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## Mandel & Faster than Light Communication ?

 Quote by ppnl2 Consider a laser, you have an excited gas to which you introduce a photon. This photon excites the emmision of more photons. So far we just have an expensive monochrome flashlight. But the key is that the emmited photon has the same phase as the exciteing photon so all the photons are in phase.
Individual photons are not "in phase" with others. What happens in a laser is stimulated emission which means that an n-photon state becomes an n+1 photon state (and ok, somehow you can say that the n+1 photons are "in phase" with eachother, but they cannot be a "bit out of phase" which doesn't make sense: they just belong to an n+1 photon state |k1,k1,k1,...k1>).
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).

 My understanding is that a down converter works the same way. The outgoing downconverted photons will have half the wavelength but will preserve the phase of the original photon. Look at figure 6 and you see the path length V1+S1 is equal to the pathlength V2+S2. That is all you need to preserve the phase relation between the two paths.
It is not that simple: V1 has a higher frequency and so "counts off" its phase (classically speaking) much faster than S1 which has a lower frequency. So depending on exactly WHERE in NL1 the conversion takes place, the phase of S1 will change. And I take it that NL1 is much thicker than the wavelength of S1 and V1 (or their difference) so that according to the conversion point, you can have several cycles through 360 degrees.

 It makes no difference where in the crystal the down conversion happens because you have the same total path length.
It does because you change the frequency (or better, the wavenumber). The wavenumber is the number of 1 radian phase changes per unit length. So if you change the wave number at different points, over a fixed distance interval, you'll change the total accumulated phase.
All this is of course "classical speak", but relates to the phase relationship between the two downconverted pairs.

 But this gives me the clue I need to maybe fix it. Look at fig 6 and the first thing that bothered me (And I gather it bothered you as well.) is the way I1 passes through NL2. It seemed clumsy. But think about it, that is the only place you could use I1 to prevent a two photon state from ever forming in the first place! Remember I1 is created before I2 so you put I1 at that point in the correct phase and I2 is never a distinguishable state. You are protected from FTL signaling by the fact that I1 must combine with I2 at its point of creation in NL2. Thus the fastest you can send a signal to Ds is the light travel time between NL2 and Ds.
That's entirely correct. And why is light speed signalling allowed for, and not FTL ? Because at light speed you cannot exclude INTERACTIONS. And that's what I claim: you have an interaction in NL2 between i1 and the others.

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)

cheers,
Patrick.

I think we have major confusion here.

 Quote by ME It makes no difference where in the crystal the down conversion happens because you have the same total path length.
 Quote by YOU It does because you change the frequency (or better, the wavenumber). The wavenumber is the number of 1 radian phase changes per unit length. So if you change the wave number at different points, over a fixed distance interval, you'll change the total accumulated phase. All this is of course "classical speak", but relates to the phase relationship between the two downconverted pairs.
First you only have one downconversion event. It is in superposition of haveing happened in NL1 and NL2. Now all you have to do is send I1 to NL2 with the exact same beam length as V2 to NL2 You have the ghost of a photon in V2 and a ghost of a photon in I1. When you combine these two paths you can no longer say which downconverter the downconversion happened in. In fact this isn't just ignorance but rather the basic nature of reality is that the event cannot be said to have happened in either.

Once you combine these two paths you have a one photon state and so an interference pattern at Ds.

 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 there isn't an interaction in any classical sense. You don't have to refere to the physics of the downconverters to find a mechanism to phaselock them. You just have to join two photon ghosts to form one real photon. To do that all you have to do is put them at the right place at the right time and going in the right direction. It turns out that when you do this you cannot send signals FTL.

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 Quote by ppnl2 First you only have one downconversion event. It is in superposition of haveing happened in NL1 and NL2. Now all you have to do is send I1 to NL2 with the exact same beam length as V2 to NL2 You have the ghost of a photon in V2 and a ghost of a photon in I1. When you combine these two paths you can no longer say which downconverter the downconversion happened in. In fact this isn't just ignorance but rather the basic nature of reality is that the event cannot be said to have happened in either.
I agree with you here about that the downconversion happened in the two NL1 and NL2 at once, in the same way as a photon in a 2-slit experiment went through both slits at once. But that's not what I wanted to say. If you calculate the interference in a 2-slit experiment, you accumulate the phases for each possible path (slit 1 or slit 2), and it is the difference of both accumulated phases, for a certain point on the screen, which makes you have constructive or destructive interference.
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.

