Delayed Choice Bell-state Quantum Eraser

[San K]

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.

Vanesch,

Can the above (i.e. subsamples and wavevector) be extended to explain DCQE as well?

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

Specifically the fact that the pattern (obtained/filtered/subsampled via coincidence) that is formed on Ds corresponds to what we did to p (eraser or no-eraser) after s was registered at Ds.

 

[San K]

down-conversion momentum needs to be conserved.

this is a bit off topic, i am asking a fundamental question about quantum entanglement:

why does momentum need to be conserved? is it because there is no friction for dissipation of momentum thus the total momentum since the time the two pairs were created needs to remain same?

 

[StevieTNZ]

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.

I would question if its potential which-way info that causes the interference to cease.

 

[ejproducts]

I have a related question, I think. I have been reading about the Scully-Druhl experiments where the photons are directed to down-converters, which create a "signal" photon and an "idle" photon, and in which it appears that the idle-photon gives information about which path the signal photon took. This has made me curious about something.

If I have a laser, and fire a photon into a beam-splitter, then my photon can go L or R towards my photo-sensitive paper. If I can detect the path, I understand, I will get no interference pattern, but if I cannot, then I will get an interference pattern as if the photon went along both paths. I place on each path a down-converter, splitting my photon into two lower-energy photons, one - the signal photon - which goes to the photographic paper, and another - the idler photon - which does not. If I place detectors at the ends of the path of the idler photons, I can tell whether my signal photon went L or R. However, I could also direct my idler photons to a single detector, making me unable to see which path my signal photon took. The first scenario would give me no interference pattern, but the second would.

My idler photons travel down a huge length of optical fibre to a human settlement on another planet 10 light years away. At the end of their ten-year journey, they either go into separate detectors, so that I can tell if they went L or R, or into a single detector, so I cannot tell. The decision is made by my friend at the other end, whom I spoke to before conducting the experiment, and the decision is made 9.5 years after my photons hit the down-converters.

I can look at the photographic paper 10 years before it is determined that the idler photons will be able to reveal any information about what path the signal photons took. What will I see?

(Edit: I forgot to put the slits in! Each signal photon would go through a slit, and each idler photon would give information about whether the signal photon went through the L slit, the R slit, or would give no information.)

 

[DrChinese]

Down converted photons do not exhibit interference patterns unless the which path information in its partner is fully erased (as I think you are saying). Generally, it is not possible to erase the information well enough to sense this without coincidence counting. In other words, the pattern you see never changes regardless of what happens on the other planet 10 years later.

BTW: I think if you look at your diagram, the setup on the right is redundant.

 

[ejproducts]

What do you mean by "it is not possible to erase the information well enough to sense this without coincidence counting"?

You are right about the diagram, also. Cheers.

 

[DrChinese]

What do you mean by "it is not possible to erase the information well enough to sense this without coincidence counting"?

You are right about the diagram, also. Cheers.

Suppose you could erase at will. Then you could send signals FTL because Alice could make the interference pattern appear or disappear at Bob's side. But that doesn't happen. Instead, the interference pattern will only appear inside a subset of the events, and the pattern at Bob never varies at all. The subset is one which depends on coincidence counting.

 

[San K]

Suppose you could erase at will. Then you could send signals FTL because Alice could make the interference pattern appear or disappear at Bob's side. But that doesn't happen. Instead, the interference pattern will only appear inside a subset of the events, and the pattern at Bob never varies at all. The subset is one which depends on coincidence counting.

The pattern at Bob never varies at all...and ...it does not vary at Alice either?

i.e. both get a scattering of dots...till subset are created via co-incidence counting

 

[San K]

Suppose you could erase at will. Then you could send signals FTL because Alice could make the interference pattern appear or disappear at Bob's side. But that doesn't happen. Instead, the interference pattern will only appear inside a subset of the events, and the pattern at Bob never varies at all. The subset is one which depends on coincidence counting.

