Does DCQE show a shift between the two "which path" results?

[Rene Dekker]

TL;DR Summary

It is not entirely clear whether the DCQE D3 and D4 which-path results show a difference. If they indeed show a difference, isn't that quite an important result?

The Delayed Choice Quantum Eraser (DCQE) experiment attempts to show that which-path information can be "erased," and interference recovered. I'll refrain from explaining the experiment, and assume you are familiar with the setup. I refer to the Wikipedia page about it, and the original Kim et.al. research paper.

My question is: is there a difference between the results for the two "which-path" detectors? The coincidence count D0/D3 and D0/D4 both show a bulge pattern. But is the maximum of that pattern shifted between the D3 and the D4 results?
The original paper spends only one sentence on it: "There is no significant difference between the curves of R03 and R04 except the small shift of the center". It does not include the D4 detector in the experimental setup diagram, and does not show the graph for the R04 results. The casual attitude with which it is treated in the paper could indicate that it was not an important result in their mind, and that they did not pay much attention to it (for example, they may not have tested whether the shift was statistically significant).
The wikipedia page is ambivalent about it. The text only states "R03 shows a single maximum, and R04, which is experimentally identical to R03 will show equivalent results", some of the pictures show no shift, other pictures show a shift.

Hence my questions:
1) in the experimental results, was there indeed a significant shift in the patterns between the R03 and R04 results ?
2) does QM predict whether there should be a shift (sorry, my math is not good enough to understand the mathematical description in the paper).
3) if there was a significant shift, isn't that quite a profound result? After all, the patterns are for photons arriving at D0 on the signal side. On the signal side, there is no path separation at all, only combination of the paths. Doesn't the result show clearly that BOTH interference AND which-path information should be present in the wave-function for individual photons, even when the paths are combined?

 

[Cthugha]

Hence my questions:
1) in the experimental results, was there indeed a significant shift in the patterns between the R03 and R04 results ?

Depends on what you call significant. There will be a tiny shift, but you will need to detect a very large number of photons to be able to identify it.

2) does QM predict whether there should be a shift (sorry, my math is not good enough to understand the mathematical description in the paper).

Already classical optics predicts that there should be a shift. The lens in the signal path actually performs a Fourier transform, so all photons emitted under the same angle will be detected at the same horizontal position at D0. The angle between the two slits and the lens will be very slightly different, so a tiny shift of the maximum might arise, but the overlap of the two curves will be somewhere in the 99%-range or above.

3) if there was a significant shift, isn't that quite a profound result? After all, the patterns are for photons arriving at D0 on the signal side. On the signal side, there is no path separation at all, only combination of the paths. Doesn't the result show clearly that BOTH interference AND which-path information should be present in the wave-function for individual photons, even when the paths are combined?

No. Unfortunately I do not understand where you get this idea from. You can estimate the amount of which-way information you would get in such a case from the overlap of the two individual curves. If they do not overlap, you get complete which way information. If they overlap perfectly, you get no which-way information. In all other cases, you can determine the partial which-way information from the intensities of the two curves (R03 and R04) at any point. It is roughly proportional to the difference of the two intensities divided by the sum of the two intensities. A value of 1 corresponds to full which-way information. A value of 0 corresponds to no which-way information.

You can actually simulate this pretty easily yourself using scientific software (octave will work, but if this is too complicated even excel or something similar works - you just need two shifted Gaussians and need to calculate their sums and differences at each point).

 

[Rene Dekker]

Already classical optics predicts that there should be a shift. The lens in the signal path actually performs a Fourier transform, so all photons emitted under the same angle will be detected at the same horizontal position at D0. The angle between the two slits and the lens will be very slightly different, so a tiny shift of the maximum might arise, but the overlap of the two curves will be somewhere in the 99%-range or above.

In classical physics the waves pass through both slits at the same time and therefore interfere on the screen. There is no concept of determining which slit the wave passed through, when they combine afterwards. If there is, then I missed something in classical physics as well.

No. Unfortunately I do not understand where you get this idea from. You can estimate the amount of which-way information you would get in such a case from the overlap of the two individual curves. If they do not overlap, you get complete which way information. If they overlap perfectly, you get no which-way information. In all other cases, you can determine the partial which-way information from the intensities of the two curves (R03 and R04) at any point.

