Question about Other Tests of EPR Paradox

[DrChinese]

I have some questions about other possible tests of the EPR Paradox. The base of the paradox is using one entangled particle to gain information about the other. The usual setup discussed involves spin, often photon spin. But what about other measurable attributes of a particle?

If we take the same entangled photons and run them through 2 double slit setups on opposite sides of the source, they should presumably yield 2 interference patterns. Is this new setup clear? We are still observing particles in the singlet state, just not looking at the spin component.

Now, if we cover one of the slits on one side (say the Left), so we know which of the 2 slits the Left particle travels through, the interference pattern disappears on the left. Does this tell us anything about the particle on the Right? Is there some complimentary property that could collasped by such a measurement?

I am trying to imagine that the interference pattern seen on the Right would also disappear if one slit on the Left is covered. Have there been any actual experiments performed with this setup to see what would happen?

Thanks,

-DrChinese

 

[slyboy]

To get such effects the particles would have to be entangled in position-momentum space rather than in their spins. Experiments on continuous-variable entanglement have been performed, but I am no expert on them. Perhaps somebody else knows?

 

[ppnl2]

I have some questions about other possible tests of the EPR Paradox. The base of the paradox is using one entangled particle to gain information about the other. The usual setup discussed involves spin, often photon spin. But what about other measurable attributes of a particle?

If we take the same entangled photons and run them through 2 double slit setups on opposite sides of the source, they should presumably yield 2 interference patterns. Is this new setup clear? We are still observing particles in the singlet state, just not looking at the spin component.

Now, if we cover one of the slits on one side (say the Left), so we know which of the 2 slits the Left particle travels through, the interference pattern disappears on the left. Does this tell us anything about the particle on the Right? Is there some complimentary property that could collasped by such a measurement?

I am trying to imagine that the interference pattern seen on the Right would also disappear if one slit on the Left is covered. Have there been any actual experiments performed with this setup to see what would happen?

Thanks,

-DrChinese

That would imply that you could pass real information between two points dispite the fact that there is no real object or energy traveling between the two points.

As far as I know one of the experiments here do this:

http://www.fortunecity.com/emachines/e11/86/qphil.html

I'm referring to the experiment by Leonard Mandel. But I'm not a physicist and am not sure I'm understanding the thing correctly.

 

[DrChinese]

That would imply that you could pass real information between two points dispite the fact that there is no real object or energy traveling between the two points.

As far as I know one of the experiments here do this:

http://www.fortunecity.com/emachines/e11/86/qphil.html

I'm referring to the experiment by Leonard Mandel. But I'm not a physicist and am not sure I'm understanding the thing correctly.

Thanks, that is exactly in the spirit of what I am asking. I could not tell from this, but are the idler photons and the signal photons space-like separated? It seems that you could use this for FTL signalling if so. Packets of signal photons would either display intereference or not depending on the action taken at the idler photons - i.e. block or not block which would affect the interference pattern.

Anyone know about this? I'm sure Mandel must have thought of this but I did not see a specific in that source... (guess I should Google it and answer it myself :)

 

[vanesch]

Thanks, that is exactly in the spirit of what I am asking. I could not tell from this, but are the idler photons and the signal photons space-like separated? It seems that you could use this for FTL signalling if so. Packets of signal photons would either display intereference or not depending on the action taken at the idler photons - i.e. block or not block which would affect the interference pattern.

The interference pattern of the set of photons A occurs, or not, according to a subset selection which is determined by a coincidence with a test on photon B. By changing the kind of test on photon B, you get different kinds of interference, or not. What you usually have in this kind of experiments is 2 interference patterns in the entire sample of A, which smoothen each others' fringes out. When you look at the pattern of a subset of photons A (coincident with something done on B), you suddenly see (one or the other) interference pattern. But in order to find that subset, you need the classically transmitted coincidence signal. If you only look at the entire population of photons A, you always get the same picture, no matter what you do with B.
There's no way, in standard quantum theory, to do FTL information transport using entangled systems. This can be demonstrated easily using the density matrix, and "tracing out" the remote system: this "local" density matrix contains all the available information (it gives you all the expectation values of observables which act only on the local system) that you can get out locally, no matter what you do on the other system.

I discussed a similar setup that was meant to serve as an FTL phone here a while ago:

" simple ftl setup using Stern-Gerlach"

cheers,
Patrick.

 

[dlgoff]

... that is exactly in the spirit of what I am asking. ...

Maybe this paper selfAdjoint referenced in his post yesterday has some relavence here:

http://www.arxiv.org/abs/quant-ph/0501034

Regards

 

[ppnl2]

The interference pattern of the set of photons A occurs, or not, according to a subset selection which is determined by a coincidence with a test on photon B. By changing the kind of test on photon B, you get different kinds of interference, or not. What you usually have in this kind of experiments is 2 interference patterns in the entire sample of A, which smoothen each others' fringes out. When you look at the pattern of a subset of photons A (coincident with something done on B), you suddenly see (one or the other) interference pattern. But in order to find that subset, you need the classically transmitted coincidence signal. If you only look at the entire population of photons A, you always get the same picture, no matter what you do with B.
There's no way, in standard quantum theory, to do FTL information transport using entangled systems. This can be demonstrated easily using the density matrix, and "tracing out" the remote system: this "local" density matrix contains all the available information (it gives you all the expectation values of observables which act only on the local system) that you can get out locally, no matter what you do on the other system.

I discussed a similar setup that was meant to serve as an FTL phone here a while ago:

" simple ftl setup using Stern-Gerlach"

cheers,
Patrick.

I'm not sure we are talking about the same experiment. Most of the experiments do require a coincidence detector but as far as I can tell this one does not.

The signal detector has photons from two different paths hitting it and each photon is in a superposition of being in both paths. On this bases I would predict an interference pattern even without knowing about the rest of the setup. What prevents the interference pattern here? What are ther subsets of photons?

I agree that you cannot use it for FTL communication but my (very limited!) understanding is that you cannot arrange a space like separation between the blocker and the signal detector without some device (like a perfect one way mirror or something) forbiden in quantum mechanics. But I have never been able to work out the details or trace down a good explination of how this works. So I could easily be all wet.

