Quantum behavior experiment flawed?

[tistemfnp]

I would like to discuss this experiment, could it be flawed?:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Let's start here with another setup:


In the video it's shown that the field is continuous and splits 50/50 at the beam splitter, no matter how much it's attenuated. Also in the regime where (much) less than one photon per path is present, interference occurs.

So in one experiment the split ratio is 50/50 in any energy regime and in the other one it's claimed to be in either path. Does this make sense?

The flaw in the setup of Thorn et al in my opinion, is in the experimental procedure:

1) "Once this alignment is set,the alignment of the T collection optics is adjusted to maximize the coincidence rate between G and T, not the raw count rate on T."
2) "The Rdetector is connected and when counts are achieved, the window on the GR coincidence unit is set, and the GR coincidence count is maximized."

This procedure could be a recipe for minimizing the coincidence rate RT, what is of course then the outcome of the experiment. Instead of using TACs and SCAs it would be more beneficial to have data where the output of the SPCMs is combined (3bit) and subsequently sampled at 500MHz. Unfortunately the authors didn't use such a simpler setup and provide the data.

If the fibers that go to detectors R and T instead would be added on a screen there would be an interference pattern, wouldn't it?

 

[tistemfnp]

Let's self-resolve this :

The experiment is fine and valid and there would be interference. No contradiction though.

Pretty silent here...

 

[PeterDonis]

The experiment is fine and valid

Yes.

and there would be interference.

Where? The experiment isn't even testing interference.

 

[PeterDonis]

In the video it's shown that the field is continuous and splits 50/50 at the beam splitter, no matter how much it's attenuated.

Ok.

Also in the regime where (much) less than one photon per path is present, interference occurs.

Interference occurs where, though? It occurs when the two beams that were separated at the beam splitter are brought back together. But the Thorn paper experiment doesn't do that; it isn't even trying to do that. They are two different experiments.

 

[PeterDonis]

So in one experiment the split ratio is 50/50 in any energy regime and in the other one it's claimed to be in either path. Does this make sense?

Yes. In the Thorn experiment, each of the photon detectors (one in each output arm of the beam splitter) detects a photon half the time. That's the 50/50 split.

 

[berkeman]

Pretty silent here...

No, I do hear the theme music from the movie Jaws playing in the background. I wonder why that is...

 

[PeterDonis]

No contradiction though.

That's correct, there is no contradiction between the two experiments.

 

[Vanadium 50]

Could this experiment be done better? Sure - it's an undergraduate lab for heaven's sake! But does it do what it sets out to do? Certainly - take a look at Table 1. Those results match a photon theory and do not match a classical theory.

If you want to claim it is wrong, there are only three places:

1. The classical prediction is wrong.
2. The quantum prediction is wrong.
3. The authors didn't measure what they said they did.

Which is it?

 

[DrChinese]

If the fibers that go to detectors R and T instead would be added on a screen there would be an interference pattern, wouldn't it?

A common misconception. Entangled photons do not produce the interference patterns as you might expect (such as if directed to a double slit setup).

 

[tistemfnp]

A common misconception. Entangled photons do not produce the interference patterns as you might expect (such as if directed to a double slit setup).

Really? But there is no entangled photons in R/T. The entanglement is between G and R/T (, if any). R/T splits the path of one of the fields stemming from SPDC. So there should be an interference pattern. (Unlike explained in the experiment, the photon does not take one path or the other, even in a quantum-mechanical model. It would take both paths at once and only at detection the wave function would collapse and the detection only would instantiate in one path or the other. Not that I would buy that, but this is the correct QM explanation. And in case there is not two detectors but one, these paths would interfere.)

Still a question though. The video at quoted time explains that the amplitude of the EM field coming from the laser can be attenuated to an arbitrarily small value. How does this align with the statement, that a coherent state at low energy levels has higher noise (https://encyclopedia.pub/entry/30411) The video shows a constant amplitude also for the attenuated beam. Question asked alternatively: the video claims, a coherent state can be attenuated (in amplitude) to arbitrarily small, constant values. Is it or is it fluctuating, as explained in the article about coherent states? What am I missing?

