# Entanglement of photons

1. Sep 1, 2009

### alexepascual

I have read an article were it is claimed that if when two photons are entangled by parametric down-conversion, you don't observe interference in the first photon (after going through a double slit) if you don't collapse the second in such a way that which-way information is destroyed.
The reasoning is that if the second photon is still alive, in principle you can recover this which-way information.
This line of reasoning does not seem very clear to me.
What do you think?

2. Sep 1, 2009

### jambaugh

Could you cite the article?

You should note that you do not observe interference in a single photon. Only when you do many repeated experiments will an interference pattern emerge.

This question discussed in some detail in the thread:
https://www.physicsforums.com/showthread.php?t=108015"

Last edited by a moderator: Apr 24, 2017
3. Sep 1, 2009

### jambaugh

Let me add that the article is correct in its claim but I'm not sure about the reasoning as you state it. The short answer is that half of an entangled pair is not by itself in a sharp mode and so will behave "noisily" i.e. with non-zero entropy. Nothing you do to the second photon of the pair will change affect the outcome of the experiment on the first photon.

However you can select a subset of all the pairs based on measurements made on the second photon and if it is the correct type of measurement then the subset you get will demonstrate an interference pattern. Note that this is not a "spooky action at a distance" but rather a screening out of a portion of the original set of photons.

You could also for example "cause" the set of first photons in the pairs to form an image of a smiley face after passing through a double split by only selecting those pairs where the second photons land in a parallel smiley face region after going through an isomorphic double slit.

It is just a matter of masking out cases not effecting the path of the photons directly.
EDIT: (Notice the imporance then of my earlier point that you don't see interference of single photons but only for a large aggregate even though this aggregate is formed one photon at a time over many experiments.)

4. Sep 1, 2009

### DrChinese

I think he is referring to something like this. Zeilinger, page 290, figure 2, there is no direct interference pattern for entangled photons:

Experiment and the foundations of quantum physics (1999)

If there were, you know either beat the HUP or send signals FTL. So there is in fact a method which can detect an entangled particle. If only there was some way to use that information...

5. Sep 2, 2009

### Cthugha

This is a rather...media-compatible line of reasoning, maybe pretty good to get funding. ;)

A more direct line of reasoning is found in the so called Dopfer thesis (one of Zeilingers PhD students). Unfortunately as far as I know it is only available in German.

However, the argument is, that the needed prerequisites to see single photon interference and entanglement effects are complementary. The prerequisite to see single photon interference in a double slit is simple: the spatial coherence of light must be large enough so that the coherence length is larger than the slit distance. The spatial coherence is mostly determined by the spread of wavevectors present in the emitted light. A small spread corresponding to a small angular size of the light source will result in high spatial coherence.

To demonstrate entanglement effects you can use two-photon interference (like in DCQE experiments) or similar stuff. In two-photon interference experiments the situation is different: Entanglement relies on the sum of the wavevectors of the two emitted photons being well defined, while the wavevectors in a single arms have a large spread. To achieve good interference visibility in two-photon interference experiments you will need this large spread. If you use a filter in k-space and filter out some of the possible wavevectors and therefore also some entangled photon pairs, the two-photon interference pattern will become narrower and fringe visibility will go down.

Now the requirement for two-photon-interference, which is one measure of entanglement, is to have a large spread in wavevectors, while the requirement for single-photon interference is to have a small spread in wavevectors. Apparently this is a contradiction. This can be tested pretty easily. All you need to do is to vary the distance between your pdc crystal and the double slit. If you place it far away, you increase spatial coherence because the range of k-vectors, which reaches the slits is rather narrow and you will see a usual double slit interference pattern. However the range of k-vectors is now so small that entanglement does not mean much anymore and you will see no two-photon interference. If you decrease the distance the reversed situation occurs.

6. Sep 2, 2009

### alexepascual

Thanks for pointing me to that thread.
I'll think about what you said about not being able to detect interference with a single photon. It appears though that we should be able to detect the lack of interference using a single photon.

Last edited by a moderator: Apr 24, 2017
7. Sep 2, 2009

### alexepascual

Yes. That's the article I read.
Zeilinger has his interpretation of Dopfer's experiment. But I don't know if his view is generally accepted or not.