Dr Chinese, if the entire setup was placed in a noiseless environment (i.e. no random photons striking the detector other than the entangled pairs sent/created by the experimenter), would we still need coincidence counting?

 

[marksesl]

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.

I hate to say this, but I think Vanesch has a misunderstanding of the nature of the coincidence meter. It changes nothing about the general nature of the experiment. If one plots the coincidences between A and B after the fact, and certain correlations show an interference pattern while others correlations do not, it's just that same as if there was no coincidence meter at all and you did the experiment in real time. For example, in delayed erasure any B photon would have still struck before its partner A photon struck, and any state the whole apparatus is in would still produce a visible interference pattern or not (if there was a screen instead of just a detector at B). The coincidence meter is only there to filter out background noise, and if one wanted to alter the experiment at any moment, one could do so and still get valid results just by looing at correlations at some later time without having to wait a long time to get visual conformation in real time.

 

[marksesl]

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.

This is true because orthogonal waves cannot interfere. There is confusion because polarized light shone upon both slits will still behave normally, and that is what is meant when one reads that polarizing photons will not affect them in double slit experiments. Polarizing one slit differently from the other is a different matter and does affect interference, making it impossible.

However, when you now put 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.

This is where much of the confusing originates. The filter at A only polarizes about 50% of both x and y-axis A photons to diagonal. Or, you could say, it is passes 50% both x and y polarized photons (which are always being passed anyway if there is no filter), now they will all be diagonal. However, the photons at B get polarized too, to diagonal. There is no ignoring of half the hits at B ("this removes about half of the photons at B, which do not correspond to a click in A"). Since 50% of the photons at B get polarized to diagonal too, they pass through the quarter-wave plates without making the right-hand and left-hand tell-tail photons, but instead the photons leaving each side are the same, which naturally causes an interference pattern. All of the photons at B detector still coincide with their matching partners at A. There are no non-corresponding hits at B (except for noise photons which are always ignored no matter what).

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.

This is true because the idler photos at A are being polarized the same. An x photon now is diagonal, and a y is now diagonal. As mentioned above, only about 50% of horizontal or vertical photons make it through a diagonal filter, but those that do still emerge diagonal, so their associated entangled photons will be diagonal too, and only those will be counted by the coincidence counter.

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.

This seems to be giving the impression that both possibilities are always at detector B, and we just see one or the other by filtering out half of the information at B by turning the polarizer at A. In actuality (as already mentioned), the A polarizer not only polarizes the A photons to diagonal, but also the B photons to diagonal because they are entangled. Turning the A polarizer to 90 degrees or to zero degrees also does the same to the entangled photons at B, and this causes the polarizations at B to be orthogonal again, disallowing interference.

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

There is no screen at B, only a detector that moves back and forth picking up fringe patterns in that manner. Only some of the photons that enter the BBO crystal are entangled, the rest are considered to be noise. So, what affect that may have on a visual image could be taken into consideration. Any light going through the double slits should produce a fringe pattern, whether entangled photons or the noise photons. Once the quarter-wave plate filters are installed, the interference pattern should go away as long as the incident light is either horizontal or vertical polarization, as both the entangled pairs and the noise photons are (Keep in mind that the diode laser pump that produces all the photons linearly polarizes all of them). But, when a diagonal filter is placed at A, only entangled pairs at B will be polarized, so visually fringes may not arise again, since any horizontal and vertical noise photons will still result in a bar pattern. It's difficult to say what it would look like, perhaps both patterns would appear at once.

 

[Cthugha]

Any light going through the double slits should produce a fringe pattern, whether entangled photons or the noise photons.

This is obviously incorrect. It is even a prerequisite for entangled light to be incoherent (producing no first order interference pattern) in a certain experimental geometry. See Phys. Rev. A 63, 063803 (2001) for details. The article can also be found on the ArXiv: http://arxiv.org/abs/quant-ph/0112065.