The R03 result is a count of the subset of the D0 photons whose partner lands at D3. Likewise the R04 is a count of the subset of D0 photons whose partner lands at D4. Assuming that they make different patterns, that suggests to me that the wave-functions for those two subsets differ on the D0 side. If they don't differ, then how can you get different results for these subsets?

It is roughly proportional to the difference of the two intensities divided by the sum of the two intensities. A value of 1 corresponds to full which-way information. A value of 0 corresponds to no which-way information.

Do I understand correctly that you are saying that the wave-functions do indeed differ on the D0 side, and therefore which-path information is present in the wave-function (to a certain degree)?

 

[Cthugha]

In classical physics the waves pass through both slits at the same time and therefore interfere on the screen. There is no concept of determining which slit the wave passed through, when they combine afterwards. If there is, then I missed something in classical physics as well.

This feature is present indeed already in classical physics. If you consider reasonably narrow slits, you can approximate them as point sources emitting the same amout of intensity. The intensity of a point source falls of with distance r from the slit as $\frac{1}{r^2}$. If you place a detection screen in some distance, for any point on the screen not directly in the middle of the screen, the distance from slit 1 to this point and from slit 2 to this point will differ. Accordingly the intensities arriving from the two slits will be slightly different as well. This renders it more likely that a detection event on the left side of the screen originates from light going through the left slit, while a detection event on the right hand side of the screen is more likely to originate from light going through the right slit.
One usually considers a screen shifted towards a distance of infinity to make this small difference disappear. Of course, in practice this is not possible, but the effect of the finite distance can be made tiny (the lens in inserted in the path of D0 to make the effective distance effectively infinity). The longer you make the distance between the slits and the screen (and the smaller you make the distance between the two slits), the more negligible this effect becomes.

The R03 result is a count of the subset of the D0 photons whose partner lands at D3. Likewise the R04 is a count of the subset of D0 photons whose partner lands at D4. Assuming that they make different patterns, that suggests to me that the wave-functions for those two subsets differ on the D0 side. If they don't differ, then how can you get different results for these subsets?

As pointed out above: Being different is not an all-or-nothing thing, but a question of how much overlap these two distributions have. If they have an overlap of 99.99%, they are very similar, but not exactly equal. This also means that the which-way information is very close to 0.

Do I understand correctly that you are saying that the wave-functions do indeed differ on the D0 side, and therefore which-path information is present in the wave-function (to a certain degree)?

Yes, but this degree is too small to be of practical use. In most parts of the pattern it will rather be something like 50.01% probability of having gone through the left slit and 49.99% probability of having gone through the right slit or vice versa. Exact numbers depend on the geometry of the experiment.

 

[Rene Dekker]

This feature is present indeed already in classical physics. If you consider reasonably narrow slits, you can approximate them as point sources emitting the same amout of intensity. The intensity of a point source falls of with distance r from the slit as $\frac{1}{r^2}$. If you place a detection screen in some distance, for any point on the screen not directly in the middle of the screen, the distance from slit 1 to this point and from slit 2 to this point will differ. Accordingly the intensities arriving from the two slits will be slightly different as well. This renders it more likely that a detection event on the left side of the screen originates from light going through the left slit, while a detection event on the right hand side of the screen is more likely to originate from light going through the right slit.
One usually considers a screen shifted towards a distance of infinity to make this small difference disappear. Of course, in practice this is not possible, but the effect of the finite distance can be made tiny (the lens in inserted in the path of D0 to make the effective distance effectively infinity). The longer you make the distance between the slits and the screen (and the smaller you make the distance between the two slits), the more negligible this effect becomes.
As pointed out above: Being different is not an all-or-nothing thing, but a question of how much overlap these two distributions have. If they have an overlap of 99.99%, they are very similar, but not exactly equal. This also means that the which-way information is very close to 0.
Yes, but this degree is too small to be of practical use. In most parts of the pattern it will rather be something like 50.01% probability of having gone through the left slit and 49.99% probability of having gone through the right slit or vice versa. Exact numbers depend on the geometry of the experiment.

