http://stardrive.org/stardrive/index.php/blog/has-mit-s-sam-ting-found-real-wimp-dark-matter.html

## Friday, February 22, 2013

## Monday, February 18, 2013

## Monday, February 11, 2013

## Wednesday, February 6, 2013

### Shall we KISS or not?

- Jack Sarfatti Jack Sarfatti On Feb 6, 2013, at 3:49 PM, nick herbert <quanta@cruzio.com> wrote:

Again a very persuasive argument.

You are correct that the |0>|1> term is small.

But it is multiplied by a different |0>|1> term (to form the product state |0>|1>|0>|1>.

The coefficients of this different |0>|1> term are surprisingly large.

JS: Ah so, Holmes.

NH: As to your ability to make alphaxr as large as you please. Do you think you can do this

and 1) preserve normalization of the input coherent state? 2) preserve the truncation condition?

JS: This issue of the normalization of the input coherent state is non-trivial. In the literature the authors on entangled coherent Glauber state put in what looks like an observer-dependent normalization forcing the Born probability rule to be obeyed. This can always be done ad_hoc, but it is not part of the rules of orthodox quantum theory where unitary time evolution guarantees invariance of the initial normalization choice that should not depend on what future choice is made by the measuring apparatus (for strong Von-Neumann projections).

For example, for a trapped ion internal qubit +,- entangled with its coherent phonon center of mass motion z. z'+ instead of the unitary invariant choice

| > = (1/2)^1/2[|z>|+> + |z'>|->]

The Born rule trace over the non-orthogonal Glauber states gives the seemingly inconsistent result

P(+) = P(-) = (1/2)[1 + |<z|z'>|^2]

P(+) + P(-) > 1

which I say is a breakdown of the Born probability rule in the sense of Antony Valentini's papers.

The dynamics of Glauber state ground state Higgs-Goldstone-Anderson condensates with ODLRO (Penrose-Onsager) is inherently nonlinear and non-unitary governed by Landau-Ginzburg c-number equations coupled to q-number random noise. The bare part of the noise dynamics sans coupling to the condensate is of course orthodox quantum mechanical.

Now what the published paper's authors do is to use an ad-hoc

| > ' = | > = (1/2[1 + |<z|z'>|^2])^1/2[|z>|+> + |z'>|->]

giving the usual no-signaling

P(+) = P(-) = 1/2

NH: And by the way, just what is the wavefunction for the input coherent state before the beam splitter?

You are never specific about what has to go into the beamsplitter to achieve the performance you describe. - Jack Sarfatti On Feb 6, 2013, at 1:49 PM, Demetrios Kalamidas wrote:

Hi to all,

Concerning my scheme, as it appears in the paper, lets do a certain type of logical analysis of the purported result:

Let's say that the source S has produced 1000 pairs of entangled photons in some unit time interval. This means that we have 1000 left-going photons (in either a1 or b1) AND 1000 right-going photons (in either a2 or b2).

Let's say we have chosen 'r' to be so small that only 1 out of every 1000 right-going photons is actually reflected into modes a3' and b3'. So, 999 right-going photons have been transmitted into modes a2' and b2'.

In my eq.6, we observe that the 'quantum erasure' part is proportional to 'ra'. Let's say we choose 'ra' such that '|ra|squared', which gives the probability of this outcome, is 10 percent.

This means that roughly 100 right-going photons have caused 'quantum erasure', for their 100 left-going partners, by mixing with the coherent states in a2' and b2'.

Thus, "fringes" on the left will be formed that show a variation of up to 100 photons, as phase 'phi' is varied, between the two outputs of beam splitter BS0.

Now, for this total batch of 1000 right-going photons, ONLY ONE PHOTON, roughly, has made it into a3' or b3' and mixed with the coherent states over there.

So, even if that ONE PHOTON contributes to "anti-fringes" on the left, it could only produce a variation of, roughly, up to 1 photon, as 'phi is varied, between the two outputs of BS0....and that is nowhere near canceling the "fringe" effect, but can, at most, cause a minute reduction in the "fringe" visibility.

