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On Feb 1, 2013, at 8:10 PM, nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:

Demetrios--

I'm already assembling a thought experiment in my head.
The nicest thing about thought experiments is that
all the sources and detectors are ideal
and work perfectly every time.

If we can't find a flaw using thought experiments
then physicists in every optics lab on Earth
will stampede
to be the first to observe
the Kalamidas Effect.

And reap the rewards.

Nick


On Feb 1, 2013, at 7:30 PM, Demetrios Kalamidas wrote:

Nick, you state:
"Although all of the parts of this experiment are
possible the whole experiment itself would be quite difficult."
  It would indeed be a technically challenging experiment, on the order of complexity of Zeilinger's recent Canary Islands teleportation stuff, IF the required distance to achieve the superluminality condition is sought for....
HOWEVER, if this bizarre effect is observed in just a table-top version, on the order of one meter, it will be extremely strong evidence that the same effect will be seen even if we stretch out the left and right wings to 10s or 100s of miles....there is no change in the nature of the set-up by doing this.
   The superluminality condition in my set-up is achieved when the distance is large enough for an observer, on the left, to statistically distinguish a "1" from a "0" before a classical and luminal signal gets there (and that is just a function of the efficiency of my scheme and technology).
Demetrios



On Fri, 1 Feb 2013 19:02:04 -0800
nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
Well it isn't going to work.
But we may learn something
by seeing where it fails.
Although all of the parts of this experiment are possible
the whole experiment itself would be quite difficult.
Thought experiments are easier and cheaper
and don't need any hardware
except the human mind
and some paper and pencils.
So real experiment is premature.
On Feb 1, 2013, at 6:55 PM, JACK SARFATTI wrote:
Nose to the grindstone Nick!
I await your penetrating analysis.
If this worked it would be a Brave New World. ;-)
I have been preoccupied with Jim Woodward's Star Ship book and have  only been giving this partial attention.

On Feb 1, 2013, at 6:44 PM, nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:

Demetrios--

Clever. And of course--if it works -- there exist an optimum  product alpha x r that maximizes the Kalamidas Effect.
I can't offhand refute it but now that I understand what you're doing
I will certainly try.

Thanks, Jack, for sending me this clever FTL scheme.

Nick



On Feb 1, 2013, at 6:00 PM, Demetrios Kalamidas wrote:

Hi Nick,

  Yes....you got the main point of what I'm trying to do.
In Mandel's experiment, the "two halves" in which an idler photon  can exist are collapsed into a single path such that the origin  of the idler is "in principle" impossible to determine....we  don't even need any detectors in that idler path to destructively  register a photon.
  I am doing an analogous action by "blurring" each of the two  halves (modes a2 and b2),in which a right-going photon can exist,  with an indefinite photon number so that again, albeit in a less  efficient and more noisy way, we cannot SOMETIMES tell, even in  principle, if that right-going photon existed in mode a2 or in  mode b2.
  The "sometimes" part is, namely, the outcome |1>a2'|1>b2'  since it could be that: the photon in a2' came from the entangled  pair while the photon in b2' came from a weak coherent state !OR!  the photon in a2' came from a weak coherent state while the  photon in b2' came from the entangled pair.
  We DO NOT NEED ANY DETECTORS on the right wing of the experiment, as in Mandel's set-up. In my scenario, the possible  outcomes in modes a2' and b2' (in terms of photon number) are:  01,10,11,02,20,12,21 which are all "in principle"  distinguishable, with the only caveat being that the outcome "11"  has the special effect of erasing the path information of a left- going photon (which in turn leads to a small amount of  interference on the left).
Demetrios



On Fri, 1 Feb 2013 15:56:00 -0800
nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
Demetrios--
I am trying to understand your device.
You seem to be trying to erase the "which path" info
without combining the two possible paths.
How are you doing this?
For clarity I assume your detectors are perfect
and measure the Number of Photons in
each two-photon entangled event.
In any ordinary experiment that number (or either side a or b)  must  be One.
And where that One Photon ends up can indicate
Which Path or Which Interference Pattern depending on design.
Both of these designs involve coincidence detection.
If I understand your proposal
you attempt to erase the which-path info
by adding (via a biased beam splitter)
a coherent state to each possibility channel.
Since coherent states possess an indefinite photon number
the number of photons that appear at the detectors is also  indefinite
and the observer cannot decide
which path the photon took
no matter what the reading of the detectors.
Is this how your device works?
Nick
On Jan 30, 2013, at 4:51 PM, nick herbert wrote:
Each single photon of the pair is produced in a SUPERPOSITION
of a and b directions. Observation of "which path" can collapse  the
superposition into either a or b but (in conventional experiments)
these collapses (in the absence of coincidence signals) appear
to occur at random.

