What destroys interference: Possibility or actual measurment?

[greypilgrim]

Hi.

If we place two perpendicularly oriented linear polarizers at a double slit, there will be no interference pattern. In the wave picture, this is trivial: Two waves with perpendicular polarization cannot interfere.

In a quantum picture, I've often seen the interpretation that the sheer possibility of finding out which path a photon took (by measuring its polarization) destroys the interference, even if this measurement hasn't been performed.

Now there's those infamous delayed choice quantum eraser experiments: Apparently one can measure the path information (thereby destroying interference) and erase it to restore the interference patter (using some rather shady post-processing). This appears to me as if the actual performance of a measurement of the path information destroys interference, and apparently it can be undone.

So what exactly destroys interference: Is it the sheer possibility of measuring the path information or does it need an actual measurement?

Is the "measurement and erasure" procedure in a quantum eraser experiment even a measurement in the orthodox (Kopenhagen) sense, i.e. is it a projection, or is it more of a unitary time evolution?

 

[DrChinese]

Hi.

If we place two perpendicularly oriented linear polarizers at a double slit, there will be no interference pattern. In the wave picture, this is trivial: Two waves with perpendicular polarization cannot interfere.

In a quantum picture, I've often seen the interpretation that the sheer possibility of finding out which path a photon took (by measuring its polarization) destroys the interference, even if this measurement hasn't been performed.

Now there's those infamous delayed choice quantum eraser experiments: Apparently one can measure the path information (thereby destroying interference) and erase it to restore the interference patter (using some rather shady post-processing). This appears to me as if the actual performance of a measurement of the path information destroys interference, and apparently it can be undone.

So what exactly destroys interference: Is it the sheer possibility of measuring the path information or does it need an actual measurement?

Is the "measurement and erasure" procedure in a quantum eraser experiment even a measurement in the orthodox (Kopenhagen) sense, i.e. is it a projection, or is it more of a unitary time evolution?

There must be the possibility of obtaining the information. In this sense, the measurement can be performed and then the information discarded - or not. There will be no interference.

On the other hand: If the measurement is performed and then "erased" - by eliminating any and all traces of the results - then there is interference (i.e. it is as if the measurement never happened). Typically this involves restoring the quantum state prior to measurement.

 

[A. Neumaier]

So what exactly destroys interference: Is it the sheer possibility of measuring the path information or does it need an actual measurement?

It is the presence of an interaction with the environment. This means that some information is passed to the environment, even when it is impossible to actually measure it.

If the measurement is performed and then "erased" - by eliminating any and all traces of the results - then there is interference (i.e. it is as if the measurement never happened).

''Erasure'' of true measurements is impossible; in the experiments that carries this name the measurement is not actually performed. Performed measurements must leave a macroscopic, on the microscopic time scale permanent record and are therefore always irreversible, by definition.

 

[vanhees71]

The interference terms vanish, because the partial beams of particles running through the slits with perpendicularly oriented polarization filters is in the formalism due to the fact that the corresponding states are orthogonal to each other. Physically it describes a strict entanglement between which-way information and polarization of the particle that has run through the double-slit setup since the polarization filters impose the corresponding polarization state on each particle running through it. That means that by measuring the polarization of the particle you know through which slit it has come, and whether or not you measure it the which-way information is carried by the particles' polarization state. So there is no interference pattern, because you can distinguish through which slit each particle came with certainty.

If you slight distort the polarization filters from not exactly being perpendicularly oriented, you get an interference pattern but also loose the certainty through which slit it has come. The more uncertain the which-way information gets (at the extreme you orient both polarizers exactly parallel, and then you cannot distinguish through which slit each particle came at all), and the interference pattern gets it's highest contrast.

There's the fascinating possibility thanks to the possibility to create polarization-entangled photon pairs to "erase" which-way information after all measurements are done. You get then subensembles of the full ensemble of measured photons which show an interference pattern, while the total ensemble refers to one where you have which-way information (in principle), and thus there's no interference pattern. For details, see

https://arxiv.org/abs/quant-ph/0106078

 

[DrChinese]

''Erasure'' of true measurements is impossible; in the experiments that carries this name the measurement is not actually performed. Performed measurements must leave a macroscopic, on the microscopic time scale permanent record and are therefore always irreversible, by definition.

