Mandel & Faster than Light Communication ?

In summary, Mandel et Al's experiment shows that interference patterns at the signal detector only appear or disappear when the idler beams are allowed to interfere. The time gap between blocking the idler beam and noticing the signal interference pattern disappearing is equal to the direct distance between the blocking point and the signal detector. This is because the effect of blocking the idler beam travels at the speed of light and cannot exceed this speed. Bell's non-local universe concept suggests that entanglement may cause FTL communication, but this communication cannot be used for sending information.
  • #36
I don't understand.

We're spending billions of dollars on super powerful particle accelerators, to try and understand the fundamental make up of our quantum mechanical particles.

And yet, the problem described here is an even more fundamental gap in our understanding, with the prize of a significantly deeper understanding of quantum mechanics, far beyond what we know today, and no-one is interested in discovering what is in front of our noses ?

Why ?


ShalomShlomo
 
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  • #37
ShalomShlomo said:
How can we experimentally measure the sought time delay between the blocking of ‎the Idler 1 beam, and the corresponding disappearance of the Signal interference ‎pattern (the cause/effect time delay) ?‎

...

Who will be the first to carry out this or a similar experiment, which is crying-out to be completed ‎‎? Has it been carried out already ?‎

ShalomShlomo

When there is a serious suggestion that there is a superluminal signal device involved, you can bet it will be tested. This particular experimental setup precludes such result, since the idler travels (more or less) "alongside" the signal photon during much of the journey - at least that is my interpretation. :smile: QM itself does not predict any FTL signalling possibilities, at least at this time.

There is in fact a lot of experimentation going on in the area of parametric down conversion and entangled photons. Desktop versions of EPR tests are appearing in undergraduate labs as the price is dropping (currently under $50,000). See this slide presentation by Mark Beck (Whitman College) showing about 12 variations on these tests. I think it is safe to say that a lot more permutations will be tested in the near future. In fact, I suspect there is a lot of stuff already in the preprint archives that is up this alley. If only I weren't so lazy...
 
  • #38
ShalomShlomo said:
And yet, the problem described here is an even more fundamental gap in our understanding, with the prize of a significantly deeper understanding of quantum mechanics, far beyond what we know today, and no-one is interested in discovering what is in front of our noses ?

Except for the specific experiment of which I'm convinced the explanation in the paper is not correct, and which must have something to do with the properties of non-linear optical materials, these experiments don't add much to our understanding of quantum mechanics per se: they just ILLUSTRATE where quantum mechanics is at work. Most of these experiments have very clear predictions in quantum theory and are also confirmed, so they only indicate that quantum mechanics is ALSO valid in those particular cases.
What they learn us is where we already DON'T have to look for deviations from quantum mechanics (although the strange predictions of quantum mechanics would make us think that this "cannot be"). They only add to our understanding by EXCLUDING possible semiclassical explanations.

Most of these experiments are confirming extremely simple quantum mechanical models which can be worked out on 2 or 3 pages, as a simple exercise in quantum theory. (the i1 through NL2 xtal experiment is an exception to it, that's also why I think that some more complicated physics must be going on there). There IS a fundamental issue to solve in QM, but these experiments only show us that they are NOT exploring it, and that QM is still "alive and kicking" at the level where they explore nature.

So although it is always good to do experiments, and to check them against QM predictions, especially when they are "spectacular", they don't learn us much in fact. On the other hand, big accerator experiments try to explore NEW stuff, of which we are NOT sure that we know what is going on (unfortunately, for about 20 years, they all agree with the predictions of the standard model in as far we can make predictions). Hopefully LHC will show unexpected things ; in any case it WILL learn us something (supersymmetry, or not, Higgs, or not).

cheers,
Patrick.
 
  • #39
I guess the dispute between us is - Is there anything in this experiment beyond our current understanding of Quantum Mechanics ?

I say yes, a major hole in our knowledge is staring us in the face, and we are fools to ignore it and not grasp the opportunity.

You both say no, the experiment is indeed described by our current knowledge of Quantum Mechanics.

History will show who was right (If there is only one history :!) ).

ShalomShlomo
 
  • #40
Theres not much else to do is there. For us to answer your question of what time it takes for interference to disappear we must consult QM and let it give us the answer and than you want to use that answer to figure out if QM is wrong or not ? Seems to be a kind of logical circle don't you think ? It would be better to search for experiments that show that there is FTL communication.
 
