|Jun5-05, 12:47 AM||#52|
Mandel & Faster than Light Communication ?
Now, I agree with you that in EACH NL1 and NL2, the "stimulated pair" |s1> |i1> has an underlying "phase relationship" in that they both belong to the same 2-photon state. But what happens in NL1 and in NL2 can have a different phase relation from event to event depending on the conversion points in NL1 and NL2. In fact, it is even more complicated: one should in fact integrate over all points (Feynman path integral).
All this is of course "classical speak", but relates to the phase relationship between the two downconverted pairs.
My argument about no FTL had in fact nothing to do with FTL, it had to do with locality: in that whatever you measure PURELY on system 1 cannot be influenced by whatever you do on purely system 2 (a measurement, a change, anything). The only exception of course being that if system 2 INTERACTS with system 1 (which is excluded when they are space-like separated, for instance).
So quantum theory is "reasonable" here in that expectation values of observables purely relating to system 1 (here, the signal photons) cannot change anything whatever we do to system 2 (the idler photons): destroy them, stop them, mix them etc... EXCEPT OF COURSE IF WE CHANGE THE INTERACTION BETWEEN SYSTEM1 AND SYSTEM2 in doing so.
But as long as the only things that happen to H1 and H2 are:
1) measurements of observables of the style A x 1 or 1 x B
2) interactions, but not with eachother (unitary evolution U1 x U2)
then <A> is independent of what happens to H2 (and <B> is independent of what happens to H1).
I'm not inventing this: this is a theorem in quantum mechanics, which I tried to illustrate in some previous posts, with the density matrices.
Now, where they can interact of course, you do not have U1 x U2 but an overall U which cannot be written as U1 x U2 and the theorem is not valid anymore.
So, according to this theorem you cannot change an expectation value of the signal photons (say, the degree of interference pattern) by doing stuff only to the idler photons UNLESS this "doing stuff" interacts with the signal photons of course. That's why I am convinced that the explanation in the paper is wrong: it goes against a theorem in QM. (of course, QM could be wrong, but then the explanation is wrong too of course)
|Jun8-05, 12:42 PM||#53|
I think we have major confusion here.
Once you combine these two paths you have a one photon state and so an interference pattern at Ds.
|Jun8-05, 02:14 PM||#54|
If we apply the same reasoning to the two downconverters, and we consider one path, with a down conversion point x in NL1, and then another path, with a down conversion point y in NL2, and (ignoring for the moment even the fact that an idler beam is generated) we accumulate the total phase that s1 reaches at the screen, and the total phase that s2 reaches at the screen, you will see that this crucially depends upon the exact location x and y, (even if the total path is of course the same: the part done by V1 plus the part done by s1 for instance, is independent of x, but the total ACCUMULATED PHASE is not). Indeed, up to point x, the "phase counter" ran with a wavenumber of k_v1 radians per meter ; after point x, the phase counter ran with wavenumber k_s1 radians per meter. So the total accumulated phase of s1 at the screen, starting from the splitter of V into V1 and V2, is:
(l1 + x) k_v1 + (l2 - x) k_s1 = constant + x (k_v1 - k_s1).
In the same way, the total accumulated phase of s2 at the same point of the screen will be constant' + y (k_v2 - k_s2).
Now, of course k_v1 - k_s1 = k_v2 - k_s2 = delta_ks.
So the phase DIFFERENCE at the screen point where s1 and s2 meet (both indeed part of the same photon in two different states) is a constant" + delta_ks(y - x) (this is the same kind of reasoning as in a Young experiment (when we neglect the extra complication of i1 and i2).
So even when i1 and i2 weren't there, we would get a difference in phases for the two different paths (which will give us our constructive or destructive interference pattern) which is dependent on x and y. QM tells us that we should then INTEGRATE over x and y, washing out all interference, because delta_ks (x - y) rotates several times through 2Pi when we integrate over all possible positions x and y in NL1 and NL2 respectively.
Now clearly, the signal photon exhibiting interference or not is an observable that only depends upon the signal photon (s1 and s2). So if the idler system (i1 and i2) is NOT interacting in the classical sense (meaning: with an interaction hamiltonian), then normally, whatever you do to i1 and i2 (measurements, sending them onto mirrors, combining them, etc...) should not affect the observables pertaining only to the signal photon.
Now, if we also postulate that interaction hamiltonians are local (in the relativistic sense) then it follows that FTL is not possible. But what is shown is in fact stronger: it means that with non-interacting systems, expectation values of system 1 cannot be affected by things (interactions, measurements) you do to system 2. And that theorem is violated here if there is no interaction of i1 in NL2, that's why I think there must be some interaction.
