Organization: Internet Science Education Project
Andrew Gray is no flake. He presents a practical design for n 2q-bit quantum computers that are able to do reliable (almost zero error-rate), reproducible, precognitive remote-viewing arbitrarily far into the future. The reliability increases exponentially with n. Attempts to make a time-travel paradox fail because the machine's error-rate increases to compensate. This machine can be built today at small cost.
Of course, quant-ph/980408 seems to violate Eberhard's theorem forbidding nonlocal communication in orthodox quantum mechanics. It also seems to bypass the no-cloning a photon theorem that defeated Nick Herbert's similar FLASH gedankenexperiment many years ago. But there is no obvious error I can detect on a quick, but not careful, first reading. So if any of you pundits out there, like Nick Herbert, for example, or Henry Stapp, detect a flaw, please let us know. :-)
Subject: photon down-conversion
Date: Sun, 5 Apr 1998 10:16:37 -0700 (PDT)
From:
nick herbert
To:
sarfatti@well.com
Jack--
Photon down-conversion does not violate Zurek's no-cloning rule because it
is duplication of a KNOWN polarization state. It is possible to make a
machine that will just clone, say, H photons. The ZNC rule forbids the
cloning of an arbitrary UNKNOWN polarization state. It is impossible to
make a machine, for example, that will clone both R and H photons.
Nick
This is clearly a quantum model for
precognitive remote viewing without any appeal to
post-quantum nonlocal communication. We need to consider it
very carefully. Click here an example of modern parapsychological
research. It
proposes a simple device that can be built. Gray's paper is
to remote-viewing, what Paul Hill's book "Unconventional Flying Objects" is to NASA's goal
of "breakthrough propellantless propulsion".
There are no autocidal anomalies. Just as I said in my Sept
1991 Physics Essay. Any attempt to make such an anomaly
simple increases the error rate of the precognitive device.
Gray uses standard Feynman path quantum theory to get his
results. His device (his Fig. 2) uses standard off-the-shelf
optical equipment consisting of a single "primary-photon"
source, a "first beam splitter", etc.. Each of the two paths
from the splitter goes to a down-converter. Each
down-converter outputs two EPR-entangled photons
("interference" and "measurement") of half the frequency of
the original primary single photon input to the splitter.
One "upper" path out of each down converter for the
"interference-photon" goes to the same interferometer
consisting of a fully reflecting mirror that will catch only
one path from one down-converter and a half-silvered mirror
that reflects and transmits from each relevant
down-converter "upper" path with 50-50 probability going to
detectors A and B. The remaining two "down" paths from each
down-converter for the "measurement photon" go to the same
delay device, are then reflected by fully silvered mirrors
into the same "measurement-photon detector". A "blocker" can
be inserted, or not, into one possible path of the
measurement-photon on the output side of the delay device.
The delay device means we can arrange for the measurement
photon to be detected long after its twin interference
photon has been detected either by detector A or by detector
B. "If both possible paths of the measurement-photon are
made to converge again at a screen there will be
interference between the two histories corresponding to each
route the primary photon can take at the first beam
splitter." It is easily arranged to have destructive
interference for each interference photon detected at A and
constructive interference for each interference photon
detected at B. Therefore, A's detection probability is 0 and
B's detection probability is 1. We have the "delayed choice"
in the time-like future of the interference-photon detection
event of putting, or not putting, the blocker into one of
the two possible paths of the twin measurement-photon. This
future delayed choice represents one classical macroscopic
"c-bit" of Shannon information. Standard quantum theory
(i.e., "one-way" Q -> P in the Bohmian mechanical version,
where Q is the quantum potential and P is the "hidden
variable" i.e. "system point") implies that if the blocker
is inserted into one of the two relatively future paths of
the measurement-photon, then its twin interference-photon,
to which it is nonlocally connected has a 50-50 chance of
being detected at A or B. In other words, the relatively
future delayed choice to block one possible path of the
measurement photon, will act backward-in-time to destroy the
destructive interference of the twin photon at A and its
alternative constructive interference at B in the relative
past!