8. Sep 2, 2009

### alexepascual

Yes, I had read Zeiliger's paper. It appears that Dopfer interpreted her own experiment in a slighty different way.
I'll give some thought to what you say in your post and look into the DCQE experiments. Thanks for your help.

9. Sep 2, 2009

### alexepascual

Reading your post again, it looks that you imply that Zeilinger's view is the one that is media-compatible and Dopfer's view is more direct. As you say, Dopfer's thesis is in German, so I have only read articles about her view but nothing actually written by her.

Now, here is what Zeiliger says (pg 290) and I am wondering there is something objectable in it:

"Will we now observe an interference pattern for particle 1 behind its double slit? The answer has again to be negative because by simply placing detectors in the beams b and b' of particle 2 we can determine which path particle 1 took.
Formally speaking, the states |a>1 and |a'>1 again cannot be coherently superposed because they are entangled with the two orthogonal states |b>2 and |b'>2 .
Obviously, the interference pattern can be obtained if one applies a so-called quantum eraser which completely erases the path information carried by particle 2.
That is, one has to measure particle 2 in such a way that it is not possible, even in principle, to know from the measurement which path it took, b or b'."

Note that in the obove I corrected two mistakes in the original article. My version here seems to make more sense and I think this is what Zeilinger intended to say.
I changed "|b'>2 and |b'>2 ." to: "|b>2 and |b'>2"
I also changed ( a' or b' ) to ( b or b' )
If I made a mistake in correcting Zeilinger please let me know. On the other hand I think that this idea may be objectable.

10. Sep 2, 2009

### alexepascual

I'll have to think about this. Thanks jambaugh.

11. Sep 2, 2009

### alexepascual

As Dr. Chinesse listed in his post, this is the article:
http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf" [Broken]

With respect to the link you included to another thread in this forum, I wish I had seen it before starting this thread. That thread was abandoned in 2006, but what we are talking about here is the same thing. It would be nice if both threads could be merged.
Or perhaps we could abandon this thread revive that one?
What do you think?

Last edited by a moderator: May 4, 2017
12. Sep 2, 2009

### DrChinese

I say proceed from here...

By the way, there are experiments that have been done which demonstrate the effect. They show that there is NO interference for entangled photons in the "normal" case.

http://grad.physics.sunysb.edu/~amarch/Walborn.pdf [Broken]

I think it is figure 3 for above.

Last edited by a moderator: May 4, 2017
13. Sep 2, 2009

### alexepascual

What do you mean by "proceed from here?"
I'll take a second look at the Walborn experiment but I don't think it adds anything. As far as I remember, it adds another level of complexity by giving you fringes and anti-fringes.

14. Sep 2, 2009

### DrChinese

I meant: ignore the old thread and discuss in this one.

I thought one of the figures had the effect of showing fringes and anti-fringes added together so that you are essentially seeing without coincidence counting (you are effectively looking at 100% in other words). I.e. you are not looking at a subsample. But on further review, I am not sure. Makes my head hurt looking at them all...

15. Sep 2, 2009

### alexepascual

Maybe you are now under hypnotic trance and I can brain-wash you!
It takes time to think about these things. I'll see if I have a chance tonight. I think I had it printed out.
In Dopfer's experiment, the coincidence counter is needed because a lot of photons going towards the double slit in the lower path get absorved by the plate that contains the slits so that many of the photons found in the upper path don't correspond to any of the photons detected in the lower path. But I have seen another proposed design of this experiment, where the upper path has a double-slit that corresponds exactly to the path that only the entangled photons take. So it screens-out the unentangled photons. In that case you don't need the coincidence counter.
Some people claim that the coinidence counter is the reason you see interference, but I can't see a reason for this. At least in principle. It may be that something in the setup was not considered carefully and the coincidence counter has this side-effect though.
But before getting into an analyisis of the influence of the coincidence counter, which I think could be ommited anyway, I would like to analyze the case where a photon at a time is sent (no counter) and the upper photon is not measured. (Let's send it on it's way to Andromeda).

16. Sep 2, 2009

### Cthugha

Well I think both statements are correct (and I am in no position to tell Zeilinger how he should present his results ;) ).

