# Resolution to the Delayed choice quantum eraser?

jarekd
marksesl, the coincidence-counter is essential: it usually distinguishes between different interference patterns, such that their sum is without interference (like fringe and anti-fringe).
It is especially important in Walborn's setup as the erasure is not made randomly there, but we can control it: https://www.physicsforums.com/showthread.php?t=710357

About understanding quantum retro-causality here or in Wheeler's experiment, we have to remember that against our "evolving 3D" intuition, all fundamental physics we use is Lagrangian mechanics - with time/CPT symmetry, where the present moment is action optimizing equilibrium between past and future.
In this "4D spacetime" action optimizing picture, photon is its 1D trajectory - the measurement is kind of mounting this trajectory in the future, what in Lagrangian mechanics influences ensemble of paths, among which physics finds the action optimizing one.

San K
Re: 1. why future is not effecting the past in the DCQE and
2. why information cannot be transferred FTL via DCQE or similar experiments such as Wheeler's delayed...etc.

Two things that may be helpful to keep in mind

1. The blob/dots on the screen don't have information extracted yet. Information is created/built via use of co-incidence counter.

All the time when DCQE experiment is being conducted no information has been extracted/transferred.

The status - as far as information (processing) is concerned -- has not changed.

No progress (in obtaining information) has been made via the DCQE experiment, prior to use of the co-incidence counter.

An interference pattern is seen only after use of co-incidence counter, which help build/create the information via filtering.

As an example:

Let's say a map has been torn into two pieces.

The DCQE has simply transferred the two pieces (separately) from two locations to two different locations.

The DCQE has not joined the maps, till the co-incidence counter is used.

Thus the DCQE has made no progress - as far as extracting/getting information is concerned - until, and unless, the co-incidence counter is used.

To join the two pieces of the map, i.e. to get information, a co-incidence counter is required.

2. Entangled photons don't self interfere i.e. don't show single particle interference (because one photon and two photon interference are complimentary)

Thus doing the DCQE in a noiseless environment (and trying to use which-way or no which-way) won't work either because entangled photons don't "self-interfere". i.e. entangled photons don't exhibit single particle interference (because they are no longer coherent).

However partial entangled and partial interference is possible but it, again, cannot be used to send information.

Essentially, in this case, the complimentarity prevents use from sending information FTL.

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marksesl
Two things that are helpful to keep in mind:

1. The blob/dots on the screen don't have information extracted yet. Information is created/built via use of co-incidence counter.
All the time when DCQE experiment is being conducted no information has been extracted/transferred. An interference pattern is seen only after use of co-incidence counter, which help build/create the information via filtering.

2. Entangled photons don't self interfere.

Thus doing the DCQE in a noiseless environment and trying to use which-way or no which-way) won't work either because entangled photons don't interfere.

However partial entangled and partial interference is possible but it, again, cannot be used to send information.

Essentially, in this case, the complimentarity prevents use from sending information FTL.

I fully understand no.1. The patterns are seen only after incidence pairs are plotted. But, if entangled photons don't interfere, how can there ever be an interference pattern made, with or without incidence counting?

San K
I fully understand no.1. The patterns are seen only after incidence pairs are plotted. But, if entangled photons don't interfere, how can there ever be an interference pattern made, with or without incidence counting?

Good question marksesl because I had the same question, a while back.

There are two kinds of interference being discussed here.

1. Single particle interference i.e. the photon interfering with itself such as in a single photon being sent through a double slit or a mach-zehnder.

i.e. photon A interfering with itself
or
(it's entangled partner) photon B interfering with itself

2. Two-particle interference (two-photon interference) -- interference between the two entangled photons A & B.

i.e. photon A interfering with (it's entangled partner) photon B

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marksesl
Ok, so in the Bell state quantum eraser configuration an entangled pair is produced by the BBO crystal. One is shot through the double slits, while the other is shot at the other detector. So are you saying this is a one photon or a two photon experiment. After all, one is going through the double slits. If that entangled photon cannot interfere with itself, how can there be an interference pattern made?