  • #42
ShalomShlomo said:
History will show who was right (If there is only one history :!) ).

ShalomShlomo

:rofl:

Anyway, entangled photons via PDC give us a great opportunity to probe reality. I am sure there are exciting experiments yet to come, perhaps som with unexpected results!
 
  • #43
vanesch said:
Except for the specific experiment of which I'm convinced the explanation in the paper is not correct, and which must have something to do with the properties of non-linear optical materials...
Patrick.


Why? Is it violating some principle of QM?

Ok what if we change the experiment a little. What if we made the thing symmetric across V1 and V2 so that we are generating one interference pattern with S1 and S2 and another interference pattern with I1 and I2. This should be possible as long as the path lengths are equal. Now if we block I1 it will stop interference at Di. But it should also stop interference at Ds. But I1 never goes through NL2 here.

I would really like to get an understanding of this experiment nailed. I have been trying since I first saw it in scientific american years ago.
 
  • #44
ppnl2 said:
Why? Is it violating some principle of QM?

Not if something is changed in NL1. But yes if not, for the reason I will line out after your next comment.

Ok what if we change the experiment a little. What if we made the thing symmetric across V1 and V2 so that we are generating one interference pattern with S1 and S2 and another interference pattern with I1 and I2. This should be possible as long as the path lengths are equal. Now if we block I1 it will stop interference at Di. But it should also stop interference at Ds. But I1 never goes through NL2 here.

By blocking i1 somewhere, you should, according to QM, not be able to change the result of s1 and s2 (except if i1 DID something, like interacting in NL2 or so of course).
The reason in QM is the following: if rho is the global density matrix, and rho1 is Tr_2(rho) (the density matrix, traced over the space H2 which corresponds to the i1 and i2 degrees of freedom), then ALL observables A1 which are local to H1 (the s1 and s2 degrees of freedom) have the following expectation values: <A1> = Tr(A1 rho1).
In the same way, observables which are local to H2 (only measure things about i1 and i2) are given by <A2> = Tr(A2 rho2).
That means that, no matter what measurement we do on i1 and i2, this should not influence any outcome of measurement (expectation value of operator) at s1 and s2. Only CORRELATIONS can see influences (because these are expectation values of observables that are NOT local to H1 and H2 respectively). Now, Interference, or not, of s1 and s2 is clearly a local observable of H1, and should hence be described by Tr(A1 rho1), IRRESPECTIVE of what is measured at i1 and i2, because (in Copenhagen view) applying a weighted projection in just any basis (any measurement) in H2 does not change the trace over H2, and hence leaves rho1 invariant.
This fundamental property of QM makes that superluminal information transfer is not possible in principle (though Bell violating *correlations* are possible).

Now, if your setup (or the explanation in the paper) were correct, then you could use this setup to do some FTL signalling of course, by deciding, or not, to block i1 (very late after emission).
That's why I think that it is not possible (and I'm SURE it is not possible within quantum theory). It might of course, in which case we found a deviation from QM. But Occam's rasor points us first to consider that i1, traversing NL2, simply DOES something in NL2, like locking in s2.

cheers,
Patrick.
 
  • #45
vanesch said:
By blocking i1 somewhere, you should, according to QM, not be able to change the result of s1 and s2 (except if i1 DID something, like interacting in NL2 or so of course).
The reason in QM is the following: if rho is the global density matrix, and rho1 is Tr_2(rho) (the density matrix, traced over the space H2 which corresponds to the i1 and i2 degrees of freedom), then ALL observables A1 which are local to H1 (the s1 and s2 degrees of freedom) have the following expectation values: <A1> = Tr(A1 rho1).
In the same way, observables which are local to H2 (only measure things about i1 and i2) are given by <A2> = Tr(A2 rho2).
That means that, no matter what measurement we do on i1 and i2, this should not influence any outcome of measurement (expectation value of operator) at s1 and s2. Only CORRELATIONS can see influences (because these are expectation values of observables that are NOT local to H1 and H2 respectively). Now, Interference, or not, of s1 and s2 is clearly a local observable of H1, and should hence be described by Tr(A1 rho1), IRRESPECTIVE of what is measured at i1 and i2, because (in Copenhagen view) applying a weighted projection in just any basis (any measurement) in H2 does not change the trace over H2, and hence leaves rho1 invariant.
This fundamental property of QM makes that superluminal information transfer is not possible in principle (though Bell violating *correlations* are possible).

