Entanglement vs. Coherence in SPDC

[Erik Ayer]

Hello all, I have read that these two properties are complimentary, similar, at least in a sense, to uncertainty between momentum and position. John Cramer used what he referred to as a Sagnac Source to vary the balance between entanglement and coherence. What, physically, has to be done to downconverted light to make it more coherent?

In a type II downconversion, the light comes out in two cones. I imagine that, since the pump has a finite cross-secional area and the nonlinear optic has some thickness, where downconversion happens can be anywhere within the intersection of these two things. Hence, an infinite number of overlapping sets of cones would be generated, making the output non-coherent, spatially. Feeding that to a double-slit would result in varying phase shifts between the waves of the photons, shifting the interference pattern around such that it all adds up to a big mess.

It would seem like, to get coherent light out, the laser beam would need to be infinitely concentrated into a geometric line and the optic would need to be infinitely thin. Obviously that does not constitute a Sagnac Source, so how can coherence be increased?

Thanks,
Erik

 

[Cthugha]

One has to be a bit precise here. Entanglement in one property and first-order coherence in the corresponding property are complementary. This has been shown e.g. here: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.62.043816. This means that entanglement in momentum and spatial coherence are complementary. If you come to think of it, this is somewhat trivial. Spatial coherence is a measure of how point-like a light source is. If the distribution of momenta is narrow, the degree of spatial coherence is high. For momentum entanglement, you want a broad distribution of momenta. If the two photons always leave the light source under exactly the same angle, there is no entanglement at all. You want a broad distribution, where the sum of both momenta adds up to the same value every time. As you obviously cannot have a broad and a narrow distribution simultaneously, these two properties are complementary.

The light source Cramer has in mind is this one: https://www.osapublishing.org/oe/abstract.cfm?&uri=oe-15-23-15377. If you rotate the half-wave plate inside the Sagnac interferometer, you may end up with either a well defined Bell state or a mixture of several. Although Cramer puts it this way, you do not really modulate coherence this way. To increase coherence, the easiest thing you can do, is to simply filter the light field. Place a narrow pinhole somewhere and spatial coherence will be enhanced and entanglement will be reduced. You can even just place the double slit further away from the light source. In effect, this is not different from filtering. Fewer photons will arrive at the slits and if you draw a line from the light source to the slits, the range of angles will become narrower. At some point, the subset of photons arriving at the slits will be coherent, but not show any signs of entanglement anymore. This was one of the earliest experiments done in Zeilinger's group.

 

[Erik Ayer]

Okay, this makes a some sense, thank you! The different momenta refer to photons moving at different angles, correct? So a photon could have been downconverted on the left side of the nonlinear crystal, on the right side, or in between so that it's partner would have a correlated momentum/angle - assuming that whatever position they cam from, the angle is such that they arrive at the same point. Does this make some sense, kind of?

Imaging a laser beam of some diameter being downconverted by a BBO cut for colinear downconversion, then sent into an infinitely long tube with the same diameter as the laser beam. Any light that was not going perfectly straight through would hit the walls of the tube. I would think that the pairs of photons that did go straight through would be in superposition of being all across the beam, so if they were then to go through a double-slit, they would go through both slits and form an interference pattern that would be visible/detectable, yet if the actual position within the cross section of the beam were measured for one photon, it should determine an actual location for both photons - they would both be, say, on the right side of the beam. That's kind of what "entanglement" means, but the infinitely long tube should destroy the entanglement (and give great coherence).

Somewhere in here I'm being stupid :) Could you point out where?

 

[Cthugha]

Yes, different momenta mean different angles. Well, they will not necessarily arrive at the same point. If you image the real space somewhere, you will most usually image them to the same point. If you do Fourier space imaging, you will image them to a position depending on their momentum. For a double slit you usually use the latter. The optical element that performs the Fourier transform to go from real to momentum space is a simple lens.

For the same reason, I am not sure I get your scheme. The long tube effectively performs some momentum filtering. If you do a measurement in the Fourier plane, you have no information about the initial position anyway. Due to the good Fourier space filtering, the position information should be lost.

 

[Erik Ayer]

I've been pondering this for a couple days...

The Fourier transform, in this case, is just mapping the different angles that photons are moving to locations, after the lens, correct? It happens that angle and wavelength are related, or at least the angle relative to the center of the cones of light after downconversion.

 

[Cthugha]

Well, yes. The standard "job" a lens does is that a parallel beam arriving at a normal angle at the lens gets focused to a single point in the focal plane. Now, if you do the same for a parallel beam that does not arrive at 90° to the lens surface, but at a different angle, the beam also gets focused to a single spot in the focal plane, but it will be at a different lateral position. Essentially different wavevectors get mapped to different positions.

