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Gordon Watson wrote:

"Even though we have introduced λ as the state of a specific single system, the assumed objectivity of the system described by this state allows us to consider an ensemble of these, physically identical to the extent that they are all characterized by the same λ. The probabilities are to be associated with this ensemble. Clearly, this procedure is conceptually sound, even in cases where we cannot in practice prepare the pure λ ensemble," CH74, p.534, footnote 11 (with my underlining).

I'd welcome any and all comments on this footnote; even to the point of suggesting a new thread: "Bell's first error revealed!" But that would be conditional on a few others seeing the gross boo-boo and oops in the cited footnote!

My thanks again; Gordon

FrediFizzx wrote:Yeah, I guess minkwe, that basically started this thread, got busy for the holidays. And I guess Joy doesn't care about the CH inequality. Anyways I think our discussion bore some fruit. I finally figured out that the maximum result of QM for the CH74 string is about 0.207. According to Ballentine in "Quantum Mechanics: A Modern Development", P1 = P2 = 1/2 and P12(θ) = (cos(θ)^2)/2 and the CH74 string can be expressed as,

3*P12(θ) - P12(3*θ) - P1 - P2

According to Ballentine a maximum result occurs at 22.5 degrees which I think is for a photon state so plugging in the values we get,

1.28 - 0.073 - 1 = 0.207

So I think this confirms my result of the independent absolute bound of CH being 1/2. We could have for independent terms,

1/2 - 0 + 1/2 +1/2 - 1/2 -1/2 = 1/2

Yeah, finally!!!!!!!!
****

FrediFizzx wrote:

Gordon Watson wrote:

FrediFizzx wrote:No. It is easy to figure out that the maximum probability for ++ counts for P(b+|a+) is 1/2. And it is just common sense that say out of 10000 trials, you would only be able to get 5000 ++ counts maximum.

Fred, ?????; maybe try writing and analysing "+|+" when you relate "++" [sic] to P(b+|a+)!!

For sure, one of us is missing something very simple! Let's see:

Using CH74 notation and limitations but avoiding their errors:

Expectation; E(AB) = 2P(b|a)-1; etc. (1)

Marginal probability; P(a) = P(b) = 1/2. (2)

Joint probability; P(ab) = P(a)P(b|a) = P(b)P(a|b). (3)

Thus: P(a|b) = P(b|a) = 2P(ab). (4)

THUS: Conditional probability; 0 ≤ P(b|a) = P(a|b) ≤ 1 !! (5)

Thus, using (1): Expectation; -1 ≤ E(AB) = 2P(b|a)-1 ≤ 1; etc. (6)

So, as far as I can see: my results for CH74, CH74' and CHSH follow, error free.

(i) That is what I told you. Your P's in CH74' are now expectations that have to range from -1 to +1.

(ii) Look, if b = -a for every iteration, then there is a maximum 1/2 probability that you will get ++ counts and the same for no counts. The total probability is 1 but only a max of 1/2 for ++.

(iii) So your CH74' ≤ 2 is impossible unless the terms are expectation terms since the maximum absolute bound you could get if they are not expectation terms is 1.5 not 2.

(iv) It is your problem to solve what you are doing wrong. Not mine. I gave you the clues so figure it out.

****

Gordon Watson wrote:

FrediFizzx wrote:No. It is easy to figure out that the maximum probability for ++ counts for P(b+|a+) is 1/2. And it is just common sense that say out of 10000 trials, you would only be able to get 5000 ++ counts maximum.

Fred, ?????; maybe try writing and analysing "+|+" when you relate "++" [sic] to P(b+|a+)!!

For sure, one of us is missing something very simple! Let's see:

Using CH74 notation and limitations but avoiding their errors:

Expectation; E(AB) = 2P(b|a)-1; etc. (1)

Marginal probability; P(a) = P(b) = 1/2. (2)

Joint probability; P(ab) = P(a)P(b|a) = P(b)P(a|b). (3)

Thus: P(a|b) = P(b|a) = 2P(ab). (4)

THUS: Conditional probability; 0 ≤ P(b|a) = P(a|b) ≤ 1 !! (5)

Thus, using (1): Expectation; -1 ≤ E(AB) = 2P(b|a)-1 ≤ 1; etc. (6)

So, as far as I can see: my results for CH74, CH74' and CHSH follow, error free.