JS: This seems to be a plausible rational intuitively understandable informal argument. Very nice. However, words alone without the math can be deceiving.

DK: Please note that we can choose 'r' to be as small as we desire, i.e. we can arrange so that one out of every billion right-going photons can be reflected into a3' and b3' WHILE STILL MAINTAINING the '|ra|squared'=10percent value (by just cranking up the initial coherent state amplitude accordingly).

I wrote this logical interpretation of my proposal in order to show that Nick's analysis goes wrong somewhere in predicting equal amplitudes for the "fringe" and "anti-fringe" terms.

Demetrios

JS: I do hope Demetrios will prove correct of course. Even Nick Herbert desires that. Is young Demetrios the new Arthur? Has he pulled the Sword from The Stone?

Using my number uncertain state |U> = x|0> + y|1> instead of your truncated coherent state,

I calculate that these two outcomes have exactly the SAME AMPLITUDE.

This fact is the essence of the refutation..

agreed that is the question.

On Feb 6, 2013, at 10:41 AM, nick herbert <quanta@cruzio.com> wrote:

>>>>In other words, even though the |0>|1>|0>|1> outcome may produce "anti-fringes", it has nowhere near the amplitude to cancel the "fringes" caused by the |1>|0>|1>|0> outcome....since the former outcome describes a right-going photon being reflected (extremely rare due to vanishing 'r') while the latter outcome describes a right-going photon being transmitted (very likely due to 't' approximately equal to 1).<<<<

A very plausible argument

But restore the missing term, Demetrios,

Do the calculation.

Then see if you still believe

that the |1>|0>|1>|)> term and the |0>|1>|0>|1>

have different amplitudes.

Using my number uncertain state |U> = x|0> + y|1> instead of your truncated coherent state,

I calculate that these two outcomes have exactly the SAME AMPLITUDE.

This fact is the essence of the refutation..

Nick--

I was up all night calculating these terms

and I am pretty sure your scheme is refuted.

Using the Feynman rule the probabilities for these two distinguishable processes are indeed equal

and do not cancel but one process is linked to fringes in Alice's detectors

and the other process is linked to anti-fringes in Alice's detector.

An incoherent equally-weighted sum of fringes and anti-fringes = no interference.

Your error consists of dropping a term that seems to be harmlessly small.

When you restore this term, the scheme becomes an ordinary coincidence-triggered distance interference device.

Since you are more familiar with these sorts of calculations than I am,

I urge you to restore the missing term and recalculate.

I would be surprised if you do not agree

that KISS is refuted.

However your measurement scheme -- ambiguating the Fock states by mixing with states of uncertain photon number --

is very clever and may find some use in less-preposterous applications.

I really have enjoyed interacting with your and your KISS scheme.

Nick

On Feb 6, 2013, at 9:38 AM, Demetrios Kalamidas wrote:

Hi Nick,

It is both a pleasure and an honor that you have analyzed my scheme to this extent and, thankfully, so far your hard analysis has not disproved it....and may have even generalized and strengthened the argument.

If my idea is described in a mathematically valid way then, as you seem to point out, the experimental proposal is also a powerful test of the strength of "The Feynman Dictum", which, so far, has never failed.

Thanks Nick

Demetrios

On Wed, 6 Feb 2013 04:32:09 -0800

nick herbert <quanta@cruzio.com> wrote:

Demetrios--

I have been calculating my own version of your KISS proposal

using the state |U> = x|0> + y|1> instead of a coherent state with alpha amplitude

as input to the beam splitter which you use. Using this state allows me to avoid

approximations. But yours is a robust proposal and should be immune to approximations.

Indeed I get the same result as you, making the approximation rx ---> 0 to eliminate a small |0>|1> term

as do you. I calculate the amplitude of the quantum erasure term |1>| 0>|1>|0> to be -trxy.

Hence my result for the probability of the FTL effect is 1/2 |etrxy|^2

which is comparable to your 1/2|etralpha|^2.

So far so good. The KISS and KISS(U) calculations give compatible results. FTL signaling seems secure.