San K
Ok, so in the Bell state quantum eraser configuration an entangled pair is produced by the BBO crystal. One is shot through the double slits, while the other is shot at the other detector. So are you saying this is a one photon or a two photon experiment. After all, one is going through the double slits. If that entangled photon cannot interfere with itself, how can there be an interference pattern made?

The double slit is prior to the BBO crystal...:)...not after as you mention above.

http://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser

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marksesl
The double slit is prior to the BBO crystal...not after as you mention above.

Not in the Bell-state configuration. You are referring to the other one where you have the B0 detector and the mirrors that lead to the other four detectors. I actually don't understand how interference takes place there at all, unless the signal photos from both slot A and slot B are interfering with one another. This implies though that a single photon passes through both slits and each half of that photon's wave function each makes an entangle pair, so you wind up getting 4 photons from only one. Surly that can't be the case.

San K
Not in the Bell-state configuration. You are referring to the other one where you have the B0 detector and the mirrors that lead to the other four detectors. I actually don't understand how interference takes place there at all, unless the signal photos from both slot A and slot B are interfering with one another. This implies though that a single photon passes through both slits and each half of that photon's wave function each makes an entangle pair, so you wind up getting 4 photons from only one. Surly that can't be the case.

1. There is (conceptually) only one configuration for the DCQE as inserted in post 41 above.

2. The wave function does not go-on to make an entangled pair (It's just a mathematical construct yet)
The photon does.

The single photon strikes the BBO. The photon's energy is transferred to the two photons that emerge from the BBO.

(Note: keep in mind that an entangled pair emerges only roughly in 1 in a trillion tries/times)

Hence you have two photons (from one photon) not four.

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msumm21
EDIT: Below I'm referring to an experiment which produces entangled photons and sends one of them to a double slit.

I didn't see which particular experiment you were referring to above, but hopefully the same concept applies to your discussion as in
http://arxiv.org/abs/quant-ph/0106078
in which there IS interference. The photons are entangled AND one of them is shown to demonstrate an interference pattern. I did not read all the above posts, but the reason this can happen without causing a lot of problems might be a question. The answer is that you will not "see" the interference patterns unless you obtain some extra information from the other entangled partner. Let me explain a bit more below.

In a simple way that may miss many details but provide the basic concept, let's say the other photon A is in one of 2 states |0> or |1>. If it is in state |0> then it's partner B will show an interference pattern P0, but if it is in |1> then B will show a different interference pattern P1. The overall net pattern (P0+P1)/2 is just Gaussian-looking, no interference pattern. So if you don't look at A you don't _see_ the interference patterns, but 2 are indeed buried in there. However, if you measure A and then you only look at the B-photons in which A was in |0> only you will find an interference pattern (and the remaining B-photons will form the other interference pattern). If you haven't read the paper I linked above I'd recommend that one and http://arxiv.org/abs/1007.3977

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marksesl
As shown, there is a double slit, followed by the BBO crystal. A photon's wave function supposedly traverses both slits in a double slit, which, in conventional experiments results in the interference pattern. Two sets of entangled pairs are made in this particular configuration, a pair from slit A and from slit B. I do not know if they are made simultaneously, but if so, than one would get 4 photons from one, and if they are not, then how can there ever be an interference pattern made at detector D0? To get an interference pattern at D0 one needs, I would assume, for the two signal photons to be trying to get to the same final state at the same time.

San K
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I fully understand no.1. The patterns are seen only after incidence pairs are plotted. But, if entangled photons don't interfere, how can there ever be an interference pattern made, with or without incidence counting?

