Does a beam of entangled photons create interference fringes?

[JesseM]

To add to my last post, I just came across http://grad.physics.sunysb.edu/~amarch/ which describes the experiment that Edgardo linked to in terms that are more accessible to a laymen.

diagram: http://grad.physics.sunysb.edu/~amarch/PHY5656.gif

To make the "which-way" detector, a quarter wave plate (QWP) is put in front of each slit. This device is a special crystal that can change linearly polarized light into circularly polarized light. The two wave plates are set so that given a photon with a particular linear polarization, one wave plate would change it to right circular polarization while the other would change it to left circular polarization.

With this configuration, it is possible to figure out which slit the s photon went through, without disturbing the s photon in any way. Because the s and p photons are an entangled pair, if we measure the polarization of p to be x we can be sure that the polarization of s before the quarter wave plates was y. QWP 1, which precedes slit 1, will change a y polarized photon to a right circularly polarized photon while QWP 2 will change it to a left circularly polarized photon. Therefore, by measuring the polarization of the s photon at the detector, we could determine which slit it went through. The same reasoning holds for the case where the p photon is measured to be y. The following table provides a summary.

[I couldn't reproduce the table here, but go to the page to see it]

The presence of the two quarter wave plates creates the possibility for an observer to gain which-way information about photon s. When which-way information is available, the interference behavior disappears. It is not necessary to actually measure the polarization of p and figure out what slit s passed through. Once the quarter wave plates are there, the s photons are marked, so to speak.

So, if you know the linear polarization of the s photon before it went through the quarter-wave plates, then by measuring its circular polarization at the detector Ds (which functions as the 'screen' in the double-slit setup), you can retroactively tell which slit it went through. But circular polarization and linear polarization are noncommuting variables like position and momentum, so once you measure the circular polarization of s, its linear polarization should be scrambled; however, you can tell what the linear polarization of the s photon was before it hit the quarter-wave plates by measuring the linear polarization of the p photon that it was entangled with. But by putting a polarizer in the p photon's path before you measure its linear polarization, you can lose the information about what its original polarization was, and thus also lose the information about the linear polarization of the s photon:

diagram: http://grad.physics.sunysb.edu/~amarch/PHY5657.gif

Increasing the strangeness of this scenario, the next step is to bring back the interference without doing anything to the s beam. A polarizer is placed in the p beam, oriented so that it will pass light that is a combination of x and y. It is no longer possible to determine with certainty the polarization of s before the quarter wave plates and therefore we cannot know which slit an s photon has passed through. The s photons are no longer marked. The potential to gain which-way information has been erased.

In this case you regain interference if you do a coincidence count between the two detectors. But because the linear polarizer is in the path of the p photons, I'm guessing that only half as many p photons would make it to the detector with the polarizer in place, so that many of the s-photon detections do not have a corresponding p-photon detection to include in the coincidence count; presumably the total pattern of s photons would still show no interference, because if they did in this case, then as always this would lead to the possibility of FTL or backwards-in-time signalling by changing the total pattern of s photons by removing or replacing the polarizer in the path of the p photons.

 

[DrChinese]

If you don’t get interference I’ll eat my hat...

You already know you won't get interference, because then you could perform FTL signalling if there were. So the real question is: why no interference?

The answer, in case it wasn't clear, is that entangled photons will act differently than photons which are not entangled. There are different constraints - i.e. different degrees of freedom.

Or seen a different way: for there to be interference, a photon needs to be able to able to traverse all possible paths so it can interfere with itself. As Mandel has stated in his writings, "On the other hand, if there is some way, even in principle, of distinguishing between the possible photon paths, then the corresponding probabilities have to be added and there is no interference."

This behavior occurs in the quantum world, but has no classical counterpart. In the classical world, there would always be interference.

 

[vanesch]

Then without changing the beams just change the test on beam A to a double slit test, rotating that test to any or all polar alignments as you like. And made as sensitive to “entangled degrees of freedom” (whatever that means) as you like.
Just as long as no part of beam B is brought into the experiment to convert it into some kind of entanglement or erasure test. (beam B is discard, redirect it to empty space or a blank wall whatever).

First of all, I'd like to point out that interpretations of QM have nothing to do with this, it is all standard quantum theory (different interpretations all agree on extracted probabilities of observation ; if they don't, they are not *interpretatons* but different theories). So, or I'm making elementary errors in using the quantum formalism, or the result is independent of any interpretational scheme.

If you have an entangled system, and you dump the partner (as you suggest), then you can only do measurements on the first beam. So all observables we're going to discuss are determined by the *reduced density matrix* of the first system.

Now, what is this "interference, sensitive to the entangled degrees of freedom" ?

First of all, what is interference ? You have interference when you have a 2 states |a> and |b> and you have (at least) 2 observables A+ and A-, which measure (|a> + |b>) and (|a> - |b>). If the incoming state is |a>+|b>, then A+ clicks 100% sure, and A- doesn't click, 100% sure. You can have more than 2 observables, for instance, A(x), where x is the position on a screen. A(x) clicks 100% sure if the state is |a> + exp(i x) |b> and doesn't click if the state is |a> - exp(i x)|b>. A+ then corresponds to x = 0 and A- corresponds to x = 180 degrees (pi).