As far as I know, you are right about experiments labeled as "erasure". However, there are others that have long been postulated - I can't say if they have actually been performed or not.

J.H. Eberly, Bell inequalities and quantum mechanics (2001)
http://www.pas.rochester.edu/~advlab/Eberly_Bell_Inequalities_AJP.pdf
See Figure 1, itself taken from A.P. French and E.F. Taylor, An Introduction to Quantum Mechanics (1979).
See also Figure 2.

You can see that the idea is that a PBS measures the polarization of a photon. Then the PBS outputs are recombined by a second "reverse" PBS, before anything is recorded indicating which path the photon took. The original photon state, prior to entering the first PBS, is restored.

So in fact it would be reversible measurement, if they are correct. I believe that an entangled photon, run through a suitable version of Figure 1, would in fact remain polarization entangled afterwards. Thoughts?

 

[A. Neumaier]

See Figure 1

In Figure 1, nothing is measured at all, and the author's don't claim that anything is measured.

 

[DrChinese]

In Figure 1, nothing is measured at all, and the author's don't claim that anything is measured.

Sure something is measured by the first PBS: polarization. Of course it is the entire measurement context that matters. The second PBS reverses the initial measurement, so it is as if it never happened. Net: nothing is measured. My original statement on which you commented:

"If the measurement is performed and then "erased" - by eliminating any and all traces of the results - then there is interference (i.e. it is as if the measurement never happened)." But something did happen at a quantum level, it just happens to net out in the example.

It is obvious that the first PBS does in fact split the photon into 2 distinct paths. These components can be manipulated independently. You could rotate one or both of them, for example. You could alter their path lengths. And you can even recombine them into something close to their original. When you do the last operation, you have the potential to restore the original input entangled state. That entangled state could not be evidenced from either component alone, must first be recombined.

So depending on your semantics: there is measurement, and then "erasure". Or you can say there was no measurement at all. I say the first is more descriptive.

 

[PeterDonis]

So what exactly destroys interference

One answer that seems to be increasingly popular with quantum physicists: decoherence.

So depending on your semantics: there is measurement, and then "erasure". Or you can say there was no measurement at all. I say the first is more descriptive.

Another possible description would be to say there was measurement, but not decoherence. The lack of decoherence is what makes the erasure possible. But not all physicists would agree with this definition of "measurement", since one such definition is that measurement requires decoherence, so the process going on in this experiment would not be properly termed a "measurement", just an interaction between the photon and the beam splitter.

 

[greypilgrim]

There must be the possibility of obtaining the information. In this sense, the measurement can be performed and then the information discarded - or not. There will be no interference.

On the other hand: If the measurement is performed and then "erased" - by eliminating any and all traces of the results - then there is interference (i.e. it is as if the measurement never happened). Typically this involves restoring the quantum state prior to measurement.

I don't quite understand. In the second paragraph you say there is interference if the system is brought back into the state prior to measurement. I assume in this state there is the possibility of obtaining the path information (otherwise the measurement would have been pointless anyway). But in the first paragraph you say that there will be no interference in the presence of such a possibility, even if no measurement is performed?

 

[DrChinese]

I don't quite understand. In the second paragraph you say there is interference if the system is brought back into the state prior to measurement. I assume in this state there is the possibility of obtaining the path information (otherwise the measurement would have been pointless anyway). But in the first paragraph you say that there will be no interference in the presence of such a possibility, even if no measurement is performed?

In my first paragraph: I draw the distinction of a situation where a measurement is clearly made, but the result is discarded into the environment. If you choose to ignore that information (or look at it), the outcome is the same. There is NO interference.

In my second paragraph: I draw the distinction where a state is returned to the original. There is no lingering information available, either to look at or ignore. There IS interference.

 

[DrChinese]

One answer that seems to be increasingly popular with quantum physicists: decoherence.

Another possible description would be to say there was measurement, but not decoherence. The lack of decoherence is what makes the erasure possible. But not all physicists would agree with this definition of "measurement", since one such definition is that measurement requires decoherence, so the process going on in this experiment would not be properly termed a "measurement", just an interaction between the photon and the beam splitter.