This was published in SCIAM about ten years ago and I have been trying to get a handle on it ever since.

 

[Hans de Vries]

It seems that you could use this for FTL signalling if so. Packets of signal photons would either display intereference or not depending on the action taken at the idler photons - i.e. block or not block which would affect the interference pattern.

I was reading this much more detailed description of a class-room
table-top quantum eraser here thinking the same:

http://people.whitman.edu/~beckmk/QM/qe/qe.pdf

It switches an Interference Pattern on and off at one place
by manipulating a $\lambda/2$ Half-Wave plate at another place,
in another beam, that went the other way, to never to
go back to the place were the interference happens.

So, if indeed it works this way then It seems almost unavoidable
that what you suggest should occur. It's a simple 1:1 relation with
a defined binary input at one place causing a defined binary output
(yes or no interference pattern) at another place.


Regards, Hans

 

[DrChinese]

I discussed a similar setup that was meant to serve as an FTL phone here a while ago:

" simple ftl setup using Stern-Gerlach"

cheers,
Patrick.

Thanks, I found it at https://www.physicsforums.com/archive/topic/t-46015_simple_ftl_setup_using_Stern-Gerlach.html .

 

[DrChinese]

I was reading this much more detailed description of a class-room
table-top quantum eraser here thinking the same:

http://people.whitman.edu/~beckmk/QM/qe/qe.pdf

It switches an Interference Pattern on and off at one place
by manipulating a $\lambda/2$ Half-Wave plate at another place,
in another beam, that went the other way, to never to
go back to the place were the interference happens.

So, if indeed it works this way then It seems almost unavoidable
that what you suggest should occur. It's a simple 1:1 relation with
a defined binary input at one place causing a defined binary output
(yes or no interference pattern) at another place.


Regards, Hans

From the reference: "Measurements made on one beam affect the
visibility of interference fringes in another, spatially separated, beam."

If I am understanding what they are doing (and it really is doing what they say), then the setup could be used as a building block to send a FTL signal. Clearly, a single signal photon in this case does not transmit enough information when it arrives at a detector to be a full bit of information. But a suffiiently large packet of them - even if from different source apparati - would contain enough information to constitute a bit.

The part I don't understand is that FTL is not mentioned in the reference, and you would expect them to have considered this already and commented upon it.

 

[Caroline Thompson]

From the reference: "Measurements made on one beam affect the
visibility of interference fringes in another, spatially separated, beam."

Am I right in thinking that this refers to a 1999 experiment by Kim et al?

Yoon-Ho Kim, R Yu, S P Kulik, Y H Shih and Marlan O Scully, “A delayed choice quantum eraser”, quant-ph/9903047 (1999)​

If I am understanding what they are doing (and it really is doing what they say), then the setup could be used as a building block to send a FTL signal. Clearly, a single signal photon in this case does not transmit enough information when it arrives at a detector to be a full bit of information. But a suffiiently large packet of them - even if from different source apparati - would contain enough information to constitute a bit.

Sigh! It's amazing what paradoxes you can come up with if you insist on modelling light with photons! I maintain that experiments such as Kim's can all be explained once you recognise a few facts about the real, perfectly ordinary, correlations in the properties of the beams emitted simultaneously in the process of (degenerate) PDC. I have to admit that, despite some correspondence with Kim, I never quite got to the bottom of it -- I could not find out quite enough facts about, e.g. the orientation of the various bits of apparatus -- but I am confident that if I'd been there myself I'd have been able to give a complete explanation.
Caroline

 

[DrChinese]

Am I right in thinking that this refers to a 1999 experiment by Kim et al?

No, it is a different one: Gogo, Snyder and Beck, 2004. You might like this one, Caroline. Beck's interest is on tabletop experiments that can be done in an undergraduate setting. To me, that means a much lower cost to perform EPR tests. One thing that you and I agree on is that more tests should be done. I want to see all of the details of EPR tested! My questions are simply different than yours. So perhaps your ideas can be tested in the near future anyway if there are more labs doing this kind of research.

 

[Caroline Thompson]

No, it is a different one: Gogo, Snyder and Beck, 2004. You might like this one, Caroline. Beck's interest is on tabletop experiments that can be done in an undergraduate setting. To me, that means a much lower cost to perform EPR tests. One thing that you and I agree on is that more tests should be done. I want to see all of the details of EPR tested! My questions are simply different than yours. So perhaps your ideas can be tested in the near future anyway if there are more labs doing this kind of research.

I haven't seen this particular article, but please do, if you've a moment, look at a letter I wrote relating to what must have been a similar paper back in 2002:

http://freespace.virgin.net/ch.thompson1/Letters/AJP.htm

which concerns:

C H Holbrow, E Galvez and M E Parks, "Photon quantum mechanics and beam splitters", Am. J. Phys. 70 (3), 260-265 (2002)

I wrote to Holbrow (and had a little correspondence) with CC to the journal.

I want to see both the double slit and Bell test investigated, but not by students supplied with ready-made apparatus and instructions on how to obtain the "correct" results! It is all too easy to produce the appearance of agreement with QM if you only look at a very limited range of the experimental parameters.

What I should like to see is comprehensive sets of experiments, investigating what happens if you vary such things as the make of beamsplitter and the make of photodetector. You can also investigate the effect of different detector settings, but some of the characteristics of the apparatus will have been "programmed" into it in the course of manufacture and calibration.

Caroline
http://freespace.virgin.net/ch.thompson1/

 

[DrChinese]

I was reading this much more detailed description of a class-room
table-top quantum eraser here thinking the same:

http://people.whitman.edu/~beckmk/QM/qe/qe.pdf

It switches an Interference Pattern on and off at one place
by manipulating a $\lambda/2$ Half-Wave plate at another place,
in another beam, that went the other way, to never to
go back to the place were the interference happens.

So, if indeed it works this way then It seems almost unavoidable
that what you suggest should occur. It's a simple 1:1 relation with
a defined binary input at one place causing a defined binary output
(yes or no interference pattern) at another place.


