Mandel & Faster than Light Communication ?


by ShalomShlomo
Tags: communication, faster, light, mandel
Hans de Vries
Hans de Vries is offline
#55
Jun8-05, 09:41 PM
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Quote 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 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
ppnl2
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#56
Jun9-05, 08:52 AM
P: 28
Quote 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?
ppnl2
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#57
Jun9-05, 09:07 AM
P: 28
Quote 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.
Hans de Vries
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Jun9-05, 12:33 PM
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Quote 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.
vanesch
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Jun9-05, 02:12 PM
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Quote 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.


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