So to review. Not putting the blocker in, in the future,
means zero probability for detector A to fire, and 100%
probability for detector B to fire in the past of that
future delayed choice. Putting the blocker in, in the
future, means probability 1/2 for either A or B to fire for
each primary photon input into the "quantum time machine",
which is here really a mechanical precognitive
remote-viewing machine that the CIA would love to
manufacture. Actually there is nothing stopping them, or any
one else, with enough money and competence to put one
together in a very short time indeed. This is not a fancy
device. Whoever has such a machine will have a decisive
military and economic advantage if its competitors do not
have such capability. This is not science fiction, but
science fact, unless one of you finds a flaw in Andrew
Gray's argument. It's a very clear paper.
There are four possible decoherent classical histories in
the Gell-Mann/Hartle sense to consider.
History 1: the A-B detectors of the interference photon in
the past show the incoherent 1/2-1/2 response, and the
actual future state of the blocker for the twin measurement
photon is actually in the beam (blocker is "on"). This is,
therefore, an error-free history.
History 2: the A-B detectors of the interference photon in
the past show the incoherent 1/2-1/2 response, and the
actual future state of the blocker for the twin measurement
photon is actually out of the beam (blocker is "off"). This
is, therefore, an error-history.
History 3: the A-B detectors of the interference photon in
the past show the coherent 0-1 response, and the actual
future state of the blocker for the twin measurement photon
is actually in the beam. This is, therefore, an
error-history.
History 4: the A-B detectors of the interference photon in
the past show the coherent 0-1 response, and the actual
future state of the blocker for the twin measurement photon
is actually out of the beam. This is, therefore, an
error-free history.
These four histories are for a single n = 1 primary photon.
Now go to a bulk ensemble of n of these individual 2 qu-bit
quantum computers from n primary photons. We have arranged
it so that "all of the resultant interference photons arrive
at the interference apparatus before the first of the
measurement photons arrives at the measurement apparatus."
The heart of the argument for this device is on page 4 of
Gray's paper where he argues how to combine the possible
Feynman histories for n primary photons with the proper
interference factors. It is the interference factors that
have causality violation here allegedly amplified to the
macroscopic level in apparent defiance of Eberhard's
theorem.
Table 2 has:
History 1 decoherent classical probability = (1 - 2^-n)Pb
History 2 decoherent classical probability = 0
History 3 decoherent classical probability = 2^-n Pb
History 4 decoherent classical probability = 1 - Pb
Therefore, the classical probability that the bulk ensemble
of n double q-bit quantum computers will correctly remote
view the future state of the blocker is the sum of the
classical probabilities for histories 1 and 4. This is 1 -
2^-nPb. The probability that the bulk quantum-computing
ensemble decodes the message from the future incorrectly is
2^-nPb.
Therefore the reliability for precognitive remote viewing of
this machine, where no attempt is made to make a paradox, is
1 - 2^-nPb. This rapidly approachs 100 % reliability as n
increases. There is nothing mystical or supernatural here.
Parapsychology has been reduced to quantum mechanics. In a
sense, disputes over parapsychology statistics has become
irrelevant in terms of a practical no-nonsense purely
mechanistic precognitive technology. Fred Alan Wolf and I,
of course, precognitively remote-viewed all this 25 years
ago "through the glass darkly" in our comic book Space-Time
and Beyond. So much for those orthodox Rabbis who say we
cannot know the future - or should not! :-)
Note that the interference photon provides the past decoding
q-bit. The measurement photon provides the future reference
q-bit. The two photons from the same primary photon are
nonlocally connected to form a 2 entangled q-bit quantum
computer. But there is a third classical Shannon c-bit that
we can call the future encoding c-bit. So one c-bit controls
the message. That is, a local future interaction of the
encoding c-bit on the reference q-bit communicates a message
backwards in time to the decoding q-bit that is entangled
with the reference q-bit. This is like a transistor switch
acting backward in time. This is a new kind of
advanced-action computing "gate".
What about time paradoxes? There aren't any. The above
analysis assumed that the probability of the blocker being
on in the future was independent of any reading of the A-B
detectors in the past. If you set up a Rube Goldberg
situation to make a paradox, the above probabilities change.
In the most extreme case the error probability is 100% so
the machine totally fails. "The time machine gives the wrong
answer if we try to trick it."
Go to http://www.stardrive.org or write to sarfatti@well.com
Nick Herbert wrote on April 14, 1998:
Jack
Yes, I can see why. But that is because you are rushing to
judgement on this most important issue and somehow you have
not given as clear a refutation as Gray gives in his
argument for the machine that decodes the future. It would
help if you actually read his paper, which you say you
haven't. Gray writes like an Englishman -- very clearly and
convincingly. I, myself, have not read his first paper prior
to the quantum time machine one we are debating. I will.
You know how Bohr warned of changing the "total experimental
arrangement". It is not obvious to me that the changes you
have made have not thrown the superluminal baby out with the
bathwater. Your argument that the photons come out 50-50
randomly really, in itself, says nothing about coherent
interference. Sure if you do something to detect which
channel each photon comes out, then you will destroy the
coherence. Gray is well aware of that. I think Gray's
gedankenexperiment is sufficiently interesting that it
should be tried as a real experiment. I am not convinced
that the "isomorphism" you claim is real. I still do not
see, in any clear and convincing way, from your argument why
the coherence is always destroyed for the interference
photon whether or not the blocker is placed in one of the
alternative paths of the twin measurement photon.
I don't believe it nor do I disbelieve it. I hang suspended
coherently in both parallel universes, one where it works
and one where it doesn't. My post-quantum computation has
not "halted" yet. If we believe many worlds, there is no
collapse, only recombination. However, in the post-quantum
many-worlds there is a new kind of intrinsically-sentient
objective collapse that Penrose calls "orch OR" rather than
"R". Oh well, this is getting off the present point. :-)
For another view:
Subject:
Figuring out Gray's Box
Date:
Tue, 14 Apr 1998 15:28:09 -0700
From:
Fred Alan Wolf
To:
Jack Sarfatti
Jack,
a) either measure which Down Converter the photon came from
(particle measurement, blocker in place)
or
b) both paths (interference wave measurement at the
measurement detector).
If the blocker is in, the path of the measurement photon is
determined, and there cannot be any interference in the
"interference" photon--it goes into A or B with equal odds.
If the blocker is out, there must be interference in the
"interference" photon channel and, as I see it, only one
detector fires .....
I think Nick Herbert argues that b never can happen ever.
That is Nick believes that A and B will fire with equal odds
independent of the blocker being on or off. If that is not
the case, then Eberhard's theorem is violated. So, Nick
thinks there is some violation of unitary time evolution
between collapses, if you believe in collapses that is, in
Gray's scheme even though he has not read Gray's actual
paper on this.
My premise is is that nothing happens at the A or B detector
until the measurement photon is measured. Only then is the
whole experiment completed. If the blocker is in or out
(and it must be one or the other) there is an effect at A or
B regardless and always. Thus I don't believe the
experimental setup can be a time machine as suggested.
Well this is strange. What happens to the energy in the
interference photon? It is supposed to propagate at the
speed of light.
I am not sure what you mean. There is a pair of photons
leaving one of the down-converters. There are two
alternatives, and the issue is whether or not the two
alternatives coherently interfere. Nick says they never do.
1) There will be no detection of the photon at A or B until
the experiment is completed at the blocker site. This is a
kind of backward-through-time measurement, but I don't see
how it can be used to send anything, including information,
back through time (I'm still thinking about that). Simply,
nothing happens at A or B until the delayed photon emerges
and then and only then will a detection event occur at A or
B. This is perhaps bizarre, but it is consistent.
But that IS a local retroactive decoding of active quantum information from
the future! Isn't it? I mean you say there will be a
"strange silence" that neither A nor B fire even though the
photon energy has already been delivered. This is like
inhibition in a nerve cell. But it is clearly a signal.
Remember we can shoot in n pairs fast, and then our computer
at the A and B receiver will note the clear anomaly that nothing happens when
it should happen! This is definitely a signal from the
future (a signal without a signal), if Nature were, in fact, to follow your hypothetical
scenario. I mean, the strange silence would tell us that the
twin measurement photons were not yet detected, because
ordinary retarded physics would say that, subject to small
fluctuations, and imperfect detection efficiencies, either
n/2 photons should reach both A and B and fire them, OR, all n photons should reach
B in the short time it takes classical light to get there and fire it.
The inhibition you propose here is in fact a precognitive
signal - a precognitive "null" signal where no measurement
is a measurement. I mean if it works this way, Intel
engineers can use it to advantage in queing their bit
streams in the quantum computing chip.
Let us review some salient points of Andrew Gray's paper.
History 1 (error-free): receiver A detects n/2 photons and
receiver B detects n/2 photons and the blocker will be on in
the future. This happens with probability
in the past according to Gray, where Pb is the probability that the
blocker will be on at the appropriate time in the future. In contrast, Nick disagrees and says this
probability is 1/2 always.
Now Gray says that the A and B events happen in the past of
the blocker being switched on. You say they happen
simultaneously, but even that will be a signal, because
neither A nor B firing in spite of n energy packets arriving according to the classical flight time
is clearly an objective anomaly that can be detected.
History 2 (error): receiver A detects n/2 photons and
receiver B detects n/2 photons in the past and the blocker will be off
in the future. Gray says the probability for this is always
absolutely zero for all n and for all Pb.
Here Gray is again in dramatic disagreement with Nick Herbert who
says that the probabilities for histories 1 and 2 are always
equal to 1/2 and that the probabilities for histories 3 and
4 below are zero.
History 3 (error): receiver A never fires, and receiver B
detects n photons and the blocker is on . Gray's probability
for this is
History 4 (error-free): receiver A never fires, and receiver
B detects n photons and the blocker is off . Gray's
probability for this is
Note that
That is, classical probability is conserved. Remember these are coarse-grained classical histories as in
the Gell-Mann Hartle theory.
Note that these probabilities come from the random
combinatorics of the branching of n events each event is
statistically independent of the others. The fact that two
of the histories have errors is from quantum randomness.
Again here is Gray's argument for the random statistically
independent branching.
This is clearly from statistical independence among the
photons where each of them has a 50-50 chance of going to A
or B in the absence of any nonlocal influence. Gray never
invokes any kind of Hanbury-Brown Twiss "photon clumping"
from Bose-Einstein quantum statistics as Nick mentioned in
one message.
The condition "independent of the result at the interference
apparatus" means no attempt to make an autocidal causal
anomaly i.e. time-travel paradox. Gray's precognition
machine only works for globally self-consistent loops in
time. Attempts to generate inconsistent loops reduce the
precognitive reliability of the machine. Now we come to the
essentially, and perhaps controversial debatable point, the
"interference factors".
Remember B is defined as that detector for which there is
constructive interference when the path lengths are adjusted
properly. That means the two Feynman
amplitudes are each
and, if the blocker will be off so that there is no measurement of which path the primary photon took, they add in phase
coherently, before squaring, to make 2^1/2. Now square to
get the relative factor 2. It is assumed that the process of down-conversion is not itself a "which route?" measurement of the primary photon, or, at least, some kind of quantum erasing is done.
In the case of the blocker on
above we square before adding to get 1/2 + 1/2 = 1 for a
relative interference factor of 1.
Finally, for each
interference photon arriving at A there is destructive
interference when the blocker is off given 1/2^1/2 - 1/2^1/2
= 0. So this makes perfect sense.
However, the kicker is the
basic assumption Gray makes that blocker off in future means
coherent interference in the past of the primary photon,
hence of the interference photon AND blocker on in the
future acts back in time, in the Wheeler "delayed choice"
mode to measure which path the primary photon took in the
past. This destroys coherence of the interference photon
from the down-converter. You need to buy this basic physical
idea. Nick Herbert doesn't buy it, but he does not give IMHO
a convincing intuitive or mathematical argument why there
can never be coherent interference at the receiver in this
set up. If Gray's machine works then it also explains
retro-PK ,as in the Schmidt experiment considered by Stapp,
at least in a qualitative way.
