A question about density matricies

[Erik Ayer]

TL;DR

The density matricies for H/V mixtures are the same as for D/A mixtures. Somewhere, my reasoning is wrong...

The density matricies for statistical mixtures of horizontal and vertical polarizations are the same as for mixtures of diagonal and anti-diagonal mixtures, so there should be no experiment that behaves differently for each mixture. I had an idea about it that's got to be wrong and am hoping someone can point out the flaw(s) in my reasoning.

If the H/V mixture is sent to a double-slit experiment with quarter wave plates in front of the slits, one with the fast axis horizontal and the other vertical, H will keep its polarization but be delayed at one slit, V will also keep its polarization but be delayed at the other slit. This results in the interference patterns overlapping and becoming indistinguishable. Diagonal and anti-diagonal polarizations will get converted into left- and right-circular polarization at each slit and not interfere at all.

If a lens is in the zone where H and V interference happens (or lack of it for D/A), if will focus the light to images of the two slits. For H/V, the light at each will be a mixture of H and V. For D and A, they get converted to circular and there ends up being a mixture of left- and right-circular polarization at each image.

If beam blocks, like wires or 3D-printed "jail bars" are put into the interference area such that they are where the bright fringes of one of the interference patterns are at, they will block most of that polarization in the H/V mixture but evenly block the circular polarization from D/A. At the images, they will get a lot of one polarization and little of the other for H/V but get an even mixture of left- and right-circular polarization for D/A. Putting H and V polarizers in the images and detecting light levers would then give different amounts of light for H/V and equal amounts of light for D/A.

That can't be. What am I doing wrong?

 

[SergejMaterov]

The quarter-wave plates reduce the interference to zero, and both mixtures behave identically. Any further manipulations (blocking the "bands", polarizing filters on the images) will not show the difference between HV and DA - because you simply do not have any coherence between the slits left.

 

[Erik Ayer]

In the MZ with QWPs below, the V input would be changed to left and right circular, marking which way photons went, and thus preventing interference. If, however, the input was diagonally polarized, it would remain diagonal but be slowed down in one path vs. the other. The arm lengths could be adjusted such that the output was constructive at one face of th final beam splitter and destructive at the other. Anti-diagonal light would be constructive at the other face, so the combination of D and A would have light coming out both faces...

And with H and H polarized light, there would be no interference and again, light coming out both faces. The path would definitely be marked so no interference, and the light would split at the final beam splitter.

What I was thinking is similar, and here's a link to a diagram:



Instead of an MZ, it would use a double slit and get the two, overlapping interference patterns. On the other hand, if the input is D/A, the QWPs mark the path and there is no interference:



The advantage of a double-slit over an MZ is that the interference patterns overlap and beam blocks can be place in one of them, but if there is no interference, one of those non-interference patterns isn't completely blocked like it would be with the beams coming out of the MZ.

 

[SergejMaterov]

$$D = \frac{1}{\sqrt{2}}(H + V) \xrightarrow{\text{QWP}(H)} \frac{1}{\sqrt{2}}(H + iV) = \text{RCP}$$
$$D = \frac{1}{\sqrt{2}}(H + V) \xrightarrow{\text{QWP}(V)} \frac{1}{\sqrt{2}}(iH + V) = \text{LCP}$$
That is, diagonal polarization also turns into a right/left circle, in the same way "marking" the path through the left or right sleeve. There is no interference. You incorrectly assumed that QWP would "slow down" D/A, leaving them coherent, and H/V would be marked forever. In fact, any input polarization in such a scheme is converted into different elliptical (or circular) polarizations in the two arms and completely loses coherence between them. Unfortunately, you will not return interference by any clever adjustment of the length or polarizers at the output.
Intensity on screen: $I(x) = I_1(x) + I_2(x)$
and it should be: $I(x) = \left| \psi_1(x) + \psi_2(x) \right|^2$
ADD:
The double slit wins because of the overlapping patterns, while the MZ scheme does not.
– In the double slit, you actually get two interference patterns (H-polarization vs. V-polarization), which are superimposed in space. There you can physically close one and see the other.
– In the MZ scheme with QWP, you never get two superimposed interference patterns, because there is no interference for either H/V or D/A.

 

[Erik Ayer]

I'm not sure I understand. D/A polarized light will be turned into left- and right-circular polarization (or vice-versa) and will not interfere - the paths are marked. For H and V, they align with the fast or slow axes of the QWPs, so there will be a shift in the phase through one slit or the other. So for those, they stay H or V polarized, not converted into circular, so they should create interference. H and V each create interference patterns that are shifted one way or the other such that the bright and dark parts from each overlap.

What am I missing?

 

[SergejMaterov]

You are absolutely right that each pure H or pure V beam will create an interference pattern by itself, once you set up the quantum photons to act only as phase shifters. But you missed one subtlety: when you feed an incoherent HandV mixture (rather than a single pure state), the two interference patterns are out of phase with each other, and so they sum to form a perfectly flat intensity distribution. Bottom line: any incoherent mix of two complementary fringe patterns gives you no visible modulation. That’s why you can’t recover a “hidden” H/V interference by blocking peaks or shifting phases: the two sets of fringes already sum to flat.

 

[Erik Ayer]

Right, H and V will both create interference patter but they will be shifted relative to one another. The light and dark parts add up to a uniform (or Gaussian) profile, but the interference patterns are still there, just obscured by the overlap.

 

[Erik Ayer]

You are absolutely right that each pure H or pure V beam will create an interference pattern by itself, once you set up the quantum photons to act only as phase shifters. But you missed one subtlety: when you feed an incoherent HandV mixture (rather than a single pure state), the two interference patterns are out of phase with each other, and so they sum to form a perfectly flat intensity distribution. Bottom line: any incoherent mix of two complementary fringe patterns gives you no visible modulation. That’s why you can’t recover a “hidden” H/V interference by blocking peaks or shifting phases: the two sets of fringes already sum to flat.

Here are some better diagrams to hopefully illustrate how the beam blocks work: