CH Inequality

[minkwe]

I think it pays to revisit the derivation of the CH inequality step by step.
http://journals.aps.org/prd/abstract/10 ... evD.10.526

See Appendix A.
It has the exact same problems like the CHSH we've talked about before. It too cannot be violated without mathematical error.
It's comparison to experiments and QM involves the same bait and switch tactic we've documented at great length in several threads here.

[FrediFizzx]

I think it pays to revisit the derivation of the CH inequality step by step.
http://journals.aps.org/prd/abstract/10 ... evD.10.526

See Appendix A.
It has the exact same problems like the CHSH we've talked about before. It too cannot be violated without mathematical error.
It's comparison to experiments and QM involves the same bait and switch tactic we've documented at great length in several threads here.

Thanks. If you have access to that journal, please email me a copy of the paper and I will get the step by step started. Otherwise, go ahead and lay the steps out here.

[FrediFizzx]

Here is the proof of the CH inequality in Appendix A. It is for eq. (4) in the paper. I am going to suppress using underline for some of the subscripts.

-1 =< p_12(a, b) - p_12(a, b') + p_12(a', b) + p_12(a', b') - p_1(a') - p_2(b) =< 0 eq. (4)

We prove the following thereom: Given six numbers x1, x2, y1, y2, X, and Y such that

0 =< x1 =< X,

0 =< x2 =< X, (A1)

0 =< y1 =< Y,

0 =< y2 =< Y,

then the function U = x1y1 - x1y2 + x2y1 + x2y2 - Yx2 - Xy1 is constrained by the inequalities

-XY =< U =< 0 . (A2)

To establish the upper bound, consider two cases. First assume that x1 >= x2 and rewrite (A2),

U = (x1 - X)y1 + (y1 - Y)x2 + (x2 - x1)y2 .

We have thus assumed the last term to be nonpositive. Inequalities (A1) require the first two terms to be nonpositive, and the validity of the upper bound is demonstrated for this case. Next assume the other alternative, i.e., that x1 < x2, and use this assumption to bound U, thus:

U = x1(y1 - y2) + (x2 - X)y1 + x2(y2 - Y) =< x1(y1 - y2) + (x2 - X)y1 + x1(y2 - Y)
= (x2 - X)y1 - x1(y1 - Y) =< 0 .

Thus, the upper bound is established in general.
The proof of the lower bound follows from a consideration of three cases.

I don't think we are concerned with the lower bound here so I don't think we need to follow that. Correct me if I am wrong.

So a dependency comes into the CH inequality here. Where exactly?

[Gordon Watson]

Fred, I've no wish to spoil the fun to be had with CH (nor the lessons therein) but:

(i) Your last equation (I suspect) should read = (x2 - X)y1 + x1(y1 - Y) =< 0.

(ii) But this "typo" is irrelevant (see next).

(iii) Rest assured: CH contains typical Bellian errors!

(iv) Here's a clue: the same error is found in Don's essays!

(v) All of which comes with the usual guarantee: I could be wrong.

(vi) Season's greetings to all; with special thanks, on behalf of many of us, to you, Fred!

(vi) Dear Santa: How about CH, free online, and in full; or delivered privately:
.

[FrediFizzx]

Fred, I've no wish to spoil the fun to be had with CH (nor the lessons therein) but:

(i) Your last equation (I suspect) should read = (x2 - X)y1 + x1(y1 - Y) =< 0.

The minus sign is correct from their paper. Perhaps you could elucidate why you think it should be + ?

Anyways, we can see that the dependency is in the function U = x1y1 - x1y2 + x2y1 + x2y2 - Yx2 - Xy1. We can set X = Y = 1 and easily see that it will always be equal to or less than zero. But I believe the function U also contains the "bait and switch". What does x1y1 have to do with p_12(a, b)? Is the x1 in the x1y1 term actually going to be the same as the x1 in the x1y2 term experimentally? I don't think so. IOW, the a from p_12(a, b) could be the same as the a from p_12(a, b') but it is not the same thing as the x1 in the x1x2 term. So basically once again they are mixing up A and B outcomes with a and b angle settings.

[Gordon Watson]

Fred, I've no wish to spoil the fun to be had with CH (nor the lessons therein) but:

(i) Your last equation (I suspect) should read = (x2 - X)y1 + x1(y1 - Y) =< 0.

The minus sign is correct from their paper. Perhaps you could elucidate why you think it should be + ?