Thanks for your solid responses, Cthuga. But I admit that I am still a bit puzzled.
I can imagine that, the RO3 photons all have a wave-function that describes for an interference pattern that has its peak in front of one of the slits, and the RO4 photons all have a wave-function that show an interference pattern with its peak in front of the other slits. That would explain the complete experiment behaviour nicely. Is that how I should imagine it?

When you talk about a large overlap between the patterns (let's say 90%), do you mean that the RO3 photons all have a different probability pattern from the RO4 photons, but these two patterns overlap by 90%. Or do you mean that 90% of the photons have the same probability pattern, and 10% of them differ from that?

For me it surprising that there is any difference at all (even if small) between the individual photon's wave functions, and that the "which-way" information is encoded in them to some degree. I always thought the wave-function was always the same for all of them. But that might be just a gap in my knowledge, and might be less profound for the experts in the field.

 

[Cthugha]

Thanks for your solid responses, Cthuga. But I admit that I am still a bit puzzled.
I can imagine that, the RO3 photons all have a wave-function that describes for an interference pattern that has its peak in front of one of the slits, and the RO4 photons all have a wave-function that show an interference pattern with its peak in front of the other slits. That would explain the complete experiment behaviour nicely. Is that how I should imagine it?

Just as a disclaimer: Talking about wave-functions for photons is difficult as they are relativistic particles without rest frame. For example you cannot really put them into eigenstates in the sense that you perform a measurement and if you perform it again and again you will always find it in the same state. The first measurement will already destroy the photon.

In quantum optics you instead typically talk about probability amplitudes (which may look similar to wave-function from a non-expert point of view). You consider some initial situation (photon is moving towards the double slit), some final situation (a detection event has taken place at some part of the screen) and can consider probability amplitudes for different intermediate events (photon goes via the left slit or photon goes via the right slit).
In order to get the total probability of the final detection event, you sum up all the probability amplitudes corresponding to indistinguishable paths from the initial situation to the final event (in a phase-correct manner - these are complex values) and calculate the absolute square value of this sum. This is very similar to what one does with wave-functions.
In that sense, you get a probability amplitude for the photon going through the left slit and one for the photon going through the right slit.

If you have some way to mark which photon went through the left slit (e.g., by placing a polarization-rotating optical element there), the probability of finding the photon at some point at the detection screen will be given by the sum of absolute squared value of the probability amplitude for the left slit and the
absolute squared value of the probability amplitudes for the right slit. You do the absolute square first and then calculate the sum.

If the paths are indistinguishable, you instead do the sum first and then calculate the absolute squared value.

Now the probability amplitudes will have more or less the properties you want to ascribe to wavefunctions. The squared absolute of the probability amplitude for the left slit is a Gaussian and the one for the right slit is a slightly shifted Gaussian. The squared absolute of the sum of the probability amplitudes will show the interference.

When you talk about a large overlap between the patterns (let's say 90%), do you mean that the RO3 photons all have a different probability pattern from the RO4 photons, but these two patterns overlap by 90%. Or do you mean that 90% of the photons have the same probability pattern, and 10% of them differ from that?

See above. It is maybe better to think of it in terms of geometric constraints. Consider the pattern you get if only the left slit was open and the pattern you would get if only the right slit was open and consider the overlap of that. When both slits are open, the total distribution will be the same for all of them, but for any point on the detection screen you might end up with the probability of a photon having gone through the left slit being slightly higher than going through the right slit (or vice versa) simply due to the geometry of the problem.

For me it surprising that there is any difference at all (even if small) between the individual photon's wave functions, and that the "which-way" information is encoded in them to some degree. I always thought the wave-function was always the same for all of them. But that might be just a gap in my knowledge, and might be less profound for the experts in the field.

The probability distribution is the same for every photon arriving at the double slit.

That just does not exclude imbalances in the probabilities of a photon having gone through the left slit or right slit, respectively, when looking at the subset of photons that arrive at a single point on the screen.

 

[Rene Dekker]

In quantum optics you instead typically talk about probability amplitudes ...

I think I am familiar with the way probability amplitudes work, and your description confirms that.

If the paths are indistinguishable, you instead do the sum first and then calculate the absolute squared value.

That's the point I am trying to make: on the signal side, the paths ARE indistinguishable, therefore I would expect the probability distribution to be the same for all photons, or at least not have any "which-path" bias in them.