Destroying the path information by conventional means
(say, combining a and b in a beam splitter) does not
produce interference by itself but can do so if coincidence
signals are introduced.

DAK claims that by adding coherent states to the separated
halves of the superposition, that he can destroy "which path"
information in a manner that produces "weak interference"
without resorting to coincidence signals.


On Jan 30, 2013, at 2:30 PM, Demetrios Kalamidas wrote:

Hi guys,
....and thanks for the interest in my idea....and SORRY! Fred  for  not getting back to you, I've been traveling all last  week and  this week for my job....I'm responding from an MIT  computer right  now (as I'm working).

Let me try to quickly clarify some points:
  The source S produces only SINGLE PAIRS of photons, with a   photon pair created in modes a1a2 !OR! b1b2.
   In Mandel's experiment, it is the overlap of the two idler  modes causes erasure of the 'which-way' info for a  signal photon.  I wanted to find an 'unfolded' version of this  concept so that  space-like separation could be achieved.
  The method that, I purport, does the job of erasing the  'which- way' info for a left-going photon (that could be in  EITHER mode a1  OR in mode b1) is that the corresponding modes, a2 and b2, are  'mixed' with weak coherent states (each  having at most one photon)  such that, sometimes, we'll get  one photon in each of the two  output modes, a2' and b2', and  this makes it impossible to tell  where each of these two  photons came from.  If the math is valid,  this procedure  leads to a small amount of 'pure state' on the left wing of  the experiment....as opposed to the completely mixed state   that would arise if the coherent states were absent and only  the  two-photon state from S was present.
  I'll try to keep up with any further comments, questions,  and  discussions.
Demetrios



On Wed, 30 Jan 2013 13:03:37 -0800
JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
PS
OK the two coherent state inputs replace Mandel's idler photons.  So when you include a3 & b3 with the original pair  from S you  have 4-photon states in the Hilbert space two of  them are Glauber  states and the original pair are Fock states.
Begin forwarded message:
On Jan 30, 2013, at 12:56 PM, JACK SARFATTI   <This email address is being protected from spambots. You need JavaScript enabled to view it.> wrote:
Wait a second, he has 4 photons s1, i1, s2, i2 - at least in  the  Mandel experiment
However, you & Fred are right, Kalamidas's picture is  confusing  it seems to show only two photons, but he cites  Mandel, so does  he actually have 4 photons - two signal &  two idler like Mandel?  On Jan 30, 2013, at 12:41 PM, nick  herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.>  wrote:
Fred Wolf is right. Like the original EPR this is a TWO- PARTICLE experiment -- one particle going to the left and  one  particle going to the right in each elemental  emission. If  DAK's argument depends on seeing this as a 4- particle  experiment, then DAK is certainly WRONG.
Nick Herbert
On Jan 29, 2013, at 10:22 AM, JACK SARFATTI wrote:
Thanks Fred.
I hadn't thought to check out his starting point Eq. 1 I only  looked at Eq. 6. These experiments are tricky. I  have not yet  understood the details. Hopefully Nick &  others will chime in.  Begin forwarded message:
From: "fred alan wolf" <This email address is being protected from spambots. You need JavaScript enabled to view it.>
Subject: RE: PPS Demetrios A. Kalamidas's new claim for superluminal entanglement communication looks obvious at second sight
Date: January 28, 2013 11:11:31 PM PST
To: "'JACK SARFATTI'" <This email address is being protected from spambots. You need JavaScript enabled to view it.>
            Of course it is wrong for some serious and perhaps not so obvious reason.  He has confused a four  photon  state with an entanglement of two entangled (two)  particle  states. He approached me and I explained why it  was wrong.  Here is my explanation sent to him to which he has not  responded:
“Thanks for the paper.  Following Zeilinger’s paper   (attached) I am having some trouble understanding your  eq. 1.  If I understand it correctly you are using a  path  entanglement scheme similar to the one illustrated in  Zeilinger’s attached paper (p S290).  Therefore I  think you  should have  a1 entangled with b2 and a2  entangled with b1.  We would get e.g., (|a1>|b2>+ |b1>| a2>)/Ö2. Given that |a1> =  (|0>+exp(iphi)|1>)/Ö2, and  similarly for a2, b1, and b2, I  fail to see how you get your eq. 1, which seems to be some  kind of mixed four  photon state.”     Best Wishes,
Fred Alan Wolf Ph.D.  aka Dr. Quantum ®

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