🚀 go-pugleaf

I wrote:
||Therefore,
||
||p(V',OFF) = |<b(V')|<h|a.b>'|^2
|
|This is the probability that V' is observed *given that the photon is
|detected in state |h> at the transmitter end*. This is *not* what you
|want. How many times do we have to go over this particular point
|before you absorb it? You *must* count the other photon pairs, or you
|are not counting anything which can be determined at the "receiver"
|end of the apparatus.

In article <Bx632J.5Eo@well.sf.ca.us> sarfatti@well.sf.ca.us 
(Jack Sarfatti) writes:
|Ramsay, I am saying there is not anything else if all the light goes into
|one detector symbolized by |h>. 

This is precisely what does not make sense. There is not just one
state describing a photon hitting the detector. There are many
possible states a photon could have, while hitting the detector. (Many
different momentum states, for example.) So you cannot simply choose
to "symbolize" the detector by |h>.

Perhaps you don't realize that there is a Heisenberg uncertainty
relation involved here. If you reduce the uncertainty of the position
of your photons, the uncertainty of the conjugate momentum has to
increase. This prevents you from both focussing the beams on a small
spot, and having the phase vary slowly across that spot.

What you say is also completely contrary to standard quantum
mechanics:

According to standard quantum mechanics, observations correspond to
self-adjoint operators. When an observation is carried out, the state
of the system is projected onto one of the eigenspaces of the
observable (and the projected vector is then normalized). The
probability of each such projection occurring (and the corresponding
eigenvalue being the observed value of the observable) is proportional
to the square of the magnitude of the projection.

For whatever observation you make, the probability of a state |f>
being projected to a fixed state |h> is at most |<h|f>|^2. You cannot
make both |<h|f>|^2 and |<h|g>|^2 be nearly 1, since |f> and |g> are
orthogonal.

|If a little is lost it should not make a
|big difference.  Almost everything collapses in a non-unitary process into
|the ket |h> - there are no other photon pairs. 

This is definitely not standard quantum mechancics-- you cannot simply
assert that there is just one state involved and no other possibility.

Here's another way to see it: Suppose that we call the two possible
values of the variable observed by the detector "detect" (+) and "does
not detect" (-). You have not mentioned the detector observing
anything else about the photon, just whether it strikes the given
spot. If you can (almost) guarantee that a photon in state |f> will be
detected, then |f> must be (almost) in the (+)-eigenspace of the
operator corresponding to the observable. Therefore when (if) it is
detected, it will remain the same (be changed by only a little bit).
Likewise for |g>. So |f> and |g> cannot both be (almost) guaranteed to
be collapsed to the same state |h> unless they are (almost) the same
as |h> to begin with, since neither one can be changed (by more than a
small deviation-- one of small norm). *That* is what standard quantum
mechanics says.

| If there are other photon
|pairs, they are not counted at the transmitter nor at the receiver
|counters. 

Why not? This seems totally arbitrary.

|So that what is counted will show the "fringe modulation".  Now
|if there is a density matrix so we do a classical average over several
|phases we may get zero signal in the mean- but we can pick it up in a
|Hanbury Brown Twiss correlation of intensity fluctuations from photo-
|currents of the two receiver counters. This is simple math and I have
|already done it more than once.

The math does not say what you are claiming that it says.

Here again is an piece of text where you do this "simple math":

|Again: the V' receiver counter photo-current I(V') will be proportional to
|
|I(V') = x + ysin(2theta)cos(phi)<cos(arg<h|g> - arg<h|f>)>
|
|the H' counter photo-current proportional to
|
|I(H') = x - ysin(2theta)cos(phi)<cos(arg<h|g> - arg<h|f>)>
|
|where <  > is a classical statistical average over set of spots or set of 
|kets |h>. The different macro spots of finite resolution each represent 
|distinguishable Feynman non-interfering histories.
|
|Even if 
|
|<cos(arg<h|g> - arg<h|f>)> = 0
|
|<cos^2(arg<h|g> - arg<h|f>)> is not zero.

So? This is not the "root mean square" variation of anything you can
detect at the receiver. I think you must be misapplying some
statistical formula, but since you're not specifying what formula it
is, it is hard to tell why you think this is obviously correct.

|Therefore the root mean square fluctuations in the receiver counter 
|electric currents will be modulated by sin(2theta)cos(phi).
|
|This is a simple argument. Why don't you get it? Sarfatti

Because you do not give a solid justification. Elsewhere, you describe
how you are going to measure these "fluctuations":

|Let J be flux of photons arriving at a receiver V' counter. The count rate
|(clicks per unit time) will be proportional to J times p(V'). Plot the
|count rate as a function of time over scale bigger than duration required
|to determine count rate. This time series will fluctuate about a mean. 

So let N be the total number of photons which arrive over the interval
of time required to determine the count rate. The probability of
counting m of those photons at the V' counter is C(N,m) p(V')^m
(1-p(V'))^{N-m}, since the probabilities of each photon being counted
V' are mutually independent. Thus the likelihood of getting a certain
rate m/N of detection over that period of time is dependent only on
p(V'). Correlations with this extra variable |h> are simply
irrelevant. So how can "fluctuations" possibly enter into it, unless
you are able to affect p(V') in the first place?

It *appears* that you want to calculate these deviations in a more
devious way as follows:

Calculate the variance of the number of photons which arrive over the
interval (out of N) by partitioning them up according to the
associated values of |h>. Take the mean of each group. Calculate the
variance of the resulting distribution of means. This corresponds to
what you have calculated above. Unfortuately for you, this procedure
is simply wrong. So what are you in fact thinking?

                       "In no way, shape or form did Kevin represent
Keith Ramsay            a viable alternative to mental illness."