It is essential to many variants of the quantum eraser experiment that the whole light field is incoherent and only a subset picked by a spatially narrow coincidence counter is coherent. Or as Stephen Walborn formulated it at CQO X: It is a matter of bookkeeping.

The coincidence meter is only there to filter out background noise, and if one wanted to alter the experiment at any moment, one could do so and still get valid results just by looing at correlations at some later time without having to wait a long time to get visual conformation in real time.

This is also incorrect. It is not the sole purpose of the coincidence counter to remove noise. While the counters do not change any detection, it is very important that the counters do not sample the whole field, but just a subset of them. The nature of this subset (width in real space or momentum space) has a significant impact on the pattern seen in the coincidence counts. They filter out more than just background noise.

 

[marksesl]

This is obviously incorrect. It is even a prerequisite for entangled light to be incoherent (producing no first order interference pattern) in a certain experimental geometry. See Phys. Rev. A 63, 063803 (2001) for details. The article can also be found on the ArXiv: http://arxiv.org/abs/quant-ph/0112065.

It is essential to many variants of the quantum eraser experiment that the whole light field is incoherent and only a subset picked by a spatially narrow coincidence counter is coherent. Or as Stephen Walborn formulated it at CQO X: It is a matter of bookkeeping.

Light must be "coherent" to produce interference patters. The light from the diode laser pump that produces the photos obviously produces coherent light. Any photons not involved in entanglement is still coherent, and any entangle pair is coherent and can produce interference. If not, it would be utterly impossible to preform even the first stage of the experiment - produce interference. The entangled pairs are oppositely polarized, so perhaps that is what you mean, they are incoherent in respect to one another.

 

[marksesl]

This is obviously incorrect. It is even a prerequisite for entangled light to be incoherent (producing no first order interference pattern) in a certain experimental geometry. See Phys. Rev. A 63, 063803 (2001) for details. The article can also be found on the ArXiv: http://arxiv.org/abs/quant-ph/0112065.

It is essential to many variants of the quantum eraser experiment that the whole light field is incoherent and only a subset picked by a spatially narrow coincidence counter is coherent. Or as Stephen Walborn formulated it at CQO X: It is a matter of bookkeeping.



This is also incorrect. It is not the sole purpose of the coincidence counter to remove noise. While the counters do not change any detection, it is very important that the counters do not sample the whole field, but just a subset of them. The nature of this subset (width in real space or momentum space) has a significant impact on the pattern seen in the coincidence counts. They filter out more than just background noise.

Anyone can do a very similar quantum eraser experiment at home using two linear polarizing filters and one diagnal filter, a laser pointer, and a piece of foil with two slits cut in it. There is no need for an incidence counter. Just how to proceed with this experiment should be obvious. After producing an interference pattern with the double slits, sticking on the polarizing filters over the slits will make the interference pattern vanish. The placing the diagnal filter in front of the two slits will bring the interference pattern back "erasing which-way information," but it's really just the geometry of the light. Great science project for the kids.

 

[Cthugha]

Light must be "coherent" to produce interference patters.

Yes, but there are different orders. Two-photon interference requires coherence in terms of the relative phase of a photon pair. Single-photon interference requires single-photon coherence.

The light from the diode laser pump that produces the photos obviously produces coherent light.

Of course it is second-order coherent (which the SPDC light is not) and of large spatial coherence (which the SPDC light is also not), but the SPDC light is relevant.

Any photons not involved in entanglement is still coherent, and any entangle pair is coherent and can produce interference.

No, entangled photons are never first-order coherent for the typical pumping scheme used here. Also note that first order coherence (what is tested in a simple double slit) is not a property of the source, but also of the experimental geometry. You can increase it by increasing the distance between the light source and the double slit. The reason is simple: spatial coherence is inversely proportional to the spread in momentum space. A large spread in momentum space is equivalent to a large range of angles under which light is emitted. This translates into a path length difference and therefore a phase difference which reduces the visibility of the interference pattern seen. For entangled light you need a large spread in momentum space. If you do not have it, you cannot violate Bell's inequalities.

