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by **Esail** » Mon Feb 05, 2018 4:48 am

Heinera wrote:
but still have

$\text{Corr}(A(10^\circ, \lambda), B(55^\circ, \lambda)) = \text{Corr}(A(0^\circ, \lambda), B(45^\circ, \lambda))$

If it wasn't possible, we could dismiss LHV models right there, and would have no need for Bell's theorem.

But Esail hasn't shown that his model has this property.

With P1 setting =0° and delta(alpha=0°) =0 and P2 setting = 45° delta(beta=45°) = 45°-0°-90°=-45° ,see my manuscript after equation 16, we get cos(2delta)= 0.
Then we get A(0,lambda) = 1 and B(-45°,lambda)= -1 for -1<lambda<=0 and B(-45°,lambda)= +1 for 0<lambda <1.
With P1 setting =10° and delta(alpha=10°) =0 and P2 setting = 55° delta(beta=55°) = 55°-10°-90°=-45° ,see my manuscript after equation 16, we get cos(2delta)= 0.
Then we get A(0,lambda) = 1 and B(-45°,lambda)= -1 for -1<lambda<=0 and B(-45°,lambda)= +1 for 0<lambda <1.

As the values for A and are identical in both cases so are the correlations.

by **Heinera** » Mon Feb 05, 2018 3:18 am

minkwe wrote:
Heinera wrote:I mean he must show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

gives the same correlation value as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

since he simply evaluates all correlations with the first polarizer stuck at $0^\circ$ .

I'm saying that does not make sense to expect that. What's the point of the functions A, B if the angle setting is irrelevant?

Well, it is possible for some local models to have

$A(10^\circ, \lambda)\neq A(0^\circ, \lambda)$ and $B(55^\circ, \lambda)\neq B(45^\circ, \lambda)$

but still have

$\text{Corr}(A(10^\circ, \lambda), B(55^\circ, \lambda)) = \text{Corr}(A(0^\circ, \lambda), B(45^\circ, \lambda))$

If it wasn't possible, we could dismiss LHV models right there, and would have no need for Bell's theorem.

But Esail hasn't shown that his model has this property.

by **minkwe** » Sun Feb 04, 2018 2:57 pm

Heinera wrote:I mean he must show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

gives the same correlation value as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

since he simply evaluates all correlations with the first polarizer stuck at $0^\circ$ .

I'm saying that does not make sense to expect that. What's the point of the functions A, B if the angle setting is irrelevant?

by **Heinera** » Sun Feb 04, 2018 5:08 am

I mean he must show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

gives the same correlation value as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

since he simply evaluates all correlations with the first polarizer stuck at $0^\circ$ .

by **minkwe** » Sat Feb 03, 2018 5:34 pm

Heinera wrote:The best way for you to prove just one example of rotational invariance of your model is to show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

give the same values as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

This cannot be right because it implies that
$A(10^\circ, \lambda) = A(0^\circ, \lambda), =A(a, \lambda) = A(\lambda)$
and, the angle is irrelevant, and same for B. But the lambda's are defined according to a given basis, which means changing the basis changes the definition of the lambdas. Rotational invariance means: "rotating the basis does not change the outcome", it does NOT mean "rotating the setting, within the same basis, does not change the outcome". Perhaps you mean

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

give the same values as

$A(0^\circ, \lambda'), B(45^\circ, \lambda')$
Where $\lambda'$ are defined according to a new basis rotated by -10 degrees.

by **Esail** » Fri Feb 02, 2018 2:43 am

Heinera wrote:
The best way for you to prove just one example of rotational invariance of your model is to show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

give the same values as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

A(alpha, lambda) and B(beta, lambda) are not unambiguous functions of alpha or beta respectively.
The function B(beta, lambda) is defined as Beta(delta,lambda) where delta= beta-polarization angle of incoming photon 2. Thus B(beta, lambda) depends on the setting alpha of P1. delta(beta) = beta-alpha-pi/2.
The rotational invariance comes from the fact that the polarization of photon 2 is perpendicular to the setting of the polarizer P1 for any setting of P1.

Esail wrote:If everything were rotated polarizer P1 were at angle 0° and Photon 1 had the polarization - alpha and photon 2 had the polarization - alpha + pi/2. This would not reproduce the probability sin**(beta-alpha).

This is an error. Please drop this sentence.

by **Joy Christian** » Thu Feb 01, 2018 11:31 am

***
None of the above issues matter because Esail's model contradicts at least one of the most elementary predictions of quantum mechanics, as I pointed out above.

***

by **Heinera** » Thu Feb 01, 2018 11:29 am

Esail wrote:If everything were rotated polarizer P1 were at angle 0° and Photon 1 had the polarization - alpha and photon 2 had the polarization - alpha + pi/2. This would not reproduce the probability sin**(beta-alpha).

Wait wait wait.. are you saying that if everything were rotated by alpha you would get different results? This would be sensational (and immediately refuted by experiment). Any laboratory on the surface of the earth rotates full circle every 24 hours.

Esail wrote:In order to proof rotational invariance we need to show that the polarization of photon 1 is equal to the setting of the polarizer P1 for any setting of P1 and thus the polarization of photon 2 is perpendicular to the polarization of photon 1.

