Heinera wrote:Plus the information of what polariser exit they took, according to yourself.
Matching events can only occur if both polarizers had detected a photon simultaneously. This is not additional information.
Heinera wrote:Plus the information of what polariser exit they took, according to yourself.
Esail wrote:Heinera wrote:But Bell's matching does not in any way depend on the properties of the particles, only on the information that they were created at the same time. That is not exactly your setup. And it makes all the difference.
Matching on the information that the photons were created at the same time is my setup indeed.
Esail wrote:Heinera wrote:Plus the information of what polariser exit they took, according to yourself.
Matching events can only occur if both polarizers had detected a photon simultaneously. This is not additional information.
Heinera wrote:The point is that for your model to work, some photons must go unmatched, and thus excluded from the computation of the correlations. This is not necessary with QM, or in Bell's ideal experiment.
Esail wrote:Heinera wrote:The point is that for your model to work, some photons must go unmatched, and thus excluded from the computation of the correlations. This is not necessary with QM, or in Bell's ideal experiment.
I do not see this. Can you explain?
Esail wrote:I have to correct myself. Changing the angle of P1 does not change the ensemble detected by P2. Only the fraction of photon 2 whose peer was detected at P1 changes.
Heinera wrote:
If only a fraction is matched, the rest must go unmatched (i.e., discarded).
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.
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.
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.
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)
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.
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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.
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)
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.
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 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.
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?
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.
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).
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.
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