This is the quintessence of the experiment. The whole ensemble of SPDC photons is first-order incoherent and will not create any interference pattern (or equivalently a superposition of many of them resulting in no pattern at all) in a simple double slit experiment. You can get two-photon coherence, though. This means that when you pick a certain subset of entangled photons on one side (typically with a small spread in momentum space), the coincidence counts will also correspond to another subset of photons with well defined momentum. This subset is spatially coherent and can produce an interference pattern, but is is a two-photon interference pattern as you cannot see it without "cherry-picking" by doing coincidence counting and selecting a proper subset showing the properties you ask for (here narrow momentum distribution).

If not, it would be utterly impossible to preform even the first stage of the experiment - produce interference. The entangled pairs are oppositely polarized, so perhaps that is what you mean, they are incoherent in respect to one another.

No, this is absolutely not, what I mean. I have given you a good reference to understand why spatial coherence is important in any SPDC experiment. Zeilinger also gave a good discussion on that in an older paper, where he calculates the optimal distance between SPDC crystal and experiment for a given pump spot size as he needs to stay away from the far field to avoid spatially coherent light at the setup. I do not know the reference by heart, but I might be able to dig it up.

edit: Just to make sure that we do not misunderstand each other: When using entangled light, there is no "bare" interference pattern without using coincidence counting.

Anyone can do a very similar quantum eraser experiment at home using two linear polarizing filters and one diagnal filter, a laser pointer, and a piece of foil with two slits cut in it. There is no need for an incidence counter.

Well, yes, you can do a lot of stuff without delayed choice. That is indeed a trivial experiment. The delayed choice part is the interesting part.

Wow...I just noticed that this topic is 2 years old. Do you have any deeper interest in that particular setting? Otherwise it might be easier to just let the thread die. We have plenty of topics on DCQE and similar stuff on these forums.

 

[marksesl]

Yes, but there are different orders. Two-photon interference requires coherence in terms of the relative phase of a photon pair. Single-photon interference requires single-photon coherence.



Of course it is second-order coherent (which the SPDC light is not) and of large spatial coherence (which the SPDC light is also not), but the SPDC light is relevant.



No, entangled photons are never first-order coherent for the typical pumping scheme used here. Also note that first order coherence (what is tested in a simple double slit) is not a property of the source, but also of the experimental geometry. You can increase it by increasing the distance between the light source and the double slit. The reason is simple: spatial coherence is inversely proportional to the spread in momentum space. A large spread in momentum space is equivalent to a large range of angles under which light is emitted. This translates into a path length difference and therefore a phase difference which reduces the visibility of the interference pattern seen. For entangled light you need a large spread in momentum space. If you do not have it, you cannot violate Bell's inequalities.

This is the quintessence of the experiment. The whole ensemble of SPDC photons is first-order incoherent and will not create any interference pattern (or equivalently a superposition of many of them resulting in no pattern at all) in a simple double slit experiment. You can get two-photon coherence, though. This means that when you pick a certain subset of entangled photons on one side (typically with a small spread in momentum space), the coincidence counts will also correspond to another subset of photons with well defined momentum. This subset is spatially coherent and can produce an interference pattern, but is is a two-photon interference pattern as you cannot see it without "cherry-picking" by doing coincidence counting and selecting a proper subset showing the properties you ask for (here narrow momentum distribution).




No, this is absolutely not, what I mean. I have given you a good reference to understand why spatial coherence is important in any SPDC experiment. Zeilinger also gave a good discussion on that in an older paper, where he calculates the optimal distance between SPDC crystal and experiment for a given pump spot size as he needs to stay away from the far field to avoid spatially coherent light at the setup. I do not know the reference by heart, but I might be able to dig it up.

edit: Just to make sure that we do not misunderstand each other: When using entangled light, there is no "bare" interference pattern without using coincidence counting.



Well, yes, you can do a lot of stuff without delayed choice. That is indeed a trivial experiment. The delayed choice part is the interesting part.