The best way for you to prove just one example of rotational invariance of your model is to show that

$A(10^\circ, \lambda), B(55^\circ, \lambda)$

give the same values as

$A(0^\circ, \lambda), B(45^\circ, \lambda)$

because from (16) to (17) you are simply substituting one for the other. Angles are in the fixed coordinate system of the laboratory. Feel free to just compute it the hard way.

(Hint: They are not the same).

by **Esail** » Thu Feb 01, 2018 10:52 am

Heinera wrote: He needs to show that the results are invariant when only the polarizers are rotated, but the rest of the system stays unrotated. And in this case the results are not invariant.

If everything were rotated polarizer P1 were at angle 0° and Photon 1 had the polarization - alpha and photon 2 had the polarization - alpha + pi/2. This would not reproduce the probability sin**(beta-alpha). The situation of entanglement is different from a mixture of photon 1 and photon 2. This is reflected by model assumption M3.

In order to proof rotational invariance we need to show that the polarization of photon 1 is equal to the setting of the polarizer P1 for any setting of P1 and thus the polarization of photon 2 is perpendicular to the polarization of photon 1.

Those photon 1 which pass P1 at alpha are a selection with p-state alpha and have by means of M3 also the polarization alpha. The peer photon at side 2 are in p-state alpha+pi/2 and have by means of M3 also the polarization state alpha+pi/2.

If we, this in mind, then rotate the coordinate system by alpha we get the setting of P1 at 0° and polarization of the selected photon 1 at 0° as well and the polarization of the corresponding photon 2 at pi/2 and the polarizer setting of P2 at beta -alpha.

by **Heinera** » Thu Feb 01, 2018 9:47 am

Heinera wrote:Or maybe you mean that if you rotate the polarizers by an angle alpha, you also rotate tho photon by the same angle?

Yes, this is the problem. By "rotational invariance" Esail means that (quote) "we can rotate the coordinate system by an angle alpha" and get the same results. This means that the results are the same if we rotate everything by an angle alpha: polarizers, source, photons, everything is rotated. It is correct that the results will then be the same. But this is of course not the kind of invariance he needs to go from (16) to (17). He needs to show that the results are invariant when only the polarizers are rotated, but the rest of the system stays unrotated. And in this case the results are not invariant.

by **Heinera** » Thu Feb 01, 2018 6:24 am

Esail wrote:
Heinera wrote:The whole derivation from (16) to (17) in your paper relies on the assumption that your model is rotationally invariant. But you haven't shown that. You have only shown that is invariant under rotations of 90^o, but you must show it is invariant for any angle.

Rotational invariance is shown for any angle alpha between equation 11 and 12.

No, alpha = phi_1 there (the state of photon 1), which means it cannot be any angle.

Or maybe you mean that if you rotate the polarizers by an angle alpha, you also rotate tho photon by the same angle?

by **Joy Christian** » Thu Feb 01, 2018 5:43 am

Joy Christian wrote:The bottom line is that quantum mechanics predicts P(+/-, 0) = 0 = P(0, +/-). That is, the probability of a photon detected at one station, but not detected at other station is exactly zero according to quantum mechanics. The question you have to ask yourself is: Is that probability predicted to be zero or non-zero in your model?

Esail,

So is the probability P(+/-, 0) = 0 or not in your model? If it is not, then your model contradicts at least one of the elementary predictions of quantum mechanics.

It has been well known since 1970's that with non-zero P(+/-, 0) (i.e., by contradicting quantum mechanics) it is very easy to reproduce the cosine correlations.

If you don't believe me, then you can look that prediction up in this classic 1970 paper by Philip Pearle: https://journals.aps.org/prd/abstract/1 ... evD.2.1418.

***

by **Esail** » Thu Feb 01, 2018 5:23 am

Heinera wrote:The whole derivation from (16) to (17) in your paper relies on the assumption that your model is rotationally invariant. But you haven't shown that. You have only shown that is invariant under rotations of 90^o, but you must show it is invariant for any angle.

Rotational invariance is shown for any angle alpha between equation 11 and 12.

by **Heinera** » Thu Feb 01, 2018 4:11 am

Esail wrote:
Heinera wrote:But this is refuted by QM. In QM, the number of coincidences does not depend on the detector settings. And to my knowledge, none of the experiments performed to date have shown such a dependence.

If you count as a matching event a photon pair whose photon 1 has passed P1 at alpha and whose photon 2 has passed P2 at beta the number of matching events depend on the polarizer setting. Suppose alpha = beta then there are no matching events. Only if you also have a detector on the other exit of P2 at beta + pi/2 then you can count all photon 2 which have a peer photon 1 detected at alpha. The probability (photon 2 passing P2 at beta and photon 1 passing P1 at alpha/ all photon 1 passing P1 at alpha) = sin**2(beta-alpha)

OK, you seem to have a definition of "matching" that is different from my. "coincident" I don't think we get any further here, so let's discuss something else:

The whole derivation from (16) to (17) in your paper relies on the assumption that your model is rotationally invariant. But you haven't shown that. You have only shown that is invariant under rotations of 90^o, but you must show it is invariant for any angle.

by **Joy Christian** » Thu Feb 01, 2018 3:31 am

Esail wrote:
Joy Christian wrote:The crucial question is: What is the probability predicted by your model for photon passing P1 at alpha and photon not-passing P2 at beta, whether or not P2 or a detector exists at station 2? Quantum mechanics predicts that probability to be exactly zero: P(+/-, 0) = 0.

