"in a manner that produces "weak interference" without resorting to coincidence signals."
Yes Nick, but is it true? - is the 64 trillion dollar question. ;-)
On Jan 30, 2013, at 4:51 PM, nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it.> 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, $ 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