Anyways, we can see that the dependency is in the function U = x1y1 - x1y2 + x2y1 + x2y2 - Yx2 - Xy1. We can set X = Y = 1 and easily see that it will always be equal to or less than zero. But I believe the function U also contains the "bait and switch". What does x1y1 have to do with p_12(a, b)? Is the x1 in the x1y1 term actually going to be the same as the x1 in the x1y2 term experimentally? I don't think so. IOW, the a from p_12(a, b) could be the same as the a from p_12(a, b') but it is not the same thing as the x1 in the x1x2 term. So basically once again they are mixing up A and B outcomes with a and b angle settings.

(i) Fred, from you (above) and CH74:

U = x1(y1 - y2) + (x2 - X)y1 + x2(y2 - Y) =< x1(y1 - y2) + (x2 - X)y1 + x1(y2 - Y) = (x2 - X)y1 - x1(y1 - Y) =< 0 [sic]

Penultimately, you (AND CH74) have -x1y1 + x1Y but the prior expression has +x1yi - x1Y! Hence my corrective PLUS: + x1(y1 - Y)

(ii) New clue re CH74 error: Look at the unnumbered equation (the old Bellian trick) on p.528; call it (3a). Now look at their (4), same page.

(iii) I think now you can see what you truly call bait and switch! [Young minkwe will love it!]

(iv) Let me know if any of this is unclear; I'm flat out clearing projects before year-end so errors may linger hereabouts.
.

[FrediFizzx]

(ii) New clue re CH74 error: Look at the unnumbered equation (the old Bellian trick) on p.528; call it (3a). Now look at their (4), same page.

I don't see anything wrong in going from eq. (3a) to eq. (4) in the CH74 paper. Looks like pretty standard math stuff.

[Gordon Watson]

(ii) New clue re CH74 error: Look at the unnumbered equation (the old Bellian trick) on p.528; call it (3a). Now look at their (4), same page.

I don't see anything wrong in going from eq. (3a) to eq. (4) in the CH74 paper. Looks like pretty standard math stuff.

Fred, with apologies (and aware that my response is overdue), this note is conditional under "E and OE" due to current time pressures. NB: For clarity, I use P instead of their p:

So, seeing nothing wrong, we note that (4) in CH74 is wholly amenable to analysis under ordinary probability theory (OPT). In particular, the following dissection can never be wrong under OPT and CH74; ie, from CH74 eqn. (4):

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') - P1(a') - P2(b) ≤ 0. (4a)

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

P1(a) = P1(a') = P2(b) = 1/2. (4b)

So from (4a) we have: -1 ≤ (1/2) [P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') - 1 - 1] ≤ 0. (4c)

So: CH74' = |P2(b|a) - P2(b'|a) + P2(b|a') + P2(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 questions arise: (i) Where does CH74 go wrong? (ii) What definition of realism do they rely on? (iii) What alternative definition of realism is warranted under QM?

And since these are good questions under the excellent new topic viewtopic.php?f=6&t=231 , I'll take them there as well.

[FrediFizzx]

Fred, with apologies (and aware that my response is overdue), this note is conditional under "E and OE" due to current time pressures. NB: For clarity, I use P instead of their p:

So, seeing nothing wrong, we note that (4) in CH74 is wholly amenable to analysis under ordinary probability theory (OPT). In particular, the following dissection can never be wrong under OPT and CH74; ie, from CH74 eqn. (4):

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') - P1(a') - P2(b) ≤ 0. (4a)

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

P1(a) = P1(a') = P2(b) = 1/2. (4b)

So from (4a) we have: -1 ≤ (1/2) [P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') - 1 - 1] ≤ 0. (4c)

So: CH74' = |P2(b|a) - P2(b'|a) + P2(b|a') + P2(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 questions arise: (i) Where does CH74 go wrong? (ii) What definition of realism do they rely on? (iii) What alternative definition of realism is warranted under QM?

And since these are good questions under the excellent new topic viewtopic.php?f=6&t=231 , I'll take them there as well.

You have a mistake on (4c). It should be,

-1 ≤ (1/2) [P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') - 1/2 - 1/2] ≤ 0. (4c)

That will get you (4d). I suppose that is one way to go from CH to a CHSH type arangement. But probably not right since CHSH is expectations that can range from -1 to +1, not probabilities that range from 0 to 1. (4e) is not correct based on (4d). It probably should go more like this.