 

[DrChinese]

Really? But there is no entangled photons in R/T. The entanglement is between G and R/T (, if any). R/T splits the path of one of the fields stemming from SPDC. So there should be an interference pattern. (Unlike explained in the experiment, the photon does not take one path or the other, even in a quantum-mechanical model. It would take both paths at once and only at detection the wave function would collapse and the detection only would instantiate in one path or the other. Not that I would buy that, but this is the correct QM explanation. And in case there is not two detectors but one, these paths would interfere.)

No interference for entangled photons, as you can see from the 1999 Zeilinger reference. See figure 2 and equation (4).

https://courses.washington.edu/ega/more_papers/zeilinger.pdf

There is of course one and only one entangled photon going into the R/T splitter. That is the basis for the experiment, which is to demonstrate it goes one way or the other (but never both). The number of detectors is not relevant for demonstrating interference either way.

 

[vanhees71]

Fig. 3 is about the Dopfer experiment, realizing a quantum version of the famous analysis of "wave-particle duality" by von Weizsäcker. Here you have to be careful again to really explain the full details. You can choose whether you want to see full-contrast double-slit interference fringes for the photons detected by D2, you have to put D1 in the focal plane of the "Heisenberg lense", which refers to a momentum measurement of the corresponding photon, which excludes to know through which way each of these photons went, and due to the entanglement you see the full-contrast interference pattern for the photons observed at D1 in coincidence with the photons observed at D2.

On the other hand, if you put D1 at the "image plane", where you depict the two slits sharply, you know precisely through which of the two slits the photon registered in coincidence by D2 in coincidence with the photons registered at D1. Then of course there's no double-slit interference picture at all.

Of course you can have all situations in between these extreme ones, by putting D2 somewhere else behind the lens. Then you have a more or less sharp or blurred double-slit interference pattern. The thesis (in German) can be downloaded here:

https://people.isy.liu.se/jalar/kurser/QF/assignments/Dopfer1998.pdf

 

[tistemfnp]

Here you have to be careful again to really explain the full details.

Exactly. The condition to have no interference in the other photon is to have a configuration where in path 1 the location of origin (and thus its path) is revealed, which needs a certain setup (said lens and a placement of the detector at 2f). This isn't the case in the experiment with detectors G, R/T. G does not reveal any "which way"-information about G (and so R/T). Hence there is interference between R/T if these paths would interfere at a screen (path 2 in the Dopfner setup). As the matter of fact, even in case there would be such a setup on path, interference would occur. Why?: -> There is a multimode fiber involved, so no "which-way"-information on where the particle is from (destroyed by the multimode-fiber passage!), whatever theater is performed in path G.

Anyone bother about the other question on attenuating a coherent state?

 

[DrChinese]

Fig. 3 is about the Dopfer experiment...

Oops. I wrote 3, but meant figure 2. I corrected post #11.

Again, entangled photons do not produce the usual double slit type interference. As vanhees71 correctly points out, the Dopfer type experiments demonstrate something different.

 

[tistemfnp]

I wrote 3, but meant figure 2.

Same thing. Only by placing detectors to determine the "which-path" information on beams b, b' the interference on a, a' is suppressed. Otherwise not.

 

[tistemfnp]

Again, entangled photons do not produce the usual double slit type interference.

Unless the "which-path" information is revealed, they do exactly that. In the G, T/R experiment "which-path" information is not available --> interference. (That's explicitely written in the Zeilinger paper you provided, mentioning the eraser part.)

 

[DrChinese]

1. Only by placing detectors to determine the "which-path" information on beams b, b' the interference on a, a' is suppressed. Otherwise not.

2. Same thing. Only by placing detectors to determine the "which-path" information on beams b, b' the interference on a, a' is suppressed. Otherwise not.

3. There is a multimode fiber involved, so no "which-way"-information on where the particle is from (destroyed by the multimode-fiber passage!), whatever theater is performed in path G.