The coincidence counter is indeed the reason, why you see an interference pattern. Without it, there will be no pattern. In the Dopfer experiment for example you will never see an interference pattern in D1 alone because there is no double slit in that path. Also you will never see an interference pattern in D2 because it is a bucket detector and not position sensitive, so there is no possibility to get an interference pattern here. Of course you could exchange this bucket detector for a small detector like D1. Now, however, you are in a regime similar to the Walborn experiment, where you can have fringes and anti-fringes showing up in coincidence counting. In fact, you will then notice that the situation is symmetric: You will se an interference pattern if you move D1 and keep D2 fixed and you will see another pattern if you move D2 and keep D1 fixed.

The reason for that is not too difficult. To see any kind of interference pattern you need a fixed phase relationship. In the original Young double slit experiment this is achieved by putting a single slit in front of the double slit. The single slit acts as a point source. There is a fixed phase relationship between the fields at the two slits and accordingly also a fixed phase relationship between the two fields originating from the slits at each point of the detection screen.

Entangled photons do not have this property. The phase in a single arm changes extremely fast due to the PDC process. If such light passes a double slit, there will be no fixed phase relationship and accordingly no interference. But the two-photon state has a very well defined phase. This means that although the phase of a single field changes pretty randomly, the phase difference between the two fields resulting from the PDC process is very well defined. So if you put D1 in a position, where the detection of a photon gives you information about the momentum of that photon two things happen:

1) By doing coincidence counting out of all the photons arriving at D2 only those having the same small momentum range as measured at D1 are filtered out. So the photons at D2 are also in a momentum eigenstate and which way information can not be present. This assures that that the photon at D2 could have gone through both slits of the double slit, which is the first prerequisite for interference to occur.

2) Due to the fixed two-photon state phase relationship this also means that there is now a fixed phase relationship between the fields at both slits of the double slit and accordingly also of the two fields originating from the two slits and arriving at D2. Interference can occur.

If you did not use coincidence counting, you will measure the superposition of all interference patterns of all possible momentum eigenstates instead of just the pattern of a single eigenstate. As a consequence the superposition of all of these patterns cancels out leaving you with no interference pattern at all. Coincidence counting is not just a means of reducing erroneous counts occuring due to stray light or other error sources, but indeed elementary.

17. Sep 3, 2009

### alexepascual

Do you have a copy (electronic) of the paper?. I read that Zeilinger had it in his university website but it looks like it has been removed now. If you have it I would appreciate it if you attached it to this thread or sent it to me by email.

I didn't know this.

I guess you mean "distance between slits" right?
I haven't gone back to take a look at my optics book, but it looks to me that this is an approximation. With the screen at a reasonable distance from the slits, the coherence length should be able to be smaller than the distance between slits and be limited by the difference in path. (in order to get interference, of course)

I would guess you are talking about longitudinal coherence right?

I assume here you are referring exclusively about the SPDC process. The angle of emission for each photon is related to the magnitude of it's momentum. Therefore, to small difference in angle corresponds small spread in momentum. Did I interpret you correctly?

Looking at Scully's paper, I can see (as you say) that this is different from Dopfer's experiment because you are mixing photons comming from different atoms.
I think this is a more complicated setup. I have printed out the paper and I will continue to think about it, but I would rather concentrate first on a Dopfer's kind of setup. I am not sure that an analyisis of Scully's setup will help in understanding Dopfer's experiment.

18. Sep 3, 2009

### alexepascual

According to Dopfer, you do see an interference pattern in D1 when you place the detector at the focal distance of the lens. You loose which-way information. On the other hand, if you place the detector at a distance of 2f, it is as if you were looking directly at the slits on the other arm. Then you get which-way information and interference is lost.
The reason you do see interference on D1 is that you are looking at the photons that are entangled with those that went through the slit on the other side. If you wanted, you could screen out all the photons in the upper arm that are entangled with photons in the lower arm that did not make it through the slits.

I'll try to skip Walborn's experiment for the moment. But I'll keep it in mind and come back to it if I feel it is necessary.

OK, I understand this.

The frequency is proportional to the photon's energy. Now, if we consider both photons (entangled) as a single thing; should the frequency be that obtained by adding the energy of both photons?. I don't think so, unless the assumption that each photon's wavelength is twice that of the original photon is wrong. Now, if the frequency of each photon is what we would expect from its energy, why would the phase in a single arm change very fast?