 But there isn't an interaction in any classical sense. You don't have to refere to the physics of the downconverters to find a mechanism to phaselock them. You just have to join two photon ghosts to form one real photon. To do that all you have to do is put them at the right place at the right time and going in the right direction. It turns out that when you do this you cannot send signals FTL.
As I said, FTL is not really what is shown in QM. What is shown in QM is that system 1 (the signal photon ghosts) and system 2 (the idler photon ghosts) have INDEPENDENT expectation values for observables which only depend upon one of the systems, if there is no interaction hamiltonian.
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.

cheers,
Patrick.

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 Quote by vanesch This was also my first idea, but it cannot simply be partial reflection, given that the frequencies of idlers and signals are different,
It seems that, in the case of the used Lithium Iodate PDC's, the idlers
and signals have the same frequency (702 nm) according to this:
(see page 3)

http://arxiv.org/PS_cache/quant-ph/pdf/0310/0310020.pdf

It also states here that both photons are emitted simultaneous (within
femto seconds)

 Quote by vanesch BTW, I also agree fully with you that the author makes quite some errors by considering the s1 and s2 photon states as "independently generated" by a "spontaneous process" : in that case, in no way ever there could be any interference. They are produced in superposition (by the pump photon), but, as you also point out, with random phase relation due to different points of conversion. cheers, Patrick.
When both photons have identical frequencies then stimulated emission
(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.

Regards, Hans

 Quote by vanesch I agree with you here about that the downconversion happened in the two NL1 and NL2 at once, in the same way as a photon in a 2-slit experiment went through both slits at once. Patrick.
You are still talking as if there were two downconversion events at different depths and so out of phase with each other. There was only one event and it by definition can't be out of phase with itself.

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?

 Quote by Hans de Vries 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. Regards, Hans
Do the excited states last long enough to be released by the I1 photons? In any case something is inverted here. When we have a photon in I1 we should NOT have a down conversion event in NL2. But if I1 photons are stimulating the decay of the excited states then an I1 photon means we DO have a downconversion event in NL2.

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.

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 Quote by ppnl2 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.
You'll loose the interference because the EM components can not cancel each
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
emission.

Often it's not properly understood why the PDC's do what they do.

Regards, Hans.

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 Quote by ppnl2 You are still talking as if there were two downconversion events at different depths and so out of phase with each other. There was only one event and it by definition can't be out of phase with itself.
I agree with you that we have one 2-photon system (signal and idler), which happens to be in a superposition of a downconversion in NL1 and in NL2. But I don't see why I cannot first analyse my "first path" completely, and then my "second path" completely. After all, in a 2-slit experiment, one analyses the first path (slit) completely, giving rise to a certain total phase at the point of arrival, then one analyses the second slit, and in the end one adds together the two unit vectors (if the slits are inifinitely small).

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.

 The total phase contribution is the sum over ALL paths (so all paths over NL1 and over NL2)
Exactly. That's what I tried to explain: there are so many paths possible, so that for each path through x in NL1, you will find paths y in NL2 that are constructively interfering, and paths y' in NL2 that are destructively interfering. Another point on the screen will give you, for the same x, about the same amount of constructive and destructive interfering paths y and y', even though the exact paths doing so in NL2 have shifted. But overall, for each point on the screen, you get about the same amount of constructive and destructive interference, which means: flat image and no interference pattern.

 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 happns 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.
No, you don't NORMALIZE of course !!

 Surely there cannot be disagreement over whither or not a coherent beam into a down converter produces a coherent beam out.
Careful. A downconverter is a strange thing! For instance, each outgoing angle has a different wavelength, so you can never get 2-slit interference from the rainbow coming out of a downconverter, for instance.

cheers,
Patrick.