In a setup such as the double slit experiment, |a> = |through-the-left-slit> and |b> = |through-the-right-slit>

We say that we have interference on an incoming state if A+ clicks always, and if A- doesn't click at all, or vice versa. So in our example, if the incoming state is |a>+|b>, we have interference, and if the incoming state is |a>-|b>, we have (complementary) interference. However, if the state is |a>, we don't, and if the state is |b>, we don't: A+ and A- both have 50% chance to click, in both cases.

It should be clear that interference occurs when the incoming state is |a>+|b>, or |a>-|b>. However, when the incoming state is a STATISTICAL MIXTURE of 50% |a> and 50% |b>, then you DON'T have interference.

Now, what did I mean when I said that we cannot have interference if the *relevant* degrees of freedom are entangled ? It means that we have an entangled state |a>|u> + |b>|v>. That the degree of freedom (in this case, the degree of freedom, or hilbert space, in which |a> and |b> live) of which we are going to take "sum" and "difference" for A+ and A-, occur in an entangled way.

Note that |a>|u> + |b>|v> = 1/2 [(|a>+|b>)(|u>+|v>) + (|a>-|b>)(|u>-|v>)]

Clearly, the |u> and the |v> states in the second beam contain the "which slit" information.

If we are going to perform an interference experiment on the second beam, and we are going to trigger only on (|u> + |v>) (using a U+ detection of interference on the second beam), then the selected states of the first system are |a> + |b>, and, as we know, they interfere, because we now have selected the states |a>+|b> for the first beam.

If we trigger on the |u>-|v> state (using U-), then we will trigger on the events with |a>-|b>, having the complementary interference pattern. Clearly, we cannot perform AT THE SAME TIME a measurement for, say, |u> AND a measurement for U+ or U-, which makes it impossible to have at the same time the 'which path' information (|a> or |b>, or: |u> or |v>) AND the "which interference pattern" (U+ or U-, hence A+ or A-) information.

But now you also see why, when only looking at ONE beam, you cannot get this interference pattern: the reduced density matrix of the one beam gives us an equivalence of the beam with a statistical mixture of 50% |a> and 50% |b>. A+ and A- will have 50% chance to trigger for this mixture, so no interference.

You can of course also consider this mixture as 50% (|a>+|b>) (which will have A+ trigger, and not A-), and 50% (|a> - |b>), which will have A- trigger, and not A+. So also in this view, we get 50% A+ triggering, and 50% A- triggering.

How can you have "interference" then from an entangled pair ? Well, simply by using a degree of freedom that is NOT entangled. For instance, we could have as initial state:

(|a+> + |b+>) |u> + (|a-> + |b->)|v>

We have added here an extra degree of freedom, which is the +/- (say, the spin of the particle. It is the SPIN of the particle that is now entangled with |u> and |v>, NOT the |a>+|b> quality. So if |a> is still "the left slit" and |b> is still "the right slit", then the "which path" degree of freedom (a/b) is NOT entangled. So we CAN get interference now, if that interference is NOT depending on the entangled degree of freedom (+ or -). But the entanglement is then also useless to determine "which path" information - because the SAME combination is present in both terms (and can hence be factored out).

If you don’t get interference I’ll eat my hat, or whatever stakes you care to put on it (and I do have a hat).

Bon appetit, as they say around here

 

[RandallB]

or I'm making elementary errors in using the quantum formalism,

I assume this is the case

If you have an entangled system, and you dump the partner (as you suggest), then you can only do measurements on the first beam. So all observables we're going to discuss are determined by the *reduced density matrix* of the first system.

Not my suggestion that was the point of the opening post.

Now, what is this "interference, sensitive to the entangled degrees of freedom" ?
... |a> = |through-the-left-slit> and |b> = |through-the-right-slit>
...
But the entanglement is then also useless to determine "which path" information - because the SAME combination is present in both terms (and can hence be factored out).

Bon appetit, as they say around here

What is this nonsense? No one is trying to figure out “which way” here. Are you just making this up to bail out of a gross error? Don’t be a wimp own up.

I notice you still have not detailed what it will mean to you if the experiment described provides interference patterns. On that ‘wild chance’ what will it mean to you?
Nor have you put up anything, what no got hat? – I’ll send you some shorts.

If you have contacts to a research facility or at a decent school lab this is a reasonably simple experiment to have run. Let us know when you find someone to give you real results.

 

[RandallB]

Did you actually bother looking over the delayed choice quantum eraser experiment?

I know how a delayed choice quantum eraser experiment works. What’s that got to do with any experiment that only uses one side of the PDC output?
DID YOU read this problem??
We are not sending the A side to one slit and the B side of the PDC to the other slit.
We are sending the A side to the double slit, then to the film to record the pattern - that’s it. Doesn’t get much simpler.