Regards, Hans

I received some clarification on the experiment. I didn't see it at first, but the interference is not the kind where it appears a la a double slit experiment. The pattern appears on a graph of coincidences, so that makes it take on the characteristics of a Bell test.

You might be interested to learn that it violated a Bell Inequality by over 20 standard deviations. It does use 2 channel detectors so it has some similarity to the experiments Caroline denies. But to the rest of us, it is another setup which yields a value which miraculously just happens to match the QM predicton.

Most importantly, this experimental setup is designed for operation at the undergraduate level. This is a great breakthrough; I couldn't understand previously why Aspect-like tests were not being used to probe many different permutations of EPR and generally provide a flood of confirming tests.

Aspect had to use time varying analyzers to rule out the possibility of some kind of signal being sent from one part of the measuring apparatus to another. The results were the same with or without the time varying enhancement to the experiment. A reasonable person would conclude this does not need to be a feature of every Bell test. That simplifies the development of EPR probes.

The Gogo/Snyder/Beck setup, like other quantum erasers, uses parametric down converters (PDCs) to create the source beams. This too should make further tests cheaper, easier such that they should proliferate.

 

[vanesch]

The pattern appears on a graph of coincidences, so that makes it take on the characteristics of a Bell test.

EXACTLY ! Hence my remark that "quantum eraser" is a strange name for it. These experiments are indeed CORRELATION MEASUREMENTS, a la Aspect.
Only, what's very nice is that instead of a single number, you get out a whole pattern of correlations which take on the aspect (:tongue:) of an interference pattern. That's much more difficult to explain away as the result of inefficient detectors and so on. How the hell can you, by coincidence, get an entire pattern with wobbles, if the two photons were not entangled ?
Some work for Caroline

cheers,
Patrick.

 

[DrChinese]

EXACTLY ! Hence my remark that "quantum eraser" is a strange name for it. These experiments are indeed CORRELATION MEASUREMENTS, a la Aspect.
Only, what's very nice is that instead of a single number, you get out a whole pattern of correlations which take on the aspect (:tongue:) of an interference pattern. That's much more difficult to explain away as the result of inefficient detectors and so on. How the hell can you, by coincidence, get an entire pattern with wobbles, if the two photons were not entangled ?
Some work for Caroline

cheers,
Patrick.

We both know that it will be no difficulty for Caroline... just another wave of the hand .

I agree that "quantum eraser" is a strange name for it. It's snappy, but not very descriptive.

 

[Hans de Vries]

I received some clarification on the experiment. I didn't see it at first, but the interference is not the kind where it appears a la a double slit experiment. The pattern appears on a graph of coincidences, so that makes it take on the characteristics of a Bell test.


You might be interested to learn that it violated a Bell Inequality by over 20 standard deviations. It does use 2 channel detectors so it has some similarity to the experiments Caroline denies. But to the rest of us, it is another setup which yields a value which miraculously just happens to match the QM predicton.

Most importantly, this experimental setup is designed for operation at the undergraduate level. This is a great breakthrough; I couldn't understand previously why Aspect-like tests were not being used to probe many different permutations of EPR and generally provide a flood of confirming tests.

Aspect had to use time varying analyzers to rule out the possibility of some kind of signal being sent from one part of the measuring apparatus to another. The results were the same with or without the time varying enhancement to the experiment. A reasonable person would conclude this does not need to be a feature of every Bell test. That simplifies the development of EPR probes.

The Gogo/Snyder/Beck setup, like other quantum erasers, uses parametric down converters (PDCs) to create the source beams. This too should make further tests cheaper, easier such that they should proliferate.

It seems to be an ideal experiment in the sense that it has non of
the disadvantages of the usual EPR experiments:

- No math at all needed to convince, No dependence on the
interpretation of mathematical formula's. No distraction with
toy theories which seem arbitrary and possibly irrelevant.

- A very clear yes or no signal (yes or no interference) already at
this stage. No issues with efficiencies or accuracies.

- No unknowns with raw data or selected data.

- An undergraduate laboratoria version will result in wide spread
access (and thus verification) of the experiment.

Regards, Hans

http://people.whitman.edu/~beckmk/QM/qe/qe.pdf

 

[DrChinese]

(OK, this probably makes no sense whatsoever, but...)

The EPR Paradox applies to any quantum properties which are measured on one particle of an entangled pair. That should apply to a position measurement as well. If you learn something about the position of one particle, say the Right, haven't learned something about the position of the one on the Left?

In our attached figure, one photon goes Left and the other Right. We measure the path/position of the photon on the Right by means of a detector. That leads to a loss of interference per the standard Double Slit experiment, and no interference is seen on the Right screen. We now know the which-path information of the Left photon as well. Should that affect what we see on the Left screen, so that its expected interference pattern is also erased?

-DrChinese

 

[Caroline Thompson]

EXACTLY ! Hence my remark that "quantum eraser" is a strange name for it. These experiments are indeed CORRELATION MEASUREMENTS, a la Aspect.
Only, what's very nice is that instead of a single number, you get out a whole pattern of correlations which take on the aspect (:tongue:) of an interference pattern. That's much more difficult to explain away as the result of inefficient detectors and so on. How the hell can you, by coincidence, get an entire pattern with wobbles, if the two photons were not entangled ?
Some work for Caroline

cheers,
Patrick.

OK, I might take up the challenge! I've printed the paper out, had a quick read, and checked out the authors. It might be more worthwhile, though, to concentrate on some other similar experiment, since it does not look as if they intend submitting to a journal. If I'm going to analyse something like this in detail, I might as well produce a paper and try and publish.

Anyway, as I supected, it is the usual fun and games, where what is really going on is the use of one interference pattern to enable (by looking at coincidences) selection of another, the whole trick being shrouded in mysticism with stories of photons whose path you know, ones you could know if you tried, and extra information that washes out previous information!

