Reversing a series of polarizers

[kurt101]

TL;DR Summary

What happens when you reverse a series of polarizers?

If light at a known polarity goes through a beam splitting polarizer and then goes through the reverse orientation of that polarizer it will exit with the same polarization that it entered with.

What happens if you put light through a series of beam splitting polarizers and then through the same series in reverse order with reversed orientation? Will the light exit with the same polarization as it entered? For example, if polarized light at +45 degrees goes through polarizer A, then B, then B reversed, then A reversed. Will the light exit with +45 degree polarization?

 

[StevieTNZ]

Do you intend to count all photons? If so, you'd most likely use wave plates rather than polarization
filters.

EDIT: at what angles do you propose setting A, B, B reversed and A reversed at?

 

[kurt101]

@StevieTNZ yes I would verify that the exit beam contained only +45 photons and no -45 photons.

 

[PeterDonis]

beam splitting polarizers

If you are using beam splitters then "reversing" one requires you to bring the two output beams back together in just the right way (normally you would use mirrors to reflect the beams, as in a Mach-Zehnder interferometer, for example). Using wave plates, as @StevieTNZ suggested, seems like a much better idea if you want to reverse operations you are doing on the beams, since a wave plate has only one input beam and one output beam.

 

[StevieTNZ]

@StevieTNZ yes I would verify that the exit beam contained only +45 photons and no -45 photons.

What angles are polarizer B and B reversed (as well as A and A reversed) set at? That's if you're happy with some loss of photons but nonetheless endeavour to achieve your end goal of photons ending up as |45>.

 

[kurt101]

What angles are polarizer B and B reversed (as well as A and A reversed) set at? That's if you're happy with some loss of photons but nonetheless endeavour to achieve your end goal of photons ending up as |45>.

I don't understand how the angles of A and B impact the end result of the experiment. Are you saying they don't in principle, but in practice based on what angles I choose for A and B I will get photon loss and if I use wave plates instead to do a similar experiment that I can avoid most of the photon loss problem?

I am most interested in the principle of the experiment, but certainly the practical aspects are interesting and important as well.

 

[DrChinese]

I don't understand how the angles of A and B impact the end result of the experiment. Are you saying they don't in principle, but in practice based on what angles I choose for A and B I will get photon loss and if I use wave plates instead to do a similar experiment that I can avoid most of the photon loss problem?

I am most interested in the principle of the experiment, but certainly the practical aspects are interesting and important as well.

If you start with a stream of light oriented at angle A; run that light through a polarizer oriented at +theta and a second oriented at -theta: All exiting light will be oriented at -theta. For most A and theta, there will be a reduction in intensity and the output stream will not be polarized at the angle of the input stream. Hence why you are being asked about the angles.

 

[kurt101]

If you start with a stream of light oriented at angle A; run that light through a polarizer oriented at +theta and a second oriented at -theta: All exiting light will be oriented at -theta. For most A and theta, there will be a reduction in intensity and the output stream will not be polarized at the angle of the input stream. Hence why you are being asked about the angles.

I should probably draw a picture. I thought I could explain it well enough so others understand what I mean, but that usually does not work for me.

So I don't disagree with what you just said, other than I think you are not understanding the experiment I am running. When I say light first goes through a beam splitting polarizer and then goes through that polarizer in the reverse orientation, I mean that it first goes into a polarizer where there is 1 input and 2 outputs and then goes into a polarizer where there is 2 inputs and 1 output. Or you might call one a splitter and the other a combiner.

I am pretty certain that my first assertion is true that with 1 polarizer in splitter configuration and 1 polarizer in combiner configuration it will not change the polarization of the incoming light because I read it from what I believe is a credible source. What I am not sure about is when you start to create a cascade of these, does it do the same thing.

 

[kurt101]

@DrChinese here is a picture of the experiment that I want to know the result for. I believe the polarization of the incoming light will be the same as the polarization for the outgoing light, but I don't know.

 

[PeterDonis]

When I say light first goes through a beam splitting polarizer and then goes through that polarizer in the reverse orientation, I mean that it first goes into a polarizer where there is 1 input and 2 outputs and then goes into a polarizer where there is 2 inputs and 1 output. Or you might call one a splitter and the other a combiner.

What you are describing is a Mach-Zehnder interferometer. I suggest looking up references that discuss this device and what it is used for.

 

[kurt101]

What you are describing is a Mach-Zehnder interferometer. I suggest looking up references that discuss this device and what it is used for.

The Mach-Zehnder Interferometer is different than what I am describing and have drawn. It measures interference and I am measuring polarization. It splits the beam at the beginning and then at the end where as the most simple version of my experiment splits the beam and then combines the beam. Here is a picture of the most simple version. I think the technical name for this is loop analyzer:

 

[PeterDonis]

The Mach-Zehnder Interferometer is different than what I am describing and have drawn. It measures interference and I am measuring polarity.

You do not appear to understand what the MZI actually does. The MZI has a first beam splitter that splits one beam into two, and a second beam splitter (or "recombiner"--the same kind of device does both jobs) that recombines two beams back into one--if the beams have no additional interactions applied to them in between. That is what you are describing.

If you are trying to change the polarization (not "polarity") of the beams, that is an additional interaction applied to the beams in between the first and second splitter, and of course that will mean there will be two output beams after the second splitter instead of one (because there will no longer be complete destructive interference in the second output beam, the one that is dark if the beams have no other interactions applied to them).

It splits the beam at the beginning and then at the end where as the most simple version of my experiment splits the beam and then combines the beam.

That is an MZI. See above. Your picture is simply failing to include all of the relevant information, like what directions the beams travel and how their directions are changed so that they can recombine again (in an MZI the direction changes are done with mirrors).

