From: scerir (scerir@libero.it)
Date: Fri May 10 2002 - 00:32:16 MDT
D3 D2
| |
| |
| source of |
D4 -- -- -- / <-- two entangled --> \ -- -- -- D1
photons
interferometer 2 interferometer 1
John K Clark wrote:
> > if we detect photon 1 in D2, we destroy
> > interference in the interferometer 1
> Yes, but the key word is "if". You have no way of knowing
> or influencing the outcome, it's a 50-50 crap shoot.
Hmmm. Let us say this way. "If" I put a detector in D2 this
detector can tell if the photon #1 is there or if the photon
#1 is in D1. But in both cases (100%) the interference
inside the interferometer 1 is destroyed. So, "if" I put
a detector in D2, even if I do not read it, I destroy the
interference.
Now the next question, imo, is: if a put a detector in D2,
I also destroy the interference inside the interferometer 2,
where the entangled photon #2 is passing through?
This question has, maybe, something to do with the time-energy
(Heisenberg's) relation, of course for both the entangled photons,
bounded in a 'singlet' state.
If the symmetry of the system is perfect (interferometer 1
& interferometer 2 are identical, photon # 1 & photon #2
have the same timing) we can destroy interference inside
interferometer 2 if we put a detector inside interferometer
1 in D2. At least this is my bet! That is because each of
the two entangled photons behaves, inside his interferometer,
exactly like the other.
(Still thinking about the rest of those questions).
s.
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