There are a couple of basic facts that I challenge, in addition to the ludicrous way they word their "explanation". I strongly suspect that the true explanation will turn out to depend on the fact that they don't get either vertical or horizontal polarisation emitted from their pair of nonlinear crystals but (usually) both at once. [If they didn't, how would the two components interfere at the beamsplitters as they say they do?] The key "hidden variable" is likely to be, as in my "rotational invariance" paper (quant-ph/9912082), the phase difference between the two.

And those half wave plates don't just flip the polarisation direction, though sometimes this may be the net effect. If you look into how they work, you find that the name gives the game away: they delay one polarisation component by half a wave relative to the other. They only do this exactly, though, if the frequency of the light is exactly that for which the plate is designed. If the frequency can vary from one pulse to the next (as is apt to happen with PDC output) the delay will not always be exactly half a wavelength, with the effect that input polarisation at 45 deg is not exactly flipped to -45 deg. I don't know yet whether or not this will prove important here ...

Oh, and incidentally, they may have got the wrong idea about what polarisation directions they are really dealing with. They might be better off not assuming they know what effect half wave plates and the other apparatus has on polarisation direction and concentrating on their effects on relative phases of the two components.

That's a start. You may hear more from me on this in a week or two.

Caroline

 

[Hans de Vries]

Anyway, as I supected, it is the usual fun and games, where what is really going on is the use of one interference pattern to enable (by looking at coincidences) selection of another, the whole trick being shrouded in mysticism with stories of photons whose path you know, ones you could know if you tried, and extra information that washes out previous information!

Maybe you should not be put of by terms like "measurement",
"knowledge of paths" et-cetera. These terms are indeed vague
and tend to lead to philosophers metaphysics. Concentrate on
the physics.

A defining moment is the absorbtion of the photon. The only thing
we could measure in the past, causing the term "measurement"
to be associated with the absorbtion of the photon.

A bit further in the future we may be able to do much more sensitive
measurements. For instance with nano-antennas that measure the
properties of the EM field of single photons passing by non-destructive.

The optical paths of the entangled photons in the quantum eraser
experiment have dozens of stages with different optical properties
between the interferometer and the detector of the idler photons.

The photons push billions of atoms and electrons around in the
optical dielectra without loosing their properties, the entangle-
ment or the superposition. So it doesn't seem to far fetched that
we will be able to make measurements as well without disturbing
these properties.

Regards, Hans

 

[Hans de Vries]

I should add this to the above post:

I don't want to suggest that we might be able to make simultaneous
measurements of non-commuting physical quantities (location-
momentum et-cetera) with future non-destructive measurements.

On the contrary.

I would expect that these quantities are fundamentally not there
simultaneously at unlimited precision just because of the idea of
non-destructive measurements some were in the future.

Many scientist (Including the EPR people) think these quantities
are still there somehow simultaneously and that we only have a
measurement issue. (teleportation would copy both non-commuting
parameters exact and simultaneously)

I think that this is a sort of denial of Heisenberg's Uncertainty
Principle and that both quantities simply do not exist exact
simultaneously.

There's nothing magical here. If momentum is associated with a
wavelength then it follows from Fourier Analyses that it fundamentally
can't be determined with high precision by looking at it at a small
interval.

One can find Heisenberg's Uncertainty Principle in many all day
situations like in an audio-set's spectrum meter. It's not possible
to determine the frequencies in an arbitrary short amount of time.

There is mathematically no way of determining the frequencies
here at unlimited precision. It is not a measurement issue.

Regards, Hans

 

[vanesch]

OK, I might take up the challenge!

Well, after some reflection, I came to the conclusion that indeed, a big part (but not everything !) of the quantum eraser experiment can be explained by classical optics.

The part that can be explained is the following (and formally, follows extremely closely the quantum formalism, btw).

If you assume that the initial "pump beam" consists of wave train pulses (necessary to have simultaneous clicks by photodetectors: you need "bumps in amplitude"), which are split evenly and excite, at A and B, the generation of double wave trains ("idler" and "signal" wavetrains), in such a way, that there is no phase relation between what happens at A and at B, but that there is a strict phase relation between the idler and signal wavetrain generated at each spot, then, throughout the "mixer" setup, you will get constructive interference at D1 for the idler pairs which have a certain phase relation (hence fixing, for these cases, the phase relation between A and B generations), and you will have constructive interference at D2 for the opposite phase relation.
As such, the correlation between clicks at D1 should favor a certain interference pattern at D0 (because there's a phase relation between A and B now, we've subselected that), and we should find the 180 degree shifted phase relation with a correlation at D2.
Without correlation, the random phases between A and B wash out any interferences at D0, so without this subselection you ain't got any interference pattern.

So the main observation is indeed classically explainable.

However, ANTICOINCIDENCES (not published for THIS experiment, but very analogous to, say, Thorn's experiment) will not be explained:
for instance, there's a finite probability that D1 and D2 will click together, which can be calculated. Also for D3 and D4.
QM predicts that only ONE of D1, D2, D3 and D4 can click at a given instant.

cheers,
Patrick.

 

[Caroline Thompson]

Well, after some reflection, I came to the conclusion that indeed, a big part (but not everything !) of the quantum eraser experiment can be explained by classical optics.

The part that can be explained is the following (and formally, follows extremely closely the quantum formalism, btw) ...

Yes, this is almost always true. The exceptions are important though.

...So the main observation is indeed classically explainable.

However, ANTICOINCIDENCES (not published for THIS experiment, but very analogous to, say, Thorn's experiment) will not be explained:
for instance, there's a finite probability that D1 and D2 will click together, which can be calculated. Also for D3 and D4.
QM predicts that only ONE of D1, D2, D3 and D4 can click at a given instant.

I'm afraid we're talking about slightly different experiments, though I think the one you're discussing is essentially the Kim et al one that I have looked at:

Yoon-Ho Kim, R Yu, S P Kulik, Y H Shih and Marlan O Scully, “A delayed choice quantum eraser”, http://arxiv.org/abs/quant-ph/quant-ph/9903047

This has the two pairs of PDC outputs, but the one DrChinese mentioned and which I have now read is the Gogo et al group's experiment:

Ashifi Gogo, William Snyder and Mark Beck, “Quantum eraser using polarization entangled photons”, http://people.whitman.edu/~beckmk/QM/qe/qe.pdf

This is simpler, with only the one.