Ok this is mostly over my head. I don't really understand what a density matrix is. But from what very little grasp I have of it I'm suprised its applicable here. I'm stuck trying to visualize the unitary evolution of a single photon as it goes through the experiment. From this perspective if you have "which way" information then you have no interference pattern. If you combine I1 and I2 then you don't have that information and should have interference. As you have pointed out this may not be takeing into accout the physics of the down converters.

As you can see my understanding of QM is less than adequate.



vanesch said:
Now, if your setup (or the explanation in the paper) were correct, then you could use this setup to do some FTL signalling of course, by deciding, or not, to block i1 (very late after emission).
That's why I think that it is not possible (and I'm SURE it is not possible within quantum theory). It might of course, in which case we found a deviation from QM. But Occam's rasor points us first to consider that i1, traversing NL2, simply DOES something in NL2, like locking in s2.

cheers,
Patrick.

As you say my setup would allow ftl signaling. A real bad thing that I will have to think about.

(Hmm, my setup is a two photon state. But in the real experiment when you combine I1 and I2 you are reduceing back down to a one photon state. Aren't you? Wouldn't this make a difference?)

But in the real setup there is no place that you could block I1 that would make the signal FTL. You may be able to change it in some way to do it but it isn't clear how. You could use mirrors to route I1 around NL2 but then you would need to make I1 and I2 colinear again and in phase.

I understand that it has been proved that you cannot use QM to signal FTL but has it been proved that you cannot use QM to signal at light speed in some similar way?
 
  • #46
ppnl2 said:
Ok this is mostly over my head. I don't really understand what a density matrix is. But from what very little grasp I have of it I'm suprised its applicable here.

The reason is simply that a density matrix describes a statistical mixture, which can also be a pure state of course. Now, when you do a measurement, then you transform a density matrix of the situation before measurement (written in the eigenbasis of the observable of which we are performing the measurement) into a density matrix with the same elements on the diagonal, and 0 elsewhere. That's the Born rule in "density matrix" language.

For a pure state, you can understand this. Imagine that we have an initial pure state given by |psi> = a |e1> + b|e2> and we are going to measure the E quantity with eigenvectors |e1> and |e2>.
The density matrix corresponding to psi is simply
|psi><psi| = (a |e1> + b |e2>) (<e1| a* + <e2|b*) which gives the 4 terms corresponding to the 4 elements of the density matrix:
{ a a* ; a b*}
{ b a* ; b b*}

After measurement, we know that we have a MIXTURE, given by:
probability a a* to be in state |e1> and probability b b* to be in state |e2>, so the expectation value of A is aa* e1 + bb* e2.

This state is represented by a density matrix
{ a a* ; 0 }
{ 0 ; b b*}

So you see, keeping the diagonal elements and putting 0 everywhere else is the effect of a measurement (the Born rule).
In this basis, of course, it is simple to see that A is a diagonal matrix with e1 and e2 on the diagonal, so that rho A equals:

{ a a* e1 ; a b* e2}
{ a* b e1 ; b b* e2}

and the trace of it, Tr(rho A) = aa* e1 + bb* e2, the expectation value of A.

This seems trivial of course, because we were working in the right basis. But a property of the Trace is that it is invariant wrt a change of basis, so this expression is valid in EVERY basis !

Now, in the case of a composite system H1 x H2, the states can be written in a kronecker product basis (local basis). If an observable A only observes on system 1 then A can in fact be written as A x 1 (1 = unit operator in H2). In the same way, if B only observes a property of H2, it can be written as 1 x B (here, 1 is unit operator in H1).
Now, let us consider that the overall state |psi> = a |e1> |f1> + b |e2> |f2> where e1 and e2 are eigenvectors of A (but f1 and f2 are not basis vectors of H2, they just happen to be vectors in H2). Clearly |e1>|f1> is an eigenvector of A x 1 and so is |e2> |f2> (and both are orthogonal, even if f1 and f2 aren't). The probability of finding e1 is a a* and the probability of finding e2 is b b*.
As a density matrix, it can be written as:

{ a a* |f1><f1| ; a b* |f1><f2|}
{ b a* |f2><f1| ; b b* |f2><f2|}

where the |f><f| stand for submatrices written out in a basis of your choice in H2.