What are you putting up for your lunch?

 

[RandallB]

You already know you won't get interference, because then you could perform FTL signalling if there were.
... "On the other hand, if there is some way, even in principle, of distinguishing between the possible photon paths, then the corresponding probabilities have to be added and there is no interference."

What FTL? Where can you sending anything FTL here?

And exactly where in this problem is there any possibility of distinguishing between possible photon paths.

 

[vanesch]

I know how a delayed choice quantum eraser experiment works. What’s that got to do with any experiment that only uses one side of the PDC output?
DID YOU read this problem??
We are not sending the A side to one slit and the B side of the PDC to the other slit.
We are sending the A side to the double slit, then to the film to record the pattern - that’s it. Doesn’t get much simpler.

I UNDERSTAND THAT. And what I'm telling you is that IF the setup is such that the entanglement cannot discriminate between the left slit and the right one, then YES YOU CAN HAVE AN INTERFERENCE PATTERN. But IF the entanglement is going to have something to do with the left slit or the right one, then IT IS NOT POSSIBLE TO HAVE AN INTERFERENCE PATTERN.

Now, as I recognized (post #10 for instance), I didn't immediately consider the first possibility, because it would be a silly experiment, given that you are testing an interference condition on a non-entangled degree of freedom. So YES YOU CAN HAVE INTERFERENCE with ONE SINGLE beam of a PDC xtal ON THE CONDITION that the relevant degree of freedom does not play a role in the entanglement. Normally, when you go through great pains of making an entangled beam, you're interested in studying the quantity that is entangled (for instance, spin). I assumed that the interference experiment was going to test that quantity.

About the rainbow: given that the entanglement of the beams coming out of the xtal are such, that there is a relationship between the exit angle and the wavelength (and thus, the other beam has the complementary relationship), the entanglement of these two angles makes that you are not supposed to be able to get an interference pattern from the "rainbow" light, because using a prism on the other side would let you find out what wavelength (and hence, what angle, and hence what slit) you were hitting. The relevant degree of freedom that is entangled here, is wavelength and direction, and this allows you to find back the "which way" information. So this would be a situation where you would have no possibility of an interference pattern.

Now, you will always be able to set up SOME interference experiment with double slits in such a way that you will see an interference pattern. Only, then you are looking at a quantity that IS NOT ENTANGLED, period.

 

[vanesch]

And exactly where in this problem is there any possibility of distinguishing between possible photon paths.

As I said, it was implicitly assumed when people talk about interference experiments and entangled systems, that the "which path" information was in principle recoverable from the partner beam.

It was my error (in the beginning) for not having stated that, but I corrected that (several times) later, by saying that the interference to be discussed should be using the degree of freedom (left slit or right slit: because that's the two states between which we are going to look for interference) that is entangled.
In this case, (when the entangled degree of freedom is the "relevant one" for the interference, so when it can distinguish between left and right slit), no interference will be observed.

In the case that the entangled degree of freedom CANNOT distinguish between left slit and right slit, and hence is NOT the relevant degree of freedom tested with the interference pattern, then you can have all the interference you want. As such, it will remain so, when looking ONLY at the first beam, or when applying all kinds of coincidences on all kinds of measurements on the second beam ; in other words, this interference pattern has nothing to do with the entangled degree of freedom.

Take the typical quantum eraser as an example:
http://grad.physics.sunysb.edu/~amarch/

For instance, here, in the first experiment, an interference pattern HAS been found with one beam (before the quarter wavelength plates were introduced). So YES you can have an interference pattern. But it has nothing to do with the entangled degree of freedom (which is, in this case, spin). You could now do all kinds of experiments on the second beam, and subsample this interference pattern, and you'd get out the same interference pattern.
However, when the quarter-wavelength plates are introduced before the slits, the interference becomes now DEPENDENT UPON SPIN, an as such, is now dependent on the entangled degree of freedom. In this case, you CANNOT get an interference pattern.

The day that you can present me an interference experiment, where I can recover the which path information potentially from the second beam, and you succeed in obtaining an interference pattern from the first beam only when you dump the second beam, I eat MY HAT.

 

[Sherlock]

I'm not sure I grasp yet the QM accounting for no interference pattern emerging via detector D_s when the s-beam is unfiltered prior to passing through the double-slit. I want to try to get some physical conceptualization of this -- if that's possible.

Using the diagram that JesseM linked to in a recent post: http://grad.physics.sunysb.edu/~amarch/PHY5657.gif

... and removing the qwp's and running the s-beam through just the double slit device (the scenario that RandallB has been talking about), then IF the s-beam was a classical superposition of red and blue frequencies, then these frequencies would combine prior to passing through the double-slit and you would expect to get some sort of discernible interference pattern at detector D_s. Is that correct?

