# Delayed Choice Bell-state Quantum Eraser

1. May 28, 2006

### Burnsys

http://www.joot.com/dave/writings/articles/entanglement/spookiness.shtml

http://www.joot.com/dave/writings/articles/entanglement/quantum-eraser.png

1. a laser fires photons into a Beta Barium Borate (BBO) crystal;
2. the crystal entangles some of the photons; and then
3. entangled photons travel to two different detectors: A and B.

Placed between the crystal and detector B is a double-slit, like in the previous experiments. Immediately in front of detector A is a polarizing filter that can be rotated.

Each slit is covered by a substance that changes the polarization of a photon. Consequently, the left-hand slit will receive photons with a counter-clockwise polarization, and the right-hand slit will pass photons with a clockwise polarization.
-----------------------------------
If we measure polarization at detector A then we have wich-way information and the interference pattern disapear at B, if not we have interference pattern at detector B

Now suppose we place a 1 millon km long fiber optic between the BBO crystal and detector A so that each photon will arrive 3 seconds later at detector A.

Now we have 3 seconds to decide if we want to get WichWay information or not after the photons hited detector B.

So, if i see an interference pattern at B then i set detector A to measure Polarization while the photons are still traveling.. What would be the results of this experiments?

Last edited: May 28, 2006
2. May 29, 2006

### vanesch

Staff Emeritus
This comes back and back and back. You will not see an interference pattern at B. You only find an interference pattern at B *IN COINCIDENCE* with A. The "interference pattern" is in fact a "correlation pattern": if you look at all clicks at B WHICH CORRESPONDED TO A CLICK IN A, then the hits at B show an interference pattern. If you look at all the hits at B, irrespective of what might happen at B, then you'll see nothing.

This is the usual misunderstanding of these experiments, but I can understand why: very often the publications "over-sell" the result, and make it sound AS IF there was a bare interference pattern at B only, which then triggers (rightly) all these questions about paradoxes or FTL communication.

3. May 29, 2006

### Burnsys

Then detector B is not a screen and i will not see somenthing like this :

http://www.physics.utoledo.edu/~ljc/interfer03.jpg

???

Just trying to understand. Thanks

4. May 29, 2006

### vanesch

Staff Emeritus
The erroneous statement is this:

Of course polarization affects interference patterns !

What happens, is simply this: when you put the perpendicular polarizers in front of each slit at B, you DO NOT GET AN INTERFERENCE PATTERN.
However, when you put now a polarizer at 45 degrees in front of detector A, and you PICK THE COINCIDENCES of A and B (this removes about half of the photons at B, which do not correspond to a click in A), then it turns out that this SUBSAMPLE shows an interference pattern.
But given that you don't know the polarization of the pair (given that your A-click was after a polarizer at 45 degrees), you will not be able to say through which slit its partner went.
However, if you put the A polarizer to 90 degrees, or to 0 degrees, AND ASK COINCIDENCE AGAIN, you will have a subsample at B that will NOT show interference. This is because knowing the click at A, you know what polarization its partner had, and hence through which slit it went at B.

But in no case, by doing something at A, you see something change at B WHEN ONLY LOOKING AT B.

5. Nov 14, 2010

### Kyle-DFW

Vanesch:

I agree with your "subsample" comments. Since you've earned credit in my book as someone who is well-versed on the details, I'm wondering if you have a theory or explanation for why an interference pattern is observed in the simple double-slit experiment where regular (non-entangled) photons are used and the photons are shot at the slits one at a time?

It seems to me that one must conclude that individual photons are interfering with themselves, and therefore an individual photon must truly be a wave which is passing through both slits simultaneously. Therefore, the particle-like properties of a photon (like position) may not actually be present until the moment of observation where the wave of potential locations collapses to a single measured location. The possible measured locations are then constrained to the interference pattern.

I suppose another possibility is that our belief that we can emit individual photons may be wrong.

6. Nov 15, 2010

### DrChinese

Well, actually, that would simply be quantum theory. In sum: if you have the potential for which-slit information, there is no interference pattern.

So I think what you are asking is: what is the physical mechanism by which this result occurs? That is presently unknown, even though the quantum description appears complete.

7. Nov 22, 2010

### Kyle-DFW

When I first read this thread, I tended to agree with the comments from vanesch that the interference pattern was an artifact of looking only at a subsample of photons which are detected at B after a corresponding photon is detected at A.

This would seem to make the experiment not seem so mysterious because it would suggest that only photons that come through the slits with opposite polarizations will generate an interference pattern.