The subtrace of this matrix with respect to H2 is obtained by tracing out each of the 4 submatrices:

rho1 =
{a a* Tr(|f1><f1|) ; a b* Tr(|f1><f2|) }
{b a* Tr(|f2><f1|) ; b b* Tr(|f2><f2|) }

Now, it is a property of Tr(|a><b|) that it is equal to <b|a>.
Remember that f1 and f2 are not basis vectors of H2, they are just 2 arbitrary vectors in H2, so <f1|f2> is not necessarily 0. However, <f1|f1> = 1 because they are supposed to be normalized.

So rho1 =

{ a a* ; a b* x*}
{a* b x; b b*}

with x = <f1|f2>

and we see that our expectation value for A is given by:

<A x 1> = Tr( A rho1)

in the same way as before.

Now, the important part is that the basis in H2 in which we did this (worked out the trace) DIDN'T MATTER. Also, it wouldn't matter if we applied a measurement in H2 or not, which would have (as we saw before) transformed the submatrices |f1><f1| and so on into their DIAGONAL matrices in the basis corresponding to the measurement. Indeed, the only quantities needed to calculate a trace are the diagonal elements, and they don't change, with or without measurement.
So we can conclude 2 things:
1) the expectation value of A doesn't change, whether we apply, or not, a measurement in H2 (full submatrices, or only diagonal)
2) if we apply a measurement in H2 it doesn't matter which one (the eigenbasis in which we write out f1 and f2).

I'm stuck trying to visualize the unitary evolution of a single photon as it goes through the experiment. From this perspective if you have "which way" information then you have no interference pattern. If you combine I1 and I2 then you don't have that information and should have interference.
It is not because you don't have the information that you necessarily have interference ! It is if you USE that information to subselect a sample and then you are working out expectation values of a CORRELATION, which is not a local observable anymore acting only on H1 or H2.


(Hmm, my setup is a two photon state. But in the real experiment when you combine I1 and I2 you are reduceing back down to a one photon state. Aren't you? Wouldn't this make a difference?)

No, the 2-photon state is |signal> |idler>
and we have a SUPERPOSITION of 2-photon states:
|s1>|i1> + exp(i phi) |s2>|i2>
(this superposition is due to the pump photon being in a superposed state: one state going to NL1 and the other going to NL2)
So s1 and s2 are part of one and the same "photon" (in 2 different states).
The trick of combining the i1 and i2 beams into one single beam is to evolve unitarily |i1> and |i2> into the same state (in order to keep unitarity then this must in fact result into 2 different states but in such a way that they are indistinguishable for one detector: |i1> goes to |u> + |v> and |i2> goes to |u> - |v> and we detect |u> and |v> ; this is typically done with a beamsplitter).
By doing so, we rewrite the wave function as

(|s1> + exp(i phi) |s2> ) |u> + (|s1> - exp(i phi) |s2>) |v>

If we now look at |u> detections of the idler we find an interference pattern of the two different states of the "signal" photon, and if we look at the |v> detections, we find a complementary interference pattern.

But normally, WITHOUT subsampling (tagging on u or on v) an ENTANGLED state can never give rise to interference in the whole sample (that's the entire content of decoherence!).
However, if some physics in the NL2 converter makes that the |u> state (it is there if i1 gets into the converter) can LOWER the |s2> phase (that's possible, a unitary phase rotation), and the |v> state can INCREASE the |s2> phase, then both interference patterns are NOT perfectly complementary anymore, and the overall sample will show a partial interference pattern. But that is due to an interaction of the idler photon states (u and v) with the signal photon state s2. This is possible because the 3 states are PRESENT, LOCALLY, IN NL2. Mind you that it cannot do anything to s1 (which is far away, in NL1).

cheers,
Patrick.
 
  • #47
ShalomShlomo said:
I don't understand.

We're spending billions of dollars on super powerful particle accelerators, to try and understand the fundamental make up of our quantum mechanical particles.

And yet, the problem described here is an even more fundamental gap in our understanding, with the prize of a significantly deeper understanding of quantum mechanics, far beyond what we know today, and no-one is interested in discovering what is in front of our noses ?

Why ?

But it IS technically easy enough (and cheap!) to make timing measurements
to distinguish between mere Correlation and actual "Action on a Distance"

Analyzing detection times and actuator times afterwards should reveal
the propagation speeds you're after in the "Action on a Distance" case.
(Apart from the question if one can transmit information FTL) If one can
detect correlation afterwards then it is also possible to analyze the timing
of the correlated events.