I am not sure what you mean by "related". In your last scenario you applied strong momentum filtering. Which scenario do you have in mind, exactly?

 

[Erik Ayer]

What I mean by "related" is this: The downconverted light is being emitted into cones in which the wavelength varies from the center to the outside - longer wavelengths at the inside of one cone and at the outside of the other. The wavelength is a function of the angle between the center line of the cone and the angle that light is coming from, so a given wavelength, other than exactly at the center, will form a ring. A lens should still focus light of a given wavelength on a ring, although ideally that ring would be thin.

What I'm trying to get a feel for is in what ways the downconverted light is in superposition and entangled. When a standard laser beam is used with a double-slit to produce interference, the photons are in a superposition with respect to where in the cross-section of the beam they are, such that the wave function for each photon goes through both slits to interfere with itself. When light is downconverted, there would also be a superposition as to which angle the new photons were emitted relative to the direction of the pump beam - hence the downconverted light is not coherent and doesn't for interference if it goes through a double slit. More precisely, there should be many interference patterns, overlapping.

Your comment about a lens doing a Fourier transform is interesting in that it provides a new way to look at Birgit Dopfer's experiment. Different wavelengths of light would be moving at different angles, which would shift the interference patterns they created from the double-slit. The lens on the idler side would do the Fourier transfrom - each wavelength would have its own position along the x-direction, so the detector and coincidence circuit would essentially be filtering out a specific wavelengths and hence, a specific interference pattern.

 

[Cthugha]

Your comment about a lens doing a Fourier transform is interesting in that it provides a new way to look at Birgit Dopfer's experiment. Different wavelengths of light would be moving at different angles, which would shift the interference patterns they created from the double-slit. The lens on the idler side would do the Fourier transfrom - each wavelength would have its own position along the x-direction, so the detector and coincidence circuit would essentially be filtering out a specific wavelengths and hence, a specific interference pattern.

The same goes for different momenta. Plane waves arriving at the double slit at different angles will create different interference patterns. Due to the phase difference at the slits, the position of the central peak (and all other peaks) will be shifted. The superposition of all of these patterns will not look ike an interference pattern at all, but one may single out a visible pattern by doing momentum filtering.

 

[Erik Ayer]

This makes sense. So the downconverted light could be filtered with a lens or even an opaque material with a hole at some distance, in order to get rid of some or most of the overlapping interference patterns. That would destroy the momentum entanglement between pairs of photons. However, it should not affect polarization entanglement, so could this filtering be applied to the quantum eraser? The coincidence detection is, effectively, filtering specific momenta to get a single interference pattern, so it seems obvious that this would be another way to get the filtering. Since it's obvious, I'm sure it has been done, but I haven't run across it yet.

 

[Cthugha]

Actually, this is how some of the quantum erasers work. If you have a look at the version by Walborn et al, the detector in the arm without the double slit uses an iris with a length of 600 microns placed 2 meters away from the BBO. This is exactly the kind of momentum filtering you just mentioned. If you move that detector around, the interference pattern seen in coincidence counts will change. Of course you still need to get the polarization settings right to see interference, but I guess you are aware of that.

 

[Erik Ayer]

If the iris is filtering out certain momenta in order to get interference (assuming the which-way marking has been erased), why is coincidence detection needed? I will read up on Walborn at lunch - and will likely sneak in some reading not at lunch - but I think this question gets at the fundamental thing I am missing.

 

[Cthugha]

You do the filtering in the arm, where the double slit is not located. All of the photons and momenta still arrive at the double slit. Let me put it this way. In the original double slit experiment using sunlight, one needs to put a pinhole in front of the double slit to filter the light and make it spatially coherent. In the quantum eraser, you simply put this pinhole in the path of the twin photon and perform coincidence counting to get the same result.

 

[Erik Ayer]

Ok, that makes sense. So there's still a mess on the far side of the double-slit, but momentum-filtering on the other arm combined with coincidence filters that mess into a single interference pattern, again assuming that which-way marking has been erased with the linear polarizer.

What happens if the aperture is places right before the double-slit in its arm? That should filter the momenta and such that there is a single interference pattern that can be observed without any coincidence filtering. Which-way marking could then be added with the quarter wave plates, then erased in the other arm as usual with a linear polarizer.

I have this overwhelming feeling that I'm missing something really stupid :) Is it that, if the aperture is far enough away to sufficiently filter momenta, that there somehow isn't enough light to create interference, or something like that?

 

[Cthugha]

What happens if the aperture is places right before the double-slit in its arm? That should filter the momenta and such that there is a single interference pattern that can be observed without any coincidence filtering. Which-way marking could then be added with the quarter wave plates, then erased in the other arm as usual with a linear polarizer.