A. Neumaier might quite agree with this approach. I don't really object, and probably most would say the same as you. I still think it comes back to your definition. If you don't see a measurement and subsequent erasure (or reversal), there is no objective way for me to say it is there. We're back to the idea that speculating on the activities of quantum particles (when we are not obtaining information about them) is a bad idea.

But I still see measurement and reversal.

 

[Nugatory]

In my first paragraph: I draw the distinction of a situation where a measurement is clearly made, but the result is discarded into the environment. If you choose to ignore that information (or look at it), the outcome is the same. There is NO interference.

In my second paragraph: I draw the distinction where a state is returned to the original. There is no lingering information available, either to look at or ignore. There IS interference.

In both cases there is an interaction. But not all interactions are measurements, so we need an agreement about which interactions are measurements. If we consider that a measurement is a statistically irreversible interaction (a position that interacts nicely with decoherence) then the second case is not a measurement.

So another question: Can the interference pattern be maintained across a statistically irreversible interaction? (Before the pattern itself forms on the screen - that's pretty clearly irreversible).

 

[atyy]

Measurement is irreversible.

Passing through a PBS, although colloquially may mean a measurement - strictly speaking it is not, since considering a larger system, the projection of the PBS can be modeled as unitary evolution. https://arxiv.org/abs/quant-ph/0305007v2

 

[OCR]

Sorry DrChinese, but I am completely baffled by this statement...

Typically this involves restoring the quantum state prior to measurement.

Is it even possible to know what the prior quantum state is ?

Wheeler's delayed choice experiment.
Delayed choice quantum eraser experiment.

 

[atyy]

Another possible description would be to say there was measurement, but not decoherence. The lack of decoherence is what makes the erasure possible. But not all physicists would agree with this definition of "measurement", since one such definition is that measurement requires decoherence, so the process going on in this experiment would not be properly termed a "measurement", just an interaction between the photon and the beam splitter.

Probably the other way - measuring and throwing away information is equivalent to decoherence which is unitary evolution (in principle reversible). It is only when a measurement is made and the information retained that state reduction is necessary.

 

[A. Neumaier]

Sure something is measured by the first PBS: polarization

No. As long there is no measurement result (communicated by the interaction to the environment) there is no measurement in any meaningful sense of the word.

You could rotate one or both of them, for example. You could alter their path lengths.

But these are all unitary transformations, not measurements. The state remains completely unknown to the observer (unless it was known initially, in which case it remains completely known). But without information flow no measurement!

I'd be interested in your definition of measurement that leads you to the opposite conclusion, and to a justification why what you call so deserved to be called a measurement although nobody can know anything about the measurement result.

 

[vanhees71]

I don't know, about which concrete experiment you guys are talking right now. In the quantum-erasure experiment by Walborn et al that I mentioned above, of course very specific measurements of the polarization of photons are done and measurement protocols irreveribly fixed. These measurement protocols then allow to choose subensembles of the total ensembles such that putative which-way information is "erased" for this subensemble. It's a pretty easy to understand delayed-choice experiment without any surprise, if you are familiar with quantum mechanics, and it's very clear, how the interference pattern is restored for the subensemble but missing for the total ensemble. For a simple explanation in German, see

http://theory.gsi.de/~vanhees/faq/qradierer/qradierer.html

 

[DrChinese]

Measurement is irreversible.

Passing through a PBS, although colloquially may mean a measurement - strictly speaking it is not, since considering a larger system, the projection of the PBS can be modeled as unitary evolution. https://arxiv.org/abs/quant-ph/0305007v2

Nice reference, thanks. And yes, I agree that strictly speaking, a measurement is a measurement is a measurement.

 

[DrChinese]

No. As long there is no measurement result (communicated by the interaction to the environment) there is no measurement in any meaningful sense of the word.

But these are all unitary transformations, not measurements. The state remains completely unknown to the observer (unless it was known initially, in which case it remains completely known). But without information flow no measurement!

I'd be interested in your definition of measurement that leads you to the opposite conclusion, and to a justification why what you call so deserved to be called a measurement although nobody can know anything about the measurement result.