So clearly there is statistical indepedence for each part of
the process with n photons, each leaving each down converter
with 50-50 random chance, and obeying the Ergodic theorem
and any other red herring placed in the path of Gray's
cogent rational argument. Nick Herbert and Larry Crowell
have so far failed IMHO to show rationally, clearly and
convincingly that there is no coherent interference when the
blocker is off in Gray's actual setup. It is against the
rules of relevance IMHO, to change the total experimental
arrangement, as Nick has done.
If we accept Gray's idea, then we can treat History 4 as a base line. The local decodability of the nonlocal message from the future really depends on the relative probabilities p(1) and p(3) corresponding to signal S and noise N for the classical message bit over the quantum communication channel. Therefore, the proper measure of the signal to noise ratio of Gray's conjectured nonlocal communicator is
See the graph of this above. Note for n =1, S/N = 1, which is the random baseline of Eberhard's theorem. That is, we cannot locally decode the message from the future using only a single photon pair at a time, but we can do it using several redundant pairs at once. For n =2, S/N = 3, for n = 3, S/N = 7 etc.
Fred Wolf continued ( his "2" omitted for now as I did not
understand it) :
4) If the delayer is capable of holding the coherent
superposition, it would appear we may have something. Then
I suggest that what probably happens is case 1). A or B
will not fire until the experiment is completed, even if it
takes a long time to complete. This leads to the
possibility of ghost-like apparitions or manifestations
suddenly appearing for no apparent reason in detection
devices wherever they happen to be. Perhaps this is how Sai
Baba does it, or perhaps this is the cause of some noise in
circuits or even the appearance of images on films, thoughts
from the blue, and so on. In other
words this could explain any sudden "miraculous"
occurrence.
Still another perspective from John Gibbs and Tony Smith.
Subject:
More time machine discussion
Date:
Tue, 14 Apr 1998 22:33:05 EDT
From:
JMTSGibbs
To:
sarfatti@well.com
More regarding Gray's quantum time machine. First, let
me state for the
record that I did read Gray's paper thoroughly before
responding and I did
understand it. After discussions with Tony Smith, we agreed
that the crucial
thing in the experiment is the exact nature of the
down-converters.
First, as I stated before, there is no process for
down-converting a
single photon which will not transfer both linear and
angular momentum to the
down-converter, however, it can be done with two or more
photons. This should
not be too big a problem because it is a nonlinear device so
there should be a
large number of photons passing through at all times.
This would support Nick Herbert's contention that there is
never interference at A and B. That is Histories 3 and 4
have probabilities zero. But I thought Aephraim Steinberg
said this was not so? Maybe I misunderstood? However, your comment that it can be done with at least n = 2 is consistent with my S/N formula above.
This again supports Nick Herbert's estimation of the
probabilities.
--- Michael Gibbs "
Intel, IBM, Motorola, AMD, Cyrix ... are you listening? :-)
Nick Herbert on April 15, 1998 wrote:
nick herbert wrote:
here's the gist of my calculations re Gray
(see figure Gray.gif for info on naming conventions)
OK
OK
OK
OK
E(A) is the relative phase shift in the upper (A) interferometer
E(B) is the relative phase shift in the lower (B) interferometer
This is OK, so far, but what is not yet clear is that this
change in Gray's total experimental arrangement that you
have made by adding E and F, is "isomorphic" to Gray's.
OK
= [aF(C) + bF(D)]F(E) +[cF(C) + dF(D)]F(F)
where
From this equation much follows but a few facts are particularly interesting:
1 Adjust the phases E(A), E(B) so that z = 1, then
when upper photon exits channel C
lower photon always exits channel E
when upper photon exits channel D
lower photon always exits channel F
Cool! :-)
independent of z.
so photons always exit upper combiner in 50/50 mixture of C & D channels
nothing you can do to lower beam changes the statistics of the upper beam.