I think that to explain Kim's results, which are based on anticorrelations, you'll find you need my hypothesis that the phase differences between the A and B pairs are not random but restricted to one of two possibilities: 0 or 180 deg. As I said, I can't work out the full details because the information supplied about orientations of the various bits of apparatus is confusing. With the Gogo et al experiment I can probably work out from the published coincidence graphs what they must have done. If not, I shall write. I'll probably sketch a classical explanation and send it to them anyway.

Perhaps you could have a look at Kim's paper and see if is the same as the Am J Phys article? I can't readily get hold of a copy of the latter.

Cheers
Caroline

 

[DrChinese]

Please... will someone (other than Caroline) please comment on my post #18 above (go up 6 posts)!

I have an attached diagram you should be able to view in your browser. It is using the EPR paradox setup with double slits. Your comments are appreciated!

 

[vanesch]

In our attached figure, one photon goes Left and the other Right. We measure the path/position of the photon on the Right by means of a detector. That leads to a loss of interference per the standard Double Slit experiment, and no interference is seen on the Right screen. We now know the which-path information of the Left photon as well. Should that affect what we see on the Left screen, so that its expected interference pattern is also erased?

-DrChinese

Ok,I'll have a go. I'll throw some ugly standard quantum mechanics onto it.

I guess that the correlation between the photons is, that if the left photon goes up, then the right goes down, and vice versa. If it is the opposite, swap the sense of "up" on the right side .

So your initial state is:

|psi0> = 1/sqrt(2) ( |+>|-> + f |-> |+>)

Here, the first state in the tensor product is the left photon, and the second state is the right photon. + means up and - means down (in position, here).
f is the phase factor between the two possibilities, taking into account path lengths ; usually this will be a fixed factor and we can take f = 1.

The image on the left is found by looking at each position along the left screen, and the corresponding left eigenvector (left photon seen) is |x> = |+> + g |->. The eigenvector (photon not seen) is the orthogonal one: |~x> = -g*|+> + |-> (up to sqrt(2))

g is a rotating phase factor along the image.

Inverting: |+> = |x> - g* |~x>
|-> = g |x> + |~x>

writing psi0 in the detector basis:

|psi0> = 1/2 ( (|x> - g* |~x>)|->
+ ( g |x> + |~x>) |+> )

The case of a detector at the low right end gives us a picture on the left (coincidence with detector)

<x|<-| psi0> = 1/2, so the probability is 1/4

This is independent of g, so no interference in the correlation.

The case of a detector at the upper right gives us
<x|<+| psi0> = 1/2 g*, so probability also 1/4

The full density matrix is given by |psi><psi| If you then trace out the second basis (the |+> and the |->) you get the (left) local density matrix, which comes down to a 1/2 - 1/2 mixture of |x> and |~x>. So also no interference.

full matrix:

1/4 g*/4 g*/4 - g*^2/4

g/4 1/4 1/4 -g*/

g/4 1/4 1/4 - g*/4

-g^2/4 -g/4 -g/4 1/4


local matrix (traces of the 4 (2x2) blocks)

1/2 0

0 1/2



We NEVER see any interference on the left, with or without detector on the right, with or without coincidence.

cheers,
Patrick.

 

[seratend]

Ok,I'll have a go. I'll throw some ugly standard quantum mechanics onto it.

I guess that the correlation between the photons is, that if the left photon goes up, then the right goes down, and vice versa. If it is the opposite, swap the sense of "up" on the right side .

So your initial state is:

|psi0> = 1/sqrt(2) ( |+>|-> + f |-> |+>)

...

We NEVER see any interference on the left, with or without detector on the right, with or without coincidence.

cheers,
Patrick.

Hi patrick! I am suprised, because I think we should always see an interference pattern on the left screen if we assume that the size of the slits allow them (in accordance with the wavelenght of the light beam). However, I am not sure that I have well understood Dr Chinese experiment thought and all your demonstration.
So I will add my own interpretation (do not hesitate to correct me if I am wrong):

We do not need to detail very much the right slits + detector: It is, in fact, a particular measurement apparatus (i.e. we may consider the right slits+detector as a right measurement apparatus with given measurement results):
The “right detector+slits” will give results, which we model with a set of coarse states:

** for a received right photon |+> before the right slits, we have the result |detector+>
** for a received right photon |-> before the right slits, we have the result |detector->

I agree we can detail the slits interaction + detector result (we may increase the number of Hilbert spaces, but it will not change this “macroscopic” behaviour, i.e. we can describe the diffraction of the photons and thereafter the detection on a given location on the right part, but it is not really important for the result on the left beam, as we have local interactions for the measurements).

Thus, if we accept this basic approach, we can model the right measurement apparatus by the following "projector":
P_right= |R:+><R:+||detector+><RIDS|+
|R:-><R:-||detector-><RIDS|
Where RIDS means “right initial detector state” (in order to avoid extra complications)

So, if we take your initial state outputted by the PDC source and the initial right detector state together with the initial left screen state, we have (removing the normalisation coefficients and imperfections):

|psi0> = (|L:+>|R:-> + |L:-> |R:+>)(x)|RIDS>(x)|LISS>

( |LISS> is the left initial screen state)

After the interaction with the right measurement apparatus, we have:

|psi>=P_right.|psi0>=|L:+>|R:->|detector->|LISS>+|L:->|R:+>|detector+>|LISS>

And if we model the left slit interference pattern on the left screen, we have (it is also a measurement):
** for a received |L:+> left photon state, the interaction of the left slits product the state |interference+> (the interference pattern of spin + photons on the left screen)
** for a received |L:-> left photon state, the interaction of the left slits product the state |interference-> (the interference pattern of spin - photons on the left screen)

i.e. we can model the left screen interference result by a "projector":

P_left=|L:+><L:+||Interference+><LISS|+
|L:-><L:-||Interference-><LISS|

Where LISS means “Left initial screen state”.