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') - P1(a') - P2(b) ≤ 0. (4a)

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

P1(a) = P1(a') = P2(b) = 1/2. (4b)

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') ≤ P1(a') + P2(b). (4c)

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') ≤ 1. (4d)

|P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a')| ≤ 1. (4e)

Then your (4e) above makes sense. However I think your (4f) (not labelled) should equal π/4 not 3π/4.
***

[FrediFizzx]

I suppose the following expression could be considered the probability version of CHSH.

|P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a')| ≤ 1. (4e)

But it is easy to see that with independent terms we could have,

1 - 1 + 1 + 1 = 2 not 1,

so the absolute bound is 2. And with (a,b) = (a',b) = (a',b') = (a,b')/3 = π/4 we would have for QM,

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

so no violation of the dependent form.

[Gordon Watson]

Fred, with apologies (and aware that my response is overdue), this note is conditional under "E and OE" due to current time pressures. NB: For clarity, I use P instead of their p:

So, seeing nothing wrong, we note that (4) in CH74 is wholly amenable to analysis under ordinary probability theory (OPT). In particular, the following dissection can never be wrong under OPT and CH74; ie, from CH74 eqn. (4):

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') - P1(a') - P2(b) ≤ 0. (4a)

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

P1(a) = P1(a') = P2(b) = 1/2. (4b)

So from (4a) we have: -1 ≤ (1/2) [P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') - 1 - 1] ≤ 0. (4c)

So: CH74' = |P2(b|a) - P2(b'|a) + P2(b|a') + P2(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 questions arise: (i) Where does CH74 go wrong? (ii) What definition of realism do they rely on? (iii) What alternative definition of realism is warranted under QM?

And since these are good questions under the excellent new topic viewtopic.php?f=6&t=231 , I'll take them there as well.

You have a mistake on (4c). It should be,

-1 ≤ (1/2) [P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') - 1/2 - 1/2] ≤ 0. (4c)

That will get you (4d). I suppose that is one way to go from CH to a CHSH type arangement. But probably not right since CHSH is expectations that can range from -1 to +1, not probabilities that range from 0 to 1. (4e) is not correct based on (4d). It probably should go more like this.

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') - P1(a') - P2(b) ≤ 0. (4a)

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

P1(a) = P1(a') = P2(b) = 1/2. (4b)

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') ≤ P1(a') + P2(b). (4c)

-1 ≤ P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a') ≤ 1. (4d)

|P1(a)P2(b|a) - P1(a)P2(b'|a) + P1(a')P2(b|a') + P1(a')P2(b'|a')| ≤ 1. (4e)

Then your (4e) above makes sense. However I think your (4f) (not labelled) should equal π/4 not 3π/4.
***

Fred,

I'm still time-poor (so still E and OE), but I see no mistakes on my part. Did you maybe miss the 1/2 on LHS (4c)? Also: 3π/4 is OK.

However, since my goal was (4f), I took a shortcut after (4c). Here I take no shortcut:

From CH74 and my (4c) above, we have:

0 ≤ P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') ≤ 2. (4d*)

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

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

But a, b, a', b' are unrestricted! So let (a,b) = (a',b) = (a',b') = (a,b')/3 = θ.

So, via CH74, (4e*) becomes: 0 ≤ 3sin^2(θ/2) - sin^2(3θ/2) ≤ 2. (4f*)

For comparison, also under EPRB, CHSH69 is: |cos(3θ)-3cosθ|≤ 2. (4g*)

For Bell's supporters: (4f*) and (4g*) are identically absurd for more than 75% of the range -π < θ < π!

So the same questions arise: (i) Where do CHSH69 and CH74 go wrong? (ii) What definition of realism do they rely on? (iii) What alternative definition of realism is warranted under QM?

PS: Sorry for the E and OE again. Just wanted to quickly/hopefully before Xmas correct some of your statements and encourage others to check the consequences of our work for "realism".

[FrediFizzx]

Fred,

I'm still time-poor (so still E and OE), but I see no mistakes on my part. Did you maybe miss the 1/2 on LHS (4c)? Also: 3π/4 is OK.

However, since my goal was (4f), I took a shortcut after (4c). Here I take no shortcut:

From CH74 and my (4c) above, we have:

0 ≤ P2(b|a) - P2(b'|a) + P2(b|a') + P2(b'|a') ≤ 2. (4d*)

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

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

But a, b, a', b' are unrestricted! So let (a,b) = (a',b) = (a',b') = (a,b')/3 = θ.