The total absence of any such experimental timing results after all these
years points strongly towards Correlation and against "Action on a Distance"



Like Patrick, I don't agree with the author's conclusion on the experiment
depicted by figure 6. Partial reflection of i1 to the s2 path would simply
explain the results.

Worse, The author in fact assumes One photon interference with a single
path! First he excludes simultaneous down conversion by both PDC's as
"very improbable" He then excludes an i1 to s2 path and goes on with:

"one might not expect to see ONE-photon interference at Ds , but, ...,
interference fringes were actually observed"


Interference with a single path? Later then he talks about interference
again between s1 and s2 which he first ruled out. This is TWO photon
interference... from two different PDC's which are unlikely to be always
coherent with the same phase since the down conversion can take place
anywhere in the PDC resulting in phase differences.


Regards, Hans
 
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  • #48
Hans de Vries said:
Like Patrick, I don't agree with the author's conclusion on the experiment
depicted by figure 6. Partial reflection of i1 to the s2 path would simply
explain the results.

This was also my first idea, but it cannot simply be partial reflection, given that the frequencies of idlers and signals are different, and that there is a filter before the s1+s2 detector, filtering out any idler contribution. That's why I suspect a frequency change i2 -> s1 which is not excluded, because for that you need a non-linear interaction, and NL2 IS a nonlinear material. Moreover, the interference only appears when i2 is EXACTLY aligned in such a way, that the momentum and energy relations in the NL2 xtal are satisfied to allow for such a i2 -> s1 transition. But, as I said, I don't know enough of the detailled physics of these xtals to work out exactly what is going on.

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.
 
  • #49
vanesch said:
The reason is simply that a density matrix describes a statistical mixture, which can also be a pure state of course. Now, when you do a measurement, then you transform a density matrix of the situation before measurement (written in the eigenbasis of the observable of which we are performing the measurement) into a density matrix with the same elements on the diagonal, and 0 elsewhere. That's the Born rule in "density matrix" language.

...

Patrick.

Sorry, mostly over my head. I think I get the gest of it. But this stuff is just a hobby for me and I have always had trouble with the notation used by the people who know what they are talking about. Still, I'm going to disagree with you.

First, I have real problems believeing in your phase locking. I know less about the physics of down converters than you but still I just don't buy it.

Second, I think I have worked out a way for it to work as advertised and still protect us from FTL communication. Follow this and tell me where I'm going wrong.

We start with a single photon from the argon laser. This is split into V1 And V2. This is a one photon state, a superposition of (V1 OR V2). If we were to check now we could get a visible interference pattern with V1 and V2. Except that at this point we only have one photon.

Now we send V1 and V2 through NL1 and NL2 creating a two photon state that is in superposition ((S1 AND I1) OR (S2 AND I2)).

(Sorry, I understand boolean speak much better than that ket and bra matrix crap you guys throw around.)

Now at this point if we were to check for interference between S1 and S2 we would not see it because we have a two photon state. I assume you gave me a good explanation of why and I think I get the gest of it. Simply put it is as you say decoherence.

But that isn't the end of the story. We now combine I1 and I2 in phase creating the state ((S1 AND I2) OR (S2 AND I2)). The I2 photon is the same in both so it no longer plays a role in differentiating between the state. So we drop it leaveing us with the state (S1 OR S2). This is a SINGLE PHOTON STATE. We have interference again!

But we are protected from FTL signaling. To see why you have to see why the way we (or I anyway) have been thinking about it is wrong. Usually we think of the choice to measure one of the Idler beams causes the interference pattern to colapse. But this is exactly wrong and counter to the whole idea of decoherence. It is the choice to combine I1 and I2 thus reduceing us to a one photon state that creates the interference pattern.

And here is the key that prevents FTL signaling. The idler beams must be combined before the entangled photon in the signal beams hits the signal detector or we still have a two photon state. We must destroy entanglement before detection in order to see an interference pattern!

So what do you think?

:blushing:
 
  • #50
ppnl2 said:
We start with a single photon from the argon laser. This is split into V1 And V2. This is a one photon state, a superposition of (V1 OR V2). If we were to check now we could get a visible interference pattern with V1 and V2. Except that at this point we only have one photon.

Ok, that's the same as what I said.

Now we send V1 and V2 through NL1 and NL2 creating a two photon state that is in superposition ((S1 AND I1) OR (S2 AND I2)).

(Sorry, I understand boolean speak much better than that ket and bra matrix crap you guys throw around.)

