Probability of entangled photons passing filter independent?

  • #1
entropy1
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Consider two polarisation-entangled photons A and B fired at two polarisationfilters that are at a certain angle α. Are the probability that A is passing its filter and the probability that B is passing its filter indepedent probabilities?

I am aware that is probably an incredibly stupid question, but I need peer confirmation.

UPDATE: As far as I understand from Wikipedia, correlation (which is the case here) implies dependence. However, in this particular case the dependence only shows when the two measurement results are brought together...
 
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Answers and Replies

  • #2
Nugatory
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Consider two polarisation-entangled photons A and B fired at two polarisationfilters that are at a certain angle α. Are the probability that A is passing its filter and the probability that B is passing its filter indepedent probabilities?
For both photons, the probability that the photon passes its filter is 50%, and this is independent of the angle between them. Thus, if you're watching just one end of the experiment you will see a completely random sequence of pass/no-pass events.... You might as well be flipping an honest coin and counting heads and tails.

However, that's the answer to the question "What is the probability that a photon passes its filter?". If you're going to ask about the probability that a photon passes its filter, given that the other photon passed (or not) when the angle between the filters is ##\alpha##.... That's a different question. To see this effect we have to make measurements at both sides and then compare notes after the fact.
 
  • #3
entropy1
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If you're going to ask about the probability that a photon passes its filter, given that the other photon passed (or not) when the angle between the filters is ##\alpha##.... That's a different question.
If you look at it that way, are the measurements dependent?

Or rather: is P(A+ ∩ B+)=P(A+)P(B+) where X+ is X passing the filter?

UPDATE: I guess that would be silly, for P(A+ ∩ B+)=cos2(α), and since the setup is symmetric, it would yield P(A+)=P(B+)=|cos(α)|, which is not the case, since P(A+)=P(B+)=½. So I guess P(A+) en P(B+) are not independent then. Thanks. Figures, since there is entanglement of course.
 
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  • #4
Strilanc
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The probabilities are not independent. If the photons' polarizations are entangled into a singlet state ##\left| HV \right\rangle
- \left| VH \right\rangle## and you pass both of them through a polarizer with the same angle, then you'll find that exactly one of the photons is absorbed every time (its partner will pass through).

If you only get to see the results from one of the photons from each pair, the results will look like coin flips.

Different entangled states will show other kinds of correlation (such as both photons always doing the same thing, or the photons always doing opposite things when the polarizers are offset by 45 degrees instead of 0 degrees).
 
  • #5
entropy1
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I thought of a crazy idea and worked out some math. It all contradicts, so I guess I am not the one entitled to invent fancy theories haha! o0):biggrin::wink:
 
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