 

[PeterDonis]

I think the technical name for this is loop analyzer:

Instead of guessing, you need to go look at the literature and find a reference that actually describes the device you intend to use. If you are unable to find any reference that does that, that should be a huge red flag to you that you need to rethink your scenario.

 

[kurt101]

If you are trying to change the polarization (not "polarity") of the beams, that is an additional interaction applied to the beams in between the first and second splitter, and of course that will mean there will be two output beams after the second splitter instead of one (because there will no longer be complete destructive interference in the second output beam, the one that is dark if the beams have no other interactions applied to them).

This is why an MZI is not the same as what I am describing. What I mean by beam splitting polarizers is a calcite crystal that separates light into two plane polarized beams. Can the MZI be used to make something similar? Also, it is my understanding that the beam splitter in the MZI splits the incoming light randomly (if you were to try and detect which way the photon went) where as the crystal splitter splits the light based on its polarization.

 

[kurt101]

Instead of guessing, you need to go look at the literature and find a reference that actually describes the device you intend to use. If you are unable to find any reference that does that, that should be a huge red flag to you that you need to rethink your scenario.

I was remembering, but it was a analyzer loop, not loop analyzer. For example it is discussed in the book "An Introduction to Quantum Physics" by A.P. French and Edwin F. Taylor. chapter 7.

 

[PeterDonis]

What I mean by beam splitting polarizers is a calcite crystal that separates light into two plane polarized beams.

Ok, that helps a lot. "Beam splitter" means the kind of thing that's in the MZI. "Calcite crystal" is much better. However, as you will see below, in many respects what it does is very similar to what the beam splitter in the MZI does.

The two beams coming out of a calcite crystal still go in different directions, so you would still need mirrors or something similar to bring them back together in a second crystal. This "recombining" operation would end up outputting a single beam from the second crystal that was identical to the input beam coming into the first crystal, if no other operations were done in between to either beam, just as with the MZI, the single output beam from the second beam splitter (or "recombiner") is identical to the original input beam to the first splitter, if no other operations were done in between.

it is my understanding that the beam splitter in the MZI splits the incoming light randomly (if you were to try and detect which way the photon went)

If there is just a single input photon, then yes, the beam splitter normally used in the MZI would have a 50-50 chance of having a photon detected in either output beam, if you were running that experiment. However, it's extremely difficult to produce single photons. Usually the MZI is run with a beam of light in the usual sense, i.e., a continuous beam of light of a certain intensity; in that case, each output beam has half the intensity of the input beam.

where as the crystal splitter splits the light based on its polarization.

Yes, and if an unpolarized beam is input into the crystal, each output beam has half the intensity of the input beam, just as with the beam splitter in the MZI. Also, if you were to use a (very difficult) single photon source to fire a single photon into the crystal, and the single photon were unpolarized, there would be a 50-50 chance of detecting the photon in either output beam, just as with the beam splitter in the MZI.

The main difference with a calcite crystal as compared to the beam splitters used in the MZI is that it is possible to prepare a light source to produce a beam that will not be split by the crystal--just prepare the source to produce light in one of the two polarization eigenstates that the crystal separates (for example, +45 or -45). AFAIK there is no way to prepare such a source for the MZI beam splitter--any single input beam will be split into two.

 

[DrChinese]

The Mach-Zehnder Interferometer is different than what I am describing and have drawn. It measures interference and I am measuring polarization. It splits the beam at the beginning and then at the end where as the most simple version of my experiment splits the beam and then combines the beam. Here is a picture of the most simple version. I think the technical name for this is loop analyzer:
View attachment 293355

I think what you are looking for is this:

http://www.drchinese.com/David/EntangledFrankensteinPhotonsA.pdf

Figure 1 is from French and Taylor, 1979. Same as yours. As far as I know, no one has ever done this experiment. But this requires a very exacting setup. For full disclosure, this reference is to a paper I wrote.

 

[vanhees71]

I think it's much simpler. It's a single-photon experiment, where the photon enters a polarizing beam splitter (like the birefringent calcite crystal)

https://en.wikipedia.org/wiki/Polarizer#Birefringent_polarizers

First consider classical light (quantum mechanically that's a coherent state of not too low intensity) of arbitrary polarization. The light beam entering the crystal it's split into two beams which are polarized perpendicularly to each other.

For a single photon that means that an incoming photon of arbitrary polarization leaves the crystal with some probability either with momentum $\vec{p}_1$ and polarized in one direction (say horizontally, labelled with $H$) or with momentum $\vec{p}_2$ an polarized perpendicular to this direction (i.e., vertically polarized, labelled with $H$). In QT formulation such a polarizing beam splitter is leading to a photon state, where the polarization is entangled with the momentum of the photon.

Then in principle you can put another calcit crystal in precisely the opposite way such that the two beams are recombined again such that you get back the original single photon state, but that's very difficult to realize as Fench&Taylor stress, and I'm also a bit puzzled about what the authors want to teach with this example. I guess it's about interference, i.e., that you cannot simply use classical probabilistic arguments to analyze these experiments but need to use the "amplitudes".

It's much simpler to think in terms of the classical wave picture to get an intuition. Concerning single-photon experiments using only devices of linear optics, the only thing you have to keep in mind is that (a) there are interference effects as in the classical wave picture and (b) you can register one and only one photon. In our example after one polarizing beam splitter you find only one photon either in the state $|\vec{p}_1,H \rangle$ or $|\vec{p}_2,V \rangle$. The probabilities for either outcome are proportional to the intensity of (energy density) of the classical em. wave at the position of the detectors.