OK. JesseM.
So I want to return to your first opinition that in classicel electromagnetism there would be
no violation of Bell inequalities in a Bell-type experiment.
(This is the reason why I asked what your electromagnetic wave packet means.)
In the photoelectric effect, the light frequency is related to the energy, and the light intensity is related to the
number of emitted photoelectrons.
(This means that we can suppose the light intensity
Q is required for
one emitted photoelectron. 2Q is needed for two emitted photoelectrons ...)
So we can suppose this minimum intensity Q is equal to more than 60%
intensity of the wave packet.
(Because if you use the example of the electromagnetic wave, the intensity is related to the events at the polarizer according to Malus' law.)
JesseM said:
Well, there are no "photons" in classical electromagnetism, classical electromagnetic waves are infinitely divisible. But I imagined that the detectors were specifically designed to only go off if they received a wave packet with at least 50% of the energy of the original wave packet sent by the source, so that the classical experiment would replicate the same features as the quantum experiment (i.e. you'd always have either detector D+ or D- go off, never both).
I agree with you about this point.
And of course, as you say, the case when we detect tow photons (D+ and D- ) at the same one polarizer is meaningless (= the total number detected beomes more than 2 photons (3 or 4 photons) ).
The cases that I want to talk about are those of the
two or less than two photons (at A and B detectors).
The light intensity that passes through the filter is
[tex]I = I_0 \cos^2 \theta[/tex]
So the remaining reflection intensity is
[tex]I = I_0 \sin^2 \theta[/tex]
(Of course, a little loss exists.)
As I said, there are three patterns (pass (1) and reflect (2), and they are detected due to its
enough intensity (> Q)).
And when the light (intensity) is divided at the polarizer almost equally ( 55% + 45%, for example, in the case of near 45 degrees in the above equations.), neither pass nor reflect detector can not detect it as a photon (3).
When the two photons (A and B) with parallel poralization axis bumps into each filter of the same angle (the angle difference between two filtes [tex]\alpha = 0[/tex]),
The results (pass or reflect) of the two photons always become the same ([tex]\cos^2 \alpha = \cos^2 0 = 1[/tex]) ?
Because when the photon A (or B) passes the filter A (or B), photon A (or B) always has the polarizarion axis
near the filter A (or B) to reach the intensity detection threashold (> Q) of the dector.
In the case of equally divided lights at the polarizer as I said above, the pass or reflect light intensities can not reach the detection threashold of the detector.
This case will be ingnored, but is very important as a
underlying reality.
Sorry. I want to talk about the two photons case (not the case of ions ...).
Because the ion case uses the very
artificial condition such as Paul trap and pulse laser.
(If these artificial manipulations don't exist, the Be+ ion excitation can not occurr, which is required for entanglement condition. )