Can circularly polarized photons interfere with themselves?

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1. Dec 27, 2015

marksesl

Hello, I was studying the Bell-state quantum eraser and saw that they placed quarter-wave plates in front of each slit to mark the photons with either clockwise polarization or counterclockwise polarization that were previously either horizontally or vertically polarized. So, now I'm wondering if light polarized in opposite circular polarizations can even form a diffraction pattern. And, what about photons polarized with the same circular polarizations? Can they interfere? For instance, can two clockwise polarized photons interfere and create a diffraction pattern?

2. Dec 27, 2015

PietKuip

Photons are so complicated. It is much easier to analyze this in terms of the electric field of electromagnetic waves.

3. Dec 28, 2015

Staff: Mentor

Its got nothing to do with polarisation - its got to do with the uncertainty and superposition principle:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

Thanks
Bill

4. Dec 28, 2015

_PJ_

Polariation, to my limited understanding, is essentially just a means to filter spin varieties.
Interference occurs regardless when any two waves are measured at a singular spacetime event or area of such events.
The overall interference result is dependant on the asynchronicity/synchronicity of the phase of the #spin#.
There is no way to guarantee the polarisation unless the filter is applied across the measurement apparatus. Else the spin will gradually spread out to encompass all probabilities.

You even refer to the quantum eraser, which requires that here is no measurement made between the two, therefore, the probability of the initial polarisation is not 100% as would the case if measurement were made at that point.

5. Dec 28, 2015

zonde

I would say that it's the same way as with linearly polarized (H and V) beams of light. H polarized light does not form interference with V polarized light, even so it may seem to be coherent as we can get V polarized beam from other half of coherent H beam by converting it to V polarization using half-wave plate. You might want to look at this: http://www.scientificamerican.com/slideshow/a-do-it-yourself-quantum-eraser/
It so it seems the same way with R and L circular polarization. R and L does not form interference even when they seem coherent (say L and R polarizations are produced from two halves of the same coherent linearly polarized beam by differently placed quarter-wave plates).
Btw in experiments with GHZ entangled state L and R polarization of photons are used as well.

6. Dec 29, 2015

vanhees71

Circular polarization means that, if you deal with single photons, these photons have definite helicity $h \in \{-1,1\}$. Helicity is the component of the photon's total angular momentum in direction of its momentum and thus represented by a self-adjoint operator, and thus the eigenstates to different eigenvalues are perpendicular to each other. Consequently the interference term is absent, and there is no interference pattern formed. This is the crux of this experiment! They use the photons themselves to mark through which slit they are going. If you want to have the which-way information fully determined, you must do it in this kind of way and thus if you have full which-way information, there's no interference pattern. The mind-boggling thing, however, is that you can choose after performing the experiment different subensembles of photons, where you either have full which-way information and no interference pattern or absolutely none which-way information and an interference pattern at maximum contrast.

7. Dec 29, 2015

PietKuip

@vanhees71: But this photon helicity... does it change by interaction with a quarter-lambda plate? How do exactly photons interact with birefringent materials? Do they get flipped around?

All these horribly difficult questions are avoided when one treats the phenomenon as electromagnetic waves.

8. Dec 30, 2015

vanhees71

You gave the answer yourself. Don't think about photons as being little bullet-like objects but as what they really are, i.e., Fock states with photon number 1. You can treat the optical elements like lenses, wave-plates, beam splitters etc. quasi classically, i.e., as if they act on the em. field operators as on the corresponding classical fields, e.g., the birefringent material of a quarter-wave plate is described by an dielectric tensor. See, e.g., the textbook

M. O. Scully, M. S. Zubairy, Quantum optics, Cambridge University Press (1997)