Well, in order to get the interference pattern without coincidence detection, you would need to filter the polarization as well. Of course you can do this as well via erasing it in the other arm, but in order to get access to this polarization subset, you would need coincidence counting again.

 

[Erik Ayer]

I think this is getting closer to the fundamental thing I don't understand :)

Why does the polarization need to be filtered in order to get an interference pattern? I imagine that light is getting downconverted at the non-linear crystal, spreading out, and then hitting an opaque material that has the aperture in it. Since there is distance between the crystal and the aperture, only a limited range of directions for the light to come from would make it through the aperture and thus, could then go through a double slit to make interference. How does polarization come into the picture and why does it need to be filtered?

Looking through Walborn, there is both an interference pattern and an anti-interference pattern, selected by taking light in coincidence with different detectors on the "other" arm of the experiment. It seem like this is related to the above question on how polarization needs to be filtered. I don't understand how that anti-interference can happen. My thinking is this: with a light source that has no entangled photons - a simple laser - the wave for each photon gets to both slits, diffracts from each slit, and those two diffracted sub-waves interfere. Where the bright spots land on the screen far away from the slits depends on the relative phase of the waves between the slits. To get anti-interference, there would need to be an additional phase shift of pi between the waves at the slits. A standard laser would be, I think, made of light with random polarizations, so is there something about downconverted light where polarization matters more or differently?

 

[Cthugha]

Since there is distance between the crystal and the aperture, only a limited range of directions for the light to come from would make it through the aperture and thus, could then go through a double slit to make interference. How does polarization come into the picture and why does it need to be filtered?

Well, first of all, light beams with orthogonal polarization do not interfere, but I guess you know that already. However, there is more to it.

Looking through Walborn, there is both an interference pattern and an anti-interference pattern, selected by taking light in coincidence with different detectors on the "other" arm of the experiment. It seem like this is related to the above question on how polarization needs to be filtered. I don't understand how that anti-interference can happen. My thinking is this: with a light source that has no entangled photons - a simple laser - the wave for each photon gets to both slits, diffracts from each slit, and those two diffracted sub-waves interfere. Where the bright spots land on the screen far away from the slits depends on the relative phase of the waves between the slits. To get anti-interference, there would need to be an additional phase shift of pi between the waves at the slits. A standard laser would be, I think, made of light with random polarizations, so is there something about downconverted light where polarization matters more or differently?

A standard laser has a single well defined polarization. Downconverted light (if it is polarization entangled) does not offer that. Only the summed polarization of both beams is known, but they may be distributed among the two beams randomly. Depending on the kind of down conversion used, one may, e.g. have the situation that one beam is horizontal and the other is vertical, but you do not know which. You may also have the situation that any possible combination of orthogonal polarizations is possible. This depends strongly on the setup. If you now use quarter wave plates to mark the path information, you need to consider what wave plates actually do. They are birefringent, so that the speed of light is different for different polarizations. For a quarter wave plate, the delay between the fast and the slow axis amounts to exactly pi/2. What the quarter wave plate does to the light now depends on the relative orientation of the polarization of the light field and the wave plate. The easiest case occurs when they are aligned or orthogonal to each other: Either the whole beam gets delayed or none of it. The light field parallel to the fast axis is not delayed, while the light field parallel to the slow axis gets delayed by pi/2. The quarter wave plates at the two slits are oriented perpendicular to each other. If the polarizations are aligned with the wave plates, this means that, e.g. horizontally light may be delayed by pi/2 at the left slit and by 0 at the right slit, while vertically polarized light will be delayed by zero at the left slit and by pi/2 at the right slit. The phase difference between the slits is +pi/2 in one case and -pi/2 in the other, which gives you in total exactly the pi phase difference between the fringe and anti-fringe pattern, you were looking for.

 

[Erik Ayer]

Actually, no, I didn't know that orthogonal light beams didn't interfere. I guess I'm thinking only of a single quantum wave, such as that split between each of the two slits. Actualy upon further thought, what are you referring to? Are you thinking of two, separate light beams that are polarized orthoganally?

As for the circular polarization with quarter wave plates, I found this: https://physics.stackexchange.com/questions/247005/double-slit-with-opposite-circular-polarizers. It essentially says the same thing as you did - that the vertical components end up shifted by pi relative to each other (the horizontal components stay at a constant phase delta). Finally, it explains these anti-fringes!

I also did not know that a laser was polarized and had, in fact, heard otherwise at some point in the distant past. Since I have lasers and polarizers, I should probably try it and verify it. Without using any quarter wave plates to circularize the polarization, I don't imagine there would be a pi phase shift between the two slits, but I'm curious as to how different phases of light would then respond to a double-slit.

Thank you, you have made a dent in my thick head!