We have [P_left,P_right]=0 by construction (they act on different Hilbert spaces) that means that we can do the measurements in what order we want (left screen before the right detector or the inverse, separated by a space like interval or not, etc ...): we still have the final state:

|psi_end>= P_left|psi>=P_left.P_right.|psi0>= P_right.P_left.|psi0>=
|L:+>|R:->|detector->|interference+>+|L:->|R:+>|detector+>| interference->

This global state shows that we always have an interference pattern whatever the results on the right side are: the left screen always receives the diffraction pattern of the left photons of spin + and – whatever measurements are done on the right side of the beam.

However, I may have missed something in this discussion. So sorry, If I am out of scope.

Seratend.

 

[Caroline Thompson]

Sorry DrChinese, but I happen to recognise your idea as one of Popper's thought-experiments. Kim has done it experimentally:

Kim, Yoon-Ho and Yanhua Shih, “Experimental realization of Popper's experiment: violation of the uncertainty principle?”, Found. Phys. 29, 1849 (1999), http://arxiv.org/abs/quant-ph/9905039

I can't remember any details but presume it supporte what Popper thought QM predicted. In deference to your request, I shall refrain from looking it up and giving my usual "hand-waving" classical wave explanation.

Caroline
http://freespace.virgin.net/ch.thompson1/

 

[vanesch]

We have [P_left,P_right]=0 by construction (they act on different Hilbert spaces) that means that we can do the measurements in what order we want (left screen before the right detector or the inverse, separated by a space like interval or not, etc ...): we still have the final state:

|psi_end>= P_left|psi>=P_left.P_right.|psi0>= P_right.P_left.|psi0>=
|L:+>|R:->|detector->|interference+>+|L:->|R:+>|detector+>| interference->

You are completely right, but your conclusion that this state gives rise to an interference pattern is wrong: we are not home yet.
The problem is that you still have the R states in your expression. You should instead calculate the expectation value for a specific left operator, say the one that looks at |in a> = (|interference+> + f |interference->) which is the amplitude on the screen at a specific spot (another spot has another phase factor f).

Note that I am not very fond of the name "interference+" because a pure interference+ state gives NO interference, I hope you agree with that ; it is only when we get superpositions of interference+ and interference- that we get an interference pattern.

So we should calculate:

<psi_end | in a> <in a| psi_end>

to find the intensity at a specific spot on the left screen (determined by the phase factor f)

<in_a|psi_end> = (<interference+| + f* <interference-|)|L:+>|R:->|detector->|interference+>+|L:->|R:+>|detector+>| interference->

= |L:+>|R:->|detector-> + f*|L:->|R:+>|detector+>

and multiply this with its conjugate, to find:

1 + f f* = 2

As f is a phase factor.

We find 2 (we should find 1/2, but left out all normalizations).
Note that this is INDEPENDENT of f, and hence there is no interference pattern.

You see also why: it is the orthogonality of the entangled states (R- and R+) which makes that you will never get a dependence on f.

cheers,
Patrick.

PS:

there's a trick to DO FIND an inteference pattern at the left side: that is: put your right side detector not at one of the slits, but put it on the right screen, looking at a specific spot. If you then look at the coincidence hits, you WILL find an inteference pattern on the left side, in that correlated subsample.

 

[DrChinese]

You are completely right, but your conclusion that this state gives rise to an interference pattern is wrong: we are not home yet.

...

there's a trick to DO FIND an inteference pattern at the left side: that is: put your right side detector not at one of the slits, but put it on the right screen, looking at a specific spot. If you then look at the coincidence hits, you WILL find an inteference pattern on the left side, in that correlated subsample.

Awesome, folks! Finally some actual QM to discuss...

Vanesch,

It is interesting that you picture an interference pattern at the left after moving the detector. But as I understand what you say, the interference appears onky when you compare results from one side to the other. The interference I hoped we would see would be the ordinary kind you would see in a double slit setup... an actual set of bars and white spaces corresponding to peaks and troughs.

seratend,

My intention for the left side was as you pictured it, a double slit suitable for forming an interference pattern. I, too, was imagining that the interefernce pattern would always appear, and wondered if a measurement on the right side could destroy the interference pattern and reduce it to 2 solid bars.

My apologies if my example is not clear. But I think both of you have a good grasp of the setup, or at least what my intention was.

I plan to study Vanesch and seratend's posts more closely. Additional comments are most welcome!

-DrC

 

[vanesch]

Awesome, folks! Finally some actual QM to discuss...

Vanesch,

It is interesting that you picture an interference pattern at the left after moving the detector. But as I understand what you say, the interference appears onky when you compare results from one side to the other. The interference I hoped we would see would be the ordinary kind you would see in a double slit setup... an actual set of bars and white spaces corresponding to peaks and troughs.

Yes, you would expect classically an interference pattern. But the light coming out from a PDC xtal is not classical, because it are 2-photon states.

The reason why you don't see an interference pattern to the left (no matter what happens to the right) is that what is supposed to interfere, namely the photon state "left+" and the photon state "right+" are entangled with something else (in this case, the photon on the right).
And if you work out the expectation value of a local operator on the left, the interference terms that normally would appear if they were not entangled, are put to zero by the inproducts of the states of the entangled photons (which are orthogonal to each other), because the local operator doesn't affect them.

Let us look more explicitly at the algebra:

If you have a state (|+> + |->) |S> (a non-entangled state), and you take the expectation of a local operator (acting only on the left photon), say, O, you have as expectation value for O:

<S| (<+|+<-|) O (|+>+|->) |S> = <S|S> ((O++) + (O--) + (O+-) + (O-+))

(where Oab is a shorthand for <a|O|b> )

So you DO get interference terms O(+-) and O(-+). (<S|S> = 1)

However, if you have an entangled state (|+>|s1> + |->|s2>) (with s1 and s2 orthogonal states of the "other" system), then, the expectation value for O is:

<s1|s1> O(++) + <s1|s2> O(+-) + <s2|s1> O(-+) + <s2|s2> O(++)
= O(++) + O(--) and the interference terms are put to zero by the orthogonality of s1 and s2.

