Question on circular polarization and the quantum eraser

[Erik Ayer]

TL;DR Summary

The quantum eraser has quarter wave plates in from of each of the slits for interference, but depending on the polarization of the incident photons, which way is marked or not

Looking at the wikipedia page for the (original) quantum eraser (https://en.wikipedia.org/wiki/Quantum_eraser_experiment), quarter-wave plates in front of the slits change photons from being linearly polarized to being circularly polarized. I think this is the case when there is no polarizer in the other, entangled beam that is not going through the slits, so that would mean the light in the double slit path in unpolarized. However, looking at various pages on the 'net about creating circularly polarized light, there does have to be an initial polarization at 45 degrees relative to the quarter wave plate's fast and slow axes. My guess is that the light in the no-interference case is diagonally polarized from a linear polarizer on the entangled (non-slit) beam, so there's bot diagonal and anti-diagonal light, resulting in left-circular at one slit and right-circular at the other, marking the which-way.

When that which-way information is erased, the lower diagram shows that light from both slits is in a superposition of left and right circular polarization so as to not mark the path. Is this a fancy way to say the photons are linearly polarized? I'm wondering whether a superposition of left and right circular adds up to linear.

There is also, apparently, a linear polarizer in front of the detector after the slits. What is this doing? My guess is that the circular polarizations of the no-interference setting come out as different linear polarizations. In the case where both slits are putting out a superposition of left and right polarization, this final linear polarizer in from of the detector for the slit-path converts the photons to an indentical linear polarization.

It seems like it would be possible to make an experiment to play around with this by taking two (visible) laser beams, putting them through linear polarizers, and combining them to get either diagonal and anti-diagonal light or vertical and horizontal light. That resultant light could then be put through a double-slit or equivalent setup with the quarter wave plates to observe, visually, what pattern it produces. Would this be equivalent (except for not having an entangled beam and thus, not as nice an experiment), or am I missing something critical?

Thanks,
Erik

 

[vanhees71]

The incoming photons are prepared in an entangled state. For simplicity let's discuss the antisymmetric Bell state (corresponding to total angular momentum 0),
$$|\Psi \rangle = \frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle).$$
Here $H$ And $V$ refer to linear polarization wrt. the $x$ axis. Then you do the experiment by sending one of the photons through the double slit with the quarter-wave plates with orientations $\pm \pi/4$ relative to the $x$ axis. This implies that if the incoming photon ("signal photon") at the double slit is $H$ polarized, the other "idler photon" is $V$ polarized and vice versa.

Now you take only the case where the signal photon is $H$ polarized by using an appropriate polarization filter before the slits. Then all the corresponding idler photons are $V$ polarized.

Behind the slits IF you measure whether the signal photon is left-circular (helicity +1) or right-circular (helicity -1) polarized but then you know through which slit it came. Now you don't measure but look at all photons. You can easily check that there's no double-slit interference, because the photons going through one slit are in a polarization state perpendicular to a photon going through the other. So having the possibility to gain which-way information destroys the double-slit interference pattern.

But now you don't do any measurement with the signal photons behind the slit but rather do a measurement on the idler photon (being still $V$-polarized) whether they are left-circular or right-circular polarized (which is the case with a 50% probability for each circular polarization state). Via coincidence measurements on both places you know which photons are in the entangled two-photon state and thus looking at the part of the photons pairs, where the idler is left-circular polarized you define a new ensemble (of about half the photons) for which you are in principle not able to gain which-way information on the signal photon, and indeed the formalism shows that for these subensemble you observe a two-slit interference pattern. The other half of the original ensemble also shows a two-slkit interference pattern but of course shifted in such a way that the entire ensemble doesn't show an interference pattern. That's clear, because through the measurement on the idler photon nothing happened with the signal photons, i.e., the total ensemble just shows no interference pattern no matter whether you did or did not meaure the idler photons. But sorting according to the measurement of the idler-photons polarization in the R/L basis you "erase" the which-way information for each of the two subensembles, due to the entanglement of the photon pairs at the very beginning. It's a realization of the famous "quantum erasure through post-selection" scheme invented by Scully et al.

 

[Erik Ayer]

I get the sense that I've been thinking of this all backwards. Here's the first diagram from the wikipedia page:

How I've been thinking about it is that there would be a (linear) polarizer in the upper/idler path. It could be oriented in H or V, which would force the idler photon to be hirizontally or vertically polarized and transmitted or absorbed by the polarizer according to how polarizers work. On the lower/signal path, those photons would also be forced to be vertically or horizontally polarized, opposite of what the idler was.

Then, going through the quarter wave plates, the signal photons would be transformed to be left or right-circularly polarized depending on their initial, linear polarization and which slit they went through. Filtering with coincidence to the idler, where one polarization was absorbed and not available for coincidence, would mean that the which slit information would be present and thus, no interference.