 

[Cthugha]

Actually, no, I didn't know that orthogonal light beams didn't interfere. I guess I'm thinking only of a single quantum wave, such as that split between each of the two slits. Actualy upon further thought, what are you referring to? Are you thinking of two, separate light beams that are polarized orthoganally?

That is the easy case. Two different light beams at orthogonal polarizations do not interfere. However, if you split a single beam into two beams and rotate the polarization of one of the beams such that these two are orthogonal, they will also not interfere. In a nutshell, the heart of interference is that the modulus squared of the amplitude gives you the intensity at some point, but you need to add all amplitudes before squaring, so interference may take place. However, the two polarizations are orthogonal to each other and you need to sum up each component individually before squaring each component individually. This is like a 2d space.

As for the circular polarization with quarter wave plates, I found this: https://physics.stackexchange.com/questions/247005/double-slit-with-opposite-circular-polarizers. It essentially says the same thing as you did - that the vertical components end up shifted by pi relative to each other (the horizontal components stay at a constant phase delta). Finally, it explains these anti-fringes!

Unfortunately, this is usually not mentioned in the papers.

I also did not know that a laser was polarized and had, in fact, heard otherwise at some point in the distant past.

Let me put it this way: Most lasers are. Laser work on stimulated emission, which means that the emission into some mode becomes more probable if it is already highly occupied. As orthogonal polarizations are in fact different modes, usually one of the possible modes "wins". Many lasers are also designed such, that they are certainly polarized. However, this is not necessarily so. If two modes of similar properties, but different polarization are present, they both may lase, which may end up in a thing called gain competition. Usually, this enhances noise as the stronger mode suppresses the weaker one and is not a desired effect. However, some cheaper laser designs may not show good polarization features. This is a design question. However, good lasers are usually polarized.

Since I have lasers and polarizers, I should probably try it and verify it. Without using any quarter wave plates to circularize the polarization, I don't imagine there would be a pi phase shift between the two slits, but I'm curious as to how different phases of light would then respond to a double-slit.

Thank you, you have made a dent in my thick head!

Have fun experimenting! The easiest way to introduce a phase difference even without polarizers is just by shining the light at the double slit at some angle. Ideally, you should be able to see how the interference pattern moves around as you change the angle. Personally, I find that pretty impressive for a DIY experiment one can do at home.

 

[Erik Ayer]

I would think that in the case where a beam was split and one sub-beam rotated by 90 degrees, this would be a form of the quantum eraser (not using entanglement) and the photons would be marked as to which way they went. However, just a free rotation as opposed to making horizontal light go one way and vertical light go the other might be a different case.

I did make a double slit with a piece of 40-gauge wire and a couple razor blades. It doesn't seem to matter how the laser is oriented, but I could try polarizing it first and try that (and see whether it doesn't like certain polarizations). I'm wondering whether laser diodes are less polarized since their emission cavity is small, relative to a gas tube.

 

[Cthugha]

I would think that in the case where a beam was split and one sub-beam rotated by 90 degrees, this would be a form of the quantum eraser (not using entanglement) and the photons would be marked as to which way they went. However, just a free rotation as opposed to making horizontal light go one way and vertical light go the other might be a different case.

Yes, indeed. You can add a polarizer at +45 or -45 degrees to both polarizers afterwards and restore the interference patterns.

I did make a double slit with a piece of 40-gauge wire and a couple razor blades. It doesn't seem to matter how the laser is oriented, but I could try polarizing it first and try that (and see whether it doesn't like certain polarizations). I'm wondering whether laser diodes are less polarized since their emission cavity is small, relative to a gas tube.

Usually, the double slit should not care about polarization. Edge emitting diode lasers are usually well polarized as the cavities are pretty asymmetric. VCSELs do not necessarily show strongly polarized emission, but mostly they do. These cavities are still mostly asymmetric. Low-power multi-mode lasers are often unpolarized (or rather: several modes that may not be distinguishable lase simultaneously, so that the sum is unpolarized) and fiber lasers are pretty notorious for having unpolarized emission.

 

[Erik Ayer]

That a double-slit wouldn't care about polarization is kind of what I would expect in that the interference would be from the quantum wave function, not the electric field. So, by setting polarization to mark the which-way path, that would be carried by the field, leaving the wave function somewhat alone - the wave function has the polarization encoded in it, correct? So by "erasing" the which way information in the polarization, the quantum wave function can still interfere. Am I coming close to correct understanding, or am I off on a completely incorrect tangent? :)

I guess what I'm kind of thinking is that the double-slit experiment is interfering each photon with itself - the wave function splits into two paths, diffracts, and interferes. The EM field between multiple photons could interfere, but that isn't what's happening here. I would think that getting the EM fields of multiple photons to interfere would require forcing them to all hit the screen at the same time, like with the precision on the order of a wavelength.