However, after studying again the information I have been able to find about these experiments online, I realize vanesch made a key mistake:

He stated that the interference pattern is seen when you look at the subsample of photons at B that correspond to a click in A. IT IS THE OPPOSITE. When the path information is available at A (coincidence clicks), the which-path information is available and there is NO INTERFERENCE at B. You get two bands as you would expect if photons were classical particles. When there is no which-path information available, THIS IS WHEN YOU GET AN INTERFERENCE PATTERN. Since on each measurement, there is only a single photon fired at the slits, the presence of an interference pattern forces one to conclude that a SINGLE PHOTON at that point is in fact a wave traveling through both slits simultaneously and interfering with itself. When the photon is finally detected at B, it again looks like a particle because it is seen at a discrete location in space, but the location where it is found will never be outside of the bounds of the interference pattern.

From what I have read, it also seems reasonable to say that polarization really does NOT have an effect on the interference pattern because only coherent (in-phase) photons are required to get an interference pattern. Polarization is not required to see it, and in the experiment, the polarized photons are coherent. In other words, coherent non-polarized photons can generate an interference pattern in the absence of which-path information, but so can coherent polarized photons.

References:

http://davidjarvis.ca/entanglement/

http://en.wikipedia.org/wiki/Coherence_(physics [Broken])

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8. Nov 23, 2010

### Joseph14

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9. Nov 23, 2010

### Joseph14

To summarize the results:
1. No polarizer at A and no quarter wave plates at B------Interference
2. No polarizer at A and quarter wave plates at B---------No Interference
3. Polarizer at A and quarter wave plates at B------------Interference depending on angle of Polarizer

My interpretation
1. The photons are coherent so you get interference (even without coincidence counting)
2. QWP put the photons out of phase (by a quarter wave) so now there is no interference because the QWP messed up the coherence
3. Polarizer at A now selects out the photon polarizations such that the fast axis and slow axis of the quarter wave plate don't mess up the interference

10. Nov 23, 2010

### Joseph14

It isn't the opposite look at the experiment again. The results are always based on coincidence counts.

The spacing is close enough that you wouldn't get two bands, they would blur together.

The polarization does have an effect for the case when QWP are in place because the coherency is destroyed based on the polarization of the photon.

11. Nov 23, 2010

### Kyle-DFW

I've been re-reading and giving a lot of thought to this document:

My current belief (for today anyway lol) is that we are really not even dealing with an issue of subsamples here.

If I understand correctly, when a polarizer is present at detector A (detector p in the document), meaning which-path information is available, the position of all photons encountered at detector B (detector s in the document) is still recorded.

Because the entangled photons come through one at a time (because the BBO crystal rarely splits a photon) the computer counter knows that the photon at A should arrive with virtually no time difference to the arrival at B. If the photon at B is recorded and there is a coincident photon recorded at A (with the polarizer in place), you could say that the photon at B came through slit 1 for example. If the photon is recorded at B and within a very short time window, there is no coincident photon recorded at A (because it was blocked by the polarizer), then you would know that the photon at B came through slit 2.

When the experiment is configured this way, the positions for ALL photons encountered at B are still recorded, and there is NO interference pattern. You get two bands, which is what you would expect if each photon passed through only one slit or the other as a classical particle.

Now as I understand it, you can restore the interference pattern without changing ANYTHING about the detector B setup or the photons entering it. All you have to do is "erase" the which-path information at detector A by either removing the polarizer (whereby the photon will be absorbed at detector A with all chance to measure its polarization lost) or by rotating the polarizer at A to 45 degrees where it would be impossible to determine the x or y polarization of the photon arriving at A (because a polarizer at 45 degrees has a 50/50 chance of either passing or blocking both x and y polarized photons).

In this scenario, again, the positions of ALL photons reaching detector B are recorded, but this time the recorded positions will be constrained to the interference pattern. It is not a subsample of photons at B. It is all of them. It is true that with a 45 degree polarizer at detector A, you would only record a subsample of photons at detector A, but these photons could have been either x or y polarized (prior to the polarizer at A) which means you have no which-path information. You would see the interference pattern at B whether you examine all photons at B or even just the ones that coincide with photons detected at A.

If what I have said is correct, then it really DOES mean that you can make a change only only to the configuration of detector A and effect a change in the pattern recorded at detector B.

I believe Membrane Theory will eventually show that Quantum Entanglement is not Faster-Than-Light communication, but is the observed by-product of wave interactions on a seamless hyper-spherical 11-dimensional space-time membrane. These wave interactions are what we perceive as sub-atomic entities. All matter and energy is merely waves on the membrane, and therefore everything is connected to everything else on the membrane. The perception of separate objects or particles is an illusion.

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12. Nov 23, 2010

### Kyle-DFW

If coherency was destroyed by the polarization process (which I have read nothing to suggest it would be), then you would not be able to restore the intereference pattern simply by modifying the detector A configuration. In this case, everything is the same on the detector B side, including the QWP, but the interference pattern returns.

