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Polarized light; maximum number of disagreements?

by xb4byish
Tags: #photons
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xb4byish
#1
Nov11-12, 10:00 AM
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Starting with 2 vertical photons, how can we argue that the maximum number of disagreements and minimum number for +30 and -30 is Nmin(-30,30) ≤ N(60,0) ≤ Nmax(-30,30)? and should we also not rotate the light?
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Vanadium 50
#2
Nov11-12, 12:07 PM
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What are you talking about?
DrChinese
#3
Nov12-12, 09:36 AM
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Quote Quote by xb4byish View Post
Starting with 2 vertical photons, how can we argue that the maximum number of disagreements and minimum number for +30 and -30 is Nmin(-30,30) ≤ N(60,0) ≤ Nmax(-30,30)? and should we also not rotate the light?
Welcome to PhysicsForums, xb4byish!

I don't follow your question either. However, if by chance you are referring to entangled photons: they cannot be considered as vertically polarized.

mfb
#4
Nov13-12, 10:03 AM
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Polarized light; maximum number of disagreements?

What are Nmin(-30,30), N(60,0) and Nmax(-30,30)?

What do you mean with maximum number of disagreements? Maximal number of photon pairs where two detectors will see different orientations for both photons? That is just the number of photon pairs. The average number is more meaningful.
xb4byish
#5
Nov13-12, 11:23 AM
P: 2
perhaps its average number of disagreements. But i mean how can we prove the number of disagreements for (60,0) is between the minimum number of disagreements of (-30,0) and the maximum number of disagreements between the maximum number of disagreements of (30,0)
mfb
#6
Nov13-12, 12:24 PM
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I don't understand that question.

There is an explicit formula for the "disagreement probability", you can derive it simply by looking at the 4 different cases. That should help.


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