1. Arrrgh. As pointed out by Zeilinger, such behavior as you describe would allow FTL signaling. The idea being that if what you said were true, a decision to determine the path b, b' would instantly change the interference on a, a'. Of course, that can't happen - hopefully you can see that now.2. If you run an entangled photon into a double slit, it will produce a pattern showing no interference. This reference shows this exact result:

https://arxiv.org/abs/1304.4943
Time-resolved double-slit experiment with entangled photons, Jennewein et al (2013)
See Figure 4, the yellow triangle markers (D1 or D2, path not specified). D1 is right path/Reflected specific, D2 is left path/Transmitted specific. The yellow area is the actual pattern produced from a double slit type setup (here created via the calcite crystal instead of actual slits). It shows no interference.

That is also what is demonstrated in the Dopfer experiment, in a slightly different manner. Specifically, interference is ONLY a function of both photons together - and neither photon stream individually as you assert. The Dopfer experiment only produces a pattern consistent with an interference pattern when there is coincidence with the other entangled photon, under the specific circumstances as described.3. The presence of the fiber is not relevant in these cases, as can be seen from the reference I provided. Fiber is simply a convenience for the setup, and does not change the basic physics.

 

[tistemfnp]

hopefully you can see that now

You're right, I recall my previous statements . Basically the system of 2 entangled photons has one degree of freedom "extra" (actually not extra, but now two of them, because it's two partners), so no interference pattern. Only if you look at a specific set of detections (by fixing one photons of the pair in location or direction), the corresponding interference pattern or location can be observed (as a correlation).

Thanks a lot for your patience and perseverance to explain that.

 

[vanhees71]

Oops. I wrote 3, but meant figure 2. I corrected post #11.

Again, entangled photons do not produce the usual double slit type interference. As vanhees71 correctly points out, the Dopfer type experiments demonstrate something different.

I don't know, how you can make such a general statement. Of course entangled photons can produce double-slit interference. Some of the most beautiful experiments are about demonstrations what was called "wave-particle duality" before the discovery of modern quantum theory, e.g., the quantum erasers (some with "delayed choice") or the realization of the Heisenberg microscope in the gedanken experiment by von Weizsäcker in the Dopfer thesis, as quoted in the APS centenary RMP article by Zeilinger. Fig. 2 precisely refers to such a quantum-eraser experiment: If you look at all photon pairs, there's no interference pattern of the signal photons behind the double slit, and complete which-way information is encoded in the momentum state of the idler photon (see the text following Eq. (4)).

 

[PeterDonis]

Of course entangled photons can produce double-slit interference.

The correct solution of, for example, Ballentine Problem 9.6, disagrees with you.

 

[vanhees71]

Of course, there are experiments, where you don't see and other where you see interference patterns using one of a pair of entangled quanta. It depends on the context. My point is you cannot apodictically say "you can't produce double-slit interference with an photon entangled with another photon". The Dopfer experiment uses the fact that you can have both cases (or any in between, i.e., interference patterns with reduced contrast).

That everybody understands what we are talking about, concerning the said problem here's the excerpt from Ballentine's book:

Let's discuss the case were we do something to the particle in the left part. Of course, you can do what you want with the left particle. If you don't use the measurement result on its spin, whatsoever, nothing happens concerning the interference pattern built by the particles on the right-hand part. The right-hand particle is prepared in an unpolarized state. The distribution at the screen after the beam splitter thus is the sum of PROBAILITY DISTRIBUTIONS for going the one or the other path. There's no interference term there, i.e., you add probabilities and no interference pattern is seen.

Now If you project the left-hand particle to one of the $\sigma_z$ eigenstates. I.e., you consider in an coincidence experiment only the particles on the right-hand side if a detector in the $\sigma_z=+1/2$ beam on the left-hand side registers the particle, you don't see an interference pattern either, because the right-hand particle is in the $\sigma_z=-1/2$ eigenstate and it goes with certainty only one of the possible paths and you see the corresponding probability distribution for this path at the screen. The same is when choosing the $\sigma_z=-1/2$ beam on the left-hand side. Then the right-hand particle goes the "$\sigma_z=+1/2$ path".

