Mandel & Faster than Light Communication ?by ShalomShlomo Tags: communication, faster, light, mandel 

#55
Jun805, 09:41 PM

Sci Advisor
P: 1,136

and signals have the same frequency (702 nm) according to this: (see page 3) http://arxiv.org/PS_cache/quantph/pdf/0310/0310020.pdf It also states here that both photons are emitted simultaneous (within femto seconds) (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 



#56
Jun905, 08:52 AM

P: 28

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? 



#57
Jun905, 09:07 AM

P: 28

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. 



#58
Jun905, 12:33 PM

Sci Advisor
P: 1,136

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 halflife 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. 



#59
Jun905, 02:12 PM

Emeritus
Sci Advisor
PF Gold
P: 6,238

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 xindependent 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. cheers, Patrick. 


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