I'm not really disagreeing with everyone's final opinion. No information to the environment, no measurement. Context is king, you must look at the ENTIRE experimental setup to get the right answer.

However, it does not take any special insight to see that a PBS does "something" within that greater context. Since the net result is "no measurement" and there are 2 component apparati, what is it that the PBS does? Obviously until the passing photon is measured as going through one branch or the other, the potential is not actualized. So the irreversible detection is the space-time point of where/when the measurement occurs, and that is by definition - making it impossible to have a different opinion if we all agree on that word's exact meaning.

But as always, that means that it is the later detection (or not) that determines the nature of what happened at the PBS earlier (and that no "measuring" occurs there). This makes perfect sense to me, precisely because I agree that context is king. It does seem to me an abuse to pretend that nothing happens at the PBS or the reversing PBS. What do you call it if not a conditional measurement or the like?

Again, to me there is no disagreement about the physics. If we specify the context, we will agree on the expected outcome. I don't think the words are particularly effective in describing that, given the example and others like it.

 

[DrChinese]

Is it even possible to know what the prior quantum state is ?

You can return it to the earlier state without knowing what it is, right?

 

[DrChinese]

I don't quite understand. In the second paragraph you say there is interference if the system is brought back into the state prior to measurement. I assume in this state there is the possibility of obtaining the path information (otherwise the measurement would have been pointless anyway). But in the first paragraph you say that there will be no interference in the presence of such a possibility, even if no measurement is performed?

And to rephrase my comments so that they makes sense in terms of what everyone else is saying, since you are the OP:

A true "measurement" does not occur until the information is irreversibly released into the environment, regardless of whether a detection device or person obtains that information. In that strict sense, there is no such thing as erasure (because there was no measurement to erase). However, when you look at the setups, you can see that there are points in the quantum path which are conditional forks: which-path (L-R) or polarization (H-V) depending on the example. Those conditional forks are technically not the measurement points in the language being discussed.

So I was being loose with the word "measurement" to better match your usage. "The system is brought back into the state prior to measurement" might be more accurately phrased: "The system is brought back into the state prior to the 'conditional' measurement" and whether there is a measurement or not is conditioned on the full final context.

 

[A. Neumaier]

a PBS does "something" within that greater context. Since the net result is "no measurement" and there are 2 component apparati, what is it that the PBS does?

The question is when ''doing something'' deserves the designation ''measurement''. In general, doing something only means an activity of some sort happened, and measurement is the special activity that results in a measured value.

If a classical white ray goes through a prism, something nontrivial happens, since the ray splits into a bunch of rays of different color, but nobody in optics calls this a measurement. The measurement happens only when you look at the spectrum on a screen or a photographic plate. In a complicated traditional camera, the incoming light goes through a number of lenses - none of them is performing a measurement in any conventional terminology, but the state of the light is transformed at each stage. The measurement happens only at the photographic emulsion, which is part of the environment, and is irreversibly changed by the incoming light.

It does seem to me an abuse to pretend that nothing happens at the PBS or the reversing PBS. What do you call it if not a conditional measurement or the like?

It is called an optical transformation. There are lossless optical transformations - e.g. those that change momentum (such as a mirror, prism, lens, grating), split a beam (half-silvered mirror), rotate polarization (rotator) -, all represented by unitary operators; these are reversible and have nothing to do with measurement. There are also lossy optical transformations such as polarizers, who suppress part of the light in a state-dependent way. These are dissipative as they have an effect on the environment (energy is transmitted and can, in principle, be measured) and cannot be reversed.

To measure intensity (energy density), one counts clicks with a photocounter. To measure polarization, one passes the light through a polarizer and measures the resulting intensity at various angles of the polarizer. Thus one measures something (the energy density), and infers something else (the polarization) through a computation from the raw measurements. The combination of energy measurements and known settings of the polarizer produces the measurement. A rotator before the polarizer just affects the subsequent measurements but doesn't measure anything itself. What holds for a rotator holds similarly for any other lossless optical transformation.

