Jack SarfattiOn Jun 2, 2013, at 7:22 AM, JACK SARFATTI <adastra1@me.com> wrote:

Yes it's always the case that if the time evolution is unitary signal interference terms cancel out. That is essence of the no-signal argument.

It's what defeated my 1978 attempt using two interferometers on each end of the pair source that David Kaiser describes in How the Hippies Saved Physics that was in first edition of Gary Zukav's Dancing Wu Li Masters. Stapp gave one of the first no-signal proofs in response to my attempt.

I. However, one of the tacit assumptions is that all observables must be Hermitian operators with real eigenvalues and a complete orthogonal basis.

II. Another assumption is that the normalization once chosen should not depend on the free will of the experimenter.

Both & II are violated by Glauber states. The linear unitary dynamics is also violated when the coherent state is Higgs-Goldstone vacuum/groundstate expectation value order parameter of a non-Hermitian boson second quantized field operator where the c number local nonlinear nonunitary Landau-Ginzburg equation in ordinary space replaces the linear unitary Schrodinger equation in configuration (or Wigner phase space more generally) as the dominant dynamic. P. W. Anderson called this "More is different."

For example in my toy model NORMALIZED so as to rid us of that damn spooky telepathic psychokinetic voodoo magick without magic

|A,B> = [2(1 + |<w|z>|^2)]^-1/2[|0>|z> + |1>|w>]

<0|1> = 0 for Alice A

<w|z> =/= 0 for Bob B

Take

Trace over B {|0><0| |A,B><A,B|} = 1/2 etc.

probability is conserved and Alice receives no signal from Bob in accord with Abner Shimony's "passion at a distance".

However, probability is not conserved on Bob's side!

Do the calculation if you don't believe me.

Two more options

i. use 1/2^1/2 normalization, then we get an entanglement signal for Alice with violation of probability conservation for Alice, though not for Bob

ii Final Rube Goldberg option (suspect)

use different normalizations depending on who does the strong von Neumann measurement Alice or Bob.

Now this is a violation of orthodox quantum theory ladies and gentlemen.

Sent from my iPhone in San Francisco, Russian Hill

Jack SarfattiOn Jun 2, 2013, at 12:56 AM, nick herbert <quanta@cruzio.com> wrote:

Kalamidas Fans--

I have looked over Martin Suda's two papers entitled 1. Taylor expansion of Output States and 2. Interferometry at the 50/50 BS.

My conclusion is that Martin is within one millimeter of a solid refutation of the kalamidas scheme. Congratulations, Martin, on achieving this result and on paying so much close attention to kalamidas's arguments.

The result, as expected, comes from a very strange direction. In particular, the approximation does not enter into Suda's refutation. Martin accepts all of kalamidas's approximations and refutes him anyway.

I have not followed the math in detail but I have been able to comprehend the essential points.

First, on account of the Martin Suda paradox, either PACS or DFS can be correctly used at this stage of the argument. So martin derives the kalamidas result both ways using PACS (Kalamidas's Way) and then DFS (Howell's Way). Both results are the same.

Then Martin calculates the signal at the 50/50 beam splitter (Alice's receiver) due to Bob's decision to mix his photon with a coherent state |A>. Not surprisingly Martin discovers lots of interference terms.

So Kalamidas is right.

However all of these interference terms just happen to cancel out.

So Kalamidas is wrong.

Refutation Complete. Martin Suda Wins.

This is a very elegant refutation and if it can be sustained, then Kalamidas's Scheme has definitively entered the Dustbin of History. And GianCarlo can add it to his upcoming review of refuted FTL schemes.

But before we pass out the medals, there is one feature of the Suda Refutation that needs a bit of justification. Suda's formulation of the Kalamidas Scheme differs in one essential way from Demetrios's original presentation. And it is this difference between the two presentations that spells DOOM FOR DEMETRIOS.

Kalamidas has ONE TERM |1,1> that erases which-way information and Suda has two. Suda's EXTRA TERM is |0,0> and represents the situation where neither of Bob's primary counters fires.

Having another term that erases which-way information would seem to be good, in that the Suda term might be expected to increase the strength of the interference term.

However--and this is the gist of the Suda refutation--the additional Suda term |0.0> has precisely the right amplitude to EXACTLY CANCEL the effect of the Kalamidas |1,1> term. Using A (Greek upper-case alpha) to represent "alpha", Martin calculates that the amplitude of the Kalamidas |1,1> term is A. And that the amplitude of the Suda |0,0> term is -A*.

And if these amplitudes are correct, the total interference at Alice's detectors completely disappears.

Congratulations, Martin. I hope I have represented your argument correctly.

The only task remaining is to justify the presence (and the amplitude) of the Suda term. Is it really physically reasonable, given the physics of the situation, that so many |0,0> events can be expected to occur in the real world?

I leave that subtle question for the experts to decide.