Ok, maybe a little explanation of the language of quantum optics is in order. "...photons do not interfere" is some kind of unfortunate wording. Coherence is a statistical concept and should be investigated first and foremost at the field level. Whenever you have two well defined fields originating from a double slit, you will find some kind of interference pattern. For a fixed geometry, the actual shape of the pattern will depend on the relative phase between the fields at the two slits. A different phase difference will give a completely different interference pattern with peaks located at different positions. This is always the case. However, the meaning of interference and coherence is defined in an ensemble averaging manner. If you repeat the experiment again and again, you can imagine two situations: Either the relative phase difference at the slits is similar for each repetition of the experiment or it is different. In the first case, you will always get the same interference pattern, in the second case, you will get a completely different pattern every time. This means that when averaging over all these measurements, you will only be able to see a pattern on average in the first case, while in the second case many different patterns will overlap, resulting in a broad Gaussian distribution.

When people were able to repeat this experiment with very low mean photon numbers, the "photons do/do not interfere"-terminology was born, but it still means the same: if photons are non-interfering in a double slit setting, this means that the ensemble of photons is not prepared in a way to give the same detection probability distribution (still determined by the phase difference at the slits) behind the double slit. If they do, the ensemble is prepared in a way that always gives the same distribution. It is important to realize that simple coherence is a question about averaged quantities.

As shown, there is a double slit, followed by the BBO crystal. A photon's wave function supposedly traverses both slits in a double slit, which, in conventional experiments results in the interference pattern. Two sets of entangled pairs are made in this particular configuration, a pair from slit A and from slit B. I do not know if they are made simultaneously, but if so, than one would get 4 photons from one, and if they are not, then how can there ever be an interference pattern made at detector D0? To get an interference pattern at D0 one needs, I would assume, for the two signal photons to be trying to get to the same final state at the same time.

You only get two photons, but you will get probability amplitudes for all four events. These are the quantum equivalents of em fields. For high intensities you would get four fields in this kind of setup. For very low intensities, you move over to describing the situation with probability amplitudes. These pretty much behave like fields, but only give the probability to detect a photon at some given position and time under the constraint of energy conservation instead of describing a quantity with direct physical reality. You get detection probabilities for all four positions, but there will only be two real detection events each time you perform a single run of the experiment.

San K
if photons are non-interfering in a double slit setting, this means that the ensemble of photons is not prepared in a way to give the same detection probability distribution (still determined by the phase difference at the slits) behind the double slit. If they do, the ensemble is prepared in a way that always gives the same distribution. It is important to realize that simple coherence is a question about averaged quantities.

Another good post Cthugha. Very Informative.

So, in an entangled pair...

Is the ensemble of photons not prepared in a way to give the same detection probability distribution (still determined by the phase difference at the slits) behind the double slit?

or alternatively

what is the behavior of the entangled photons with respect to interference patterns (say on passing from a double slit)?

marksesl
Ok then the photon pairs are not produced simultaneously. Correct? So then what is doing the interfering at D0? A single photon can't just go there by itself and form an interference pattern.

And getting back to the double slits. I still say that the photon's wave function must be incident on the BBO crystal at two locations at the same time. So, you must believe that the wave function of that photon must decohere and reform into a discrete photon inside the BBO crystal where it is absorbed, and then two photons of half wavelength are reemitted, the signal and the idler. So, again, what is the signal photon interfering with if it's alone and not itself going through a double slit?
You see, in the Bell-state delayed quantum eraser experiment, the signal photon goes on through a double slit. But, I'm being told that entangled photons cannot interfere with themselves, so what's the purpose of the double slit if entangled photons can't pass making an interference pattern?

San K
So, you must believe that the wave function of that photon must decohere and reform into a discrete photon inside the BBO crystal where it is absorbed, and then two photons of half wavelength are reemitted, the signal and the idler.

agreed, though there will be variation depending upon which interpretation of quantum mechanics is considered

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Gold Member
Ok then the photon pairs are not produced simultaneously. Correct? So then what is doing the interfering at D0? A single photon can't just go there by itself and form an interference pattern.