But because the photons in the s-beam are EACH in a superposition of red and blue, then what has passed through the double-slit photon-by-photon is either a wavetrain of red OR a wavetrain of blue, one after the other in random order (tending toward 50% of each as you record more detections). Each of these frequencies, after passing through the double-slit, produces its own interference distribution at D_s, but the TOTAL distribution at D_s (after a large number of detections at D_s) is not an interference pattern because the areas of maximal detection of one frequency more or less overlap with the areas of minimal detection of the other frequency.

Whoops! Ok, I just read the first experiment in the link provided by Vanesch in the preceding post: http://grad.physics.sunysb.edu/~amarch/

Apparently, I've been thinking about this incorrectly. From what I see at that link, the BBO created photons, s and P, aren't entangled in wavelength (they're the same, each 702.2 nm) -- so, of course you'd see an interference pattern emerge by putting the unfiltered s-beam (or P-beam) through a double-slit.

Or, am I still missing something?

 

[JesseM]

I know how a delayed choice quantum eraser experiment works. What’s that got to do with any experiment that only uses one side of the PDC output?
DID YOU read this problem??
We are not sending the A side to one slit and the B side of the PDC to the other slit.
We are sending the A side to the double slit, then to the film to record the pattern - that’s it. Doesn’t get much simpler.

Did YOU actually read about how the delayed-choice quantum eraser works? Both members of the entangled pair, A and B, are not sent to different slits in the same double-slit setup. The signal photon is sent to the screen through a double slit, while the idler photon goes in the opposite direction towards a separate detector, never going through either slit (or in Scully's version of the experiment, both photons go in opposite directions from one of two possible locations which function as the 'slits', the signal photon going towards the screen and the idler going towards a set of beamsplitters/mirrors/detectors). The pattern on the screen will consist solely of the signal photons. However, by making the right kind of measurement on the idler, you can reconstruct the which-path information of both photons. Complementarity says if you know the which-path information of the signal photons, you shouldn't see any interference on the screen--agreed? And yet if the pattern on the screen depended on whether or not you chose to measure the idler in a way that reconstructed its which-path information or in a way that erased it, then this would lead to the possibility of FTL or backwards-in-time signalling.

 

[Edgardo]

No ! Because you've lost the essential subselection information. Nature is not going "to check whether you could eventually...". It wants you to prove that you cannot restore the which-path information, by USING the complementary information. (or vice versa). If you can't show anything, you get nothing.

Yes, this makes sense to me. So, in order to erase the which-path information of the upper photon one has to get "complementary information" (the interference pattern).

However...

In the original poster's setup, no interference occurs you claim. But I wonder how exactly is the which-path information erased when you use coincidence counting with the upper detector? Where do you gain "complementary information" in coincidence counting?

 

[Sherlock]

In message 1, bruce2g asked:

Here's the setup: shine a laser beam on a crystal which performs SPDC (spontaneous parametric down conversion). This produces two beams, A and B. The photons in beam A are entangled with the photons in beam B.

Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?

In message 2, vanesch answered:

Short answer: no.

In message 38, vanesch answered:

Take the typical quantum eraser as an example:
http://grad.physics.sunysb.edu/~amarch/

For instance, here, in the first experiment, an interference pattern HAS been found with one beam (before the quarter wavelength plates were introduced). So YES you can have an interference pattern.

So, it seems that, after a bit of confusion, which nevertheless resulted in me learning some stuff, the answer to the original question, as stated, is either a qualified yes or a qualified no.

The SPDC photons A and B are not entangled in wavelength. That is, beam A (or beam B) is not composed of photons that are in a superposition of two different wavelengths (which was the misapprehension that I was originally considering). A and B are however characteristically entangled in space-time and momentum. If this entanglement is assumed (because of the source), but nothing is done to reveal it, then simply directing beam A through an unobstructed double-slit will result in interference banding.

Now if you outfit side B with a double-slit device and put a detector on the upper slit, then pair (via coincidence matching) the B slit-detected results with the appropriate subset of the A detections, then you will still see an interference pattern on A's detection screen --- unless you select out that subset from A's data.

So it seems that we can have which way information and an interference pattern at the same time. But not really, because the subset of A's data that corresponds to the B slit-detected data has been produced by only one slit, so via data (or signal) processing this subset of A's data will not produce interference banding.

And on the B detection screen there is of course no interference pattern because the slit detector has effectively closed one of the slits on B's double-slit device.

Is this ok or am I still missing something?

 

[JesseM]

Now if you outfit side B with a double-slit device and put a detector on the upper slit, then pair (via coincidence matching) the B slit-detected results with the appropriate subset of the A detections, then you will still see an interference pattern on A's detection screen --- unless you select out that subset from A's data.

So it seems that we can have which way information and an interference pattern at the same time.