------------

Next I decided to include the small term we both threw away. This means calculating the amplitude for

the detector response |0>|1>|0>|1>.

Imagine my surprise when I discovered that this (also a quantum erasure term by the way) amplitude is also trxy

with a plus sign!!!!!!!!!!!!

One might think that this term will exactly cancel your former quantum erasure term and refute your KISS proposal.

But I do not think that's the way it works. According to the Feynman rules you add amplitudes for indistinguishable paths,

and add probabilities for distinguishable paths. Since the |1>|0>|1>| 0> result is distinguishable from the |0>|1>|0>|1> result,

it seems that the proper thing to do here is add probabilities rather than amplitudes. So not only do these two processes

not cancel but THEY DOUBLE THE SIZE OF YOUR FTL EFFECT.

At least that's the way I see it right now.

Seems like my attempt to refute your proposal is traveling in the opposite direction.

Thanks for the fun.

Nick

Part 2

The more I think about this, the more I am convinced that this calculation refutes KISS.

The amplitude for |1>|0>|1>|0> is "-trxy"and for |0>|1>|0>|1> is "+trxy".

If you coincidence-trigger on the detector result |1>|0>|1>|0> you get fringes.

If you coincidence-trigger on the detector result |0>|1>|0>|1> you get anti-fringes.

If you do not coincidence-trigger you get an equal mixture of fringe and anti-fringe.

QED: No FTL signaling.

*On Feb 6, 2013, at 10:07 AM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:*

Hi to all,

I stated that in my previous message that "....thankfully, so far your hard analysis has not disproved it" but forgot to include the text of why I believe this:

The only way for "fringes" on the left wing of the experiment (caused by the |1>|0>|1>|0> term on the right) to be canceled by "anti-fringes" (caused by the |0>|1>|0>|1> term on the right) is if BOTH the |1>|0>|1>|0> term and the |0>|1>|0>|1> term had the SAME AMPLITUDE, and therefore the same probability of happening.

HOWEVER, in my scheme, the |1>|0>|1>|0> outcome is HEAVILY FAVORED when compared to the |0>|1>|0>|1> outcome because of the high asymmetry of the two beam splitters on the right.

In other words, even though the |0>|1>|0>|1> outcome may produce "anti-fringes", it has nowhere near the amplitude to cancel the "fringes" caused by the |1>|0>|1>|0> outcome....since the former outcome describes a right-going photon being reflected (extremely rare due to vanishing 'r') while the latter outcome describes a right-going photon being transmitted (very likely due to 't' approximately equal to 1).

Demetrios

Hi to all,

I stated that in my previous message that "....thankfully, so far your hard analysis has not disproved it" but forgot to include the text of why I believe this:

The only way for "fringes" on the left wing of the experiment (caused by the |1>|0>|1>|0> term on the right) to be canceled by "anti-fringes" (caused by the |0>|1>|0>|1> term on the right) is if BOTH the |1>|0>|1>|0> term and the |0>|1>|0>|1> term had the SAME AMPLITUDE, and therefore the same probability of happening.

HOWEVER, in my scheme, the |1>|0>|1>|0> outcome is HEAVILY FAVORED when compared to the |0>|1>|0>|1> outcome because of the high asymmetry of the two beam splitters on the right.

In other words, even though the |0>|1>|0>|1> outcome may produce "anti-fringes", it has nowhere near the amplitude to cancel the "fringes" caused by the |1>|0>|1>|0> outcome....since the former outcome describes a right-going photon being reflected (extremely rare due to vanishing 'r') while the latter outcome describes a right-going photon being transmitted (very likely due to 't' approximately equal to 1).

