Does the Quantum Eraser Experiment Truly Erase Information or Alter Causality?

[Cemre]

Hello,

info about quantum eraser experiment is at wikipedia
and at http://grad.physics.sunysb.edu/~amarch/

I want to express my view of last part of this experiment and want to discuss whether there really is an "erasure" or a breakdown of causality even in the delayed case. the way the coincidence counter measures the interference patterns really gets me thinking, cause i think there are some photons we ignore to measure which we shoudn't... please read ahead:

scenario 1 -> WITHOUT the polarizer on p-way :
-> one photon with 2f frequency hits BBO and two entangled photons get out each with f frequency. one partner goes through slits ( s-way ) goes through QWP1 and QWP2, which-path information becomes known, pattern is gone.

scenario 2 -> WITH the polarizer on p-way :
-> partner photon on p-way hits the polarizer, by %50 chance gets x-polarized and goes through. with %50 chance gets y-polarized and doesn't get through ! ( absorbed or reflected back ) *** the "doesn't get through" *** thing got me thinking:

if photon on p-way goes through the polarizer it will hit the detector on p-way, other entangled photon on s-way will also hit the its detector and "coincidence counter" will "tick"

BUT if photon on p-way doesn't go through, the other photon on s-way will STILL hit the detector and "coincidence counter" will NOT "tick", it will just ignore the photon on s-way and it will not be counted.

I think if we were to count ALL the photons on s-way, interference patterns would still be "gone", "not present", even in the case the polarizer exists on p-way.

the reason that the interference pattern SEEMS to come back is that;
-> ALL the s-way photons do NOT form a pattern. ( since which-way info is known due to the presence of QWPs )
-> some specially selected (*) subset of photons on s-way DO form an interference pattern picture.

the polarizer on p-way just tells the coincidence counter;
-> which photon on s-way to count ( the partners of p-way photons that go through polarizer -> the specially selected subset )
-> and which to ignore, ( the partners of p-way photons that doesn't go through the polarizer )

polarizer on p-way just effect the coincidence counter and not the photon on s-way so there is no communication to s-photon here... and the interference pattern doesn't really come back ! so nothing is erased here. the interference pattern, that we observe after putting the polarizer, is NOT an actual interference pattern, it is just a picture of selected subset of all the photons on s-way, which do not form an interference pattern.

am i right? any ideas?

 

[DrChinese]

the reason that the interference pattern SEEMS to come back is that;
-> ALL the s-way photons do NOT form a pattern. ( since which-way info is known due to the presence of QWPs )
-> some specially selected (*) subset of photons on s-way DO form an interference pattern picture.

the polarizer on p-way just tells the coincidence counter;
-> which photon on s-way to count ( the partners of p-way photons that go through polarizer -> the specially selected subset )
-> and which to ignore, ( the partners of p-way photons that doesn't go through the polarizer )

polarizer on p-way just effect the coincidence counter and not the photon on s-way so there is no communication to s-photon here... and the interference pattern doesn't really come back ! so nothing is erased here. the interference pattern, that we observe after putting the polarizer, is NOT an actual interference pattern, it is just a picture of selected subset of all the photons on s-way, which do not form an interference pattern.

am i right? any ideas?

I have to admit I am confused a bit myself, and I have looked at this quite a bit previously.

Case I: if no polarizer on p-way, no quarter wave plates on s-way, then total* s-way pattern is interference.

It has always been my understanding that a beam of entangled photons does NOT produce an interference pattern (for reference see Zeilinger, s290, fig. 2). So this seems to stand in direct contradiction to Case I.

Case II: if no polarizer on p-way, quarter wave plates added on s-way, then total* s-way pattern is NO interference.

(*Question: Why bother with coincidence counting in cases I and II? In cases I and II, there clearly is no subset to consider in the first place; there is nothing sub-selecting the p-way stream in any way I see. So I think the total pattern and the coincidence pattern are the same.)

