Well, as you can see here:


He isn't very experienced yet. I know of some experienced  professors of
have made worse mistakes :)

So, what goes wrong? Well, you don't get an interference pattern at one end
even if you don't detect the photon at the other end. To see this, just
write down the two particle state and add the phase shifts. If the detectors
on the other sides are off, then the two contributions corresponding to the
photon being detected at some position z consist of two orthogonal terms;
one term correpsonds to the other photon in pipe 1 and the other for that
photon in pipe 2

 Suppose that you add a plate to detect the photon on that side as well.
Then the probability that the photon at one end is detected at position z1
and the other is detected at z2 does contain an interference term of the

Cos[delta1(z1) + delta2(z2)]

If you don't detect where photon 2 is absorbed you have to integrate over z2
and the interference term vanishes. To see an interference term you must
keep  z2 fixed. This means that you must consider only those photons for
which the entangled partners were detected at at some fixed z2. But this
means that this value must be communicated by the observer there.

----- Original Message ----- 
From: ""Hal Finney"" <[EMAIL PROTECTED]>
To: <everything-list@eskimo.com>
Sent: Thursday, October 13, 2005 03:59 AM
Subject: RE: Quantum theory of measurement

> Now that you are experts on this, try your hand on this FTL
> signalling device, <http://arxiv.org/abs/quant-ph?0204108>.
> The author, Daniel Badagnani, is apparently a genuine physicist,
> <http://cabtep5.cnea.gov.ar/particulas/daniel/pag-db.html>.
> Hal Finney

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