See above. It is maybe better to think of it in terms of geometric constraints. Consider the pattern you get if only the left slit was open and the pattern you would get if only the right slit was open and consider the overlap of that. When both slits are open, the total distribution will be the same for all of them, but for any point on the detection screen you might end up with the probability of a photon having gone through the left slit being slightly higher than going through the right slit (or vice versa) simply due to the geometry of the problem.

I cannot get my head around this. On the signal side, both slits are open, so the distribution should be same for them, as you say. The signal photons arrive at D0 before their partners arrive at one of the detectors on the idler side. Unless you subscribe to super-determinism or retro-causality, when the photon arrive at D0, it is not known yet whether its partner will arrive at D3/D4 or D1/D2.
Yet when we look at the R03 subset of the photons afterwards, then we see a shifted pattern when compared to the R04 subset (assuming that is indeed the outcome of the experiment). Those (subset) patterns are made before their entangled partner arrives D3 or D4.
I don't see any other explanation for this, except that the "which-path" information is already in the wave-function from the start. That is, half of the photons have probability distributions that (when combined) form a Gaussian shape shifted to the D3 side. The entangled partners of those photons will never arrive at D4 (but can still arrive at D1/D2). The other half of the photons form a Gaussian shape shifted to the other side, and their entangled partners never arrive at D3. In a sense, the photons really "go through" one of the slits, and that information is reflected in their wave-functions.
Of course, the entangled partners of those photons could still arrive at D1/D2, so their individual probability patterns cannot be a pure Gaussian, but should be an interference pattern with its peak to one or the other side.

I thank you for your answers, but I was not able to derive from those answers how these (subset) patterns on D0 can be made before the information of whether the entangled partner arrives at D3 or D4 is available, unless there is some encoding in the wave-function for them.
Does that make sense? Did I miss something in your answers?

The probability distribution is the same for every photon arriving at the double slit.

I understand that. I am concerned with the probability distribution of the photons arriving at D0.

 

[Cthugha]

I think I am familiar with the way probability amplitudes work, and your description confirms that.

Great. Thanks for letting me know.

That's the point I am trying to make: on the signal side, the paths ARE indistinguishable, therefore I would expect the probability distribution to be the same for all photons, or at least not have any "which-path" bias in them.

The issue remains the same: It is not meaningful to attribute the which-path bias to the photons. You need to attribute it to the detection positions. While the set of all photons has a 50/50 probability to go through either slit, the subset of photons that end up on the very left of the detection screen will be more likely to have passed through the left slit, while the subset of photons that end up on the very right will be more likely to have passed through the right slit. Imagine a double slit with very large separation between the slits to make this more transparent.
The paths to a single point on the screen are only indistinguishable if the mean intensity getting to the point from each individual slit are exactly the same. This is ONLY possible, if both slits are at exactly the same position. Obviously this cannot be the case as you would then only have one slit.

Really, this is simple: Take any single point on the detection screen and calculate the distance from the left slit to this point ($r_1$) and the distance from the right slit to this point ($r_2$). What are the distances you get?
Now just imagine that both slits get the same intensity $I_0$. The mean phase-averaged (meaning that the relative phase between light fields at the two slits is randomized) intensity arriving at that point via the left slit will be $\frac{I_0}{r_1^2}$. The mean phase-averaged intensity arriving via the right slit will be $\frac{I_0}{r_2^2}$. This is completely unrrelated to the photons, but ONLY a consequence of the distances being different. These different intensities give you a slight asymmetry in which-way probabilities. Just calculate it. This can be done in 2 minutes. It is very simple.

I cannot get my head around this. On the signal side, both slits are open, so the distribution should be same for them, as you say. The signal photons arrive at D0 before their partners arrive at one of the detectors on the idler side. Unless you subscribe to super-determinism or retro-causality, when the photon arrive at D0, it is not known yet whether its partner will arrive at D3/D4 or D1/D2.

No, the distribution relative to the middle of each slit should be the same. As the slits are at different positions, the distributions will be shifted by the difference in the position of the slits. When the photon arrives at D0, you can predict a probability whether it will rather arrive at D3 or D4 (the same with D1/D2). You do not need retrocausality for that.