This is an example of Eberhard's theorem (forbidding nonlocal communication) in action.
3
This is comparing apples to oranges. Gray does not have an E
and F channel at all! I have no doubt that this new design
you give here works exactly like you say it does. Yes, what
you have here is definitely NOT a time machine! I agree with
your very nice clear analysis. But what is still not clear
to me is your intuitive leap that your new design is
"isomorphic" to Gray's? For example, would a good patent
clerk think so, given the profound context-dependence of
quantum machines made clear in the Einstein-Bohr debates? I
think Gray can fairly say that you changed his argument to
get what you wanted to prove. So, I have no doubt that your
design will work as you say it will, but is it really
equivalent to Gray's design? You have an extra interference
stage with new reflections and transmissions that Gray does
not have. You have two output channels for the measurement
photon where Gray only has one. This could be a significant
difference invalidating your claim of "isomorphism"? I agree
that maybe Gray is doing post-quantum theory not quantum
theory. That is, his really new step is how he derives the
n-dependence and this really does violate Eberhard's
theorem.
4
Well if it is so easy, why not do it explicitly that way at
the beginning so that I could not, in fairness to Gray, make
the above objections?
5
After calculating the quantum mechanical probabilities for this
device--either as above or using some equivalent method--anyone who still
believes that this device can send signals FTL is motivated not by
scientific logic but by muddleheaded and pathologically wishful thinking.
So what do you think, Jack? Where's my Bohm Prize?
So far you are in the running for the Grand Bohm Prize. In
the meantime I will send you $100 honorable mention with a
$50 bonus if you do the "simple calculation" rather than
leaving it to the reader. :-)
Is it POB 251, Boulder Creek, CA 95006 (from memory)? Please
confirm address.
A believer in quantum theory
OK Nick, here I have a go at refuting what you just did. I am acting
as a proxy for Andrew Gray who with his large head, liquid black eyes and spindly arms and legs, is at this moment (on some spacelike surface fixed to the Hubble flow) , a few
light years distant from us looking for an appropriate
traversable wormhole to get back here. You know these Grays
can't take too much Earth-time without some R & R at The
Pleasure Dome. Imagine if you had to wear a gray kevlar
form-fitting fiber-opticked transistorized nano-suit with night-vision contact lenses all the time. :-)
I will cut to the quick of your ill-fitting argument. :-)
nick herbert wrote:
1 Adjust the phases E(A), E(B) so that z = 1, then
when upper photon exits channel C
lower photon always exits channel E
when upper photon exits channel D
lower photon always exits channel F
Gray does not have any E and F, instead he only has M. Gray
argues that when the blocker is off, the correct wave
function is not what you wrote, but, rather, it's, (for the
n = 1 case only, without the branching for n pairs)
Because there is destructive interference at C and
constructive interference at D. Therefore, the contribution from C is zero.
On the other hand, when the blocker is on, Gray argues that
the wave function is
where @ is a random phase, like in Bohm's "Quantum Theory"
text. This is F(on) is a mixed (impure) state. Now there may
be a hidden post-quantum nonunitary evolution required to
make this work. But there is something peculiar and
non-standard about your purported demonstration. You are
ambiguous about the distinction between wave functions at a
single spacelike instant, and global Feynman path
amplitudes. I mean when you write
What do you mean, wave function or amplitude? They are not
the same in Feynman's path quantum mechanics. If you mean a
wave function, then you have a non-standard "two-time" wave
function
Instead of the standard form
One could perhaps do this using a density
matrix or some kind of propagator formalism, but you did not
do that when you did your step-by-step retarded algorithm as
if you had a sequence of wave functions on advancing
spacelike slices (snapshots).
"the accuracy of the time machine doubles with each extra
photon used, and can in principle be made as accurate, and
see as far into the future, as one desires."
"We could potentially send several photons into this
apparatus one after the other, and have all of the resultant
interference photons arrive at the interference apparatus
before the first of the measurement photons arrives at the
measurement apparatus. In this case we could we could see if
there was interference, with an arbitrarily large set of
photons, before it is decided whether or not to observe
which route each primary photon took. We would thus know in
advance the future state of the blocker."