Wow...I just noticed that this topic is 2 years old. Do you have any deeper interest in that particular setting? Otherwise it might be easier to just let the thread die. We have plenty of topics on DCQE and similar stuff on these forums.

Your remarks are quite informative. As I understand it now, the photons from the BBO crystal are no longer coherent because entangled pairs do not have to be equal in frequency, but just have frequencies that add up to their parent photons. The entangled pairs represent kind of rainbow of all colors, thus being incoherent, cannot interfere. Is that correct? How can the detectors detect entangled pairs that are coherent though from the total where most are not? And, what exactly what happens when a diagonally polarized photon goes through a quarter-wave-plate? Also, how can I edit my post? I edited it before, but I no longer see an edit button.

 

[DrChinese]

This is obviously incorrect. It is even a prerequisite for entangled light to be incoherent (producing no first order interference pattern) in a certain experimental geometry. See Phys. Rev. A 63, 063803 (2001) for details. The article can also be found on the ArXiv: http://arxiv.org/abs/quant-ph/0112065.


(and other comments)
...

Cthugha, a question for you. I see that when you have a smaller width source, we end up with something that cannot produce momentum entanglement because the delta p grows as delta q shrinks. But it seems to me that the output of the PDC crystal would still be polarization entangled. Is that correct? I know that must be wrong, but cannot figure out why.

 

[Cthugha]

Your remarks are quite informative. As I understand it now, the photons from the BBO crystal are no longer coherent because entangled pairs do not have to be equal in frequency, but just have frequencies that add up to their parent photons. The entangled pairs represent kind of rainbow of all colors, thus being incoherent, cannot interfere.

This would be the case for energy entanglement vs. temporal coherence (measured in a Mach-Zehnder interferometer), yes. For many typical PDC sources it is rather momentum entanglement vs. spatial coherence (measured using a double slit), but the principle is the same. The total off-axis momentum (corresponding to emission angle from the normal) must add up to that of the parent photon, but the momentum distribution in each arm is broad, resulting in a wide emission cone and low spatial coherence.

Is that correct? How can the detectors detect entangled pairs that are coherent though from the total where most are not?

The way to get back coherence is filtering. In the case of entanglement in energy you could place a narrow spectral filter in one arm and only pick a narrow energy range. The corresponding photons in the other arm will then also feature a narrow spectral range and you will see some longer coherence time when doing coincidence counting.

For momentum entanglement and spatial coherence, filtering is much simpler. You just need a narrow detector which is so small that it only detects photons emitted under some specific angle, preferably placed in the Fourier plane. All of these photons will have similar off-axis momentum. So will the corresponding photons in the other arm, which are then coherent enough to show an interference pattern.

And, what exactly what happens when a diagonally polarized photon goes through a quarter-wave-plate? Also, how can I edit my post? I edited it before, but I no longer see an edit button.

I think editing is only possible for a certain period of time after writing the initial post. What happens when a diagonally polarized photon passes a quarter wave plate depends on the relative angle between the slow/fast axis of the wave plate and the direction of polarization.

Cthugha, a question for you. I see that when you have a smaller width source, we end up with something that cannot produce momentum entanglement because the delta p grows as delta q shrinks. But it seems to me that the output of the PDC crystal would still be polarization entangled. Is that correct? I know that must be wrong, but cannot figure out why.

Excellent question. To be honest, I am not exactly sure about what happens in that case. The paper I linked is an example for mutually exclusive requirements for spatial coherence and momentum entanglement. I think that most properties which can be entangled are linked to some other property via an uncertainty relation and you can always find some kind of coherence which is incompatible with entanglement.

For example energy entanglement requires a large spread in frequencies, while temporal coherence requires the opposite. Polarization is a bit complex, but you can at least relate the degree of circular polarization to the uncertainty relation between angular momentum and angular position (see, e.g. New J. Phys. 6 103, 2004: http://iopscience.iop.org/1367-2630/6/1/103, I hope it is open access). So entanglement in circular polarization and "angular coherence" should be mutually exclusive. I am not sure about more general and arbitrary polarizations though.