Now if you do the projection to a $\sigma_x$ eignstate of the particle on the left-hand side and again only register the particle on the right-hand side if you find the first particle to have $\sigma_x=1/2$. Then the particle on the right-hand side is in the eigenstate with $\sigma_x=-1/2$. Since then the way this particle goes is undetermined, you'll get an interference pattern. Now if you do the experiment also with the left-hand particle being found in the state $\sigma_x=-1/2$. Then the right-hand particle is in the state $\sigma_x=+1/2$ and again you get an interference pattern.

Now, an ideal beam splitter and a mirror is always described by a unitary transformation, which implies, no matter how it works in detail, a relative phase shift of $\pi$ between the different $\sigma_z$ states. This means that the two interference patterns in the $\sigma_x$ experiment are shifted against each other, and when you not look at them together you'll again see no interference pattern. So you get an interference term $\propto +1$ times the product of the position wave functions in one case and $\propto -1$ times this product in the other case. Adding the two probability distributions cancels thus again the interference term.

This resolves the apparent paradox.

I hope you agree with my solution ;-)).

This is common to all these kinds of experiments. The minute details don't play to much of a role. You only need basic quantum-mechanical principles like unitarity of beam splits etc. There's never a contradiction or a paradox when carefully analyzing each experiment.

As I said above you can NOT make a general statement "entangled particles/photons don't make double-slit interference patterns" or interference patterns in the here discussed beam-splitter experiment. In short: Whether there is an interference term or not when superimposing states depends upon the selection of the ensemble being considered. That's precisely the quantum properties you test with such entangled particle or photon pairs.

 

[DrChinese]

Of course, there are experiments, where you don't see and other where you see interference patterns using one of a pair of entangled quanta. It depends on the context. … The distribution at the screen after the beam splitter thus is the sum of PROBAILITY DISTRIBUTIONS for going the one or the other path. There's no interference term there, i.e., you add probabilities and no interference pattern is seen.
As I said above you can NOT make a general statement "entangled particles/photons don't make double-slit interference patterns" or interference patterns in the here discussed beam-splitter experiment.

As I referenced, individual entangled photons do not generally produce interference patterns. As you correctly state (and there’s not any debate about this), coincidences between pairs of entangled photons can demonstrate interference patterns.

Generally, to get a single stream of entangled photons to produce an interference pattern, you must first eliminate the entanglement. You might do that, for example, by running that stream through a single slit to produce the usual coherence required to see the double slit pattern.

 

[tistemfnp]

Guys, the reason for not seeing interference from entangled photons is ultra short : conservation of momentum.

 

[tistemfnp]

Somehow the other question of the original post got lost in the discussion. The video claims, a coherent state can be attenuated to a "arbitrarily low value", that could "penetrate polarizers only considering the cosine-law".

So my current understanding, that a coherent state at least would need to contain 1 photon is wrong? And the statistics is none-Poissonian, as can be concluded from the video?

To detail the question: Let's assume a laser that would emit pulses of 10µs with a power of 1mW within the pulse. (The laser is only locked within that pulse, there is no coherence between pulses and the pulses are not created by chopping a continuous laser but by switching the laser on and off and locking it at the beginning of each pulse. For simplicity, assume locking happens in a very short time, to not overcomplicate the discussion with irrelevant details.) Each pulse contains about 8x10¹⁰ photons (feel free to calculate the wavelength, not important here). Then the pulses get attenuated to counts of few photons (or less).

Now consider you are confronted with an experiment for which the outcome is (it's not the question if the outcome could be or not or if the experiment has a flaw or how this could be done, the outcome is reality by definition here and consider to be confronted with that reality):
- The statistics is not Poissonian. For example, if the attenuated beam contains 4 photons per pulse (in average), the experiment shows, each pulse contains 4 photons (or their energy equivalent)
- The number can be fractional. The attenuation can be adjusted, that in each pulse you can detect a fractional number of photon energy.
- The number can be less than 1.
- The statistics within a pulse is not Poissionian. For example, if the complete pulse has 4 photons, it's 2 photons in the first 5µs of the pulse and 2 photons in the second measurement period.

What would such experimental outcome mean for QM?

