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.
No comments:
Post a Comment