I am also not sure what happens when you have hyperentangled states which are entangled in more than one property like momentum and polarization. I do not know whether several kinds of entanglement necessarily "break" when one kind of entanglement is broken. Intuitively I would say no or at least not totally or just to the minimal degree making sure that ftl information transfer is impossible, but maybe someone else on these forums knows better. If not, I hope people like Kwiat, Boyd and Padgett know and have written papers about that and we might be able to dig them up.

 

[Zafa Pi]

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.

You are referring to: Quote by Joseph14
2. No polarizer at A and quarter wave plates at B---------No Interference

1st of all you'll see this was tested by checking the list at the end of the article.
2ndly, you are correct in saying no path info implies interference. BUT, about half the photons make one interference pattern and the other half make another interference pattern 180º out of phase with respect to the first batch. What you see at B is the combination of the two; they wash each other out so you don't see any interference pattern. Whether a photon falls into the 1st or 2nd batch is a 50/50 crap shoot. What the detector at A does is let you find the members of the 1st (or 2nd) batch so you can see that interference pattern (or the shifted pattern from the 2nd batch).

Does this help?

 

[marksesl]

The way to get back coherence is filtering. In the case of entanglement in energy you could place a narrow spectral filter in one arm and only pick a narrow energy range. The corresponding photons in the other arm will then also feature a narrow spectral range and you will see some longer coherence time when doing coincidence counting.

But, there is no mention of this being done in the delayed choice quantum experiments.

 

[Cthugha]

But, there is no mention of this being done in the delayed choice quantum experiments.

Of course not. At least not in the majority of experiments as they involve double slits. Temporal coherence does not matter for a double slit. A double slit measures spatial coherence. The quantity of interest in this case is momentum. The filtering in momentum space is mentioned. Just have a look at the size of the detectors given in the manuscripts.

edit: As it turns out, I was wrong about this. They do indeed also use a spectral filter, too. From the Walborn paper (PHYSICAL REVIEW A, VOLUME 65, 033818 (2002)):

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

 

[marksesl]

Of course not. At least not in the majority of experiments as they involve double slits. Temporal coherence does not matter for a double slit. A double slit measures spatial coherence. The quantity of interest in this case is momentum. The filtering in momentum space is mentioned. Just have a look at the size of the detectors given in the manuscripts.

edit: As it turns out, I was wrong about this. They do indeed also use a spectral filter, too. From the Walborn paper (PHYSICAL REVIEW A, VOLUME 65, 033818 (2002)):

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

Ok, thanks. I have that paper printed out actually. I'll check it out. So, then this is differently one reason for doing incidence counting, pick out coherent light from the mass of incoherent light made up of the incoherent entangled photons, which are comprised of a spectrum of different wavelength emitted by the BBO crystal.

 

[marksesl]

Has anyone read the description of the double slit experiments on this site? They claim that a detector if left on, but with the recording tape removed will still result in the fringe pattern being restored. I don't believe it. It just can't be.

http://genesismission.4t.com/Physics/Quantum_Mechanics/double_slit_experiment.html

 

[Cthugha]

This is just another try to push new age claims by misinterpreting the experiment. If there is an irreversible interaction at the detector giving which-way information, the fringe pattern will not be visible. It does not matter, whether you destroy the record, do not look at the data or even keep the detector switched off.

 

[rkastner]

I address quantum eraser experiments in some detail in my book, in Chapter 5. I note that these experiments are often hyped and misconstrued as, e.g., 'erasing which-way info after it's already been recorded'. This is not the case. Of course I am applying the transactional picture in my approach, but the basic point holds without that specific interpretation. There is no explicit retrocausality going on here, nothing beyond the usual EPR-type correlations.

QM is strange enough without the efforts on the part of some writers to make it even stranger than it is.