Re: FTL Quantum Comm. via 2-photon Interference?

From: Neil (paradoxer@lykose.com)
Date: 03/14/03 

From: "Neil" <paradoxer@lykose.com>
Date: Thu, 13 Mar 2003 22:01:30 -0500

"Thinh Tran" <thinhvantran@cs.com> wrote in message
news:54187164.0303130011.1836c16d@posting.google.com...
> "Neil" <paradoxer@lykose.com> wrote in message
news:<v6vq2jagoir5b3@corp.supernews.com>...
>
> > Well, does anyone have an answer yet to the question I first posed?
> >
> > Neil
> [Question] Unfortunately I have to bother you with a trivial question.
> You wanted to know if moving D in and out would change the
> distribution elsewhere? Correct?
> Thinh Tran (http://www.thinhtran.com)

Correct, and the logical distinction for measurement effects is not trivial. It
seems like it couldn't, but consider the following argument. First, assume that
two photons P1 and P2 (for tracking, so named each time of emission) are emitted
simultaneously and in a repetitive set of emissions such as to form an
interference pattern (after enough hits) when they combine. Then, look at
corrected diagram below using fixed-width font. (I put E near the corner so that
the time to reach /// and then D1 and D2 would be equal.) D1 is the constructive
interference channel, and D2 is the destructive one.

Consider what happens if the detector D3 is close to E. If it gets a hit (which
happens 50% of the time), then P1 is still in transit, and has no "competition"
when it reaches the top-right half-silvered mirror C. It then goes 50/50 into
either D1 or D2, for a subtotal of 0.25 average hits into D2 per run. If D3 does
not get a hit, then we know that both photons are headed for C. Presumably this
means complete interference, with two hits in D1 and no hits in D2. In any case,
we already have a minimum of 0.25 average hits in D2.

However, if D3 is far away or absent, the lower-left half-silvered mirror LL
attenuates the amplitude of P2 to about 0.707.. of full value. Why? Because
without an absorption yet somewhere, it is just a "split wave" with 50/50 chance
to be found in either direction, and thus has sqrt(0.5) amplitude in both
diversions from LL. Hence, we have a 0.707 wave meeting a 1.0 amplitude wave at
C all the time, instead of a zero wave meeting a full wave half the time, and a
full wave meeting a full wave the other half the time. It's the same average
input of P2, but the *amplitude* statistics are different. We must add
amplitudes, then square, then divide by two (for D1 rates), and subtract
amplitudes, square, and divide by two (for D2 rates). (We divide by two because
the output is allocated into two channels, or "twice the area."
Hence rate at D1 = [1 + sqrt(0.5)]^2 /2 = 1.457106781187,
and rate at D2 = [1 - sqrt(0.5)]^2 /2 = 0.04289321881347.
The extra 0.5 is of course due to P2 having a 50% chance of hitting D3.
Note that 0.04289321881347 is much smaller than the 0.25 minimum demanded by
the effect of knowing definitely that P1 is alone and not subject to any
interference. Once P2 is gone, P1 will act like a single photon emitted along
one path.

Hence, if we move D3 from being a critical distance from the emitter to a little
farther away, we should get a change from the first type of statistics to the
second type. This would happen because the P2 member of twin photons is no
longer detected before the waves from both can reach D1 and D2.

                       D2
                       |
    /-----------------///--D1
    | |C
    ^ |
    P1 |
    | |
    | |
    | |
    E |
    \---->--P2--------///--<var.dist.>--D3
                      LL

/// = half-silvered mirror

Neil

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