Are you arguing that by measuring which slit B went through, we can get the which-path information for A? Would this be because of momentum entanglement, or for some other reason? If there is any opportunity to obtain the which-path information for A by doing the appropriate measurement on B, then my understanding is that you will not see interference in the total pattern of A, although you will see it in coincidence-matching with any subset of B that is measured in such a way that the opportunity to determine which-path information is lost. Are you saying the same thing? I couldn't tell, it almost sounded like you were arguing that the total pattern of A would show interference but some subset would not because the which-path information was measured--but I don't think this can happen, because if there was any opportunity to obtain which-path information for A you shouldn't see any interference in its total pattern. Anyway, it shouldn't be possible for multiple non-interfering subsets to add up to an interference pattern in the total set of signal photons, because an interference pattern will have less photon hits in certain regions of the screen than a non-interference pattern.

 

[Sherlock]

Are you arguing that by measuring which slit B went through, we can get the which-path information for A?

The subset of B's total data produced by B's slit detector will correspond to a subset of A's data associated with a single slit of A's double-slit device.
Wouldn't it?

Would this be because of momentum entanglement, or for some other reason?

Momentum entanglement would be my guess.

If there is any opportunity to obtain the which-path information for A by doing the appropriate measurement on B, then my understanding is that you will not see interference in the total pattern of A,
although you will see it in coincidence-matching with any subset of B that is measured in such a way that the opportunity to determine which-path information is lost. Are you saying the same thing? I couldn't tell, it almost sounded like you were arguing that the total pattern of A would show interference but some subset would not because the which-path information was measured--but I don't think this can happen, because if there was any opportunity to obtain which-path information for A you shouldn't see any interference in its total pattern.

We already know that the total pattern of A (provided the beam is directed, unaltered, through a fully open double-slit device) will be an interference pattern.

Nothing that happens on the B side will affect that UNLESS the signals (the photons) incident on or transmitted by A's double-slit device are altered, or UNLESS A's data is altered via post-detection processing.

If you don't do any signal processing contingent on B's slit detections, then the RAW data at A will always show interference banding. Otherwise, you would have direct evidence of an FTL phenomenon --- and that hasn't happened yet, afaik.

The slit-detector detections at B ARE providing which way info about A in that there is a subset of A's data which corresponds to these detections, and this subset of A's data is associated with only one of A's slits.

But that's the point. The which way info is associated with only one of A's slits --- and one slit can't produce the interference banding ... even though you'll actually see interference banding at A in the absence of signal processing. Confusing isn't it? But not really. It's just that if you're considering the which way data, then this data isn't associated with two slits, only one. So, no violation of complementarity. In fact, complementarity can't be violated if it's applied properly. It has to do with mutually exlusive stuff, like one detection vs. many detections. Things that are, necessarily, mutually exclusive but nevertheless both necessary for the full picture are what complementarity is about. At least I think that's what it's about.

 

[JesseM]

The subset of B's total data produced by B's slit detector will correspond to a subset of A's data associated with a single slit of A's double-slit device.
Wouldn't it?

I'm not sure. I think in some cases this would be true because of momentum entanglement, but I'm not sure if all methods of producing entangled particles would produce this sort of momentum entanglement.

We already know that the total pattern of A (provided the beam is directed, unaltered, through a fully open double-slit device) will be an interference pattern.

If it is true that the total pattern of A would show interference, then I think that must mean there's no measurement you can do on B that will reconstruct A's which-path information. On the other hand, when the paper says that the pattern in the absence of the quarter-wave plates shows interference, it looks to me like they're still talking about a coincidence count between s photons that go through the slits without the plates and p photons that are detected by the other detector Dp. Maybe this other detector is placed in such a way that any photons that hit it will provide no which-path information about their entangled twins, so the subset of s photons whose p photons register at this detector show interference, but some p photons would simply miss this detector so the total pattern of s photons would not show interference? I dunno, maybe vanesch or someone can tell us the answer.

If you don't do any signal processing contingent on B's slit detections, then the RAW data at A will always show interference banding.

I think either you're wrong and the raw data shows no interference pattern (again, when the paper said that interference was detected in the absence of quarter-wave plates, it looked like they were still talking about a coincidence count), or else the photons are entangled in such a way that it's impossible to reconstruct the which-path information of the signal photons s by any measurement on the idlers p.

The slit-detector detections at B ARE providing which way info about A in that there is a subset of A's data which corresponds to these detections, and this subset of A's data is associated with only one of A's slits.

But that's the point. The which way info is associated with only one of A's slits --- and one slit can't produce the interference banding ... even though you'll actually see interference banding at A in the absence of signal processing.

I really don't think it can happen that a subset of an interference pattern can be a non-interference pattern. Your use of the term "signal processing" as opposed to "coincidence count" obscures things IMO, there's nothing at all sophisticated or technical about what's done here, it's basically just like taking the total pattern of photon hits on the screen, then coloring a certain subset one color and another subset another color, and looking at the pattern made only by photon hits of one color. The total pattern of photons must just be a simple sum of these subsets--if in a certain region of the screen there were 5 photon hits that you colored red, and 8 that you colored blue, then the total pattern must have 13 photon hits in that region. So there's no way the total pattern could have less photon hits in a region than anyone of the subsets--but wouldn't this be required, if the subsets were non-interference patterns but the total pattern was an interference pattern? I don't see how you can explain the black bands which represent the regions of destructive interference in terms of the total pattern being a sum of subsets which show no such bands.