FrediFizzx wrote:

Gordon Watson wrote:

FrediFizzx wrote:

Gordon Watson wrote:Thanks Fred. To clarify our terms under CH74: Are you saying that the maximum for P(ab) -- ie P(++|CH74, a, b) -- is 1/2? Is that what you mean by a "++" probability?
*

CH only is concerned with single channel counts. IOW there is only + and ++. There is no - and no --, +-, -+ counts in CH like there is in the expectation version of CHSH. See CH74 eq. (1) and the text around it. The maximum probability that you will get with ++ counts is 1/2 per iteration. And that is with
b - a = 180 degrees. sin(180/2) = 1 so (1*1)/2 = 1/2.

OK Fred, thanks: I take it that you mean sin^2; but what has any of this to do with me or my hasty scribblings?

I ask because, nb: I use P(b|a), etc, [= P(b+|a+), etc, implicitly] in both CH74 and chsh, and these "Ps" can certainly [I believe] occupy the whole range [0,1]?
.

No. It is easy to figure out that the maximum probability for ++ counts for P(b+|a+) is 1/2. And it is just common sense that say out of 10000 trials, you would only be able to get 5000 ++ counts maximum.

FrediFizzx wrote:No. It is easy to figure out that the maximum probability for ++ counts for P(b+|a+) is 1/2. And it is just common sense that say out of 10000 trials, you would only be able to get 5000 ++ counts maximum.

Gordon Watson wrote:

FrediFizzx wrote:

Gordon Watson wrote:Thanks Fred. To clarify our terms under CH74: Are you saying that the maximum for P(ab) -- ie P(++|CH74, a, b) -- is 1/2? Is that what you mean by a "++" probability?
*

CH only is concerned with single channel counts. IOW there is only + and ++. There is no - and no --, +-, -+ counts in CH like there is in the expectation version of CHSH. See CH74 eq. (1) and the text around it. The maximum probability that you will get with ++ counts is 1/2 per iteration. And that is with
b - a = 180 degrees. sin(180/2) = 1 so (1*1)/2 = 1/2.

OK Fred, thanks: I take it that you mean sin^2; but what has any of this to do with me or my hasty scribblings?

I ask because, nb: I use P(b|a), etc, [= P(b+|a+), etc, implicitly] in both CH74 and chsh, and these "Ps" can certainly [I believe] occupy the whole range [0,1]?
.

FrediFizzx wrote:

Gordon Watson wrote:Thanks Fred. To clarify our terms under CH74: Are you saying that the maximum for P(ab) -- ie P(++|CH74, a, b) -- is 1/2? Is that what you mean by a "++" probability?
*

CH only is concerned with single channel counts. IOW there is only + and ++. There is no - and no --, +-, -+ counts in CH like there is in the expectation version of CHSH. See CH74 eq. (1) and the text around it. The maximum probability that you will get with ++ counts is 1/2 per iteration. And that is with
b - a = 180 degrees. sin(180/2) = 1 so (1*1)/2 = 1/2.

Gordon Watson wrote:Thanks Fred. To clarify our terms under CH74: Are you saying that the maximum for P(ab) -- ie P(++|CH74, a, b) -- is 1/2? Is that what you mean by a "++" probability?
*

FrediFizzx wrote:

Gordon Watson wrote:

FrediFizzx wrote:

Gordon Watson wrote:Good try mate, but try again. It's 1 January that approaches, not 1 April. E&OE.

That is totally non-responsive. You have an error. You should figure out what it is. You might find a clue in Appendix B of the CH74 paper.

Fred, re "non-responsive" -- it was as politely responsive as I could manage at the time.

With respect; you repeatedly claim errors against me but have any (or any of significance) yet been correct?

I'm pressed for time; you again claim an error; please identify it more closely to save me reading CH junk again.

Thanks; for I'm sure you know by now that I will respond "with thanks" to the identification of genuine errors; notwithstanding E&OE!
.