I have never heard anyone in the classical (ray optics) setting using measurement terminology for lossless optical transformations. There is no reason to assume that if the same physical process is described quantum mechanically that more is measured than in the classical case - especially when there is not the slightest trace of measurement results!

If you still think that a rotator measures something, then a mirror, a prism, or a beam splitter should also measure something, since according to your reasoning, they ''do something'' to the state of light. Could you please tell me what these could possibly measure? Not even conditionally is something measured.

 

[ShayanJ]

I think the point is, there are processes that are not measurements themselves, but are important steps of a measurement. So you can either look at them as part of a measurement or as a standalone process. Now if you do one of these processes, one may think it was the beginning of a measurement and then if you reverse it, they would say the measurement was begun but was stopped. But others just say two unitary processes happened that had nothing to do with a measurement. So to have a uniform convention, we think of these processes as steps in a measurement only when there is an observation at a later point, otherwise they're just unitary processes that have nothing to do with a measurement. But maybe delayed-choice experiments challenge this way of thinking about the issue. I don't know enough to say anything so I'd appreciate any comment on this.

 

[A. Neumaier]

we think of these processes as steps in a measurement only when there is an observation at a later point, otherwise they're just unitary processes that have nothing to do with a measurement.

More precisely, if you measure the original ray then the rotator, the polarizer, and the photocess combined form the detector.

If you measure the intermediate ray then the rotator is a process unrelated to measurement (instead it is part of the preparation procedure), and polarizer and photocell combined form the detector. Whereas if you measure the final ray then the rotator and the polarizer are processes unrelated to measurement (though the polarizer is dissipative and, in principle, something could be measured there) and only the photocell is the detector.

In the three cases, although the physical setting is identical, measurement means something different, since the protocol used to go from the raw data (clicks of the photocell) to the measurement results is different in each case.

 

[DrChinese]

To measure intensity (energy density), one counts clicks with a photocounter. ... Not even conditionally is something measured.

I don't have any argument with what you are saying with the first statement. I think if you look back at the OP, you will see how we got to the point of discussing this in the first place - and how in many ways we have skirted that . The question was:

"So what exactly destroys interference: Is it the sheer possibility of measuring the path information or does it need an actual measurement?"

We could answer by saying only a measurement destroys the interference. But then you have to explain that the information being dissipated to the environment is a measurement in that sense as well. I don't think that communicates much useful.

The example is one regarding double slits and polarizers in front of each. There is interference when they are parallel, none when they are crossed. How do you explain that in terms of a measurement (note: this is a rhetorical question)? My point is that we are trying to aid the understanding of the situations in which interference occurs or does not for the OP. The variable being: in one situation there is an actual/potential measurement of which-slit, and in the other there is not.

 

[A. Neumaier]

The example is one regarding double slits and polarizers in front of each. There is interference when they are parallel, none when they are crossed. How do you explain that in terms of a measurement

There is no explanation in terms of measurement alone. Both polarizers absorb some light when light goes in that is polarized in a direction different from both polarizers, and the amount of absorbed energy (and only that) can be measured at each polarizer.

Nevertheless there is an interference pattern when the polarizers are parallel and there is none when they are at a significant angle. The reason is that in the first case (only) the electromagnetic field is transformed identically (though dissipatively) behind both slits, so that the result is identical to the result that would have been obtained if instead there was a single polarizer in front of the two slits. Since the latter produces interference when initially coherent light goes in, we have interference.

No such possibility exists in the second case. The field is transformed differently in different parts of the field, and hence coherence is (gradually, if the angle is increased) destroyed. One can actually calculate the effect of the transformations and the resulting partial interference pattern at each angle.

Note also that interference is a purely classical effect and has nothing intrinsically to do with quantum mechanics. One calculates throughout with the Maxwell equations and the intensity of light, and uses the fact (known since Maxwell, long before the advent of quantum mechanics) that the response on the screen is proportional to the intensity integrated over the period of time of observation. The pattern becomes a bit erratic when the intensity is low (a quantum property of the detector, not the light) or pulsating (when sending single photons at a time), but this has nothing to do with interference itself.

 

[DrChinese]

There is no explanation in terms of measurement alone. ...

Did I mention it was a rhetorical question?