 

[vanesch]

In message 1, bruce2g asked:

In message 2, vanesch answered:
...

In message 38, vanesch answered:
...

Yes, I recognized at least 5 times already that it was my mistake to say no in general, because I didn't consider the case where the interference experiment didn't have anything to do with the entanglement (that the degree of freedom that is entangled, is NOT related to the which-way information).

For my excuse, it was a bit like: if the transmitter antenna is not connected, can I receive something on the receiver ? And then the answer is a qualified "yes", namely if you tune in into ANOTHER broadcast station

So, it seems that, after a bit of confusion, which nevertheless resulted in me learning some stuff, the answer to the original question, as stated, is either a qualified yes or a qualified no.

Indeed. It is NO if the entanglement is related to the "which slit" degree of freedom, and it is (potentially) YES if the entanglement has nothing to do with it.

The SPDC photons A and B are not entangled in wavelength. That is, beam A (or beam B) is not composed of photons that are in a superposition of two different wavelengths (which was the misapprehension that I was originally considering).

When they come out of the Xtal, they ARE entangled. It is only because there have been FILTERS that we only take selected photons of given wavelength.

EDIT: I take the last statement back. In the proposed experiment here, there are (to my surprise) no filters. So all entanglements are possible here, depending on the specific behaviour of the xtal, and the detailled setup...

 

[JesseM]

Yes, I recognized at least 5 times already that it was my mistake to say no in general, because I didn't consider the case where the interference experiment didn't have anything to do with the entanglement (that the degree of freedom that is entangled, is NOT related to the which-way information).

But on the issue that Sherlock and I were talking about, is the experiment on http://grad.physics.sunysb.edu/~amarch/ with the quarter wave plates removed a case where the degrees of freedom that are entangled are unrelated to the which-way information, or would there be momentum entanglement such that a measurement on the p photon could potentially tell you which slit the s photon went through? There is interference in the coincidence-counting graph between the Dp detector and the Ds detector when the QWPs are absent, but I wonder if the total pattern of Ds photons without coincidence-counting would still show interference, or if you only see interference because the Dp detector is specially placed in such a way that its measurements don't give you any which-path information.

 

[vanesch]

But on the issue that Sherlock and I were talking about, is the experiment on http://grad.physics.sunysb.edu/~amarch/ with the quarter wave plates removed a case where the degrees of freedom that are entangled are unrelated to the which-way information, or would there be momentum entanglement such that a measurement on the p photon could potentially tell you which slit the s photon went through? There is interference in the coincidence-counting graph between the Dp detector and the Ds detector when the QWPs are absent, but I wonder if the total pattern of Ds photons without coincidence-counting would still show interference, or if you only see interference because the Dp detector is specially placed in such a way that its measurements don't give you any which-path information.

I don't know, honestly. They don't seem to do any pre-selection (when you look into the publication). Often, PDC experiments use a specific wavelength filter and some collimating to obtain both entangled beams, but they don't do that here. Of course, the xtal is 1 mm long, and the slits are only 200 micrometer apart, so because of the uncertainty of the emission point in the xtal (also entangled !) you cannot know, from the wavelength (and thus, the angle) alone, which slit has been hit.
So it could indeed be, as you suggest, that the specific interference pattern that they view IS already a result of the selective geometrical effect of the second detector ; or it could be that (given the uncertainty of the emission point, or some collimating optics or whatever) there is no link (entanglement) anymore between the "which slit" and the entangled states. It's because the initial state is not very well defined in this case (given that there are NO pre-selections in wavelength, emission point etc...) that all these possibilities are still open, and depend on the specific behaviour of the setup.
A test would be to use a big detector as second detector, for the coincidence counting, which has a large acceptance for both wavelength and direction/emission point.

 

[Sherlock]

If it is true that the total pattern of A would show interference, then I think that must mean there's no measurement you can do on B that will reconstruct A's which-path information.

... when the paper says that the pattern in the absence of the quarter-wave plates shows interference, it looks to me like they're still talking about a coincidence count between s photons that go through the slits without the plates and p photons that are detected by the other detector Dp. Maybe this other detector is placed in such a way that any photons that hit it will provide no which-path information about their entangled twins, so the subset of s photons whose p photons register at this detector show interference, but some p photons would simply miss this detector so the total pattern of s photons would not show interference?

You can't REconstruct something that didn't exist in the first place.

The photon distribution pattern at A, or Ds, is associated with two emission locations. The wave picture accounts for the banding that's seen on the detecting screen, and the wave picture is complementary to the particle picture. Trying to understand the double-slit distribution in terms of particles going through one slit or the other creates problems of the sort that we both seem to have encountered in this discussion.

The only way that the double-slit pattern produced on A's screen can be altered is by something that you do to the A side. Nothing done ONLY on the B, or Dp, side will have any affect on the A side whatsoever. Otherwise you'd have a FTL phenomenon --- which would be exciting, but nobody's claiming that.

