I Violation of Bell Inequality with unentangled photons

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  • #51
gentzen said:
$$(\alpha|x\rangle+\beta|p\rangle)(\alpha^*\langle x|+\beta^*\langle p|)=|\alpha|^2|x\rangle\langle x|+|\beta|^2|p\rangle\langle p|+\text{interference terms}$$
$$\text{interference terms} = \alpha\beta^*|x\rangle\langle p| +\alpha^*\beta|p\rangle\langle x|$$
Is this related to the state? to the observables or to what? I am lost here. There are clearly interference terms in whatever you are doing there.
 
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  • #52
gentzen said:
pines-demon tried to get rid of the continuous measurement outcomes, to be closer to typical Bell measurements
I agree that you can't just bin things into binary measurements. I'm not sure exact continuity is required. For example, with a momentum entangled state, I think you could have a spherical array of momentum detectors with finite angular resolution; you wouldn't need exact, infinite angular resolution. You'd just need a lot more than only two such detectors (one per hemisphere) to spot Bell inequality violations, since you'd have to make measurements at various relative angles, not just 180 degrees.
 
  • #53
pines-demon said:
Is this related to the state? to the observables or to what? I am lost here. There are clearly interference terms in whatever you are doing there.
This is an observable. Your ##S_x## and ##S_p## were observables too. Maybe to clarify, we have ##\alpha=\cos(\theta)## and ##\beta=\sin(\theta)## in your case.
 
  • #54
gentzen said:
This is an observable. Your ##S_x## and ##S_p## were observables too. Maybe to clarify, we have ##\alpha=\cos(\theta)## and ##\beta=\sin(\theta)## in your case.
Isn’t ##\alpha\beta=\cos\theta\sin\theta## different than 0 for most angles?
 
  • #55
pines-demon said:
Isn’t ##\alpha\beta=\cos\theta\sin\theta## different than 0 for most angles?
Yes, of course.

I guess what you didn't understand is that your ##|\alpha|^2S_x+|\beta|^2S_p## is missing those interference terms. That was my point, why I showed you an observable where those terms are not missing.
 
  • #56
gentzen said:
Yes, of course.

I guess what you didn't understand is that your ##|\alpha|^2S_x+|\beta|^2S_p## is missing those interference terms. That was my point, why I showed you an observable where those terms are not missing.
This is getting very cryptic. I take ##S_\theta=S_x\cos\theta+S_p\sin\theta=\alpha S_x+\beta S_p##, I do what then? ##S_\theta^2## or ##S_\theta S_\theta^\dagger##? Either way I get interference term right? Is not that different to your calculation. What am I missing?
 
  • #57
pines-demon said:
What am I missing?
You are missing
gentzen said:
If you define it like this, then ##S_x## and ##S_p## are already matrices (or rather operators), basically ...
Your sign measurement operators ##S_x## and ##S_p## are no longer representable by a single state, hence you have to represent them as matrices/operators. But forming the superposition of matrices doesn't give you interference. You only get interference, if you form the superposition of two single states, which represented two measurements.
gentzen said:
..., basically ##S_x=\int_0^\infty dx\, |x\rangle\langle x|## and ##S_p=\int_0^\infty dp\, |p\rangle\langle p|##.
If you want to be more accurate, it would be ##S_x=\int_{-\infty}^\infty dx\, sign(x)|x\rangle\langle x|##. But the crucial part is that ##|x\rangle\langle x|## got already formed.
 
  • #58
martinbn said:
In the EPR paper, the entangled particles cannot be used for Bell's inequality violations, and it was the fist paper about entanglement.
If this were correct then EPR is not a quantum experiment, but this is not the case.
 
  • #59
gentzen said:
You are missing
Who is being cryptic?
gentzen said:
Your sign measurement operators ##S_x## and ##S_p## are no longer representable by a single state, hence you have to represent them as matrices/operators. But forming the superposition of matrices doesn't give you interference.
1) Please be explicit on what is the calculation to perform to check if there is interference.
2) Why is it needed?
 
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