"To calculate the total probability for each of these
histories we must calculate the probabilities for the
possible branchings within each history, and also any
interference factors. We shall consider the general case
where n primary photons are used. For all of the resulting
interference photons to go to B gives a branching
probability of 2^-n, for any of them to go to A gives a
branching probability of 1 - 2^-n. Then we must include the
probability for the blocker to be on, which we shall assume
to be independent of the result at the interference
apparatus, and which we shall call Pb. Finally we must
include the interference factor for each of the histories.
If the blocker is on there is no interference, and the
interference factor is 1 in all cases. If the blocker is
off, then for each interference photon arriving at detector
B, there is an (constructive) interference factor of 2 when
the histories associated with both the routes the primary
photon can take reconverge. For each (destructive)
interference photon arriving at A there is an interference
factor of 0. Thus if all n photons arrive at B i.e. if the
device concludes the blocker will be off, the total product
of all the interference factors, the total product of all
the interference factors, is 2^n. If any photons arrive at
A, which will result in the device concluding that the
blocker will be on, the overall interference factor is zero.
We multiply these three probability factors together to
obtain the overall probability for each of the possible
histories. The results are shown in Table 2."
Arguing with you is like fighting the tar baby. It just gets stickier and stickier and never resolves itself.
Quantum mechanics tells us how to
calculate the results of any experiment. I've done that for Gray's setup
plus some simple generalizations of his setup and don't get any
superluminal results. To me that settles it. I'm not interested in beating
a dead horse.
I'm not interested in plowing thru Gray's paper looking for where he went
wrong. Entanglement is subtle: there are a thousand easy ways to go off the
track. As Feynman once said during a lecture on quantum gravity. "I didn't
read none of dos udder guys, 'cause dey didn't solve da problem. Here's
what I done."
I thank you for piquing my interest in FTL systems involving photon doublers. I'd never analyzed such systems before. I've convinced myself
that Gray's scheme and simple generalizations are useless for FTL
signalling. I'm not really interested in convincing others.
If you want to go on believing Gray's Box will work as a time machine,
please be my guest.
I've figured out Gray's Time Machine. It is very much like
the Reif paper I analyzed earlier. This is no more than a
two photon delayed choice experiment with the choice
existing at the "measurement" photon to:
There is apparently no time machine effect as I see it.
But this is moot. The real question is what happens at A or
B while the "measurement" photon is rattling around in the
delayer? The answer here is, perhaps, not obvious.
However, having said that consider the following. Thus,
after the photons leave both down converters.
detector A will never fire if the blocker is out and will
always fire half the time if the blocker is in, regardless
of the delay and only when the detection event with the
"measurement" photon is determined." This leads to some
unusual and interesting possibilities.
p(1) = (1 - 2^-n)Pb
p(2) = 0.
p(3) = 2^-n Pb.
p(4) = 1 - Pb.
p(1) + p(2) + p(3) + p(4) = 1
History
Gray (ET)
Nick (Contactee)
p(1)
(1-2^-n)Pb
1/2
p(2)
0
1/2
p(3)
2^-nPb
0
p(4)
1-Pb
0
For all the resulting interference photons to go to B gives
a branching probability of 2^-n.
for any of them to go to A gives a branching probability of
1 - 2^-n. Then we must include the probability for the
blocker to be on, which we shall assume to be independent of
the result at the interference apparatus, and which we shall
call Pb.
Finally we must include the interference factor for each of
the histories. If the blocker is on there is no
interference, and the interference factor is 1 in all cases.
If the blocker is off, then for each interference photon
arriving at detector B, there is an interference factor of 2
when the histories associated with both the routes the
primary photon can take reconverge.
Feynman path amplitudes = 1/2^1/2
We multiply these three probability factors together to
obtain the overall probability for each of the possible
histories.
In the case of Histories 1 and 4 the machine has correctly
predicted the future, in cases 2 and 3 it has got it wrong.