So, wrt the SPDC photon source, apparently, as long as the A side includes an unobstructed double-slit and no other filtration of any sort, then what you see on the detecting screen A will always be the banding characteristic of two waves interfering.
You can do coincidence matching with B, you can post-process the data, whatever ... it doesn't have any affect on what emerges on the A screen. At least that's my current understanding of this --- which of course could change the moment I read something from a source I trust that tells me otherwise.

I really don't think it can happen that a subset of an interference pattern can be a non-interference pattern.

If you do coincidence counting, then you get the same count for each side. But the B side photons came from one emission source, while the A side photons came from two emission sources. The B side distribution isn't a subset of the A side distribution. It's just different, because of the emission sources.

I don't see how you can explain the black bands which represent the regions of destructive interference in terms of the total pattern being a sum of subsets which show no such bands.

You can't. That's the point of complementarity. It isn't the photon detection patterns that are being combined or subtracted from. It's that you have two emission sources vs. one emission source, and these produce different distributions.

So, keeping the same setup on the A side, then if you put a double slit on the B side, and a slit detector on one of the slits of the B double slit, and you coincidence match a certain number of A detections with the number of B detections from EITHER the B slit detector OR the B screen detector, then did a calculation of the A side distribution for this numerical subset of the A detections based on one emission source, then you would get a different distribution for this numerical subset of the A detections.

 

[JesseM]

You can't REconstruct something that didn't exist in the first place.

The photon distribution pattern at A, or Ds, is associated with two emission locations. The wave picture accounts for the banding that's seen on the detecting screen, and the wave picture is complementary to the particle picture. Trying to understand the double-slit distribution in terms of particles going through one slit or the other creates problems of the sort that we both seem to have encountered in this discussion.

You're again getting into interpretational issues, but I really would prefer to avoid that entirely. It is convenient to speak of a measurement of B providing "which-path information" for A, but this does not commit you to believe that A really traveled through one slit or another, it should be understood purely as a statement about the formalism for calculating the probabilities in QM (as I suggested before, in the path integral approach it could be understood to mean you only sum over paths that go through a particular slit). My comment about "reconstructing the which-path information" was meant purely in this vein.

The only way that the double-slit pattern produced on A's screen can be altered is by something that you do to the A side.

Sure, but "altered" from what? You cannot assume that the original pattern for photons which were created in an entangled state would be the same as the pattern for unentangled photons. As I said in another post to you, this does not entail any FTL influences, because the creation of both particles in an entangled state lies in the past light cone of the event of the particle being measured. What happens to its entangled twin after the two depart cannot further alter the pattern, but the mere fact that it was created in an entangled state can be thought of as leaving an "imprint" on the photon that causes it to behave differently in the double-slit setup than an unentangled photon would (and since I don't want to get into interpretational issues this obviously isn't meant to be taken literally, it's just a way of conceptualizing the idea that entangled photons can make different patterns on the screen than non-entangled ones without them needing to be in communication with their entangled twins).

So, wrt the SPDC photon source, apparently, as long as the A side includes an unobstructed double-slit and no other filtration of any sort, then what you see on the detecting screen A will always be the banding characteristic of two waves interfering.

Nope, you are not justified in concluding that. Note that vanesch said above that he wasn't sure if the screen would show interference in the total pattern of photons on the A side, and that it would depend on the nature of the original entangled state.

If you do coincidence counting, then you get the same count for each side. But the B side photons came from one emission source, while the A side photons came from two emission sources. The B side distribution isn't a subset of the A side distribution. It's just different, because of the emission sources.

This paragraph makes no sense to me, and it makes me think that maybe you're not clear on what "coincidence counting" means. Suppose we have a bunch of distinct photon hits on the screen on the A side, and say we use a setup where the photons on the B side can end up at one of four possible detectors, as in Scully's version of the delayed choice quantum eraser. Now, since each photon on the A side is entangled with a photon on the B side, we can match photons hits on one side with photon hits on the other. For example, we can look at a particular blip at a particular location on the screen, and say "the photon that made this blip was entangled with a photon on the B side that was registered at detector 3." So if we look at the pattern of photon blips on the screen, we can divide them into four mutually exclusive subsets, one for each possible detector that their entangled twin registered at. That's all a "coincidence count" is, it's about looking at only the subset of photon hits at one location that coincided with their entangled twins being registered at another specified location. Again, you can imagine coloring the dots on the screen different colors to label them--a red dot could indicate that this photon's twin was registered at detector 1, a green dot could indicate this one's twin was registered at detector 2, and so forth. Thus you can see that the total pattern on the screen must just be the sum of these 4 subsets (assuming no photons missed the detectors in this setup).

