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While trying to understand polarization in quantum field theory, I wondered how a single photon could go through a linear polarizer. I found a paper which asked "Is a single photon always circularly polarized?"

This paper proposes an experiment to determine if a single photon can be linearly polarized, or if only pairs of photons can be linearly polarized. It suggests that there may be non-trivial consequences regarding all Bell experiments with a "linearly polarized single photon" (because such thing may not exist).

The paper is from 2014 and the experiment seems simple if you have the right equipment, so do we have the result of the experiment yet?

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    $\begingroup$ For what it's worth, that preprint is unpublished, and didn't make it into the special issue it was submitted to. That doesn't invalidate the science but it's a warning flag to be aware of. $\endgroup$ – Emilio Pisanty Oct 18 '15 at 11:43
  • $\begingroup$ @darkblue if $c_{\vec{k}}^{R/L}$ is the creation operator of a photon with momentum $\vec{k}$ and helicity $R/L$, try this creation operator now (and prove that this is a creation operator): $(c_{\vec{k}}^{L}-c_{\vec{k}}^{R})/\sqrt{2}$. I recommend you to study second quantization grounded on quantum mechanics. Second quantization is only a better way to deal with quantum systems, and some cases the only way (when the number of particles do not commute with some observables) $\endgroup$ – Nogueira Oct 20 '15 at 16:28
  • $\begingroup$ @darkblue In special relativity, the number of particles of mass $m$ does not commutes with observables distributed in small boxes ($~\frac{\hbar}{mc}$). So second quantization is needed. $\endgroup$ – Nogueira Oct 20 '15 at 16:34
  • $\begingroup$ @Nogueira About your small boxes : I would argue this is a first quantization issue. By speaking about "particles with masses in SR" you are gauge fixing yourself in momentum space. Then you talk about position space by saying space boxes. Obviously it won't commute but not because of the number of particles. Your bad choice of basis is not a reason good enough to justify needing second quantization (a conceptual jump in a higher function space). I think the experiment proposed in the paper is the analog of bell's experiments but for second quantization, a way to know if the jump is needed $\endgroup$ – darkblue Oct 20 '15 at 22:13
  • $\begingroup$ @Nogueira I think I can conceptualize quantizations correctly. Suppose we have a set S , f : S -> S , g : L2(S) -> L2(S) ,h : L2( L2 (S) ) -> L2( L2( S ) ) , then There are several interesting ways we could sample from S We start by picking a s0 in S, then f^n (s0 ) . Or we could start by picking ls0 in L2(S) then g^n(ls0) from which we sample an element which is in S : analog to first quantization Or we could start by picking lls0 in L2(L2(S)) then h^n(lls0) from which we sample an element which is in L2(S) from which we sample an element which is in S : analog to second quantization $\endgroup$ – darkblue Oct 20 '15 at 22:27
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For a single photon, the only similar physically meaningful question is whether the circular polarization is left-handed or right-handed. Quantum mechanics may predict the probabilities of these two answers. An experiment, a measurement of L/R, produces one of these answers, too. After the measurement, the photon is either left-handed or right-handed circularly polarized.

If a photon is prepared in a general state, it has nonzero probabilities both for L and R. In such a "superposition", we may perhaps say that the single photon has no circular polarization. This statement means that we are uncertain which of the polarizations will be measured if it is measured. But when the circular polarization is measured, one always gets an answer, according to the result of the measurement.

Linear polarizations are the simplest nontrivial superpositions of L and R. The absolute value of both coefficients, $c_L$ and $c_R$, is the same while the relative phase encodes the axis on which the photon is polarized.

The paper quoted in the question is completely wrong. An example of a very wrong statement is that the linearly polarized photon moving in the $z^+$ direction carries $J_z=0\cdot\hbar$. In reality, a linearly polarized photon or any photon is certain not to have $J_z=0\cdot\hbar$. A linearly polarized photon has the 50% probability to be $J_z=+1\cdot\hbar$ and 50% to have $J_z=-1\cdot\hbar$. The expectation value $\langle J_z\rangle = 0$ but it's still true that the value $J_z=0\cdot\hbar$ is forbidden.