By adding the probabilities for Histories 1 and 4 we find
that the accuracy of the machine is 1 - 2^-n Pb.
S/N = p(1)/p(3) = (1 - 2^-n)/2^-n
3) A third possibility is A and B will fire with equal
odds. Before the "measurement" photon emerges from the
delayer the "measurement" photon is on either path but not
as a coherent sum of both paths. Hence, again, the presence
of the delayer causes the decoherence to occur. Here the
delayer acts as an observer and the desired effect of the
blocker vanishes. I don't think this is the case.
Jack,
Secondly, the phase of the down-converted photons
depends sensitively on
where the down-conversion takes place. To maintain
coherence between the two
down-converters the regions of interaction must be small
compared to the
wavelength of the light (perhaps at an interface between two
crystals?). In
the devices I have heard of, the conversion takes place
inside a macroscopic
region which would completely destroy the coherence between
the two paths.
(The lengths of the paths would differ by random amounts.)
It seems to me that one would really prefer a device
which converted two
photons (from a laser) into an entangled pair of photons at
the same
frequency. This would make getting coherence much easier.
Is there an
optical device that does this?
It would be very interesting to see what would happen
if this device were
built. The amount of delay applied to the measurement
photon should be
irrelevant to the experiment and producing a delay more than
a microsecond
presents quite a technical challenge. But even a
microsecond look into the
future could be very beneficial to a computer which could
carry out 500 or
more floating point operations during the delay and transmit
the results back.
jack--
The Fs are photon wavefunctions usually written as "phi"s, but phi alas is not an ASCII symbol.
F -> F(1) + F(2) primary splitter
F(1) -> F(A,1)F(B,1) doubler #1
F(2) -> F(A,2)F(B,2) doubler #2
F(A,1) -> E(R)F(C) + F(D)
upper combiner F(A,2) -> [F(C) + E(R)F(D)]E(A)
F(B,1) -> F(E) + E(R)F(F)
lower combiner F(B,2) ->[E(R)F(E) + F(F)]E(B)
where E(R) is the photon phase shift upon reflection
All these Es have unit modulus, E(A) and E(B) can be varied by the experimenter
and E(R) is actually equal to "i" but it's more convenient to
just use symbols and put this piece of info in last. Also none of these
equations is normalized. Factors of 1/sqrt2 are left out. we will normalize
(if we have to) at the end.
A little algebra yields:
F -> [i(1+z)F(C) + (1-z)F(D)]F(E)
+ [(z-1)F(C) + i(1+z)F(D)]F(F)
z = E(A)E(B)
F -> F(C)F(E) + F(D)F(F) perfect correlation
2 Ratio of photons exiting channel C to photons exiting channel D is:
[|a|2 + |c|2]/[|b|2 + |d|2]
= 1
, The same statistics holds for one photon as for a million photons. No
matter how many photons go into this apparatus the 50/50 ratio at channels
C & D is predicted by quantum mechanics. This fact that statistics are
independent of photon number is one of the general principes of quantum
theory. If Gray gets different statistics for different photon numbers,
he's not doing quantum theory but something else
. This calculation was carried out using a slight generalization of Gray's
apparatus--two beam combiners rather than an upper beam combiner and a
lower interference screen. It is an easy exercise to specialize this
calculation to the case of Gray's device.
The conclusion is
unaffected--nothing you can do in the lower beam (putting a blocker in path B2 for instance) alters the 50/50 statistics of the upper combiner. I leave
this simple calculation to the reader.
Nick Herbert
Like Cyrano De Bergerac in a film by Jean Cocteau, Jack Sarfatti now tries to hoist Nick Herbert by his own petard! :-)
F -> F(C)F(E) + F(D)F(F) perfect correlation
F(off) -> F(D)F(M)
F(on) -> (1/2^1/2) [F(C) + F(D)e^i@] F(M)
F -> F(C)F(E) + F(D)F(F)
F(past,future) = F(Cpast)F(Efuture) + F(Dpast)F(Ffuture)
F -> F(Cpresent)F(Epresent) + F(Dpresent)F(Fpresent)