Has FTL communication really never been tested in this way?

[Andreas Most]

scerir wrote:

> And imagine we cannot see any interference pattern
> on that screen. (Apparently this is very strange
> because we have a two-slit interferometer and a beam
> of photons, and we do not see any interference
> pattern).
>
> What is the reason?
>
> The possible reason seems (to me) this one. Signal
> photons cannot cause their interference pattern on
> the screen because their momentum uncertainty is large.
> And their momentum uncertainty is large because the
> source of entangled photons we (must) use to perform
> a two-photon interference experiments has a 'large'
> size (divergence of the beam).

This has nothing to do with our poor set up of the experiment.
Imagine a quantum eraser experiment with two slits where the
left slit is covered with a horizontal polarization filter and
the right slit is covered with a vertical one. If you shoot linearly
polarized photons (at an angle of 45°) through it there will be no
interference pattern on the screen because left and right linearly
polarized electromagnetic waves do not interfere.
Now, If you chose to measure circular polarization on the idler photon
and select those events where the idler photon has e.g. right circular
polarization then you would see the interference pattern for the
selected signaling photons on the screen.
(By choosing the left circularly polarized idler photons you would
see a shifted interference pattern)

You could argue now that it is possible to use circularly polarized
photons from the very beginning in which case you would see an
interference pattern on the screen without the need of any
coincidence unit. The question is then, what is the use of the idler
photon if not to decide on the type of measurement. And, does the type
of measurement actually change the interference pattern?
The answer by QM and experiment is definitely: No.

Andreas.

 

[scerir]

Andreas wote:

> > Signal photons cannot cause their interference pattern
> > on the screen because their momentum uncertainty is large.
> > And their momentum uncertainty is large because the
> > source of entangled photons we (must) use to perform
> > a two-photon interference experiments has a 'large'
> > size (divergence of the beam).

> This has nothing to do with our poor set up of the
> experiment.

Since we were talking about the possibility of 'signaling',
at a distance, using a two-photon interference set-up,
having removed the coincidence detection unit, I was only
pointing out that the usual set-up, i.e. the usual SPDC
source itself, might not allow any single-photon interference
pattern, at the signal wing, for essential reasons (divergence
of the beam). More below.

> Imagine a quantum eraser experiment with two slits where the
> left slit is covered with a horizontal polarization filter and
> the right slit is covered with a vertical one. If you shoot linearly
> polarized photons (at an angle of 45°) through it there will be no
> interference pattern on the screen because left and right linearly
> polarized electromagnetic waves do not interfere.
> Now, If you chose to measure circular polarization on the idler photon
> and select those events where the idler photon has e.g. right circular
> polarization then you would see the interference pattern for the
> selected signaling photons on the screen.
> (By choosing the left circularly polarized idler photons you would
> see a shifted interference pattern).

Yes, I know these interesting experiments.
http://www.arxiv.org/abs/quant-ph/0106078
http://icpr.snu.ac.kr/resource/wop.pdf/J01/1998/033/R04/J011998033R040383.pd
f

> You could argue now that it is possible to use circularly polarized
> photons from the very beginning in which case you would see an
> interference pattern on the screen without the need of any
> coincidence unit. The question is then, what is the use of the idler
> photon if not to decide on the type of measurement. And, does the type
> of measurement actually change the interference pattern?
> The answer by QM and experiment is definitely: No.

Again, an important distinction is to be made.
Two-photon interference and one-photon interference
are obviously different phenomena. In the first case
you need a coincidence detection unit of some sort
(two clocks at least). In the second you do not need
any coincidence device.

It seems to me (I may be wrong of course) that
these position/momentum correlated photons
'signaling' machines are based on a sort of ... fusion :-)
of the one-photon and the two-photon interference
phenomena (you perform a specific measurement on the idler
photons and, at a distance, without checking the coincidences,
an interference pattern would appear, or disappear,
at the signal wing).

Now it is known, since long time, there is a weird
'complementarity' principle between the one-photon
and the two-photon interference. In the sense that
the more you can see the first interference, the less
you can see the second interference, and viceversa.

See, i.e., these papers:

M.A.Horne, A.Shimony, A.Zeilinger, 'Two-Particle Interferometry',
Phys.Rev.Lett. 62, 2209 (1989).

M.A.Horne, A.Shimony, A.Zeilinger,
'Two-Particle Interferometry', Nature, 347, 429 (1990).

D.M. Greenberger, M.A. Horne and A. Zeilinger,
'Multiparticle Interferometry and the Superposition Principle',
Physics Today 46 8, (1993).

and these specific experiments ...
http://www.arxiv.org/abs/quant-ph/0112065
http://josab.osa.org/abstract.cfm?id=35389

Since the 'complementarity' principles, in general,
presuppose a 'smooth' transition from the visibility
of a phenomenon to the visibility of the other,
here we can also expect (imo) a smooth transition
from the visibility of a single-photon interference
to the visibility of a two-photon interference,
and viceversa. If there is an intermediate situation
in which both interferences are (badly) visible,
and if - in this intermediate situation - it is possible
to perform 'signaling' (in principle) experiments,
I cannot say.