I tend to explain the DCQE experiment to students as a non-local pinhole. In the standard double slit experiment done by Young, it was important to increase the spatial coherence of light field used because spatially incoherent light will not create an interference pattern. This is done using a pinhole. This ensures that the light source is point-like, so the transverse momentum of the light (which corresponds to its angle relative to a slit) becomes well defined. For each angle relative to the slit, you will get a different interference pattern (because the relative phase of the light field at the two slit positions will change), so putting a pinhole effectively fixes this relative phase. If the phase is not fixed, you just get a broad distribution which is the sum over many different interference patterns.

SPDC light is inherently incoherent as the momenta (emission angles) vary strongly and the relative phase between the light fields at the two slits are undefined. Still they are correlated, which means that if you know the emission angle of the photon arriving at D0, you also know the emission angle of the other photon. By placing the lens before D0, you turn the detection at D0 into a momentum measurement. Each point on the screen will only get detection events correspond to exactly one photon emission angle. This allows you to know the emission angle of the photon on the other side as well. This is what I would call an effective non-local pinhole. Picking the subset of photons arriving at one certain position at D0 corresponds to choosing a subset of photons with a certain emission angle on the other side - which could also be achieved by using a pinhole.

Now the other side of the setup is similar to a Mach-Zehnder interferometer and you can easily simulate whether photons with a certain emission angle (which translates to a certain phase difference in the interferometer) are more likely to end up on D1 or D2. You can know this probability already when the detection at D0 has taken place and before the detection event at D1/D2 has taken place. The same goes for D3/D4, but this is more boring.

Yet when we look at the R03 subset of the photons afterwards, then we see a shifted pattern when compared to the R04 subset (assuming that is indeed the outcome of the experiment). Those (subset) patterns are made before their entangled partner arrives D3 or D4.

This is a difficult way of putting it as R03 and R04 are coincidence counts. You need both detection events at D0 and D3/D4 to actually measure the pattern. But you can predict whether an entangled partner is more likely to arrive at D3 or D4 as soon as you have recorded a detection event at D0.

I don't see any other explanation for this, except that the "which-path" information is already in the wave-function from the start. That is, half of the photons have probability distributions that (when combined) form a Gaussian shape shifted to the D3 side. The entangled partners of those photons will never arrive at D4 (but can still arrive at D1/D2). The other half of the photons form a Gaussian shape shifted to the other side, and their entangled partners never arrive at D3. In a sense, the photons really "go through" one of the slits, and that information is reflected in their wave-functions.
Of course, the entangled partners of those photons could still arrive at D1/D2, so their individual probability patterns cannot be a pure Gaussian, but should be an interference pattern with its peak to one or the other side.

What is in there "already from the start" is the sum of the momenta of both photons. However, there is absolutely no information in there about the momenta of the individual photons. The first detection at D0 is a momentum measurement and as the sum of momenta indeed is there already from the start, this allows you to know the momentum of the other photon - and the probability of the photon ending up at D1, D2, D3 or D4, which depends solely on this momentum value.

I thank you for your answers, but I was not able to derive from those answers how these (subset) patterns on D0 can be made before the information of whether the entangled partner arrives at D3 or D4 is available, unless there is some encoding in the wave-function for them.
Does that make sense? Did I miss something in your answers?

Well, there is entanglement of momenta. This IS the encoding.

 

[Rene Dekker]

SPDC light is inherently incoherent as the momenta (emission angles) vary strongly and the relative phase between the light fields at the two slits are undefined.

Yes, that is what I understood as well. That is the reason you don't see an interference pattern for all D0 photons together. Actually I understood that it is a mixture of only two relative phases that differ by 180 degrees, but I might have understood incorrectly.

Sorry for the late reply, by the way, I got caught in other stuff.

Still they are correlated, which means that if you know the emission angle of the photon arriving at D0, you also know the emission angle of the other photon. By placing the lens before D0, you turn the detection at D0 into a momentum measurement. Each point on the screen will only get detection events correspond to exactly one photon emission angle. This allows you to know the emission angle of the photon on the other side as well. This is what I would call an effective non-local pinhole. Picking the subset of photons arriving at one certain position at D0 corresponds to choosing a subset of photons with a certain emission angle on the other side - which could also be achieved by using a pinhole.