Case III: if polarizer added on p-way, quarter wave plates remain on s-way, then total s-way pattern is interference... and so is the subset which should be exactly half intensity of the total beam. Now, this MUST be true because otherwise you could perform FTL signaling by either inserting or removing the p-way side polarizer; and the s-way total pattern would change.

So the only way to reconcile this is to assume that in cases I and II above, there is a changing (conditional) subset as well - but I don't see what that is right now.

 

[DrChinese]

Hello,

info about quantum eraser experiment is at wikipedia
and at http://grad.physics.sunysb.edu/~amarch/

I think there is also an issue regarding the p-way: in some versions of the experiment, there is also a lens used to register the photon and focus it on that side's detector. This is not shown on the above link, and neither is it shown on the Wikipedia page. But the Delayed Choice versions always have that lens, and when it is present you do need to coincidence count. See also the Dopfer thesis.

 

[Cthugha]

polarizer on p-way just effect the coincidence counter and not the photon on s-way so there is no communication to s-photon here... and the interference pattern doesn't really come back ! so nothing is erased here. the interference pattern, that we observe after putting the polarizer, is NOT an actual interference pattern, it is just a picture of selected subset of all the photons on s-way, which do not form an interference pattern.

You seem to be confusing single photon interference and two-photon interference. The pattern is a real two-photon interference pattern. To this end it is a selected subset of all photons on s-way, which form several interference patterns. However the superposition of all of this patterns is no pattern at all.

So the only way to reconcile this is to assume that in cases I and II above, there is a changing (conditional) subset as well - but I don't see what that is right now.

In two photon interference there are always several subsets present due to the fixed phase of the two-photon state. Maybe it helps to picture this using a simpler model, which puts aside the nonlocalitiy issues and concentrates on the fixed phase relationship.
So consider a light source emitting two particles called A and B at once. At each run the phase of the emitted particles is random, but the same for both particles. Now imagine that the photons are directed to two small detectors. The probability to detect a photon A at some special detector position in the plane, where the detector can be moved is determined by its phase. You will have a periodic pattern of regions with high and low detection probabilities for the single photon depending on whether the wave function shows a node or an antinode there. However due to the random phase of each of the emitted photons any position of the detector will get the same amount of photons.
What gives rise to the different subsets is now the fixed phase relationship of the two-photon state. If you know the distribution of positions of high detection probabilities and low detection probabilities for single photon A on one screen, you will also know the distribution of positions of high and low detection probabilities for photon B on the other screen due to the fixed phase relationships. So therefore you get a correspondence between several points on the two screens. For each position of detector A you therefore get a probability distributions for all the possible positions of detector B telling you, whether it is likely or unlikely to detect a photon there, if a photon is also detected at this position of detector A. So therefore your several subsets are given by the possible values of the initial single photon phase, which both photons share.

 

[Sammy k-space]

Well let us try it this way: let the frequency be say 900Mhz and use a vertically polarized light source with a vertically polarized detector and a hoizontally polarized detector; or try a right handed circular polarized light source with a right handed detector and a left handed detector. What do you think?

 

[DrChinese]

In two photon interference there are always several subsets present due to the fixed phase of the two-photon state...

We know that each side has a total pattern. I believe that the total pattern (let's call this A) for the double slit side, no coincidence counting, should not show the traditional interference pattern - but rather a traditional diffration pattern (per Zeilinger reference).

Now, suppose we coincidence count with the other side, which has no double slit (let's call this B). How will the resulting subset look any different than the total pattern just mentioned? Ih other words, I would expect A=B because every time there is a click in A (which is on one side only) there will also be a matching click on both sides in B because the p-way side (no double slit) always clicks in coincidence.

So I remain cornfused!

 

[Cthugha]

Now, suppose we coincidence count with the other side, which has no double slit (let's call this B). How will the resulting subset look any different than the total pattern just mentioned? Ih other words, I would expect A=B because every time there is a click in A (which is on one side only) there will also be a matching click on both sides in B because the p-way side (no double slit) always clicks in coincidence.