So, via CH74, (4e*) becomes: 0 ≤ 3sin^2(θ/2) - sin^2(3θ/2) ≤ 2. (4f*)

For comparison, also under EPRB, CHSH69 is: |cos(3θ)-3cosθ|≤ 2. (4g*)

For Bell's supporters: (4f*) and (4g*) are identically absurd for more than 75% of the range -π < θ < π!

So the same questions arise: (i) Where do CHSH69 and CH74 go wrong? (ii) What definition of realism do they rely on? (iii) What alternative definition of realism is warranted under QM?

PS: Sorry for the E and OE again. Just wanted to quickly/hopefully before Xmas correct some of your statements and encourage others to check the consequences of our work for "realism".

Your (4d) makes no sense. You are using conditionals there and those are single probabilities. So the QM probability is 1/2 for each term. But yes, your 3π/4 was OK. I saw that later that you were using sine instead of cosine.

[Gordon Watson]

Your (4d) makes no sense. You are using conditionals there and those are single probabilities. So the QM probability is 1/2 for each term. … ...

Fred, ??? "(4d) makes no sense?"

I am knowingly using conditionals and they make sense to me: for I'm also using Ordinary Probability Theory (OPT).

Please: What do you mean by "… and those are single probabilities"? Your answer might help me understand how you arrive at this: "So the QM probability is 1/2 for each term."

At the moment, I'm standing by (4d). Thanks.

[FrediFizzx]

Gordon, what do you think P2(b|a) means?

[Gordon Watson]

Gordon, what do you think P2(b|a) means?

Fred, P2(b|a) is defined, above, mathematically! It therefore has a very clear meaning under OPT. It would help me to respond if I knew what you think it means, in words. Thanks.

[FrediFizzx]

Gordon, what do you think P2(b|a) means?

Fred, P2(b|a) is defined, above, mathematically! It therefore has a very clear meaning under OPT. It would help me to respond if I knew what you think it means, in words. Thanks.

The best I can figure out is that you have P2(b|a) = sin^2((a,b)/2). That can't be true. P2 has to be a singles probability.

[Gordon Watson]

Gordon, what do you think P2(b|a) means?

Fred, P2(b|a) is defined, above, mathematically! It therefore has a very clear meaning under OPT. It would help me to respond if I knew what you think it means, in words. Thanks.

The best I can figure out is that you have P2(b|a) = sin^2((a,b)/2). That can't be true. P2 has to be a singles probability.

Before we go beyond P2(b|a) [the sin^2 function is correct], we must agree on what P2(b|a) represents under OPT and CH74. Why do you say it has to be, and what do you mean by, "a singles probability"? [I take it that you do not mean a "marginal" probability?] Please study how and why P2(b|a) is first introduced. (NB: I am by no means saying that the CH74 notation is a good one!! Nor that I am error free.)

[FrediFizzx]

Gordon, what do you think P2(b|a) means?

Fred, P2(b|a) is defined, above, mathematically! It therefore has a very clear meaning under OPT. It would help me to respond if I knew what you think it means, in words. Thanks.

The best I can figure out is that you have P2(b|a) = sin^2((a,b)/2). That can't be true. P2 has to be a singles probability.

Before we go beyond P2(b|a) [the sin^2 function is correct], we must agree on what P2(b|a) represents under OPT and CH74. Why do you say it has to be, and what do you mean by, "a singles probability"? [I take it that you do not mean a "marginal" probability?] Please study how and why P2(b|a) is first introduced. (NB: I am by no means saying that the CH74 notation is a good one!! Nor that I am error free.)

Following the notation used in the CH74 paper for eq. (4) then it is P12(a, b) = sin^2((a,b)/2). 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.

[Gordon Watson]

Fred, P2(b|a) is defined, above, mathematically! It therefore has a very clear meaning under OPT. It would help me to respond if I knew what you think it means, in words. Thanks.

The best I can figure out is that you have P2(b|a) = sin^2((a,b)/2). That can't be true. P2 has to be a singles probability.

Before we go beyond P2(b|a) [the sin^2 function is correct], we must agree on what P2(b|a) represents under OPT and CH74. Why do you say it has to be, and what do you mean by, "a singles probability"? [I take it that you do not mean a "marginal" probability?] Please study how and why P2(b|a) is first introduced. (NB: I am by no means saying that the CH74 notation is a good one!! Nor that I am error free.)

Following the notation used in the CH74 paper for eq. (4) then it is P12(a, b) = sin^2((a,b)/2). 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.

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.

HTH.

[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).