Ok, apart from the strange boolean notation, we're on the same line...
except one point: there is a random phase shift between the first and the second pair, due to different optical lengths where the conversion v1->s1 and i1 on one hand, and v2 -> s2 and i2 on the other hand take place. These conversions can be anywhere in each of the NL xtals. This is important for what follows.

Now at this point if we were to check for interference between S1 and S2 we would not see it because we have a two photon state. I assume you gave me a good explanation of why and I think I get the gest of it. Simply put it is as you say decoherence.

Ok.

But that isn't the end of the story. We now combine I1 and I2 in phase creating the state ((S1 AND I2) OR (S2 AND I2)). The I2 photon is the same in both so it no longer plays a role in differentiating between the state. So we drop it leaveing us with the state (S1 OR S2). This is a SINGLE PHOTON STATE. We have interference again!

This is fundamentally impossible. You see, whatever you do to i1 and i2 must be a unitary transformation (any optical action, like a mirror, a lens, a beam splitter, a double slit etc... gives rise to a time evolution operator, derived from a hermitean hamiltonian).
Now, a unitary transformation CONSERVES IN PRODUCT. Clearly, <i1|i2> = 0 (they are completely different modes). So no matter what you do to i1 and i2, using a unitary transformation, you cannot have an U that transforms i1 and i2 into the same mode, because then <U i1 | U i2> would be 1.
The thing that comes closest is to transform (as I showed) i1 into something like u + v and i2 into something like u - v (in that way, <u+v|u-v> = 0) and only detect u, for instance. But that comes down to selecting a subset of the sample of s1 + s2.

But we are protected from FTL signaling. To see why you have to see why the way we (or I anyway) have been thinking about it is wrong. Usually we think of the choice to measure one of the Idler beams causes the interference pattern to colapse. But this is exactly wrong and counter to the whole idea of decoherence. It is the choice to combine I1 and I2 thus reduceing us to a one photon state that creates the interference pattern.

I agree with you that this collapse business is not correct and that one should consider unitary evolution all the way. So indeed, if you could "unify" i1 and i2 you would indeed undo the entanglement and in principle, allow for interference (if there wasn't also a random optical path length difference)...
Except that you can't do that, with a unitary transformation.
Another way to see it is that a unitary operator is reversible in principle. Now if i1 -> i2 and i2 -> i2, you cannot reverse that.

And here is the key that prevents FTL signaling. The idler beams must be combined before the entangled photon in the signal beams hits the signal detector or we still have a two photon state. We must destroy entanglement before detection in order to see an interference pattern!

Even if it were true, it would still allow for FTL signalling. The source could be 1 light year from the signal detector, and in the opposite direction, we could let the idlers travel, for say, 3/4 of a light year. You could combine them or not, according to your choice, and 3 months later, the detection would take place at the signal detector. But the decision was made at a distance of 1.75 light years, and the effect was seen only 3 months later ! That's FTL.
But as it isn't possible with unitary transformation, there is no such problem.

cheers,
Patrick.
 
  • #51
vanesch said:
Ok, apart from the strange boolean notation, we're on the same line...
except one point: there is a random phase shift between the first and the second pair, due to different optical lengths where the conversion v1->s1 and i1 on one hand, and v2 -> s2 and i2 on the other hand take place. These conversions can be anywhere in each of the NL xtals. This is important for what follows.

Ok let's handel this problem first. What you say here does not fit my limited understanding of how down converters work.

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.

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 makes no difference where in the crystal the down conversion happens because you have the same total path length.

Even if it were true, it would still allow for FTL signalling. The source could be 1 light year from the signal detector, and in the opposite direction, we could let the idlers travel, for say, 3/4 of a light year. You could combine them or not, according to your choice, and 3 months later, the detection would take place at the signal detector. But the decision was made at a distance of 1.75 light years, and the effect was seen only 3 months later ! That's FTL.
But as it isn't possible with unitary transformation, there is no such problem.

Yes I screwed that up rather badly. But I may be able to fix it.



This is fundamentally impossible. You see, whatever you do to i1 and i2 must be a unitary transformation (any optical action, like a mirror, a lens, a beam splitter, a double slit etc... gives rise to a time evolution operator, derived from a hermitean hamiltonian).
Now, a unitary transformation CONSERVES IN PRODUCT. Clearly, <i1|i2> = 0 (they are completely different modes). So no matter what you do to i1 and i2, using a unitary transformation, you cannot have an U that transforms i1 and i2 into the same mode, because then <U i1 | U i2> would be 1.
The thing that comes closest is to transform (as I showed) i1 into something like u + v and i2 into something like u - v (in that way, <u+v|u-v> = 0) and only detect u, for instance. But that comes down to selecting a subset of the sample of s1 + s2.

