On Feb 6, 2013, at 10:49 AM, JACK SARFATTI <This email address is being protected from spambots. You need JavaScript enabled to view it. > wrote:
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 <This email address is being protected from spambots. You need JavaScript enabled to view it. > 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 <This email address is being protected from spambots. You need JavaScript enabled to view it. > 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.
I thought Nick said the two amplitudes were equal and opposite. If Demetrios is correct below I will be happy to retract my Eulogy for the demise of nonlocal entanglement signaling within ORTHODOX quantum theory as opposed to post-quantum extensions.
On Feb 6, 2013, at 10:07 AM, Demetrios Kalamidas <This email address is being protected from spambots. You need JavaScript enabled to view it. > 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
Jack Sarfatti
KISS-OFF! ;-)
-
Jack Sarfatti Yes, Nick most likely the two terms cancel as you say at the end. The problem with all the attempts to derive entanglement signal nonlocality within orthodox quantum theory, is the neglect of relevant terms, which in the end as you show, cancel the result. I wrote at the beginning of this that such may happen here.
Note, that this does not affect attempts as entanglement signal nonlocality using a more general nonlinear post-quantum theory as in Steven Weinberg's, Henry Stapp's and Antony Valentini's models.
On Feb 6, 2013, at 4:32 AM, nick herbert <This email address is being protected from spambots. You need JavaScript enabled to view it. > 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.