CH Inequality

[Gordon Watson]

A: Fred, you say this: "Following the notation used in the CH74 paper for eq. (4) then it is P12(a, b) = sin^2((a,b)/2) [under EPRB]." ???

I say: "Following the notation used in the CH74 paper for eq. (4) then P12(a, b) = (1/2) sin^2((a,b)/2) [under EPRB]." Reason: for me, P12(a, b) is the joint probability that both counts are triggered under the setting combination a and b. Further (and given in CH74), P1(a) is the marginal probability of Alice's detector being triggered under setting a; so it equals 1/2. So:

P12(a, b) = P1(a)P2(b|a) = (1/2) sin^2((a,b)/2); etc., as in my original derivation (above) re CH74.

NB: I'm no fan of the CH74 notation, but P2(b|a) -- a conditional probability -- is not here the same as P2(b) -- a marginal probability.

B: You say: "So I don't know how you are getting that from P2(a|b). P12 is the pairs probability count so P2 is a singles probability count."

I say: P12(a, b) is a joint probability. Under OPT it factors into P12(a, b) = P1(a)P2(b|a) = P2(b)P1(a|b). P2 is probability function associated with Bob's detector and we have two such: P2(b), a marginal probability; P2(b|a) a conditional probability; etc.

Note that I split this discussion to a new appropriate topic.

We are discussing the CH74 paper here so we should stick to their notation. P1, P2, and P12 are defined in their eq. (1) so P2 has to be a singles count (marginal probability?). You are just confusing the discussion by introducing something like P2(b|a).

Fred, with respect: You are confusing the discussion by not understanding P2(b|a) and by pointing to "errors" that are not.

My use of P2(b|a) is licensed by OPT. Given their (1), P2(b|a) CAN NEVER BE FALSE under OPT!

If you want to stick with their silly (2')* -- which is often false under OPT: and certainly false here -- then you are stuck with their silly results!

* Read their silly paragraph below (2'). Take note of their focus on AAD and their false extrapolation from common-sense (with no mention of the consequences of correlations: (2') even being false under some elementary classical correlations).
.

[FrediFizzx]

A: Fred, you say this: "Following the notation used in the CH74 paper for eq. (4) then it is P12(a, b) = sin^2((a,b)/2) [under EPRB]." ???

I say: "Following the notation used in the CH74 paper for eq. (4) then P12(a, b) = (1/2) sin^2((a,b)/2) [under EPRB]." Reason: for me, P12(a, b) is the joint probability that both counts are triggered under the setting combination a and b. Further (and given in CH74), P1(a) is the marginal probability of Alice's detector being triggered under setting a; so it equals 1/2. So:

P12(a, b) = P1(a)P2(b|a) = (1/2) sin^2((a,b)/2); etc., as in my original derivation (above) re CH74.

NB: I'm no fan of the CH74 notation, but P2(b|a) -- a conditional probability -- is not here the same as P2(b) -- a marginal probability.

B: You say: "So I don't know how you are getting that from P2(a|b). P12 is the pairs probability count so P2 is a singles probability count."

I say: P12(a, b) is a joint probability. Under OPT it factors into P12(a, b) = P1(a)P2(b|a) = P2(b)P1(a|b). P2 is probability function associated with Bob's detector and we have two such: P2(b), a marginal probability; P2(b|a) a conditional probability; etc.

Note that I split this discussion to a new appropriate topic.

We are discussing the CH74 paper here so we should stick to their notation. P1, P2, and P12 are defined in their eq. (1) so P2 has to be a singles count (marginal probability?). You are just confusing the discussion by introducing something like P2(b|a).

Fred, with respect: You are confusing the discussion by not understanding P2(b|a) and by pointing to "errors" that are not.

My use of P2(b|a) is licensed by OPT. Given their (1), P2(b|a) CAN NEVER BE FALSE under OPT!

If you want to stick with their silly (2')* -- which is often false under OPT: and certainly false here -- then you are stuck with their silly results!

* Read their silly paragraph below (2'). Take note of their focus on AAD and their false extrapolation from common-sense (with no mention of the consequences of correlations: (2') even being false under some elementary classical correlations).
.

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?
***

[FrediFizzx]

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.

[Gordon Watson]

… … ...
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?
***

Thanks Fred, good comment and much appreciated, but I say stet for now.

I'm sure we agree that many changes would improve CH74 out of sight AND THEREBY MOVE IT ON TO OUR SIDE of the debate (see quote below; but I digress)! By using P instead of their p, I chose to make the least changes; P2 being clearer than p2 for me here. Then, based on the first paragraph in their footnote 13, I accepted that they had some knowledge of conditional probability.

Thus, based on their terms: P2(b|a) is the probability of a count being triggered at apparatus 2, conditioned on a count being triggered at apparatus 1.

PS: While I'm at it, I'd like to draw the attention of all (bellians, non-bellians and anti-bellians alike) to CH74, p.534, footnote 11 (with my underlining).

"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

[Gordon Watson]

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!

Thanks.

[FrediFizzx]

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

[Gordon Watson]

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

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

As explained above: having strictly checked that CH74 understood "conditional probability", I then went with the fact that OPT licenses me to use such. NB: Their failure to properly use "conditional Ps" is NOT due to any objection to such. Their failures and misunderstandings are conditioned by the equally silly work of Bell and CHSH: who CH thus join in ALL being wrong together!

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

Note that my CH74' with the prime (') is NOT CH74 without the prime! As I explained -- viewtopic.php?f=6&t=234#p6012 -- CH74' is the output of a shortcut (that I took at the time). I then, later, did my work without that shortcut.

"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.]

(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.

HTH; HNY.

[FrediFizzx]

(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]

(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.

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

[FrediFizzx]

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]

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]

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]

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.
****

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]

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]

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]

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.

[Gordon Watson]

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.

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.

E&OE.

[FrediFizzx]

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.

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

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 ++.

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

****

[Gordon Watson]

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.

****

(i) Wrong. (ii) Irrelevant. (iii) Wrong. (iv) Wrong (re the clues)!

Since any expectation E is a function of some P, any combination of Es and/or Ps (such as CH or CHSH) can be written as a valid combination of the related Ps!

Seeing few (if any) errors on my part anywhere near here, I suggest you talk to minkwe and Joy.
**********

[FrediFizzx]

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!!!!!!!!
****