And getting back to the double slits. I still say that the photon's wave function must be incident on the BBO crystal at two locations at the same time. So, you must believe that the wave function of that photon must decohere and reform into a discrete photon inside the BBO crystal where it is absorbed, and then two photons of half wavelength are reemitted, the signal and the idler. So, again, what is the signal photon interfering with if it's alone and not itself going through a double slit?
You see, in the Bell-state delayed quantum eraser experiment, the signal photon goes on through a double slit. But, I'm being told that entangled photons cannot interfere with themselves, so what's the purpose of the double slit if entangled photons can't pass making an interference pattern?

The reason I usually don't comment on this experiment is because it is an order of magnitude more complex than either the double slit or the entangled pair side individually. But I will pass this on:

BBo crystals take a single photon as input and outputs 2 photons. The process is not modeled as absorption and emission in the traditional sense. I wouldn't model the input as being at 2 spots either. Obviously you have a beam going in (and the beam has width), but most input photons pass through without any particular change. But occasionally one spontaneously splits. They come out more or less at the same time, with a spread of times just as their wavelengths are not always exactly equal.

Some of the photons that go through the double slit are coherent enough to self-interfere. But that is a subset for which its partner photon registers a particular way on the other side. I am not sure if this description is fully accurate, Cthugha can you comment?

So, in an entangled pair...

Is the ensemble of photons not prepared in a way to give the same detection probability distribution (still determined by the phase difference at the slits) behind the double slit?

In entangled pair production you have a huge pump spot on the non-linear crystal. The phase difference at the slits is mostly governed by pure geometry. There will be a distance in the paths from any possible point of emission on the BBO crystal to the two slits. This tiny path difference makes a huge difference in terms of relative phase and will give a very different interference pattern. Yet, this huge difference is what you need for entanglement. As the paths from tall the positions inside the pump spot to some chosen position are very different, so are the emission angles and therefore also their momentum. However, you need to have this broad momentum distribution for momentum entanglement. You cannot violate Bell's inequalities (for momentum entanglement), if this distribution is narrow. So preparing a beam in a manner that allows to see entanglement must necessarily exclude preparing it in a way that you can see a stable interference pattern in the same geometry.

You see, in the Bell-state delayed quantum eraser experiment, the signal photon goes on through a double slit. But, I'm being told that entangled photons cannot interfere with themselves, so what's the purpose of the double slit if entangled photons can't pass making an interference pattern?

Have you read my last response? As I said, interference happens on the field level and is a statistical thing which tells you how large the spread in some property is. That automatically means that this is not carved in stone at the moment of photon emission. This is a property of the FIELD and as such you can change it along the way by filtering. If you have spatially incoherent light like sunlight go through a double slit, you will not see an interference pattern. If you place a narrow pinhole in front of it, you will see a pattern, although the initial light source never changed. What people now do in delayed choice experiments is almost the same. You put incoherent light on a double slit and do not see any interference, but instead of placing a pinhole in the light path, you place it in the path of its entangled partner and filter that beam. This one now has increased coherence, but it does not go to a double slit. This is why you now do coincidence counting. You just detect these filtered photons which do not go to a double slit. Due to entanglement, their entangled partners (but only those corresponding to the filtered photons, not all of the emitted photons) on the other side will also have well defined characteristics: they will be coherent, too. This subset will therefore create a clear interference pattern. But as the filtering and the double slit happen in two different places, you will need the coincidence counting to do the "bookkeeping".

Some of the photons that go through the double slit are coherent enough to self-interfere. But that is a subset for which its partner photon registers a particular way on the other side.

That is more or less it. For a single photon and a double slit there will (most likely) always be some detection probability distribution looking like an interference pattern. The total ensemble will contain many of these subsets (for each of which the partner photon registers a different particular way on the other side) which is what is loosely translated into "photons not self-interfering".

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msumm21
I'm being told that entangled photons cannot interfere with themselves, so what's the purpose of the double slit if entangled photons can't pass making an interference pattern?