Ah yes, that makes sense. The information that the path choice depends on does not need to be present from the start but can be determined on hitting the D0 detector, and shared non-locally with the other side.

Now the other side of the setup is similar to a Mach-Zehnder interferometer and you can easily simulate whether photons with a certain emission angle (which translates to a certain phase difference in the interferometer) are more likely to end up on D1 or D2. You can know this probability already when the detection at D0 has taken place and before the detection event at D1/D2 has taken place. The same goes for D3/D4, but this is more boring.

Yes, the D1/D2 part I understood. That basically just depends on the relative phase between the paths. (Plus the non-interfering photons that happen to make it to D1 and D2 as well, but that is not relevant to the current discussion)

This is a difficult way of putting it as R03 and R04 are coincidence counts. You need both detection events at D0 and D3/D4 to actually measure the pattern. But you can predict whether an entangled partner is more likely to arrive at D3 or D4 as soon as you have recorded a detection event at D0.What is in there "already from the start" is the sum of the momenta of both photons. However, there is absolutely no information in there about the momenta of the individual photons. The first detection at D0 is a momentum measurement and as the sum of momenta indeed is there already from the start, this allows you to know the momentum of the other photon - and the probability of the photon ending up at D1, D2, D3 or D4, which depends solely on this momentum value.Well, there is entanglement of momenta. This IS the encoding.

Aha, that makes sense as well. So, the non-local "which path data" that is shared is really the momentum, determined when the signal photon hits D0. So that would imply that the detection of the signal photon would collapse the wave-function into a conditional probability amplitude, based on the momentum? For example, if the signal photon lands on D0 at one extreme side, then the associate momentum would give the idler photon a higher chance to land at D3?

One additional question, if it pleases you:
Suppose we don't send the photons directly to D0 on the signal side, but project them onto a similar setup as the D1/D2 on the idler side. That is, the light is sent through a beamsplitter, and only the output of one side of the splitter is projected onto D0 (if that is technically possible to achieve).

I assume that would mean the overall pattern on D0 of all photons arriving there would now show an interference pattern?
Would the D0-D3 coincidence count then show an interference pattern with its peak shifted slightly from the middle? And the same for the D0-D4 coincidence count, but shifted to the other side? Hence it would demonstrate both which-path and interference behaviour at the same time?

 

[Cthugha]

Yes, that is what I understood as well. That is the reason you don't see an interference pattern for all D0 photons together. Actually I understood that it is a mixture of only two relative phases that differ by 180 degrees, but I might have understood incorrectly.

Sorry, I have been a bit busy myself.
Actually, it is a mixture of all possible relative phases, but the geometry then "sorts" them such that the detection probabilities will resemble two out-of-phase patterns.

Aha, that makes sense as well. So, the non-local "which path data" that is shared is really the momentum, determined when the signal photon hits D0. So that would imply that the detection of the signal photon would collapse the wave-function into a conditional probability amplitude, based on the momentum? For example, if the signal photon lands on D0 at one extreme side, then the associate momentum would give the idler photon a higher chance to land at D3?

Calling it collapse results in a preferred interpretation, which I usually try to avoid, but yes: One may interpret the results along these lines.

One additional question, if it pleases you:
Suppose we don't send the photons directly to D0 on the signal side, but project them onto a similar setup as the D1/D2 on the idler side. That is, the light is sent through a beamsplitter, and only the output of one side of the splitter is projected onto D0 (if that is technically possible to achieve).

I assume that would mean the overall pattern on D0 of all photons arriving there would now show an interference pattern?
Would the D0-D3 coincidence count then show an interference pattern with its peak shifted slightly from the middle? And the same for the D0-D4 coincidence count, but shifted to the other side? Hence it would demonstrate both which-path and interference behaviour at the same time?

Sorry, I am not exactly sure I get the geometry you mean. A good discussion of the patterns in different geometries (for example using two double slits) is given here:
https://arxiv.org/abs/1010.1236
Maybe this more formal treatment helps.

With respect to the other question: It is possible to demonstrate which-path and interference behavior at the same time, but then you will have only partial which-way information and no full visibility of the interference pattern. A more thorough treatment can be found here (but for a different geometry):
https://arxiv.org/abs/quant-ph/0112065