Anyway, I feel we have beat this dead horse enough. Especially since the dead horse is me.

 

[vanhees71]

It's not only a measurement that destroys interference. In the eraser experiment by Walborn et al the two quarter-wave plates in relative 90-degree orientation lead to the destruction of the interference. Although the polarization state and thus the which-way information has never been measured, the interference pattern is gone. The quarter-wave plates are (when thought of being ideal) no measurement device but are described by a unitary operator in polarization space. In this case they are part of the preparation process of the photons going through the double slit. The measurement of the photons occur when they hit the CCD and are irreversibly registered (and threreby absorbed and thus destroyed).

 

[OCR]

You can return it to the earlier state without knowing what it is, right?

Yes, this state... "prior to measurement."

Typically this involves restoring the quantum state prior to measurement.

How do you know what the state is (not was, since you're " restoring the quantum state" (to now?)), prior to measurement.

No reply needed Doc... cause I know.[COLOR=#black]..[/COLOR] lol

I feel we have beat this dead horse enough. Especially since the dead horse is me.

 

[zonde]

Passing through a PBS, although colloquially may mean a measurement - strictly speaking it is not, since considering a larger system, the projection of the PBS can be modeled as unitary evolution. https://arxiv.org/abs/quant-ph/0305007v2

You need MWI to model projection of the PBS as unitary evolution. If you would attribute to photon certain position (at least up to the diameter of output beam) PBS projection would have to be considered measurement.

 

[greypilgrim]

The measurement of the photons occur when they hit the CCD and are irreversibly registered (and threreby absorbed and thus destroyed).

But in the usual delayed choice quantum eraser setups, isn't the claim that the choice can be made after the signal photon hits the detector?

 

[vanhees71]

It's not a claim, it's done, and the delayed choice is done using the fixed measurement protocol at any time after the measurement in the above sense is done! You just choose an appropriate subensemble of the full ensemble of registered photons. Please look at the very nicely written orginal paper:

https://arxiv.org/abs/quant-ph/0106078

 

[DrChinese]

It's not a claim, it's done, and the delayed choice is done using the fixed measurement protocol at any time after the measurement in the above sense is done! You just choose an appropriate subensemble of the full ensemble of registered photons. Please look at the very nicely written orginal paper:

https://arxiv.org/abs/quant-ph/0106078

That is not what is meant by "delayed choice" in this situation. From the paper (and relating to greypilgrim's question):

Delayed erasure

The possibility of obtaining which-path information after the detection of photon s leads to delayed choice [34]. Delayed choice creates situations in which it is important to have a clear notion of the physical significance of quantum mechanics. A good discussion can be found in references [11–14]. In as much as our quantum eraser does not allow the experimenter to choose to observe which-path information or an interference pattern after the detection of photon s , it does allow for the detection of photon s before photon p, a situation to which we refer to as delayed erasure. The question is: “Does the order of detection of the two photons affect the experimental results?” ... We use the term “delayed choice” loosely, in that in our experiment there is no “choice” available to the observer in the time period after the detection of photons s and before the detection of photon p.

As before, you can simply define "measurement" to mean "when both s and p have been detected" and then your view is fully descriptive (and nothing is considered as "delayed"). Or you can split the measurement into 2 parts, one for s and a separate one for p. Then the term "delayed" that the authors use is relevant, and they test that. Obviously, the order of detection of s and p does not change the results.

 

[vanhees71]

Ok, I thought the paper is clear enough, but obviously we have to discuss the details. One starts with a maximally polarization-entangled two-photon state from a appropriate parametric-down-conversion apparatus:
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\hat{x} \rangle_s \otimes |\hat{y} \rangle_p-|\hat{y} \rangle_s \otimes |\hat{x} \rangle_p),$$
where $|\hat{x} \rangle_s$ means a single photon that is polarized in $\hat{x}$ direction. The index $s$ means the "signal photon", i.e., that it is the photon that runs through the double slit (with quarter-wave plates) and is finally detected on the screen, and $p$ means the "polarizer photon", which is send to a far-distant polarization analyzer.