Did you see my previous post? Interference can still occur, but is not visible without information about the entangled partner. I think I explained a simple example of what happens in my last paragraph in my previous post here. If it's not clear just let me know I can try to explain it better. I also pointed to some papers there that explain a real experiment of this.

marksesl
If you have spatially incoherent light like sunlight go through a double slit, you will not see an interference pattern. If you place a narrow pinhole in front of it, you will see a pattern, although the initial light source never changed. What people now do in delayed choice experiments is almost the same. You put incoherent light on a double slit and do not see any interference, but instead of placing a pinhole in the light path, you place it in the path of its entangled partner and filter that beam. This one now has increased coherence, but it does not go to a double slit. This is why you now do coincidence counting. You just detect these filtered photons which do not go to a double slit. Due to entanglement, their entangled partners (but only those corresponding to the filtered photons, not all of the emitted photons) on the other side will also have well defined characteristics: they will be coherent, too. This subset will therefore create a clear interference pattern. But as the filtering and the double slit happen in two different places, you will need the coincidence counting to do the "bookkeeping".

Did you see my previous post? Interference can still occur, but is not visible without information about the entangled partner. I think I explained a simple example of what happens in my last paragraph in my previous post here. If it's not clear just let me know I can try to explain it better. I also pointed to some papers there that explain a real experiment of this.

Ok, I understand the filtering process. The entangled photons produced by the BBO crystal are no longer coherent since the pairs only need to have wavelengths that add up to the wavelengths of the parent photos, rather than being exactly half. So, in these entangled photon experiments the field must be filtered so that only photons that are coherent are counted. It is thus incorrect to simply make the blanket statement that entangled photons cannot interfere. Ones that are coherent can interfere with each other.

San K
It is thus incorrect to simply make the blanket statement that entangled photons cannot interfere. Ones that are coherent can interfere with each other.

marksesl, your statement above is correct.

For a physicist (or someone trying to understand this deeper) -- the answer (as you correctly pointed out above) is:

Photons always self interfere (as long as they are in same time-space).

The (quality and type of) interference pattern depends upon the state (phase, amplitude etc) of the interfering waves.

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It is thus incorrect to simply make the blanket statement that entangled photons cannot interfere. Ones that are coherent can interfere with each other.

Yes and no...

The tricky part is that when you just have a look at this isolated set of photons which show interference, you will not be able to show that this subset is indeed entangled in the strict sense - you will not be able to violate Bell's inequality just using this subset.

The statement about entangled photons not interfering with each other is pretty condensed. In my opinion it would be less misleading to say that it is not possible to build an experimental setup unambiguously showing or verifying both good interference and good entanglement for the same set of photons.

marksesl
Yes and no...

The tricky part is that when you just have a look at this isolated set of photons which show interference, you will not be able to show that this subset is indeed entangled in the strict sense - you will not be able to violate Bell's inequality just using this subset.

The statement about entangled photons not interfering with each other is pretty condensed. In my opinion it would be less misleading to say that it is not possible to build an experimental setup unambiguously showing or verifying both good interference and good entanglement for the same set of photons.

So, this seems to be another complementarity issue. Before tackling just how this impacts the experiment, I want to focus on the more fundamental issues: 1) How knowing, or at least being able to know, which-way information appears to cause wave-function collapse. 2) Retrocausality.

In regard to 1, it seem fairly straight forward that anything we do to try to find which-way information is the very thing that causes wave-function collapse, and then that collapse just gives the impression of which-way. It's difficult for me to see, however, in the Kim, et. al. DCQE just how we are screwing with the wave-function of the signal photons causing collapse by detection of the idlers at D4 and D3. Perhaps you can elaborate. BTW in Wikipedia, it states that the source of interference is at BSc.: "To understand the source of the derived interference pattern, one has to focus on the third beam splitter BSC, where the photon paths from both slits merge. At this point a phase difference exists between the merging paths, which is dependent on the different path lengths from slit A or B respectively to the splitter."