A different question is the polarization of an electromagnetic wave. For a wave, e.g. light, one may distinguish left-right and right-handed and $x$-linearly and $y$-linearly and elliptic polarizations of all kinds one may think of. In terms of photons, a macroscopic electromagnetic wave is the tensor product of many photons. If all these tensor factors are linearly (or circularly) polarized, then the wave may be said to be linearly (or circularly) polarized. Because the polarization of the whole wave requires some correlation in the state of individual photons, a wave may be measured not to be circularly polarized in either direction. But an individual photon is always circularly polarized in one of the directions when the answer to this question is measured.

The paper may present a proposed experiments which may be done but what is completely invalid is the author's interpretation of this experiment – even "possible interpretations" before the experiment is actually performed. The correct description by quantum mechanics isn't included among their candidate theories with which they want to describe the experiment.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Oct 21 '15 at 6:55
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Answering my own question to close a far too long debate that heated far too much in my opinion.

To sum up :

According to mainstream physics, quantum mechanics : No, a single photon isn't always circularly polarized. See Lubos's good answer if you want more details.

The paper is unorthodox science, as it proposes a test to falsify in quantum mechanics.

A little advice to any beginners to the field like me, be aware that because there has been many unsuccessful attempts to falsify QM in the past, any talking about any new experiment to falsify and you will be looked upon as crackpot.

A little QM self fulfilling prophecy joke to end on a more light tone : "Obviously because we live in a QM world, any experiment that would falsify QM can not happen" :)

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    $\begingroup$ Proposing experiments to falsify QM does not automatically mean that you're a crackpot. In fact, pretty much everyone in quantum foundations is to at least some extent unhappy with the situation, and if we found an experiment that actually broke QM then most people would be ecstatic - it would give us an upper hand on the beast. Indeed, there have been many serious attempts to falsify QM (does Bell ring a bell? when Aspect set out to do the measurements, he set out to show that common sense would hopefully triumph over QM) but what this means is that (cont.) $\endgroup$ – Emilio Pisanty Oct 24 '15 at 9:38
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    $\begingroup$ all the easy approaches have already been used, and if you want to propose a new one you had better bring something nontrivial to the table. Papers that say "I don't like QM, this other theory is better" will generally be met with "well, how does your theory deal with $X$"?, and there is a large set $\{X\}$ of situations that QM explains perfectly and your new theory needs to satisfy. $\endgroup$ – Emilio Pisanty Oct 24 '15 at 9:44
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    $\begingroup$ This particular paper makes some pretty crude mistakes (e.g. it ignores that number-resolving detectors are easy to implement now) which means that its alternate theory isn't really a workable explanation for the current experimental state of the art. That makes it hard to take seriously as an experimental proposal. $\endgroup$ – Emilio Pisanty Oct 24 '15 at 9:46
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There is a big misunderstanding about what is the spin of a photon. The orthogonal standing B- and E-field could have a left hand orientation or a right hand orientation (see last page in this elaboration).

For polarizers it is important only, how the photons E-field is orientated to the slits, in the case of 0° and 180° photons of both spin orientations are going through. (And artfully designed polarizers rotate photons with +/- 45° to the above mentioned orientations, so one get 50% transmittivity.)

For photons, going through birefringent calcite, the spin orientation play a great role. The calcite separates the two spin orientations. This is clear because even polarized light will be divided into two beams.

Circular orientated light has got a rotational momentum from the source emits it. The E- and B-field rotating together. Of course one could feel free to represent linear polarized light as a superposition of a clockwise and an anticlockwise rotating state. But this mathematics one is able to do with the state of a football too.

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    $\begingroup$ This is a classical picture. I was looking for a quantum field picture. The question I was asking is : can a quantum field with a single quanta of energy make its Left Handedness, and Right Handedness interfere to behave as a linearly polarized quanta of energy. I was looking for the physical result (real data) of the experiment proposed in the paper. $\endgroup$ – darkblue Oct 18 '15 at 20:53

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