Entangled photon polarization correlation

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Hello everyone,

I've got a quick question... Given two entangled photons, going through a polarization filter with relative angle a, what is the correlation between the two "answers" (whether the photon is blocked or let through)?
I believe it's either cos(a) or cos^2(a), but I'm not sure which of the two.

If we do the test in sequence, with the same photon, then the correlation is cos^2(a), is that correct?

Thanks in advance
 
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If analyzer A is set at an angle [itex]\alpha[/itex] and analyzer B at an angle [itex]\beta[/itex] then the probability that both photons (of the entangled pair) pass the analyzer is:

[tex]P_{AB}(\alpha,\beta) = \frac{1}{2}\text{cos}^2(\alpha-\beta)[/tex]

A derivation is given in Gregor Weih's dissertation, see page 26, Eq (1.40) and (1.41). The setup is on page 25.

I'm not sure though what you mean with the second question.
 
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If analyzer A is set at an angle [itex]\alpha[/itex] and analyzer B at an angle [itex]\beta[/itex] then the probability that both photons (of the entangled pair) pass the analyzer is:

[tex]P_{AB}(\alpha,\beta) = \frac{1}{2}\text{cos}^2(\alpha-\beta)[/tex]

A derivation is given in Gregor Weih's dissertation, see page 26, Eq (1.40) and (1.41). The setup is on page 25.

I'm not sure though what you mean with the second question.
Thank you for your answer!

My second question was more a confirmation, as I'm quite sure about it. Let's say we have a photon that went through polarization filter at 0 degrees, and we have a second polarization filter at [itex]\alpha[/itex] degrees, the chance it goes through the second polarization filter is:
[tex]\text{cos}^2 \alpha[/tex]

Right?

(I like those tex tags!)
 
699
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Thank you for your answer!

My second question was more a confirmation, as I'm quite sure about it. Let's say we have a photon that went through polarization filter at 0 degrees, and we have a second polarization filter at [itex]\alpha[/itex] degrees, the chance it goes through the second polarization filter is:
[tex]\text{cos}^2 \alpha[/tex]

Right?

(I like those tex tags!)
Yes, correct. This is Malus's law.
 
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Thanks for your help!
 
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I am confused by this whole entanglement thing - on one breathe when measurement occurs, both photons assume a definite polarisation and from this point on are no longer entangled. Yet, if we measure one photon before sending the other through a polariser orientated at a certain angle (rather than vertical or horizontal), we find the results are still correlated. Or is it that, if the photon going through the 2nd polariser had a definite polarisation, even if it were not entangled the pass/fail rate is still the same as if it were entangled?
 

DrChinese

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I am confused by this whole entanglement thing - on one breathe when measurement occurs, both photons assume a definite polarisation and from this point on are no longer entangled. Yet, if we measure one photon before sending the other through a polariser orientated at a certain angle (rather than vertical or horizontal), we find the results are still correlated. Or is it that, if the photon going through the 2nd polariser had a definite polarisation, even if it were not entangled the pass/fail rate is still the same as if it were entangled?
We don't know the moment or mechanism by which entanglement ends. We can make these statements:

a) It is "as if" both take on a definite polarization when one takes on a definite polarization. There is no sense in which the ordering of that collapse matters.

b) When collapse occurs, it is not necessary that *all* entanglement ends. Just on the related bases for the measurement. For example, they could remain frequency/momentum entangled even though they are no longer polarization entangled.
 

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