On the other hand, if the polarizer in the idler was either diagonal or anti-diagonal, the signal photon would take a complimentary state and the quarter wave plated, oriented diagonally, would not convert the signal photons to a circular polarization. Which-way information wouldn't be available, so interference wouuld be present.

Depending on whether a photon was diagonal or anti-diagonal, either the upper slit or the lower slit would have a quarter wave delay since the electric field would be in the slow axis on one or the other slit. Diagonal and anti-diagonal photons would have opposite slits delayed, so there would be a 180 degree difference between the two cases, and that should shift their interference patterns to be opposite. Thus, without coincidence to filter out one interference pattern or the other, they add up to a gaussian mess.

Is my reasoning wrong? Indicentally, here's what I was looking at for circular polarization:

Photons coming in at 45 degrees to the axes would be converted to circular polarization and light that was aligned to the axes would stay linear, although it might be delayed depending on which axis it was parallel to. Looking at the second diagram from the quantum eraser wikipedia page:

It seems to indicate that there would be both left and right-circular polarization in superposition coming from both of the slits. So I'm wondering whether light aligned to the axes and passing through without being converted to circular polarization can be thought of as that superposition of both left and right-circular polarization.

 

[vanhees71]

Have you read my posting? As I said there, to erase the which-way information, you have to measure the polarization of the idler photon in the helicity basis.

 

[Erik Ayer]

I did read you explanation multiple times, and I 've just read it again a few more times but am still not understanding several things. Starting here:

"Now you take only the case where the signal photon is polarized by using an appropriate polarization filter before the slits. Then all the corresponding idler photons are polarized."

Is this actually done? I don't see a polarizer in the diagrams before the slits, so it would seem like the photons would be unpolarized. If the idler passed through a polarization filter, it would set the polarization of the signal photons in that basis (H|V), correct? Each of the signal photons would be marked as to which slit they went through due to the quarter wave plates once they had either H or V polarization. H photons would be left-circular for the top slit and R-circular for the bottom, whereas V photons would be right-circular for the top and L-circular for the bottom (or vice versa depending on the orientation of the quarter wave plates).

Have I gotten this part correct so far?

 

[vanhees71]

I think, in my attempt to simplify the issue as much as possible, I made it more complicated. Here you find the English translation of some slides used for my "habilitation colloquium":

https://itp.uni-frankfurt.de/~hees/publ/habil-coll-talk-en.pdf

 

[Erik Ayer]

Yes, this makes sense. It being an "English translation" implies that you are a native German speaker? Cool.

I've studied the quantum eraser for more than a decade, but recently I gave a presentation to a meetup group - mostly engineers, including myself - and they asked me to write it up as a document. While creating an outline I realized that I didn't quite understand how the erasure happened. A big part of that is not understand cirrcular polarization very well.

Having read the wikipedia page on circular polarization as well as a few other pages, converting a linearly polarized photon to circular polarization is done by passing it through a quarter wave plate where its initial polarization is pi/8 to the fast and slow axes. I would think that unpolarized light passing through a quarter wave plate would NOT become circularly polarized and in fact, one web page confirms this: https://sciencedemonstrations.fas.harvard.edu/presentations/circular-polarization. The relevant paragraph is this:

"""
How it works:

The quarter-wave retardation plate is a sheet of birefringent (double refracting) material 1 of thickness such that horizontally and vertically polarized light entering in phase will emerge from the retardation plate 1/4 of a wavelength out of phase. Unpolarized light is not affected by this retardation plate (or by any thickness of birefringent material) because the retardation plate only changes the phase of each component of polarization. The situation dramatically changes when the incident light is polarized.
"""

It seems like this would be the case. Would it be correct to think of unpolarized light as being is a superposition of all polarizations? If so, would the quarter wave plate collapse the polarization into a random angle? I'm contrasting this to a linear polarizing filter where the photons collapse to either H or V and are either transmitted or absorbed. With the quarter wave plate it would seem like the polarization would take a random and and transform into elliptically polarized light between and including circular and linear polarizations.

So, I'm wondering why the s-path light is converted to circular polarization before either path is linearly polarized and why that would mark which slit it went through. Polarizing the P-beam photons would force them and the s-beam photons to be polarized H or V relative to the polarizer, so I would think having the polarizer there would be necessary.

 

[vanhees71]

Yes, I'm German, and this was one of the few occasions, where I had to give a talk in German. That's why I've translated the presentation to have it at hand for discussions :-).

The amazing thing with "entanglement" is indeed really in a nutshell in this experiment.

The entangled photon pairs have total angular momentum 0 and thus the single photon states are that of completely unpolarized photons. It's characteristic for a Bell state that the observables of the entangled parts of the system (here the polarization of the single photons making up the pair) are maximally indetermined (in the sense of von Neumann entropy).