The state of the photons immediately after the double slit is given by applying the operator describing the quarter-wave plates which are in $\pm 45^{\circ}$ direction in the $xy$ plane, which are
$$\hat{Q}_{+}=|L \rangle \langle \hat{x}|+\mathrm{i} |R \rangle \langle \hat{y},\\
\hat{Q}_{-}=|R \rangle \langle \hat{x}|-\mathrm{i} |L \rangle \langle \hat{y},$$
where $|R/L \rangle$ are the right-/left-circular polarization states (helicity eigenstates).

So when the $s$ photon would be detected right at the double slits 1 or 2, you'd have the two-photon state (we assume that there's no strange interaction of the quarter-wave plates with the $p$ photon)
$$|\Psi_1 \rangle=\hat{Q}_+ \otimes \hat{1} |\Psi \rangle, \quad |\Psi_2 \rangle=\hat{Q}_- \otimes \hat{1} |\Psi \rangle.$$
Now when the $s$ photon is observed at the far-distant screen, the state is
$$|\Psi' \rangle = \frac{1}{\sqrt{2}} (|\Psi_1 \rangle + \exp(\mathrm{i} \varphi(\vec{x}))|\Psi_2 \rangle),$$
where
$$\varphi(\vec{x})=\frac{2 \pi x d}{L \lambda},$$
with the geometry given at the figure on

http://theory.gsi.de/~vanhees/faq/qradierer/node3.html

Now if you measure all photons without taking care about anything concerning the $p$ photons, you get
$$\langle \Psi' | \Psi' \rangle=1,$$
because $\langle \Psi_1|\Psi_2 \rangle=0$. Indeed there's no interference pattern, because now you can determine uniquely through which slit each $s$ photon has gone by checking, whether they are in polarization state 1 or 2, which are true alternatives since the corresponding vectors are orthogonal (i.e., you loose the interference pattern completely).

Now to erase the which-way information to some degree or completely, you measure the polarization state of the $p$ photon with the polarizer set with an angle $\alpha$ wrt. the $x$ direction. Then you filter out only such photons for that the $p$ photon has gone through the polarizer. This is described by the projection operator $\hat{1} \otimes \hat{P}(\alpha)$, where
$$\hat{P}(\alpha)=|\hat \alpha \rangle \langle \hat{\alpha}|, \quad |\hat{\alpha} \rangle = \cos \alpha |\hat{x} \rangle + \sin \alpha |\hat{y} \rangle.$$
Applying this projector with $\alpha = \pm 45^{\circ}$ to $|\Psi' \rangle$ you get
$$|\Psi_{+45}'' \rangle=\frac{1-\mathrm{i} \exp(\mathrm{i} \varphi(x))}{2 \sqrt{2}} (|L \rangle_s \otimes |\hat{45} \rangle_p+\mathrm{i} |R \rangle_s \otimes |\hat{45} \rangle_p),$$
and thus for the corresponding partial ensemble the interference pattern
$$\langle \Psi_{+45}'' |\Psi_{45}'' \rangle=\frac{1}{2} [1+\sin(\varphi(x))].$$
The corresponding complementary part of the ensemble is given by considering only those $s$ photons whose entangle $p$ photon is not absorbed, i.e., which was polarized in direction $-45^{\circ}$. The corresponding application of the projector finally gives indeed the complementary interference pattern
$$\langle \Psi_{-45}'' |\Psi_{-45}'' \rangle=\frac{1}{2} [1-\sin(\varphi(x))].$$
As it must be both partial ensembles together give $1$ as it must be, since nothing has been done to the $s$ photons by measuring the polarization state of the $p$ photons.

If you choose any other angle $\alpha$ for the $s$ photons you get interference patterns with more or less contrast, and the complementary pattern for $\alpha'=\alpha \pm 90^{\circ}$. That's the demonstration of a delayed erasure of which-way information without ever interfering with the $s$ photon. That's made possible by the entanglement of the $s$ with the $p$ photon from the very beginning of their preparation and a specific quantum feature, i.e., here you need to really deal with the single biphotons and you have to keep the time of the measurement protocols at the screen and the polarizer (corresponding to the usual observers Alice and Bob :-)) accurately enough to know which $s$ photon was entangled with which $p$ photon.