Well, there is no rush. Another clue. The maximum a ++ probability can be is 1/2. So by simple inspection for the probability version of CHSH would be,

1/2 - 1/2 + 1/2 +1/2 = 1

For independent terms you could have,

1/2 - 0 +1/2 +1/2 = 1.5 as an absolute independent bound so it can't be 2.
****

Gordon Watson wrote:

FrediFizzx wrote:

Gordon Watson wrote:Good try mate, but try again. It's 1 January that approaches, not 1 April. E&OE.

That is totally non-responsive. You have an error. You should figure out what it is. You might find a clue in Appendix B of the CH74 paper.

Fred, re "non-responsive" -- it was as politely responsive as I could manage at the time.

With respect; you repeatedly claim errors against me but have any (or any of significance) yet been correct?

I'm pressed for time; you again claim an error; please identify it more closely to save me reading CH junk again.

Thanks; for I'm sure you know by now that I will respond "with thanks" to the identification of genuine errors; notwithstanding E&OE!
.

FrediFizzx wrote:

Gordon Watson wrote:Good try mate, but try again. It's 1 January that approaches, not 1 April. E&OE.

That is totally non-responsive. You have an error. You should figure out what it is. You might find a clue in Appendix B of the CH74 paper.

Gordon Watson wrote:Good try mate, but try again. It's 1 January that approaches, not 1 April. E&OE.

FrediFizzx wrote:

Gordon Watson wrote: (i) "Ps" with me are NEVER expectation values!!

(ii) At (4d) I did NOT "just go to" the CHSH inequality: rather, I DERIVED CH74' (again, note that distinguishing prime). CH74' is CHSH written (correctly) in probabilistic terms; call it chsh. Thus:

CH74' = chsh = |P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a')| ≤ 2.

So your equation (X) is incorrect [E&OE].

The "something [sic] that is not right" are the errors and 'somethings' in Bell, CHSH, CH74 that I've already alerted you to.

The probability version of CHSH cannot be the same bound as the expectation version of CHSH. Your CH74' can't be correct. Remember that the CH inequality is only concerned with + and ++ counts.

Gordon Watson wrote: (i) "Ps" with me are NEVER expectation values!!

(ii) At (4d) I did NOT "just go to" the CHSH inequality: rather, I DERIVED CH74' (again, note that distinguishing prime). CH74' is CHSH written (correctly) in probabilistic terms; call it chsh. Thus:

CH74' = chsh = |P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a')| ≤ 2.

So your equation (X) is incorrect [E&OE].

The "something [sic] that is not right" are the errors and 'somethings' in Bell, CHSH, CH74 that I've already alerted you to.

FrediFizzx wrote:

Gordon Watson wrote:

FrediFizzx wrote:

FrediFizzx wrote:OK, I get what you did now. However, at the point of going to your eq. (4d), you should maybe just drop the 2 off of P2. IOW, just say P(b|a) instead of
P2(b|a). Then it is less confusing. Agreed?
***

Actually I think you can just not even use the subscripts 1 and 2 at all. I don't think they are needed for your formulation since you are not strictly following CH74 anyways.

Please explain: "Not following CH74 anyways." ???

PS: Please take account of this: "I do not follow CH74 into error!" And, please: point to where you follow CH74 and I do not!

I said you are not strictly following CH74. That is because you introduced the conditional into the scheme. So just dropping the subsripts 1 and 2, you would have the following.

-1 ≤ P(a)P(b|a) - P(a)P(b'|a) + P(a')P(b|a') + P(a')P(b'|a') - P(a') - P(b) ≤ 0. (4a)

Then, since λ is a random variable, we have:

P(a) = P(a') = P(b) = 1/2. (4b)

So from (4a) we have: -1 ≤ (1/2) [P(b|a) - P(b'|a) + P(b|a') + P(b'|a') - 1 - 1] ≤ 0. (4c)

So: CH74' = |P(b|a) - P(b'|a) + P(b|a') + P(b'|a')| ≤ 2. (4d)

However, under EPRB (using Ps that are readily derived), (4d) delivers:

CH74' = | sin^2((a,b)/2) - sin^2((a,b')/2) + sin^2((a',b)/2) + sin^2((a',b')/2)| ≤ 2. (4e)

Now a, b, a', b' are unrestricted! So let (a,b) = (a',b) = (a',b') = (a,b')/3 = 3π/4.