 

[vanhees71]

As I referenced, individual entangled photons do not generally produce interference patterns. As you correctly state (and there’s not any debate about this), coincidences between pairs of entangled photons can demonstrate interference patterns.

Generally, to get a single stream of entangled photons to produce an interference pattern, you must first eliminate the entanglement. You might do that, for example, by running that stream through a single slit to produce the usual coherence required to see the double slit pattern.

Of course you need many equally prepared photons to get an interference pattern. How do you explain all the experiments with double-slit interference of photons in an entangled photon pair like the here discussed by Dopfer or the various versions of the quantum-eraser experiments, etc. if these photons don't show interference patterns, as you claim?

 

[tistemfnp]

Of course you need many equally prepared photons to get an interference pattern. How do you explain all the experiments with double-slit interference of photons in an entangled photon pair like the here discussed by Dopfer or the various versions of the quantum-eraser experiments, etc. if these photons don't show interference patterns, as you claim?

In the work of Dopfer, the interference pattern only shows if the partner is fixed in location. If the partner is not prepared, no interference pattern. Same principle applies to the quantum eraser. In fact, nothing is erased and there is no retrospectivity in the experiment.

 

[vanhees71]

As I said, it depends on the precise description of the experiment, whether interference fringes are seen or not. That's my point! You cannot apodictically say that photons in an entangled pair don't show interference fringes or not.

E.g., in the Walborn quantum eraser experiment without putting the quarter-wave plates into the slits the double-slit diffraction pattern is seen (no which-way information available -> interference), if you put them it's not seen (which-way information available -> no interference). Projection measurements on the idler photon make the fringes visible again ("erasure of which-way information" -> interference).

That's the whole point of all these experiments about the "wave-particle dualism" in context of entanglement.

 

[tistemfnp]

You cannot apodictically say that photons in an entangled pair don't show interference fringes or not.

Mhhh, I don't want to make you angry, but actually you can. Strictly, you will never be able to put a real (photo) screen and observe interference fringes. That's the statement of @DrChinese and it's valid to its full extent.

If you talk of "interference fringes" related to mentioned experiments, they are always virtual and can only be made visible by subsequently project them on a virtual screen, only displaying clicks related to clicks of the other detector that's prepared in a certain setup.

So it relates more to the context of the statements. Photo screen -> no interference fringes, virtual screen -> interference fringes under certain conditions.

 

[vanhees71]

I don't understand what you want to say. In the Walborn experiment you send one of the entangled photons through a double slit and detect them and you see interference fringes. Here's the preprint and the paper

https://arxiv.org/abs/quant-ph/0106078
https://doi.org/10.1103/PhysRevA.65.033818

If you can, read the paper, because there the figures are put nicely into the text. That's not the case with the preprint unfortunately.

Just using the setup with the QWPs taken out of the slits the signal photons show interference fringes. This is depicted on page 8 of the preprint (Fig. 2 of the paper). From the double-slit interference pattern contrast you see that the photons are pretty well coherent, i.e., they have sufficiently low momentum width to show interference.

So you can't say that you never see interference fringes from photons what are entangled with another photon.

At the next page (Fig. 3 of the paper) you see the same with the QWPs (orientied perpendicularly to each other at $\pm \pi/4$ relativive to the H-direction). There you don't have double-slit interference fringes (you still see the single-slit diffraction pattern).

For a didactic presentation of the experiment, see my habilitation colloquium slides (English version):

https://itp.uni-frankfurt.de/~hees/publ/habil-coll-talk-en.pdf

There I use idealized plane-wave photon states for simplicity.

 

[tistemfnp]

I don't get it. There is no screen in the Walborn experiment. Figure 1 shows the setup, no screen. All figures start with "coincidence", which is exactly doing what I told, relating the output to a prepared setup of the other detector. If you would count all clicks without coincidence, no interference pattern. The interference pattern is virtual.

If entangled photons would show an interference pattern, you would need to be able to have a double slit for one of the partners, followed by a screen (a lens between to project infinity to the screen is ok). No interference. Now you could argue "but the heralding rate is low" and the noise is too high. But then you just need to wait longer and average. You could take 1000 photos and average them. You will never (n e v e r) see an interference pattern with such a setup. You are doomed to coincidence counting with the other detector and coincidence means always virtual.