In Scully's version of the setup, detectors 3 and 4 allowed you to determine the which-path information for both members of the entangled pair, while detectors 1 and 2 did not. And if you look at the coincidence count graphs in Scully's paper, you see that the one for 3 shows no interference (fig. 5 in the paper--the graaph for detector 4 is not shown, but it should have the same shape), while the ones for 1 and 2 (fig. 3 and fig. 4) do show interference. But the sum of all 4, which should represent the total pattern of signal photons on the screen (again, assuming no cases where the entangled twin misses all the detectors), shows no interference, because the peaks of the graph for detector 1 line up with the valleys of the graph for detector 2, so their sum does not show interference.

But note that it isn't really possible to imagine the reverse, that the sum of two or more subsets that don't show interference would result in an interference pattern. This is because the interference pattern would have to include less photon hits in certain regions of the screen then the non-interference patterns that are supposed to sum up to produce it, which is obviously impossible.

You can't. That's the point of complementarity. It isn't the photon detection patterns that are being combined or subtracted from. It's that you have two emission sources vs. one emission source, and these produce different distributions.

Again, I think you're misunderstanding what a coincidence count is--all the coincidence counts are simply subsets of the total pattern of photon hits on the screen, you get them by looking only at the subset of photon hits on the screen that coincide with their entangled twin being registered at a particular detector, and ignoring all the other hits on the screen that don't coincide with hits at that detector.

 

[bruce2g]

I just want to thank Vanesch and all of the others who patiently contributed posts to clear up this question I had. I guess the short answer is that if the entanglement of the 'B' photons can reveal which slit the 'A' photons go through, then there will be no interference. If the 'B' photons do not reveal the which-path information, then there will be interference.

A slightly longer answer would note that you really need to do coincidence counting to be sure that you're dealing with entangled pairs, and that the geometry of the 'B' detector employed for this purpose can apparently also have an influence.

It seems pretty simple when you look at it this way. Thanks again!

 

[vanesch]

I just want to thank Vanesch and all of the others who patiently contributed posts to clear up this question I had. I guess the short answer is that if the entanglement of the 'B' photons can reveal which slit the 'A' photons go through, then there will be no interference. If the 'B' photons do not reveal the which-path information, then there will be interference.

I think that's a fair summary, yes.

A slightly longer answer would note that you really need to do coincidence counting to be sure that you're dealing with entangled pairs, and that the geometry of the 'B' detector employed for this purpose can apparently also have an influence.

It seems pretty simple when you look at it this way. Thanks
again!

Yes, exactly. And you could even add yet another caveat: EVEN if you're using entangled pairs, and EVEN if the entanglement doesn't permit you *in principle* to deduce which-slit information using photon B (so that we are in the case where A can interfere), there might always be another, prozaic reason why we don't observe interference. The only thing we can say is that interference is potentially possible (in other words, that the entanglement is not forbidding an interference ; but maybe something else is screwing it up).

But the essential statement is indeed this: if the entanglement allows you, in principle, to use photon B to find out through which slit photon A went, in a specific interference setup, then this same entanglement will make it impossible for photon A to give rise to interference in this setup.

But to the person only looking at beam A (which might even ignore that beam A is entangled with something else), the photons of beam A do not look somehow "peculiar" in that they "refuse to interfere". He just sees beam A as a stochastic mixture of "photons in a mode that only go through slit 1" and "photons in a mode that only go through slit 2", and as such, he's not suprised not to see, from this mixture, any interference pattern.
this mixture is dictated by the reduced density matrix (reduced to beam A only) of the entangled state. The reason for that is that, if the entanglement allows you to find out "which slit" A went through, it means that the mode of A entangled with this tagging state for saying "slit 1" of photon B is a mode that only goes through slit 1, and the mode of A entangled with the tagging state for saying "slit 2" of photon B, is a mode that only goes through slit 2. As such, beam A, as seen alone, is a mixture of photons that only go through slit 1, or only go through slit 2. As such, it is not surprising that there is no interference from the mixture.

 

[bruce2g]

I'm starting to think that it'd be nice to actually perform some of these experiments. As I understand it, it only costs around $10,000-$15,000 for a complete setup (laser, pdc, detectors, mounts, etc), and there have been some recent papers about setting up undergrad labs (complete with parts lists and vendor addresses!). I think I'll see if some undergrad might be interested in exploring the effect of the 'B' detector's geometry on the interference, as a lab project, and report any results back here.

 

[alexepascual]

I'm starting to think that it'd be nice to actually perform some of these experiments. As I understand it, it only costs around $10,000-$15,000 for a complete setup (laser, pdc, detectors, mounts, etc), and there have been some recent papers about setting up undergrad labs (complete with parts lists and vendor addresses!). I think I'll see if some undergrad might be interested in exploring the effect of the 'B' detector's geometry on the interference, as a lab project, and report any results back here.

I recently started another thread about the same issue without being aware obout this one.
The thread: https://www.physicsforums.com/showthread.php?t=334069

Now, I see that activity on this thread stopped in 2006. I wonder why.
I also wonder if anybody managed to do the experiments suggested by bruce.

I didn't read meticulously this thread but by looking specially at the last posts it appears that there was no consensus as to what to expect.