We have a polarizing Beamsplitter. Photons either pass at one exit, say beta or at the perpendicular exit beta + pi/2. So you say all photon 2 whose peer photon 1 would pass P1 at alpha would definitely pass P2 at beta and none at beta+pi/2. Thus with P1 and P2 set to alpha all photon 2 whose peer photon 1 would pass P1 at alpha would definitely pass P2 at alpha. This is definitely wrong.

The bottom line is that quantum mechanics predicts P(+/-, 0) = 0 = P(0, +/-). That is, the probability of a photon detected at one station, but not detected at other station is exactly zero according to quantum mechanics. The question you have to ask yourself is: Is that probability predicted to be zero or non-zero in your model?

***

by **Esail** » Thu Feb 01, 2018 3:17 am

Joy Christian wrote:The crucial question is: What is the probability predicted by your model for photon passing P1 at alpha and photon not-passing P2 at beta, whether or not P2 or a detector exists at station 2? Quantum mechanics predicts that probability to be exactly zero: P(+/-, 0) = 0.

***

We have a polarizing Beamsplitter. Photons either pass at one exit, say beta or at the perpendicular exit beta + pi/2. So you say all photon 2 whose peer photon 1 would pass P1 at alpha would definitely pass P2 at beta and none at beta+pi/2. Thus with P1 and P2 set to alpha all photon 2 whose peer photon 1 would pass P1 at alpha would definitely pass P2 at alpha. This is definitely wrong.

by **Joy Christian** » Thu Feb 01, 2018 2:57 am

Esail wrote:If you count as a matching event a photon pair whose photon 1 has passed P1 at alpha and whose photon 2 has passed P2 at beta the number of matching events depend on the polarizer setting. Suppose alpha = beta then there are no matching events. Only if you also have a detector on the other exit of P2 at beta + pi/2 then you can count all photon 2 which have a peer photon 1 detected at alpha. The probability (photon 2 passing P2 at beta and photon 1 passing P1 at alpha/ all photon 1 passing P1 at alpha) = sin**2(beta-alpha)

The crucial question is: What is the probability predicted by your model for photon passing P1 at alpha and photon not-passing P2 at beta, whether or not P2 or a detector exists at station 2? Quantum mechanics predicts that probability to be exactly zero: P(+/-, 0) = 0. But your model predicts it to be non-zero, as you noted above. Therefore your model contradicts the predictions of quantum mechanics. It is not the model for the singlet state considered by Bell or used in experiments.

***

by **Esail** » Thu Feb 01, 2018 12:29 am

Heinera wrote:But this is refuted by QM. In QM, the number of coincidences does not depend on the detector settings. And to my knowledge, none of the experiments performed to date have shown such a dependence.

If you count as a matching event a photon pair whose photon 1 has passed P1 at alpha and whose photon 2 has passed P2 at beta the number of matching events depend on the polarizer setting. Suppose alpha = beta then there are no matching events. Only if you also have a detector on the other exit of P2 at beta + pi/2 then you can count all photon 2 which have a peer photon 1 detected at alpha. The probability (photon 2 passing P2 at beta and photon 1 passing P1 at alpha/ all photon 1 passing P1 at alpha) = sin**2(beta-alpha)

by **Heinera** » Wed Jan 31, 2018 1:28 pm

Esail wrote:
Heinera wrote:
If only a fraction is matched, the rest must go unmatched (i.e., discarded).

If P1 is set to alpha and P2 is set to beta only the fraction sin**2(alpha-beta) of all detected photon 2 matches. That is to say for this fraction photon 1 is detected at P1 and photon 2 is detected at p2.

But this is refuted by QM. In QM, the number of coincidences does not depend on the detector settings. And to my knowledge, none of the experiments performed to date have shown such a dependence.

by **Joy Christian** » Wed Jan 31, 2018 12:49 pm

Esail wrote:If P1 is set to alpha and P2 is set to beta only the fraction sin**2(alpha-beta) of all detected photon 2 matches. That is to say for this fraction photon 1 is detected at P1 and photon 2 is detected at p2. No photon is discarded.The rest cos**2(alpha-beta) of photon 2 doesn't match with a photon 1 detected at P1.

This contradicts the predictions of quantum mechanics. Thus your model is not only non-local, but also simply wrong. Because quantum mechanics predicts that if any photon emitted by a source is detected at station 1, then it's twin is also detected at station 2. That is an elementary prediction of quantum mechanics, as verified in experiments. All 13 probabilities predicted by quantum mechanics for the singlet state are listed in equations (2) to (8) of this paper: https://arxiv.org/abs/1405.2355

In particular, your model violates the probabilities given in the equation (6) of the linked paper, which are elementary predictions of quantum mechanics.

***

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