So the signal-photon going through the slits is unpolarized, but its polarization state is entangled with the polarization of the idler photon.

To see, why for measuring signal photons behind the slit with the QWPs in place you have the possibility to gain which-way information, just check p. 6 of my slides. The state for the pair is $|\psi_1 \rangle+|\psi_2 \rangle$. So how can you know through polarization measurements on the signal photon behind the slits and the idler photon, through which way the signal photon went?

Obviously you have to measure the signal photon's polarization in the $L,R$ basis, i.e., determine, whether it is either left- or right-circular polarized and the indler photon's polarization in the $\hat{x},\hat{y}$ basis, i.e., determining whether it's polarized in $x$- or in $y$-direction. The possible outcomes are

$\gamma_s$ is L-polarized and $\gamma_p$ is $\hat{x}$ polarized. From the prepared state we read of that then $\gamma_s$ with certainty has come through slit 2 (because this combination in the product states is only in $|\psi_2 \rangle$)

$\gamma_s$ is R-polarized and $\gamma_p$ is $\hat{x}$ polarized $\Rightarrow$ $\gamma_s$ with certainty must have come through slit 1.

$\gamma_s$ is R-polarized and $\gamma_p$ is $\hat{y}$ polarized $\Rightarrow$ $\gamma_s$ with certainty must have come through slit 2

$\gamma_s$ is R-polarized and $\gamma_p$ is $\hat{x}$ polarized $\Rightarrow$ $\gamma_s$ with certainty must have come through slit 1.

So by measuring the polarization for $\gamma_s$ in the $L,R$ and that of $\gamma_p$ in the $\hat{x},\hat{y}$ basis you precisely know through which slit it came.

On the other hand you can measure $\gamma_p$ in the $\pm \pi/4$-linear-polarization basis (see next slide). Then only considering the photons where the $\gamma_p$ is in the $+\pi/4$ state you get a superposition, where through no polarization measurement on $\gamma_s$ it's possible to get which-way information and accordingly you get the full two-slit interference contrast back (though shifted compared to the interference pattern you'd get without the QWPs in place).

The same holds true when taking the $\gamma_s$ for whose partner $\gamma_p$ the polarization state was found to be $-\pi/4$-linear. Again you erase all possibility to gain which-way information about $\gamma_s$ and get an interference pattern, this time shifted in the opposite direction.

Of course, taking all $\gamma_s$, you always see no interference pattern, because indeed through the measurement on $\gamma_p$ you haven't influenced the $\gamma_s$ (at least if you do the measurements such that the registration events of the $\gamma_s$ and $\gamma_p$ are space-like separated), and indeed there's no contradiction between that on one hand you can gain in principle which-way information by one kind of measurement and on the other hand you can erase any possitiliby to gain such which-way information by another kind of measurement (and postselecting the corresponding partial ensembles), because if you realize the possibility to get which-way information you cannot realize the erasure measurment anymore and vice versa. That's a pretty astonishing demonstration of Bohr's "complementarity principle".

At the same time it shows that you can also do any kind of other coincidence polarization measurements on $\gamma_s$ and $\gamma_p$. Then you neither gain complete which-way information nor erasing the possibility go gain which-way information completely. Consequently the interference pattern of the corresponding partial ensembles is not completely gone but for loosing more or less of the contrast, while the uncertainty about through which slit it came is the larger the better the contrast of the interference pattern is.

 

[Erik Ayer]

Okay, this makes complete sense. I had kind of come to understand it this way earlier but didn't explain my thoughts very well such that one could follow them.

The concern about converting unpolarized light with the QWPs would not be a problem because there will always be a linear polarizer in front of Dp, and hence the signal photons will always take on a linear polarization before the QWPs do their thing. H and V polarizations will lead to the which-way information being present, diagonal and anti-diagonal polarizations will not cause photons to be converted to circular polarization and hence the interference patterns will be present, although shifted one direction or the other.

Measuring without any polarizer in front of Dp would just result in a mess.

On similar experiment, most notably this one by John Cramer:

https://arxiv.org/pdf/1409.5098.pdf

both the signal and idler beams are split into two paths:

In contrast, the quantum eraser experiment we've been talking about has only the signal being split into two paths (through the two slits). As I understand it, there is a trade-off between entanglement and coherence that these experiments where both beams are split into two paths suffer from. So in this one above (100% entanglement, 0% coherence), the waves going along the upper and lower paths would have a random phase shift that would cause light not to interfere and come out of just one face of the outer beam splitters. That phase shift would be the same between paths on the left vs paths on the right.

With the quantum eraser experiment we've been talking about and the idler beam not being split into two paths, would this keep there from being a random phase shift between the paths on the signal? It seems like it should and does, given that measuring diagonal (or anti-diagonal but not both) photons in coincidence does give an interference pattern.