I am not ignoring the rest of your post. But I would rather clarify the above issues before I go on to analyze the rest.

19. Sep 4, 2009

### Cthugha

Well, no. I always had a look at it at the Homepages of the universities of Vienna or Innsbruck. Maybe I have a copy somewhere on my office computer. I will check. Unfortunately I am attending a conference next week, so it might take some time.

Yes, distance between slits. Also maybe my usage of coherence length was a bit unclear I suppose as this is sometimes also defined has coherence time times speed of light. So to be more exact, both slits must lie within the coherence volume of the light. Btw, this does not depend on the distance between slits and screen, but between the light source and the slits.

No, longitudinal/temporal coherence are defined by the spectral width of the light. Coherence time is the decay time of the autocorrelation. The autocorrelation is the Fourier transform of the spectral power density.

Spatial coherence (or transverse coherence) are defined by the spread of wavevectors. Collecting a small solid angle of a light source therefore leads to high spatial coherence. This is foe example, why starlight has rather high spatial coherence. This property was used to measure star diameters back in the 50s.

Yes, but this is not limited to SPDC processes.

Well, you do not see a pattern in D1, you see one in the coincidence counts.
Just screening out those photons, which do not make it through the slits, is not enough. The important, but often overlooked fact is, that D1 is a small detector. It is smaller than the beam is, so there are also photons, which do not hit D1. If you replace it by a larger detector, also the interference pattern in the coincidences will vanish. In the focal plane each possible position of the detector corresponds to a small spread of momentum eigenstates. This small spread shows interference in coincidence counts. The whole spread does not (due to low spatial coherence) and therefore there is also no interference in a single arm.

I am not sure, why you bring up frequency here. However, indeed the wavelength is not necessarily exactly doubled. You get a rather broad peak of resulting wavelengths. The peak is of course indeed ad the doubled wavelength. The SPDC process can be considered as a process, which is stimulated by random vacuum fluctuations and is comparable to spontaneous emission in these terms. Therefore the phase changes just as fast as in any spontaneous emission process. In processes, where you have a well defined phase (lasing for example) most of the emission processes are stimulated emission processes. You have one well defined field and any stimulated emission process is caused by this field and "happens in phase with it". In spontaneous emission processes you have some field, but all subsequent emission processes are independent of each other because the "stimulation" happens by vacuum field fluctuations here. These happen at a random time with a random phase. As a result, the emitted light is pretty incoherent. Try for example to feed from a light bulb to a double slit and see interference. You will not see anything unless you use another single slit in front of the double slit to increase spatial coherence.

20. Sep 4, 2009

### alexepascual

Thanks, I'll appreciate it

Many of the things I said were a result of my misinterpreting the meaning of "spatial coherence" now I know it relates to the transverse component.

Let's consider the following modification to the experiment:
(1) You screen out all the photons in the upper arm that don't correspond to those that made it through the slits in the lower arm. (this is just a geometrical thing).
(2) Instead of a small detector for D1, or a small slit that moves to different positions and scans the interference area, just use a very good resolution CCD of the right size.
(3) Consider the efficieny of the detectors 100% (at least for the thought experiment)

With these modifications, the number of photons at each detector is the same. To each photon that is detected in D1 there is a corresponding entangled photon detected in D2.
If you introduce the coincience counter now, it'll allow you to know exactly which photon is entangled with which on the other side. I don't see how you can use this information to filter out some data and obtain an interference pattern after the fact. Maybe I am missing something.

My mention of energy was again due to my misinterpretation of what you had said. Still, if you consider a single photon emission, I don't see how you would get a phase difference between one slit and the other. Would the different waves that correspond to one photon be comming from different points on the crystal far appart enough to affect spatial coherence?

21. Sep 5, 2009

### Cthugha

In fact, you will not be able to filter out an interference pattern. You will be able to filter out many. If you have a look at the coincidence counts between a small portion of the CCD and the other detector, you will see an interference pattern. If you look at the coincidence counts between another portion of the CCD and the other detector, you will see another interference pattern, but the position of the fringes and antifringes will be shifted with respect to the first interference pattern. All of these patterns sum up to no pattern at all, just a broad peak. Each position on the CCD will correspond to a different momentum eigenstate. This slightly different momentum also leads to this slight shift of the patterns on the other side.