Edit: You could simulate a photo screen with the Walborn experiment. You just need to repeat the experiment with many positions of the other detector and add the outputs. As soon as you shift the other detector (perpendicular to the optical axis) the (virtual!) interference pattern would change its phase and if you overlap many of them the overall sum will not show interference anymore. This would be effectively the output without fixing the position of the other partner. And noise efficiently suppressed, using coincidence and only relating to the effective heralding rate.

Edit 2: The main flaw of thinking with the outcome of the Walborn experiment is to pretend the position of the idler arriving at detector D_P is only at position of D_P. It's not. D_P also needs to be shifted in x (as depicted for D_S). This will shift the (virtual) interference pattern in phase detected with D_S. The quantum eraser is an excellent example for the tendency in humans to confuse correlation with causality, including the authors of such experiments.

 

[vanhees71]

The screen is substituted by a photo detector which can be placed at different positions. How else do you think they made the Figures in their paper? Of course they do coincidence measurements with the two photodetectors, and that's why you can't simply use a screen for the signal photon. All this is completely irrelevant to the question whether or not there's an interference term of the partial waves going through the one or the other slit.

I don't see any flaw in the arguments of the authors. For sure they are well aware about the fact that indeed there's no retrocausality nor causal interactions between space-like separated registration events. From where in the paper to you infer the claim the authors were not aware of this basic facts about QED?

 

[tistemfnp]

The screen is substituted by a photo detector which can be placed at different positions. How else do you think they made the Figures in their paper?

Quoting @DrChinese "Arrrghh". A photo detector photo screen per se is not able to fix the position of the other partner while a detector with a coincidence counter will do exactly that.

You don't see a flaw as you obviously do not understand that fixing the other partner (x-position of D_P ) changes something in the experiment. Edit *)

A real photo-detector screen would add all virtual interference patterns and as they have random phases, related to the position in x of detector D_P, it would not show interference.

The difference between causality and correlation is well explained in this video.

Edit: Please have a look at the work of Dopfer, figure "Abbildung 4.5". Please specifically note the comment on "Quelle" D1: f i x. If you change the position of D1 in x, the virtual interference pattern on D2 will also change its position in x. If you put a real photo screen instead of a detector D2, you have no control on fixing the position of D1 in x! Hence, what you need to do in order to get the output of a real screen is to take the outcome of many such experiments, where you change the position of detector D1 in x. This will remove the interference pattern, as it adds up different phases.

 

[vanhees71]

I still don't understand your arguments. Are we discussing about the same experiment? I refer to the Walborn quantum eraser experiment.

https://arxiv.org/abs/quant-ph/0106078
https://doi.org/10.1103/PhysRevA.65.033818

In this experiment the interference pattern is measured by counting photon rates of a detector as a function of its position. The detection position is given by the position of the detector, no matter whether you use a single photon detector, a photoplate or silicon-pixel detector or whatever you want.

 

[tistemfnp]

In this experiment the interference pattern is measured by counting photon rates of a detector as a function of its position.

Half of the truth. In this experiment a virtual interference pattern is measured by counting photon rates of a detector as a function of its position coincident with counts of another detector of which the position is fixed.

The detection position is given by the position of the detector, no matter whether you use a single photon detector, a photoplate or silicon-pixel detector or whatever you want.

And how would a photoplate be able to only register detections coincident with detector D_P ? You will not see any interference pattern on a photoplate, as you won't be able to selectively look at a virtual interference pattern depending on the position of D_P.

If at detector D_S you change the position in x, and fix the position of D_P (which is the case in the Walborn experiment), you will see an interference pattern I₁. If you change the x-position of D_P, you will see an interference pattern I₂. The interference pattern I₂, is also shifted in the x-position. So if you repeat the Walborn experiment N times with N different, random x-positions of D_P, you will get N different interference patterns I_n with arbitrary phases in x-direction. Adding them up will cancel out interference and you will see approximately a Gaussian-like shape of the distribution of detections. Exactly this is what you see if you look at the screen behind the double-slit using a photoplate. Clearer?