In regard to 2, it is noteworthy that the state of the signal photons cannot be known until after idler data is known, as you have mentioned many times here and there. Nothing can be seen at the screen (even if there were one) because the information must be extracted later, once idler correspondence is determined. Now, this is a real logical "red herring." On one hand, one can say regardless if the results cannot be known until all the data is in, it can be assumed that each idler's entangled signal partner did in fact end its journey first, so there must have been retrocausality. On the other hand, one cannot ignore that the state of the signal photons always remain unknown (regardless of configuration) until the state of the idlers are known, so the game is never really over for the signal photons until all the information about the idlers is known. If one cannot know of what happed first until what happened second is known, then there isn't really any retrocausality; it is more a matter of having to wait until all pertinent data is registered. Perhaps this may lead one to the belief that the very sequence of time itself is a kind of illusion, as, I believe, Einstein himself suggested.

San K
On one hand, one can say regardless if the results cannot be known until all the data is in, it can be assumed that each idler's entangled signal partner did in fact end its journey first, so there must have been retrocausality. On the other hand, one cannot ignore that the state of the signal photons always remain unknown (regardless of configuration) until the state of the idlers are known, so the game is never really over for the signal photons until all the information about the idlers is known.

Ther is no retrocausality.

When the signal (or idler) ends it journey, the entanglement is broken. The property (on which they were entangled) becomes determinate/frozen for both the particles (simultaneously).

However we don't which photon is the entangled partner of the signal (or idler, for that matter). There are so many photons.

To find out the ex-partner we need a coincidence counter.

No information has been transmitted prior to use of the coincidence counter.

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msumm21
It's difficult for me to see, however, in the Kim, et. al. DCQE just how we are screwing with the wave-function of the signal photons causing collapse by detection of the idlers at D4 and D3.

Unless you are thinking in terms of reduced density operators (or something equivalent), you should think of the quantum state of the composite system, not just the signal photon alone. At this level, you should find that the portion of the state associated with the idler striking D3 or D4 and the signaler found at X is orthogonal to those portions in which the signaler passed through the opposite slit. This means they will not interfere. However, if the idler strikes D1 or D2 the signaler can strike X via both paths and the two associated portions of the quantum state in which this happens are NOT orthogonal (interference). I'm not sure if this explanation helps much, you may really have to do the math to understand further. Leonard Susskind has some lectures on Quantum entanglement online. Around lecture 6 or so in that class I think he goes through the math of a simple case somewhat similar to this, but maybe simpler in some respects.

marksesl

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Gold Member
This is a copy of Cthuga's detailed post number 8. I'm copying it here so I can just format it a little better to make it easier to read, before I start working through it. I hope nobody minds. It doesn't show as a quoted post normally does because I have removed the quote codes, which tend to make one lose horizontal space, which is valuable when dealing with lengthy formulas.

Beginning of reformatted Cthuga's post #8:

Here is a simplified explanation of these different patterns in the DCQE experiment as performed by Kim et al, which I posted earlier somewhere on this forum. It is a bit long, but maybe it helps you understand (and I am too lazy to write a completely new description ;) ).

You can treat this experiment as if every photon here originates from either point A or B (following the pictures in the DCQE paper by Kim). Now the setting is as following:

a) In entangled photon experiments each photon on its own behaves like incoherent light. This means, there is no or at least not much correlation between the phases (##\phi_1## and ##\phi_2##) of the fields originating from points A and B at the same moment. For the sake of simplicity I will now treat the phase of these fields as completely random and the amplitude as constant and equal.

b) The two-photon state has a well defined phase. This means that the fields of both paths (signal and idler), which originate from the same point (A or B) have a fixed phase relationship. For the sake of simplicity I will now assume, that the initial phase is the same for signal and idler.