What I did here in algebraic detail is part of the formalism of density matrices.

As an explicit example of O, I will write out O(x) with x a position along the screen on the left (and x in units of fringe phase). In that case,
O(x) takes on the form:

O(x) = |x><x| where |x> is the eigenvector corresponding to maximum intensity at x. It takes on the form:

|x> = |+> + exp(i x) |->

This means that O(x) takes on the form:

O(x) = |+><+| + |-><-| + exp(ix) |-><+| + exp(-ix) |+><-|

So O(x++) = 1
O(x--) = 1
O(x+-) = exp(-ix)
O(x-+) = exp(i x)

you see that you need the cross terms in order to have an x-dependent expectation value (which is nothing else but an interference pattern).

cheers,
Patrick.

 

[seratend]

Thanks Patrick. Before continuing the answer, I want to say that I agree with your current positions (your approach to the problem of measurement and so on: the independence of local measurements, etc …).
Therefore, my comments deal only with the result as I may have done an error in my coarse approximation of the problem.
I will add therefore some comments to the readers of this thread and to Dr Chinese, do not hesitate to correct me (and the others) if you do not agree.

To the readers of this thread: please beware that writing the state |psi_end>, we have chosen an arbitrary spin axis (the entangled photons have no specific spin axis).

First, you “enter” inside the left measurement apparatus (what I have not done): you add legitimately a measurement detector that “sees” the interference pattern (previously, my final state describes only the screen + double slits interaction). I should have done it: it is the only way to answer formally to the question “do we see an interference pattern”.

This local interference measurement has the following projector (detection of photons at position x in the left screen):

P_interf_left(x)=|in_x><in_x|
Where |in_x>= |R:+>|x>+|R:->|x>

(i.e. the detector detects undifferentiated photons, + or – at position |x>. Note that this projector approximates the real detector (to avoid complicate formulates), where we should sum to all the photons and spins of the universe that can hit the left screen at the position x.

|x> is in the same Hilbert space has |interference+> and |interference-> (we look a the position x in the interference pattern). This comes from my definition of the coarse state |interference+ or ->.

You have added an extra phase f, while the state |interference+or-> already includes it: it is the advantage to work with a coarse state (you can hide all the imperfections of the apparatus and slits within this state).
The main interest of writing |in_x> in this form is to highlight the fact that the state |R:+>|interference+> is naturally orthogonal to the state |R:->|interference-> whatever the value <interference+|interference-> is, because <R:+|R:->=0.

Thus a detector of interference in the left side will give the result:
|psi_end_interf(x)>=P_interf_left(x).|psi_end>

=<x|interference+>|x>|L:+>|R:->|detector->+<x|interference->|x>|L:->|R:+> |detector+>

And we have the norm of this vector:

<|psi_end_interf(x)|psi_end_interf(x)>=|<x|interference+>|²+|<x|interference->|²

(Note that |R:->|detector-> is naturally orthogonal to |R:+> |detector+> and we recover an orthogonal sum of interferences)

That is the pattern of the interference at the screen. Note that this result does not depend on any measurements of the right side (measurements are local).

Thus if we agree that a single beam of photons |+> produces an interference pattern on a double slit experiment as well as photons |->, a detector should see the interference pattern that is the sum of photons |+> and photons |->.

Moreover, if we agree that the spatial extension of photons |+> and |-> are the same, we should have |interference+>=|interference-> (we assume that the left slits do not interact with the polarisation of photons and the PDC source produces spatially identical photons + and -, i.e. only the polarisation is different).

Thus our divergence is located in the following problems (it is not related to the measurement, just what produces some local interactions):
** I interpret the result 2 of your previous post as:
2= Int_x <|psi_end_interf(x)|psi_end_interf(x)>dx that is a normal result:
Both vectors |interference+> and |interference-> are normalised to 1 and we have left the normalisation coefficients.

** your sentence “Note that I am not very fond of the name "interference+" because a pure interference+ state gives NO interference, I hope you agree with that ; it is only when we get superpositions of interference+ and interference- that we get an interference pattern.”
This is surely the point where we do not understand ourselves. May be I wrong with that point.
In my coarse notation, I call the state |interference+>= Int_x <x|interference+> |x> dx
Where |<x|interference+>|² is the interference pattern result on a screen of + polarised photons on a double slit experiment.
I just assume that if I have a beam of only |+> photons on a double slit experiment I will have an interference pattern on the screen. Thus I do not understand your sentence (and may be, by the way, I am saying a very stupid thing : ).

Seratend.

P.S. I agree that we can see or not see interferences by construction of measurements that only samples results (and most of the time, we need to construct non local measurements apparatuses: we need to use signals from the left and right sides of the experiment, and if we want we can describe them through non local projectors formalism: a simple way to demystify QM, however It does not prevent me to make errors!).

 

[vanesch]

P_interf_left(x)=|in_x><in_x|
Where |in_x>= |R:+>|x>+|R:->|x>

(i.e. the detector detects undifferentiated photons, + or – at position |x>. Note that this projector approximates the real detector (to avoid complicate formulates), where we should sum to all the photons and spins of the universe that can hit the left screen at the position x.

|x> is in the same Hilbert space has |interference+> and |interference-> (we look a the position x in the interference pattern). This comes from my definition of the coarse state |interference+ or ->.

You have added an extra phase f, while the state |interference+or-> already includes it: it is the advantage to work with a coarse state (you can hide all the imperfections of the apparatus and slits within this state).

No, it is absolutely necessary. |interference+> was the state that corresponded to "a photon through the upper left hole" (that's why I objected to the name). |interference-> was the state that corresponded to "a photon through the lower left hole".
But your detector of the interference pattern will combine these terms with different phases as a function of the position in the image, according to the different path lengths from both holes up to point x. An interference pattern is exactly a dependence on this phase factor!
Note that if there is ONLY |interference+>, there is NO interference pattern, because it is the illumination of a screen by a photon that went only through the upper hole. (again, that's why I thought it was a lousy name).
|interference+> is the illumination of the screen by a single source (the upper hole) and |interference-> is the illumination by the lower hole. It is only when we get combinations of both, with shifting phase factors, that we get a true interference pattern.
(OK, I understand now that you can include this phase factor in "interference" but the reasoning is the same).

