Natural Science Forum / Physics / Relativity / November 2005

On Nov 27, 2005, at 7:24 PM, Jack Sarfatti wrote:

Another Go at this:

Round 2. :-)

OK, let's look at this once more. The problem here is how to properly
apply Feynman's rules. Let's review those rules.

I. Add amplitudes before squaring if one cannot distinguish the
alternatives.

II. Square the amplitudes before adding if one can distinguish the
alternatives.

Jensen's eq. 3 is, he claims, the coherent superposition

for the total experimental arrangement in fig 3 p. 3.

Let | ) be the basis for the detectors A&B of photon 2 on the RHS of fig 3.

Jensen's eq. 4 is

Substitute (4) into (3) to get

The key physical assumption here is that (3) is a single pure state not
a mixture.

If that assumption is correct, then

p(A) = (1/2)(cos@ + sin@)^2 = (1/2)(1 + cos2@)

p(B) = (1/2)(cos@ - sin@)^2 = (1/2)(1 - cos2@)

Now this is Jensen's argument. Nick seems to be saying that his physical
assumption is wrong, i.e. that (3) is not coherent, but is a random
mixture, so that

p(A) = p(B) = 1/2 for fig 3

and the same for fig 5. Hence no signal nonlocality.

On the other hand Jensen claims that these modulated probabilities are
locally seen in actual experiments? Is that correct. If so, then really
you deny Jensen's eq. 3.

On Nov 27, 2005, at 3:13 PM, nick herbert wrote:
Hi Ray--

Your ingenious FTL scheme was forwarded to me by Jack Sarfatti.

Having constructed dozens of such schemes while writing my FTL book (none
of which ever worked) I am familiar with most of the pitfalls a would-be
EPR FTL signaller will encounter.

In your case the problem occurs with your Eq 4 where you give the
<amplitudes> for detection of photons at the two interferometer arms. What
you want is the probabilities (amplitudes squared) for each of the two
types of pure-state photons that impinge on the interferometer.

Nick, what do you mean by "two types of pure state photons"? Are you saying

as a single coherent state is wrong, and that there is a mixture of two
states

and

each with classical probability 1/2?

I assume that is what you mean?

Since these distributions are caused by separate photons (a single photon
interferes with itself) one must add probabilities--not amplitudes as you
have done. When you do this, the sin squared plus cos squared identity
wipes out all fringe variation and the two cases (0 and 1 input) give
exactly the same result at the output.

What did you expect? That FTL signalling would be easy?

Mother Nature isn't easy. She wants to be coaxed.

But thanks for trying.

warm regards
Nick Herbert
http://members.cruzio.com/~quanta
http://quantumtantra.com

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