Demetrios

## Tuesday, February 5, 2013

Jack Sarfatti This is hot. If the effect works it's the basis for a new Intel, Microsoft & Apple combined for those smart venture capitalists, physicists & engineers who get into it. This is as close as we have ever come since I started the ball rolling at Brandeis in 1960-61 & then in mid-70's see MIT Physics Professor David Kaiser's "How the Hippies Save Physics". I first saw this as a dim possibility in 1960 at Brandeis grad school and got into an intellectual fight about it with Sylvan Schweber and Stanley Deser. Then the flawed thought experiment published in the early editions of Gary Zukav's Dancing Wu Li Masters in 1979 - pictured in Hippies book tried to do what DK may now have actually done. That is, control the fringe visibility at one end of an entangled system from the other end without the need of a coincidence counter correlator after the fact. Of course, like Nick Herbert's FLASH at the same time late 70's, it was too naive to work and the nonlinear optics technology was not yet developed enough. We were far ahead of the curve as to the conceptual possibility of nonlocal retrocausal entanglement signaling starting 53 years ago at Brandeis when I was a National Defense Fellow Title IV graduate student.

Jack Sarfatti

about an hour ago near San Francisco

On Feb 5, 2013, at 12:28 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

Thanks Nick. Keep up the good work. I hope to catch up with you on this soon. This may be a historic event of the first magnitude if the Fat Lady really sings this time and shatters the crystal goblet. On the Dark Side this may open Pandora's Box into a P.K. Dick Robert Anton Wilson reality with controllable delayed choice precognition technology. ;-)

On Feb 5, 2013, at 10:38 AM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

Looking over your wonderful paper I have detected one

inconsistency but it is not fatal to your argument.

On page 3 you drop two r terms because "alpha", the complex

amplitude of the coherent state can be arbitrarily large in

magnitude.

But on page 4 you reduce the magnitude of "alpha" so that

at most one photon is reflected. So now alpha cannot be

arbitrarily large in magnitude.

But this is just minor quibble in an otherwise superb argument.

This move does not affect your conclusion--which seems

to directly follow from application of the Feynman Rule: For distinguishable

outcomes, add probabilities; for indistinguishable outcomes, add amplitudes.

To help my own understanding of how your scheme works,

I have simplified your KISS proposal by replacing your coherent states with

the much simpler state |U> = x|0> + y|1>. I call this variation of your proposal KISS(U)

When this state |U> is mixed with the entangled states at the beamsplitters,

the same conclusion ensues: there are two |1>|1> results on Bob's side of the source

that cannot be distinguished -- and hence must be amplitude added.

The state |U> would be more difficult to prepare in the lab than a weak coherent state

but anything goes in a thought experiment. The main advantage of using state |U>

instead of coherent states is that the argument is simplified to its essence and needs

no approximations. Also the KISS(U) version shows that your argument is independent

of special properties possessed by coherent states such as overcompleteness and non-

orthogonality. The state |U> is both complete and orthogonal -- and works just as well

to prove your preposterous conclusion. --- that there is at least one way of making photon

measurements that violates the No-Signaling Theorem.

Thanks for injecting some fresh excitement into the FTL signaling conversation.

warm regards

Nick Herbert

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Jack Sarfatti On Feb 5, 2013, at 1:15 PM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Nope, no refutation I can think of so far....and I've tried hard.

Demetrios

...See More

33 minutes ago · Like

Joe Ganser Jack do you know a lot of people at CUNY? I take ph.d classes there.

26 minutes ago · Like

Joe Ganser I'm interested in who may do these sorts of topics in NYC

25 minutes ago · Like

Jack Sarfatti Daniel Greenberger!

9 minutes ago · Like · 1

a few seconds ago · Like

Nope, no refutation I can think of so far....and I've tried hard.

Demetrios

On Tue, 5 Feb 2013 13:09:28 -0800

nick herbert <quanta@cruzio.com> wrote:

Thanks, Demetrios. I understand now that alpha can be large

while alpha x r is made small. Also I notice that your FTL signaling scheme seems to work both ways. In your illustration the photons on the left side (Alice) are combined at a 50/50 beam splitter so they cannot be used for which-way information. However if the 50/50 beamsplitter is removed, which-way info is present and the two versions of |1>|1> on the right-hand side (Bob) are now distinguishable

and must be added incoherently, which presumably will give a different answer and observably different behavior by Bob's right-side detectors. So your scheme seems consistent -- FTL signals can be sent in either direction.

This is looking pretty scary.