This depends strongly on the kind of detectors you wish to use. For example if you use a small sized detector to scan the x-axis at B and a bucket detector at A, which is as large or larger than the whole single slit diffraction pattern, you are right. There will be no interference pattern arising and the coincidence count pattern will look like the pattern seen before at A.

The interference pattern will only occur if you use two small detectors, so that there is no averaging over several initial phases.

 

[Dmitry67]

Probably I am missing something obvious, but why there are always coincidence couters?

What if we put a beam (intensive enough, I don't need individual photons) thru doubleslit (slits A and B) then light behind each slit is downconverted, forming A->Au, Ad and B->Bu,Bd

So Au and Bu should form an interference pattern, right or wrong?
As well as Ad and Bd?

 

[Sammy k-space]

sounds like the dancing Wu Li masters.

 

[DrChinese]

Probably I am missing something obvious, but why there are always coincidence couters?

What if we put a beam (intensive enough, I don't need individual photons) thru doubleslit (slits A and B) then light behind each slit is downconverted, forming A->Au, Ad and B->Bu,Bd

So Au and Bu should form an interference pattern, right or wrong?
As well as Ad and Bd?

You put the double slit *after* the down conversion crystal. Before doesn't really accomplish anything.

 

[DrChinese]

This depends strongly on the kind of detectors you wish to use. For example if you use a small sized detector to scan the x-axis at B and a bucket detector at A, which is as large or larger than the whole single slit diffraction pattern, you are right. There will be no interference pattern arising and the coincidence count pattern will look like the pattern seen before at A.

The interference pattern will only occur if you use two small detectors, so that there is no averaging over several initial phases.

Thanks. In the example, they use the "bucket" type for the p-way. And it seems like you could use an array of small detectors anyway and then avoid having to move the one small detector from spot to spot. (Perhaps economics was an issue.)

So that leaves me wondering in this situation, as mentioned. Perhaps I need to look at the underlying paper to the summary page. I'll see if I can find a good link, and perhaps that will explain further.

 

[DrChinese]

OK, the link was on the page itself:

http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

It says the same as the summary page referenced earlier. There is no discussion at all of the coincidence counting, other than labeling it as such.

So I remain confused. I do not see what the purpose of the coincidence counting is, and yet I know that if it is not needed then FTL signaling would be possible (which must be wrong). I guess if I saw the raw graph of the double slit side without coincidence counting, it would explain the missing piece of information. Perhaps there is a lot of noise that the coincidence counting corrects for.

 

[Dmitry67]

You put the double slit *after* the down conversion crystal. Before doesn't really accomplish anything.

I suspected that.
But could you explain why?
Is down conversion process reversible or irreversible?

 

[Cthugha]

Thanks. In the example, they use the "bucket" type for the p-way.

Are you sure? At least in the quoted paper they use a 300 or 600 micrometer slit in front of the detector.

 

[DrChinese]

Are you sure? At least in the quoted paper they use a 300 or 600 micrometer slit in front of the detector.

They say "A stepping motor is used to scan detector Ds." but not Dp. The p-way is going straight into the Dp detector and therefore does not need to be moved. There is no dispersion to speak of for that beam since it is not going through a double slit.

 

[Cthugha]

The p-way is going straight into the Dp detector and therefore does not need to be moved. There is no dispersion to speak of for that beam since it is not going through a double slit.

I disagree. You won't be able to get a perfectly parallel beam out of a BBO, therefore the emission will form a cone. The opening angle might be very small, but it will be there. Let us do a rough calculation to estimate, what happens under the conditions of the experiment. The entangled photons have a wavelength of roughly 700 nm. The distance from the BBO to the crystal is roughly one meter. Due to the small opening angle the photons, which hit the screen at a slightly different position, will have traveled a slightly different distance. Now the interesting point is, which horizontal distance D on the screen corresponds to a difference of half the wavelength in the distance the photons have traveled because as soon as the screen (or the detector) becomes this large you are averaging of all kinds of initial phases. Using Pythagoras we get:

$$1m^2 +D^2= (1m+350 nm)^2$$

This leads to $$D^2 = 7* 10^{-7}m^2 + 12,25 *10^{-14} m^2$$

If you neglect the final part you get roughly D=830 micrometers. Now the slit is smaller than this size, roughly 300 or 600 micrometers respectively. This means that you are indeed in the regime of a small detector (even if it is not moved) because your slit acts like a phase filter. If you increased the slit size to much more than 830 micrometers and performed coincidence counting, the interference pattern should vanish.