It is sufficient that a lock-in of this delta_ks(x-y) phase factor occurs, thanks to the presence of i1 (which carries the x phase information!). But there must be an interaction hamiltonian if the theorem is not to be violated.
|Jun8-05, 09:41 PM||#55|
and signals have the same frequency (702 nm) according to this:
(see page 3)
It also states here that both photons are emitted simultaneous (within
(i1 --> s2) becomes another possibility since there's no random phase
anymore depending on the conversion points.
The paper excludes this kind of stimulated emission with the argument
that the down conversion rate of the 2nd PDC doesn't change with i1
being blocked or not.
I would think that's rather logical since it's the UV laser that "pumps up"
the PDC while the i1 photons can only release exited states.
|Jun9-05, 08:52 AM||#56|
Ok lets talk about a single downconverter and the effect it has on the phase of a photon. When the down conversion event happens it happens in a superposition of all possible depths in the downconverter. To find the effect on the phase you must integrate over all possible depths and then normalize to a unit vector. It is a unitary process, all we can get is a unit vector. Unitary evolution is nothing but a rotation of a unit vector.
So given this I would assume that if we shine a coherent beam into a downconverter we will get a (actually two) coherent beam out phase shifted by an amount determined by the total thickness of the downconverter. The only way to change this is to have some device that could watch the downconverter and tell us the depths that the individual downconversion events happened. Then we would no longer have a unitary evolution and we would get decoherence.
Surely there cannot be disagreement over whither or not a coherent beam into a down converter produces a coherent beam out.
You do not have NL1 and NL2 downconverting at the same time possibly at gifferent depths. You only have one of them downconverting but it is a superposition state. And it is in a superposition of all different depths produceing an expected constant phase shift due to the fact that it is a unitary process.
What am I doing wrong?
|Jun9-05, 09:07 AM||#57|
In any case what would happen if you were to rotate the polarization of I1? If they were stimulating the release of the photons they should still do so. Yet if the polarization is wrong we still have which way information and the interference pattern should go away.
|Jun9-05, 12:33 PM||#58|
other if they are at 90 degrees That said, what is the guarantee that s2
takes over the polarization of i1? Generally even the most basic properties
of the PDC's are omitted, let alone the more detailed parameters like the
half-life of the excited state. or the polarization dependency of stimulated
Often it's not properly understood why the PDC's do what they do.
|Jun9-05, 02:12 PM||#59|
So I can just as well consider the "first path" first, with downconversion in NL1, and then the second path next, in NL2, and then add them together at the screen of s1 and s2, no ?
However, as we both agree now, the "first path" (going through NL1) is actually composed of many different paths, depending on the point of conversion. And indeed we will have to integrate over all these "first paths". But we will also have to integrate over all the "second paths".
If x is the conversion point in NL1, we have the total phase contribution:
Integral dx f(x) exp( i.delta_vs1.x) . exp(i. phi1)
where phi1 is an x-independent phase factor which depends on the position of the screen of s1 and s2 and f(x) is a kind of square function, delimiting the conversion xtal.
If y is the conversion point in NL2, we have he total phase contribution:
Integral dy f(y) exp( - i. delta_vs2. y) exp (i . phi2)
Note that delta_vs1 and delta_vs2 are slightly different and the difference depends upon the point on the screen of s1 and s2, because if this point moves, we have slightly different angles between the horizontal and s1 or s2 so slightly different effective wave vectors as a function of x or y.
Adding the two contributions together, we get something that looks a lot like the interference pattern of two LARGE slits (except that here, the phase differences of the two slits are not obtained by changing the geometrical path length sideways, but by a different conversion point x). This means in fact: almost no interference.
Intuitively, you can see why this happens: for each x in NL1 (and a chosen point on the s1 and s2 screen), you will find a corresponding point y in NL2 so that we get constructive interference, and another point y' in NL2 so that we get destructive interference (and again a point y'' with constructive and a point y''' with destructive interference etc...).
If you now change the point on the screen of s1 and s2, for the same x point in NL1, y, y',y'', y''' will have shifted a bit but you will again find about as many constructive as destructive interference paths. This means that the intensity on the s1 / s2 screen doesn't change from point to point.
|Similar Threads for: Mandel & Faster than Light Communication ?|
|Is Faster-than-light Communication Possible?||General Physics||15|
|Faster than light communication...||Special & General Relativity||8|
|Is Faster-than-light Communication Possible?||General Physics||0|
|Is Faster-than-light Communication Possible?||Quantum Physics||22|
|Faster than light communication?||Quantum Physics||2|