequence of the first, seems like a total non sequitur to me. And in any case, if you aren't even sure how to calculate P(a+, b+) and P(a+, c+) and P(c+, b+) then you obviously can't claim that you have a well-defined "model"!

OK; so we're clear on what I meant.

We appear to agree re non sequiturs.

I am sure of my calculations and reject the non sequitur.

Did I not answer this? https://www.physicsforums.com/showpost.php?p=3158191&postcount=50

The model is well-defined ... at least in so far as it agrees with your logical implication.

If by this you mean P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, then this is an automatic consequence of the definitions of P1-P8 on the Bell inequality page, if you don't think all three of these equations hold then you need to explain what your physical definitions of P1-P8 are (as I keep asking over and over) and how you calculate P(a+, b+) and P(a+, c+) and P(c+, b+).

Jesse, I'm sure that I calculate them the same way you do, using P1-P8 per the PDF.

If you refer to the calculation of each of P1-P8; they are derived from local realistic considerations. Just before this post of yours I was about to post this:

Thinking hard about our differences, does this help: Are you confusing my model with naive realism? With EPR elements of physical reality, interpreted in the naive tradition?

I AM NOT in the naive tradition in any way. My HVs are perturbed by the measurement interaction. To my mind, via spin axes following gyroscopic-like trajectories. The tricky word in EPR eprs is "corresponding".​

Don't know what you mean by "manipulation".

OK. I meant equating the 3 of them in some way.

A "local realist" model of QM would be a classical model in the sense that Maxwell's laws are classical, it would explain everything without recourse to superpositions or spooky-action-at-a-distance. Again it seems doubtful to me that you really understand the meaning of "local realism" if you don't see that local realist models should be possible to simulate to arbitrary accuracy on classical computers.

Oh! Oh! Is your requirement realistic?

I'm pretty sure that entanglement cannot be simulated on classical computers: spherical symmetry etc.

I am confident that my LR ideas cannot be simulated. You seem to be limiting LR to naive realism, in some of these phrases. My LR, primitive or cumbersome as it may seem, delivers the PDF. Your question below relates to that. LET US NOT get carried away by discussing LR now. If the PDF fails, my LR fails.

If you want to pick your strategy in the game by "encoding the PDF" be my guest. For example, on each trial you could load the flash drives with a set of predetermined results like a+,b-,c+, with the probabilities of different predetermined results on each trial being determined by P1-P8 in your PDF. Then each computer's algorithm for generating a "result" on each trial would just be to take the button pressed by the user (Alice or Bob), then look at the corresponding slot on the predetermined results on the flash drive, and give the symbol in that slot as output (for example, if Alice chooses button "b" and her flash drive contains the predetermined results a+,b-,c+, then the output would be -). But this strategy would not be a "win" for you, because the statistics over many trials would not match those of QM.

My model would require us to go beyond the encoding of gyroscopes, no two gyroscopes the correlated, except pair-wise; observers freely setting measurement orientations at random, ... Let's not digress; just yet.

Again I don't know what you mean by "manipulation", perhaps you could give an example? Anyway, my main question right now would just be how you would propose to calculate P(a+, b+) and P(a+, c+) and P(c+, b+), whether you would use the formulas P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, with P2,P3,P4,P7 all defined as in the PDF, or whether you would use some different formulas.

Manipulation addressed above.

I think I answered the next one: No difference between you and me here. Have you run a few lines using the PDF example below the table?

You do see that each result there is in total agreement with QM?

My question: Has anyone, presumably from the so-called Bell community, delivered the values for P1-P8 that the table requires? I'd be keen to see where they differ from mine.

 

[JesseM]

Jesse, I'm sure that I calculate them the same way you do, using P1-P8 per the PDF.

"Using P1-P8" is still vague, do you use all of the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, yes or no? If you do then your probabilities will obey the inequality Pab++ ≤ (Pac++) + (Pcb++)! The whole point of the proof is that QM violates this inequality, therefore it cannot be compatible with local realism! If you use the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7 then your model disagrees with QM, do you not understand this?

 

[Gordon Watson]

Your table does not give any formulas for calculating Pac++ and Pcb++.

Could you just use your formula; or the evident formula that I used in the worked example.

Why not do Pac++ and Pcb++?

Here's my answer: Sac/2, Scb/2. The QM results, I believe?

It would hold in a local realist theory where each particle pair has hidden variables that predetermine what result it would give to any of the possible polarizer angles, with P1-P8 giving the probabilities of different sets of predetermined results as shown in the wiki page's table. But Pab++ ≤ (Pac++) + (Pcb++) does not hold in the theory of quantum mechanics for certain values of a,b,c. Therefore the theory of QM cannot be explained by a local realist theory.

By predetermined; do you allow the HV to respond to each SGM that it encounters?

I'm wondering if this is the key difference between us. Explaining your focus on card-games? Card games cannot deliver the PDF table!

My calculation would yield: Pab++ = Sab/2 ≤ 1/2.

Which MUST hold in QM.

For I have Pac+– and Pbc+– added to the RHS of what you offer above.

My one question: Is it not the case that they too deliver: a = +; b = +; and so must be included in any rational study?

Thanks, as always,

JenniT

 

[Gordon Watson]

"Using P1-P8" is still vague, do you use all of the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7, yes or no? If so then your probabilities will obey the inequality Pab++ ≤ (Pac++) + (Pcb++)! The whole point of the proof is that QM violates this inequality, therefore it cannot be compatible with local realism! If you use the equations P(a+, b+) = P3 + P4 and P(a+, c+) = P2 + P4 and P(c+, b+) = P3 + P7 then your model disagrees with QM, do you not understand this?

We are cross-posting.

I use all, and maybe more, in the equation of yours above.

(This is what I meant by "manipulation, in its + sense, please understand!)

MY PDF violates this inequality, and agrees with QM, as explained in my just sent post.