Well, if you are talking about single photons from SPDC processes, you do not know much about your single photon going to the double slit. It can have a very wide range of wavevectors (and can in principle come from a lot of different positions of the SPDC crystal). All of the probability amplitudes for these processes already interfere at the slits. As the emission processes are not coherent, this leads to a rapidly changing phase of the light field in this arm of the experiment.
However, having just a single well defined photon emission process would even make the situation worse. If you really had a deterministic single photon emission process at a given time (something like a ballistic photon), you would never see any interference in this arm because you would have which-way information (the photon has to hit the slit at a certain point) and the light field would be as incoherent as it gets. Coherence is a measure of how well you can predict the field at some time t, if you know its value now. In such a ballistic emission process you would just have a delta peak now and no information about the field at later times.

22. Sep 5, 2009

### alexepascual

I didn't know you could get anti-fringes in a Dopfer-type experiment. This is type-I SPDC, isn't it?. The Wallborn experiment on the other hand uses type-II. Am I right?
If I expect to see (or not see) an interference pattern in an extended CCD D1, I don't see a reason to look at a small portion of the CCD. I should take a note of all the hits on D1 and the (x,y) coordinates on the surface of the detector, and keep track of the coincidence information. Acording to your suggestion, all the hits on D1 would not from an interference pattern. Then somehow you use the coincidence count and suddenly the interference appears. In order to do this, you should select only a portion o the hits on D2, but what criterium would you use?.

If you make the crystal small enough and set it at a reasonable distance from the slits, This shoud not be a problem. Otherwise the experiment is flawed from the start. Each photon will have a random phase with respect to other photons, but they are being produced one at a time, and each photon interferes with itself at the slits. The phase at each of the slits for a simgle photon should be considered roughly the same.

When I talk about a single photon, I mean a single quantum. I am not saying that I see the photon as a little ball. The picture I have of the photon when in transit is still that of a superposition of waves of slightly different wavelengths around a central wavelength (a wave packet).

23. Sep 9, 2009

### alexepascual

Would you say you agree with Zeilinger's analysis fo Dopfer's experiment?

24. Sep 9, 2009

### DrChinese

Absolutely.

25. Sep 10, 2009

### alexepascual

DrChinese:
According to most descriptions of entanglement, once the particles are entangled, you can't talk about the state of a single particle but about the state of the composite system. That is, you consider a Hilbert space which is the vector product of the Hilbert spaces of the separate particles.
Now, let's assume I am trying to make a computer simulation of the photon in Dopfer's lower arm going throught the slits. I want to be able to assign a value to the wave function at each point in space (..and time, which makes it an animation). I am not really planning to do this because I don't know of a good (and cheap) tool to do it, but it would definitely be something interesting to do. What factors should I consider in deriving the expressions for this wave function?. Even if there are factors that may render this problem hard to solve, I would like to make reasonable simplifications for a first approximation. Maybe we could also ignore the fact that the waves comming from each slit do not have a sharp momentum due to the finite size of the slit, which gives you a small but finite range of angles for the line-of-sight between the source and a single slit. So, I would try to ignore subtleties as much as possible and concentrate on the most fundamental problems. The main problem here I guess would be that the wave function is not that of a single photon but that of both photons. I should nevertheless be able to assign a single complex number to the wavefunction at a single point for the lower photon because I should be able to use it to calculate the probability of finding the photon at each point in space. To analyze the behavior after the slits I am OK with leaving out the portion of the wave that strikes the surface of the plate that contains the slits. I am not asking you to give me detailed expressions, that would be too much. But maybe you can give me some ideas, such as how to derive the phase for the wave function. I guess in this case there is no interaction Hamiltonian, so there should be some Hamiltonian that depends on the energy of the particles right?. I could I guess consider each combination of positions for both photons and multiply their complex coefficients (functions of time in the Schrodinger picture) from the separate Hilbert spaces, but I am not sure this is valid. Again, I would appreciate your opinions about this, as it would help me understand better the experiment and the entangled state itself.
I would also be grateful for any ideas from other posters.
Oh! one more thing.. I am assuming the photons on the upper arm have not been detected and maybe will never be detected.

Last edited: Sep 10, 2009
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