The main interest of writing |in_x> in this form is to highlight the fact that the state |R:+>|interference+> is naturally orthogonal to the state |R:->|interference-> whatever the value <interference+|interference-> is, because <R:+|R:->=0.
Thus a detector of interference in the left side will give the result:
|psi_end_interf(x)>=P_interf_left(x).|psi_end>

=<x|interference+>|x>|L:+>|R:->|detector->+<x|interference->|x>|L:->|R:+> |detector->

And we have the norm of this vector:

<|psi_end_interf(x)|psi_end_interf(x)>=|<x|interference+>|²+|<x|interference->|²

(Note that |R:->|detector-> is naturally orthogonal to |R:+> |detector-> and we recover an orthogonal sum of interferences)

That is the pattern of the interference at the screen.

Exactly ! And it corresponds to no fringes: because the first term gives you the light distribution of an illumination by the single upper hole (a blob), and the second term gives you the light distribution of an illumination by the lower hole (also a blob).

Thus if we agree that a single beam of photons |+> produces an interference pattern on a double slit experiment as well as photons |->, a detector should see the interference pattern that is the sum of photons |+> and photons |->.

Yes, but photons that come from the upper hole do NOT give an interference pattern, and so do those from the lower hole. It is only when we have superposed states that we see that, and as you calculate yourself, their combination is absent.

cheers,
Patrick.

 

[vanesch]

A caveat however. As I was thinking more about this setup, there seems to be something weird at first sight. Indeed, imagine the PDC xtal is in a box and the right-hand photon goes into a sheet of black paper, with only the lefthand photon coming out.
You can ask yourself the question: How the hell is this lightbeam different from an ordinary lightbeam and how the hell can I not produce an interference pattern with double slits on this light ??

I was a bit troubled by this question, but I think the answer resides in the following observation. Any light beam has a finite coherence length, which is correlated with its spectral bandwidth. So for any lightbeam, you should find interference if the slits are close enough to each other, and you won't get interference anymore if they are far enough apart.

Now, in order to have our initial state correctly, namely that each lefthand photon that goes through the upper slit must correspond to a righthand photon that goes through the lower slit and vice versa, these slits must be far enough apart : otherwise, some paired photons will "go through the wrong hole". If that is true, our initial state is not a pure |->|+> + |+>|-> but will also contain terms |+>|+> and |->|->, so the entanglement is not complete. As such, some interference will be possible.

I didn't look into detail, but I expect that, if you analyse in detail the PDC process, there will be a relationship between the uncertainty of the bandwidth of the pair of emitted photons and the angular precision of the "back-to-back" trajectory of the pair, in such a way that if the "back-to-back" relationship is precise enough to have essentially a right-up and a left-down correlation, then the bandwidth of the emitted beams will be such that the correlation lengths will be too short to produce an interference pattern.
In fact, you'll never have a back-to-back relationship in PDC, but another angular relationship.

cheers,
Patrick.

 

[DrChinese]

A caveat however. As I was thinking more about this setup, there seems to be something weird at first sight. Indeed, imagine the PDC xtal is in a box and the right-hand photon goes into a sheet of black paper, with only the lefthand photon coming out.
You can ask yourself the question: How the hell is this lightbeam different from an ordinary lightbeam and how the hell can I not produce an interference pattern with double slits on this light ??

I was a bit troubled by this question, but I think the answer resides in the following observation. Any light beam has a finite coherence length, which is correlated with its spectral bandwidth. So for any lightbeam, you should find interference if the slits are close enough to each other, and you won't get interference anymore if they are far enough apart.

Yes, this is what I think is strange as well. Are we saying that there is NO way to get an interference pattern out of the light emerging from a PDC? This seems so counter-intuitive to me. I am still following your analysis above, so I am a little behind you on this question. But here is my reasoning:

A single photon that goes through an ORDINARY double slit does not by itself show too much interference that is visible to the naked eye. It takes an ensemble to detect this effect. Same is true when one of the slits is blocked. There will be no interference pattern, but you cannot tell this too much from one photon alone. And yet that photon is different depending on whether which-path information is known or not.

When the PDC creates the photon pair, how do you know which type was created? I understand that you are saying it is the non-interfering type, but that doesn't obviously seem to be a consequence of the PDC source. On the other hand, the PDC is a pretty special setup so maybe that makes sense.

1. Would the same hold true for the source used in Aspects tests?
2. Would an experiment to test for visible (on one side alone) interference patterns using a PDC source be valuable?

 

[vanesch]

When the PDC creates the photon pair, how do you know which type was created? I understand that you are saying it is the non-interfering type

What I'm saying is that, probably, when the setup is such that you have a clear angular correlation between the emitted pairs which allows you to distinguish clearly through which slit the left photon went, then:
1) these slits have to be far enough apart
2) the angular correlation between the two photons of a pair must be good enough, which I think leads to a spread in frequency spectrum such that:

2) limits the coherence length of light
1) is such that you are outside of the correlation lengths in order to generate an interference pattern.

So the beam coming out on the left looks like classical light with a limited coherence length (such as all light!) and you're trying to use a Young's experiment outside of the coherence length, so you shouldn't be surprised to find no interference.

If you bring the slits closer together, you WILL see an interference pattern (you're now within the coherence length) but I suspect that the angular correlation between the two photons in the entangled pairs is now such, that the other photon will not allow you anymore to decide through which slit the first one went (they are now within the angular uncertainty on the entanglement).

As I said, I don't know too much about the details of PDC, but I know that there are relationships between the angles and the emitted frequencies. So it sounds quite possible to me that if you want high angular correlation between idler and signal beam, and you want to cover the two slits with these beams, then you will have to accept a certain bandwidth, which in turn leads to a small coherence length.
If you insist on high angular correlation and small bandwidth, then your idler and signal beams will probably be too small angularwise to cover your two slite (You will always shoot through one !).
But I didn't work out any details...

cheers,
Patrick.