Bell's test: Introducing a control experiment

San K
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If we were to introduced a set of un-entangled, but same polarized, photons as control experiment what would the results be of the control experiment?

So we have the following three cases:

Bell test Experiment: Send entangled photons

Result is that P(-30,30) is not equal to P(0,30) + P (0,-30)...hence QE proved...

(side note - with not all loopholes closed simultaneously)

Control Experiment 1: Send "un-entangled" photons, but same polarization

What would the relation be between P (-30, 30), P(0,30) and P(0,-30)?

Control Experiment 2: Send "un-entangled" photons, but random polarization

What would the relation be between P (-30, 30), P(0,30) and P(0,-30)?

I guess in the last case it would be 0.5, 0.5, 0.5
 
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San K said:
Bell test Experiment: Send entangled photons

Result is that P(-30,30) is not equal to P(0,30) + P (0,-30)...hence QE proved...
I assume by P(x,y) you mean the probability of mismatch between the results of a polarizer oriented at an angle x and a polarizer oriented at an angle y. And one correction, it's called Bell's INequality, not Bell's equation for a reason. Bell's inequality in this case, is P(-30,30)≤P(-30,0)+P(0,30), so the significant fact is that QM predicts P(-30,30) is greater than P(-30,0)+P(0,30). The significant fact is not merely that P(-30,30) is not equal to P(0,30)+P (0,-30).
Control Experiment 1: Send "un-entangled" photons, but same polarization

What would the relation be between P (-30, 30), P(0,30) and P(0,-30)?
Well, the photons aren't entangled, then you no longer get perfect correlation at identical angles, i.e. it is no longer true that P(x,x)=0 for all angles x. But this is a crucial assumption for deriving the Bell inequality, so the Bell inequality need not apply in this case.

Still, if you want to calculate the probabilities anyway it's pretty straightforward (though tedious) to compute. All you have to know is that given an unentangled photon polarized in a direction θ1, the probability that it will go through a polarizer oriented at an angle θ2 is cos2(θ1-θ2).
Control Experiment 2: Send "un-entangled" photons, but random polarization

What would the relation be between P (-30, 30), P(0,30) and P(0,-30)?

I guess in the last case it would be 0.5, 0.5, 0.5
Yes, you're right about that.
 
Thanks Lugita.

lugita15 said:
I assume by P(x,y) you mean the probability of mismatch between the results of a polarizer oriented at an angle x and a polarizer oriented at an angle y. And one correction, it's called Bell's INequality, not Bell's equation for a reason. Bell's inequality in this case, is P(-30,30)≤P(-30,0)+P(0,30), so the significant fact is that QM predicts P(-30,30) is greater than P(-30,0)+P(0,30). The significant fact is not merely that P(-30,30) is not equal to P(0,30)+P (0,-30)

agreed.

a better answer is that - the correlation is stronger than that predicted by the laws of probability...

lugita15 said:
Well, the photons aren't entangled, then you no longer get perfect correlation at identical angles, i.e. it is no longer true that P(x,x)=0 for all angles x. But this is a crucial assumption for deriving the Bell inequality, so the Bell inequality need not apply in this case.

Same polarized photons won't give same answer for polarizers that are aligned? (i.e. polarizers are same angles to each other)

However entangled photons will give the same answer for polarizers that are aligned?Bell's inequality does not apply. We are simply comparing polarized non-entangled photons with entangled photons (which necessarily are non-polarized). there is a reason for this.

lugita15 said:
Still, if you want to calculate the probabilities anyway it's pretty straightforward (though tedious) to compute. All you have to know is that given an unentangled photon polarized in a direction θ1, the probability that it will go through a polarizer oriented at an angle θ2 is cos2(θ1-θ2). Yes, you're right about that.

ok, just wanted to get the probabilities for P(-30,30), P(-30,0) and P(0,30), assuming Theta 1 is Zero degrees.
 
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San K said:
a better answer is that - the correlation is stronger than that predicted by the laws of probability...:)
It's the laws of probability plus local hidden variables.
San K said:
ok, just wanted to get the probabilities for P(-30,30), P(-30,0) and P(0,30), assuming Theta 1 is Zero degrees.
OK, based on a quick calculation in my head, P(-30,30)=37.5%, and P(-30,0)=P(0,30)=25%. EDIT:Sorry, I made a mistake the first time. Now the numbers should be right.
 
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