Likewise, changing elements in the path and or changing path lengths, does not effect the outcome of the experiment. The outcome of the experiment is determined by the conservation of momentum alone, independent on the order of events in time. Nothing is 'erased', only the state of the idler was changed. The (virtual!!) screen pattern does not change because something changed retrospective in time, but because the correlation changed and a different type of (again: virtual) coincidence-pattern is observed.

To sum that up:
1. Nothing is erased in the Quantum "eraser". Instead, different correlations are observed.
2. Entangled photons do not show (self-)interference, as this would break conservation of momentum. Only virtual interference can be observed for a fixed position in x of the partner for a certain position in z of the partner (see Dopfer), in coincidence with the detection of the partner.Btw., isn't it much more cool, that for entangled photons we don't see two maxima on the screen behind a double slit, but only one? As if it wasn't two slits but one? This has an application: you can determine if photons are entangled by looking at what happens after a double-slit: interference --> simple photons; no interference and one maximum --> entangled photons. I like that .

 

[vanhees71]

Half of the truth. In this experiment a virtual interference pattern is measured by counting photon rates of a detector as a function of its position coincident with counts of another detector of which the position is fixed.
And how would a photoplate be able to only register detections coincident with detector D_P ? You will not see any interference pattern on a photoplate, as you won't be able to selectively look at a virtual interference pattern depending on the position of D_P.

That's why you use the movable detector and not a photoplate.

If at detector D_S you change the position in x, and fix the position of D_P (which is the case in the Walborn experiment), you will see an interference pattern I₁. If you change the x-position of D_P, you will see an interference pattern I₂. The interference pattern I₂, is also shifted in the x-position. So if you repeat the Walborn experiment N times with N different, random x-positions of D_P, you will get N different interference patterns I_n with arbitrary phases in x-direction. Adding them up will cancel out interference and you will see approximately a Gaussian-like shape of the distribution of detections. Exactly this is what you see if you look at the screen behind the double-slit using a photoplate. Clearer?

Yes, it was clear all the time.

Likewise, changing elements in the path and or changing path lengths, does not effect the outcome of the experiment. The outcome of the experiment is determined by the conservation of momentum alone, independent on the order of events in time. Nothing is 'erased', only the state of the idler was changed. The (virtual!!) screen pattern does not change because something changed retrospective in time, but because the correlation changed and a different type of (again: virtual) coincidence-pattern is observed.

To sum that up:
1. Nothing is erased in the Quantum "eraser". Instead, different correlations are observed.

That's in a sense true: You select different subensembles. What's erased is the information imprinted in the corresponding states, i.e., the "preparation procedures". As Zeilinger puts it QT is foremost about "information".

2. Entangled photons do not show (self-)interference, as this would break conservation of momentum. Only virtual interference can be observed for a fixed position in x of the partner for a certain position in z of the partner (see Dopfer), in coincidence with the detection of the partner.

Of course, they show "self-interference". Otherwise you'd never see a double-slit interference patterns. I don't know, what you mean by "virtual interference". In the Dopfer experiment you observe a "real interference" in the setup, where no which-way information is present and you don't see interference, if it's not present, depending on the position of the "Heisenberg lens". That's the whole point of this experiment!

Btw., isn't it much more cool, that for entangled photons we don't see two maxima on the screen behind a double slit, but only one? As if it wasn't two slits but one? This has an application: you can determine if photons are entangled by looking at what happens after a double-slit: interference --> simple photons; no interference and one maximum --> entangled photons. I like that .

This I also don't understand.

E.g., in the Walborn experiment with no QWPs present in the slits you see an interference pattern with the signal photons. It doesn't matter that these are entangled with the other photon from the parametric down conversion. If you don't know about the idler photons you simply have unpolarized but coherent photons making a double-slit interference pattern. Without the coincidence measurements on both photons you can never figure out that the signal photons are entangled with another photon. To observe entanglement you need these two-photon observations!