Let's first consider what happens at the detector D0, which scans the x axis. Just like in an usual double slit experiment you will not detect any photons, if there is destructive interference and will detect a large number of photons, if there is constructive interference. Each point P on the x axis, which the detector scans, corresponds to a certain path difference between the paths from A to P and B to P, which can be expressed in terms of an additional phase difference ##\Delta \phi##. So you will have constructive interference at a point if ##\Delta \phi +(\phi_2 -\phi_1) = 2 \pi##.

As ##\Delta \phi## is a constant for each point this means that a detection at a certain point P means, that the phase difference between fields at point A and B had a fixed value ##\phi_2 -\phi_1## at a certain time short before. So in fact, scanning the x-axis means scanning the phase difference of the fields. As the fields change completely random and independent of each other, each phase difference will be realized and there will be no interference pattern at D0 alone.

Now let's have a look at the other side. There are two detectors, which both fields can reach and two detectors, which can only be reached by one field. Let me explain D1 first. I will assume that the distances from A and B to each of the detectors are equal. Before the field originating from A reaches the detector, it crosses 2 beam splitters (no reflection) and a mirror. This influences the phase of the field. Assuming that the beam splitters are 50-50 each transmission changes the phase by ##\frac{\pi}{2}## and each reflection changes the phase by ##\pi##. So summarizing the phase of the field originating from A will be

$$\phi_1+\pi\textrm{(reflection at the mirror)} + 2\frac{\pi}{2} \textrm{(transmission at the beamsplitters)}$$

The field reaching D1 from point B is reflected twice and transmitted once, so the phase will be ##\phi_2 + 2\pi+\frac{\pi}{2}##, so the phase difference at the detector will be

$$\Delta \phi=\phi_2+2\pi+\frac{\pi}{2}-(\phi_1+\pi+2\frac{\pi}{2})=\phi_2-\phi_1+\frac{\pi}{2}$$

Of course a detection implies again, that there is no destructive interference and most detections will occur, if ##\phi_2-\phi_1+\frac{\pi}{2}=2 \pi## (or a multiple).

Now let's check the other detector. Here the field originating from A is reflected twice and transmitted once and the field originating from B is transmitted twice and reflected once, which leads to a phase relationship of

$$\Delta \phi = \phi_2 + \pi + 2 \frac{\pi}{2}-(\phi_1+ \frac{\pi}{2}+2\pi)=\phi_2-\phi_1-\frac{\pi}{2}$$

So the ##\Delta \phi## at the two detectors are exactly ##\pi## out of phase. This means that constructive interference on one detector at some certain phase difference automatically implies destructive interference at the other. So each detector selects a set of phase differences. Let me once again stress that the phases are completely random, so there will be no interference on these detectors either. The detectors D3 and D4 are simpler. As there is only one field present, there will be no interference and the phase does not matter. The detections will be independent of ##\phi_1## and ##\phi_2##.

Now we're almost done. Now we have to consider the two-photon state, where the relative phases are not random any more. As I stated before, a certain spot on the x-axis of D0 corresponds to a certain phase difference ##\phi_2 - \phi_1##. This very same phase difference will also correspond to a certain amount of constructive (or destructive) interference on D1 and consequently also (due to the different geometries concerning transmission and reflection mentioned above) to an equivalent amount of destructive (or constructive) interference on D2. So you will see this interference pattern in the coincidence counts of D0/D1 and D0/D2 due to the fixed phase difference of the two photon state. Now it is also clear, why there is no interference pattern if you have which-way information. If you have which-way information, there is just one field present, which has a random phase. There is no interference pattern present, which corresponds to the phase difference, which is present at D0 and therefore no interference pattern can show up in the coincidence counts.

I hope this simplified scheme shows, why the choice between the interference pattern and the which-way information can be done after the signal photon has already been detected, why it does not depend on whether we have a look at the data or not and that there are no problems with causality.

End of reformatted Cthuga's post #8