Do you happen to have a refutation up your sleeve

or are you just as baffled by this as the rest of us?

Nick

On Feb 5, 2013, at 12:18 PM, Demetrios Kalamidas <dakalamidas@sci.ccny.cuny.edu> wrote:

Hi Nick,

And thanks much for your careful examination of my scheme....however, there appears to be a misunderstanding.

Let me explain:

"On page 3 you drop two r terms because "alpha", the complex amplitude of the coherent state can be arbitrarily large in magnitude."

I drop the two terms in eq.5b because they are proportional to 'r'....and 'r' approaches zero. However, the INITIAL INPUT amplitude, 'alpha', of each coherent state can be as large as we desire in order to get whatever SMALL BUT NONVANISHING AND SIGNIFICANT product 'r*alpha', which is related to the terms I retain.

In other words, for whatever 'r*alpha' we want, lets say 'r*alpha'=0.2, 'r' can be as close to zero as we want since we can always input a coherent state with large enough initial 'alpha' to give us the 0.2 amplitude that we want.

So, terms proportional to 'r' are vanishing, while terms proportional to 'r*alpha' are small but significant and observable.

You state:

"But on page 4 you reduce the magnitude of "alpha" so that at most one photon is reflected. So now alpha cannot be arbitrarily large in magnitude."

The magnitude of 'alpha' is for the INITIAL coherent states coming from a3 and b3, BEFORE they are split at BSa and BSb. It is this 'alpha' that is pre-adjusted, according to how small 'r' is, to give us an appropriately small reflected magnitude, i.e. 'r*alpha'=0.2, so that the "....weak coherent state containing at most one photon...." condition is reasonably valid.

Demetrios

On Feb 5, 2013, at 12:28 PM, JACK SARFATTI <sarfatti@pacbell.net> wrote:

Thanks Nick. Keep up the good work. I hope to catch up with you on this soon. This may be a historic event of the first magnitude if the Fat Lady really sings this time and shatters the crystal goblet. On the Dark Side this may open Pandora's Box into a P.K. Dick Robert Anton Wilson reality with controllable delayed choice precognition technology. ;-)

On Feb 5, 2013, at 10:38 AM, nick herbert <quanta@cruzio.com> wrote:

Demetrios--

Looking over your wonderful paper I have detected one

inconsistency but it is not fatal to your argument.

On page 3 you drop two r terms because "alpha", the complex

amplitude of the coherent state can be arbitrarily large in

magnitude.

But on page 4 you reduce the magnitude of "alpha" so that

at most one photon is reflected. So now alpha cannot be

arbitrarily large in magnitude.

But this is just minor quibble in an otherwise superb argument.

This move does not affect your conclusion--which seems

to directly follow from application of the Feynman Rule: For distinguishable

outcomes, add probabilities; for indistinguishable outcomes, add amplitudes.

To help my own understanding of how your scheme works,

I have simplified your KISS proposal by replacing your coherent states with

the much simpler state |U> = x|0> + y|1>. I call this variation of your proposal KISS(U)

When this state |U> is mixed with the entangled states at the beamsplitters,

the same conclusion ensues: there are two |1>|1> results on Bob's side of the source

that cannot be distinguished -- and hence must be amplitude added.

The state |U> would be more difficult to prepare in the lab than a weak coherent state

but anything goes in a thought experiment. The main advantage of using state |U>

instead of coherent states is that the argument is simplified to its essence and needs

no approximations. Also the KISS(U) version shows that your argument is independent

of special properties possessed by coherent states such as overcompleteness and non-

orthogonality. The state |U> is both complete and orthogonal -- and works just as well

to prove your preposterous conclusion. --- that there is at least one way of making photon

measurements that violates the No-Signaling Theorem.

Thanks for injecting some fresh excitement into the FTL signaling conversation.

warm regards

Nick Herbert

## Sunday, February 3, 2013

## Saturday, February 2, 2013

## Friday, February 1, 2013

http://www.physicstoday.org/resource/1/phtoad/v66/i2/p9_s1?bypassSSO=1 my letter in Feb 2013 Physics Today

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