Ok, mostly over my head. But I do agree with you. I screwed up here but I may be able to fix it.

I agree with you that this collapse business is not correct and that one should consider unitary evolution all the way. So indeed, if you could "unify" i1 and i2 you would indeed undo the entanglement and in principle, allow for interference (if there wasn't also a random optical path length difference)...
Except that you can't do that, with a unitary transformation.
Another way to see it is that a unitary operator is reversible in principle. Now if i1 -> i2 and i2 -> i2, you cannot reverse that.

My intuition tells me that this should be unitary. But you are right it is irreversable so clearly it isn't unitary. I'm still trying to wrap my mind around that.

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.

How is that?
 
  • #52
ppnl2 said:
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 each other, 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 each other (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.
 
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  • #53
I think we have major confusion here.

ME said:
It makes no difference where in the crystal the down conversion happens because you have the same total path length.

YOU said:
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.
 
  • #54
ppnl2 said:
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.
 
  • #55
vanesch said:
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 [Broken]

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

vanesch said:
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
 
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  • #56
vanesch said:
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 let's 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
Hans de Vries said:
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.
 
  • #58
ppnl2 said:
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 :smile: 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.
 
  • #59
ppnl2 said:
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 let's 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.
 
<h2>1. What is the Mandel effect?</h2><p>The Mandel effect is a phenomenon in which a large group of people remember something differently than how it actually occurred. It is named after the creator of the concept, Fiona Broome, who noticed that a large number of people believed that Nelson Mandela died in prison in the 1980s, when in reality he was released in 1990 and passed away in 2013.</p><h2>2. What is faster than light communication?</h2><p>Faster than light communication, also known as superluminal communication, is the hypothetical ability to send information or messages faster than the speed of light. According to Einstein's theory of relativity, it is impossible for any object with mass to travel at or above the speed of light, making faster than light communication currently impossible.</p><h2>3. Is it possible to achieve faster than light communication?</h2><p>At this time, there is no scientific evidence or technology that supports the possibility of faster than light communication. The laws of physics, including the speed of light being the universal speed limit, make it highly unlikely that this type of communication will ever be achievable.</p><h2>4. How does quantum entanglement relate to faster than light communication?</h2><p>Quantum entanglement is a phenomenon in which two particles become connected in such a way that any change in one particle will affect the other, regardless of the distance between them. Some theories suggest that this could potentially be used for faster than light communication, but this has not been proven and remains a topic of debate among scientists.</p><h2>5. What are the potential implications of faster than light communication?</h2><p>If faster than light communication were to become possible, it could have significant implications for the way we communicate and share information. It could potentially lead to instantaneous communication across vast distances, but it could also raise ethical concerns and disrupt our understanding of the laws of physics.</p>

1. What is the Mandel effect?

The Mandel effect is a phenomenon in which a large group of people remember something differently than how it actually occurred. It is named after the creator of the concept, Fiona Broome, who noticed that a large number of people believed that Nelson Mandela died in prison in the 1980s, when in reality he was released in 1990 and passed away in 2013.

2. What is faster than light communication?

Faster than light communication, also known as superluminal communication, is the hypothetical ability to send information or messages faster than the speed of light. According to Einstein's theory of relativity, it is impossible for any object with mass to travel at or above the speed of light, making faster than light communication currently impossible.

3. Is it possible to achieve faster than light communication?

At this time, there is no scientific evidence or technology that supports the possibility of faster than light communication. The laws of physics, including the speed of light being the universal speed limit, make it highly unlikely that this type of communication will ever be achievable.

4. How does quantum entanglement relate to faster than light communication?

Quantum entanglement is a phenomenon in which two particles become connected in such a way that any change in one particle will affect the other, regardless of the distance between them. Some theories suggest that this could potentially be used for faster than light communication, but this has not been proven and remains a topic of debate among scientists.

5. What are the potential implications of faster than light communication?

If faster than light communication were to become possible, it could have significant implications for the way we communicate and share information. It could potentially lead to instantaneous communication across vast distances, but it could also raise ethical concerns and disrupt our understanding of the laws of physics.

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