13. Nov 23, 2010

### Joseph14

A QWP provides 2 different speed to pass through the crystal corresponding to a fast polarization and a slow polarization. We will call these P1 and P2.

If light has a random polarization then half of the photons will take the slow route and half the fast route at both slit 1 and slit 2, which will mess up the interference.

Now if the polarizer at detector A forces the polarization to be P1, then photons at slit 1 will always take the fast route and at slit 2 always the slow route so there will be interference. And the opposite will happen for orrientation P2 resulting in interference as well.

14. Nov 23, 2010

### Kyle-DFW

I don't believe the polarizer at A "forces" anything. It either passes the incoming photon or doesn't, based on the photon's incoming polarization. This merely enables you to know the which-path information and should not logically have any effect on the photon behavior or results at B.

Also, since my understanding is that each measurement is taken with a SINGLE PHOTON at B along with its entangled partner at A, it seems that a single photon could not be de-coherent with itself, yet the interference pattern still emerges as these individual photons pile up.

There is nothing I have read that says anything about fast route / slow route having anything to do with whether or not an interference pattern is observed.

15. Nov 23, 2010

### Cthugha

No, it is not that easy, but the webpage you linked obscures the reason for why that is so, because the pictures are somewhat hard to read. If you have a look at the pdf linked at the beginning of the document, you will find that both detectors are not bucket detectors, but only small-area detectors. So if you detect a photon at B, but no photon at A, it does not necessarily mean that the photon was blocked by a polarizer. Your small-area detector A might also just not be at the right position to detect the photon. If you had a large-area detector in that place, you should in my opinion also get no interference pattern or - more correctly - you would integrate over many shifted interference patterns that add up to no pattern at all.

16. Nov 23, 2010

### Kyle-DFW

I have to disagree. Why would Detector A need to be a small area detector? Everything I've seen indicates that A is a bucket detector. If there is something in the PDF to the contrary, please quote it. I didn't see it.

Even if some photons were not seen at A for some sort of positional reason (rather than being blocked by the polarizer) it should be a small number that would not significantly effect the results. Detector A's only job is either absorb the photon as-is, eliminating any which-path information because the polarization was not measured, or to absorb the photon after it passes through the polarizer (if it does), at which point which-path information is available. The spatial location where the photon strikes detector A is irrelevant.

Detector B is a small area detector that moves through space across the potential range of the interference pattern and gives you a count for the number of photons detected at each spatial point in the range. Detector B sits at each position long enough to count enough photons that you get a good sense of the relative number of photons hitting each point.

17. Nov 23, 2010

### Cthugha

The paper says:
"The detectors are EG&G SPCM 200 photodetectors, equipped with interference filters (bandwidth 1 nm) and 300 micrometre x 5 mm rectangular collection slits.
A stepping motor is used to scan detector Ds ."

Both detectors are of equal size. You need to move Ds along to detect all photons. You would also need to move the other detector around to detect all photons on the other side.

Sorry, but this is not a small number. I doubt you will get more than 5% of all the photons at a fixed position of detector A. The exact position of that detector represents a measurement of the wavevector or equivalently emission angle of that photon. Entanglement relies on two photons emitted with well defined sum momentum, but variable magnitude of the momentum of the single photons. Therefore the area on the fixed-detector side which can get hit by photons is quite large. The spatial location where detector A is hit is not irrelevant. You would get an interference pattern at in the coincidence counts at each possible position of detector A, but it would be slightly shifted at each of these positions.

18. Nov 23, 2010

### Kyle-DFW

I don't believe this specific scenario was tested by these particular scientists, but everything I've read leads me to believe you would get interference in this scenario because no which-path information is available without the polarizer.

19. Nov 23, 2010

### Cthugha

The result of exactly this scenario is shown in figure 3. The result is: no interference.

20. Nov 23, 2010

### Kyle-DFW

I'll concede now that the detectors are both small and that a lot of photons are lost at A. I'll further concede to an earlier post that the experiment done by these scientists only records at B when there is a coincidence at A. However, I'm still of the opinion that this doesn't matter because I believe the lost photons don't cause a change in the pattern. They only cause it to take longer for the coincidences to be detected and longer for the pattern to build up.

The which-path information is either present or erased regardless of whether detector A was able to capture the photon. All photons at B will show interference in the absence of which-path information, and no interference in the presence of which-path information, regardless of whether you look at all photons at B or only the ones where you also capture the partner photon at A.

It may be that they only looked at the B photons which were in coincidence with A for some reason such as to eliminate noise photons that might be entering the system from other sources.

How is there going to be an interference pattern in the photons arriving at A when there are no double-slits at A? There isn't.

The guys that did this experiment seem pretty smart to me, and I'd have a hard time believing they could overlook something so seemingly basic if it was really going to somehow impact the validity of their results.