Summing up. I think WE CAN AGREE that in the two-photon
interference we need a coincidence detection unit,
and in the usual single-photon interference we do not
need such a device. I am pointing out that it is not
just about the use of the coincidence unit, or the use
of specific detectors. There is much more physics beyond,
there are many essential principles involved here.

Regards,
s.

 

[PostReplies]

On Wed, 14 Nov 2007 19:14:10 +0000 (UTC), "scerir" <scerir@libero.it>
wrote:
>
>The possible reason seems (to me) this one. Signal
>photons cannot cause their interference pattern on
>the screen because their momentum uncertainty is large.
>And their momentum uncertainty is large because the
>source of entangled photons we (must) use to perform
>a two-photon interference experiments has a 'large'
>size (divergence of the beam).
>

Reading all the posts I still don't see why there wouldn't be an
interference pattern. What causes the pattern is each photon
interferring with itself. Since each photon that registers on the back
of the apparatus had to go through a slit in order to even register,
and since we don't know which slit, then according to QM it seemingly
goes through both slits and interferes with itself. Isn't that the
crux of the dual slit experiment? That the source of the photons just
so happened to generate another entangled beam of photons I don't see
how that affects the dual slit experiment.

As far as I can see, polarization doesn't matter in Cramer's
experiment since he's only using the particle vs wave
complimentariness in the experiment--will a particle-like pattern show
up or an interference patttern.

Also, I don't think his experiment has anything to do with quantum
erasers. If you make a which-slit detection and then send the photon
through another 'which path did it take?' choice then you lose the
intereference pattern--not too surprising given what is known about
how QM works. But I don't see what that has to do with Cramer's
experiment. Neither do I see what any kind of delayed choice
experiment has to do with it since the experiment can be performed
without that.

I expect that the sender can at will turn on and off the interference
pattern on his side in the same way as the classical dual slit
experiment. Apparently, that will have no effect on the receiver side
since that would violate Relativity. Perhaps once the results of the
exxperiment are published it will be more understandable to me about
why it failed.

 

[Andreas Most]

PostReplies wrote:

> Reading all the posts I still don't see why there wouldn't be an
> interference pattern. What causes the pattern is each photon
> interferring with itself. Since each photon that registers on the back
> of the apparatus had to go through a slit in order to even register,
> and since we don't know which slit, then according to QM it seemingly
> goes through both slits and interferes with itself.

That is not quite true. It doesn't matter whether we know or we do not
know which slit the photon has taken. If we could possibly know which
path the photon has taken there will be no interference pattern.
That is a subtle but important difference and it is actually the reason
why the coincidence unit in the quantum eraser experiment is mandatory.

> Isn't that the
> crux of the dual slit experiment? That the source of the photons just
> so happened to generate another entangled beam of photons I don't see
> how that affects the dual slit experiment.
>
> As far as I can see, polarization doesn't matter in Cramer's
> experiment since he's only using the particle vs wave
> complimentariness in the experiment--will a particle-like pattern show
> up or an interference patttern.

"Polarization" is here a tool to tag the photons path. There are other
ways of doing that but polarization is probably the easiest to accomplish.

> Also, I don't think his experiment has anything to do with quantum
> erasers. If you make a which-slit detection and then send the photon
> through another 'which path did it take?' choice then you lose the
> intereference pattern--not too surprising given what is known about
> how QM works. But I don't see what that has to do with Cramer's
> experiment. Neither do I see what any kind of delayed choice
> experiment has to do with it since the experiment can be performed
> without that.

IMHO the quantum eraser experiment exhibits all the features to
understand what is going on in a delayed choice experiment and why
Cramers experiment will fail. Actually, what Cramer sets up is a
quantum eraser without a coincidence unit. Since the outcome is
already known it is clear that Cramer will fail.

> I expect that the sender can at will turn on and off the interference
> pattern on his side in the same way as the classical dual slit
> experiment. Apparently, that will have no effect on the receiver side
> since that would violate Relativity. Perhaps once the results of the
> exxperiment are published it will be more understandable to me about
> why it failed.

Andreas.

 

[Andrew Ray]

In post 23, Ben Rudiak-Gould said

... if you replace the detectors with photographic plates, there will
be no interference pattern on either plate.

Putting aside the question of FTL communication and Cramer's experiment, this appears to be an important part of the argument that coincidence detection is an essential part of Dopfer's experiment (or, at least, a show-stopping consequence of not having it).

Ben, by "either plate", do you mean 'a plate at either position of D1' (i.e. positions f and 2f beyond the Heisenberg lens), or are you saying that furthermore, there will be no fringes recorded at D2 (i.e. on the far side of the slits), regardless of where the film on the lens leg is? If the former, then do you predict there will always be fringes on the plate at D2?