Usual spot diameters of unfocused beams one encounters in optics are usually even larger than this size, so that seems resonable to me.

 

[DrChinese]

I disagree. You won't be able to get a perfectly parallel beam out of a BBO, therefore the emission will form a cone. The opening angle might be very small, but it will be there. Let us do a rough calculation to estimate, what happens under the conditions of the experiment. The entangled photons have a wavelength of roughly 700 nm. The distance from the BBO to the crystal is roughly one meter. Due to the small opening angle the photons, which hit the screen at a slightly different position, will have traveled a slightly different distance. Now the interesting point is, which horizontal distance D on the screen corresponds to a difference of half the wavelength in the distance the photons have traveled because as soon as the screen (or the detector) becomes this large you are averaging of all kinds of initial phases. Using Pythagoras we get:

$$1m^2 +D^2= (1m+350 nm)^2$$

This leads to $$D^2 = 7* 10^{-7}m^2 + 12,25 *10^{-14} m^2$$

If you neglect the final part you get roughly D=830 micrometers. Now the slit is smaller than this size, roughly 300 or 600 micrometers respectively. This means that you are indeed in the regime of a small detector (even if it is not moved) because your slit acts like a phase filter. If you increased the slit size to much more than 830 micrometers and performed coincidence counting, the interference pattern should vanish.

Usual spot diameters of unfocused beams one encounters in optics are usually even larger than this size, so that seems resonable to me.

1. You're probably right, but I think I have been denser than usual and can now explain this. Using the 3 cases described below:

Case I: if no polarizer on p-way, no quarter wave plates on s-way, then total* s-way pattern is interference.

Case II: if no polarizer on p-way, quarter wave plates added on s-way, then total* s-way pattern is NO interference.

Case III: if polarizer added on p-way, quarter wave plates remain on s-way, then total s-way pattern is interference... and so is the subset which should be exactly half intensity of the total beam. Now, this MUST be true because otherwise you could perform FTL signaling by either inserting or removing the p-way side polarizer; and the s-way total pattern would change.

The coincidence counting may not be strictly needed in the case of I and II. But that does not allow the possibility of any signaling anyway. If you modify any double slit setup so as to learn the which path information, the interference disappears. So clearly nothing strange so far.

2. Now, looking at Case II vs. Case III: The coincidence counting IS needed for case III because you are selecting the subset of the fringe pattern (or the anti-fringe pattern, which are figures 4 & 5 in the reference (8 & 9 in the delayed choice version). The subset is that which goes through the polarizer (which is erasing the possibility of learning which-path information).

The total pattern on the s-way side for all Case IIIs is always simply the same as the total for the Case II on the s-way side! To say it a different way: Total = Fringe+Anti-Fringe. That was the missing element for me, and exactly matches what the OP said now that I re-read it. So sorry for the extra time to get me to the right place.

3. Now what is left confusing for me is the idea that an interference forms in Case I, which appears to contradict what Zeilinger said in that reference provided. But looking at the diagram in his reference, (fig. 2), it is clear that there is the possibility of obtaining which path information. So that matches Case II rather than Case I. There is no interference pattern, which matches Zeilinger's statement. So that works.

But you would imagine that you could remove the double slit on one side, and it would return to the Case I. If that could happen, then the interference pattern would re-appear. And we know that doesn't happen, because FTL signaling is then possible. So perhaps the coincidence counting is strictly necessary after all in order to see the patterns. Which would agree with what you have been saying all along anyway.

But it sure makes me interested in what the Case I and Case II patterns on the s-way side look like without coincidence counting...