Then CH74' is absurd, for we find:

CH74' = 2 + (√2 - 1) >> 2. (4f) QED; E and OE!

So at eq, (4d) you just went to the CHSH inequality. And the "P's" somehow became expectation values with a range of -1 to +1. So something is not right as far as CH74 is concerned. I think the probability form of CHSH using your notation should be,

|P(b|a) - P(b'|a) + P(b|a') + P(b'|a')| ≤ 1

FrediFizzx wrote: "I said you are not strictly following CH74. That is because you introduced the conditional into the scheme."

FrediFizzx wrote:"So just dropping the subscripts 1 and 2, you would have the following. … …"

FrediFizzx wrote: "So at eq, (4d) you just went to the CHSH inequality. And the "P's" somehow became expectation values with a range of -1 to +1. So something is not right as far as CH74 is concerned. I think the probability form of CHSH using your notation should be,

|P(b|a) - P(b'|a) + P(b|a') + P(b'|a')| ≤ 1. (X)" [(X) added by GW for ID purposes.]

Gordon Watson wrote:

FrediFizzx wrote:

FrediFizzx wrote:OK, I get what you did now. However, at the point of going to your eq. (4d), you should maybe just drop the 2 off of P2. IOW, just say P(b|a) instead of
P2(b|a). Then it is less confusing. Agreed?
***

Actually I think you can just not even use the subscripts 1 and 2 at all. I don't think they are needed for your formulation since you are not strictly following CH74 anyways.

Please explain: "Not following CH74 anyways." ???

PS: Please take account of this: "I do not follow CH74 into error!" And, please: point to where you follow CH74 and I do not!

-1 ≤ P(a)P(b|a) - P(a)P(b'|a) + P(a')P(b|a') + P(a')P(b'|a') - P(a') - P(b) ≤ 0. (4a)

Then, since λ is a random variable, we have:

P(a) = P(a') = P(b) = 1/2. (4b)

So from (4a) we have: -1 ≤ (1/2) [P(b|a) - P(b'|a) + P(b|a') + P(b'|a') - 1 - 1] ≤ 0. (4c)

So: CH74' = |P(b|a) - P(b'|a) + P(b|a') + P(b'|a')| ≤ 2. (4d)

However, under EPRB (using Ps that are readily derived), (4d) delivers:

CH74' = | sin^2((a,b)/2) - sin^2((a,b')/2) + sin^2((a',b)/2) + sin^2((a',b')/2)| ≤ 2. (4e)

Now a, b, a', b' are unrestricted! So let (a,b) = (a',b) = (a',b') = (a,b')/3 = 3π/4.

Then CH74' is absurd, for we find:

CH74' = 2 + (√2 - 1) >> 2. (4f) QED; E and OE!

FrediFizzx wrote:

FrediFizzx wrote:OK, I get what you did now. However, at the point of going to your eq. (4d), you should maybe just drop the 2 off of P2. IOW, just say P(b|a) instead of
P2(b|a). Then it is less confusing. Agreed?
***

Actually I think you can just not even use the subscripts 1 and 2 at all. I don't think they are needed for your formulation since you are not strictly following CH74 anyways.

FrediFizzx wrote:… … ...
OK, I get what you did now. However, at the point of going to your eq. (4d), you should maybe just drop the 2 off of P2. IOW, just say P(b|a) instead of P2(b|a). Then it is less confusing. Agreed?
***

"Even though we have introduced λ as the state of a specific single system, the assumed objectivity of the system described by this state allows us to consider an ensemble of these, physically identical to the extent that they are all characterized by the same λ. The probabilities are to be associated with this ensemble. Clearly, this procedure is conceptually sound, even in cases where we cannot in practice prepare the pure λ ensemble," CH74, p.534, footnote 11 (with my underlining).