Talk:Photon polarization

Light has three polarization orientations, not just the two

Hi, About the first sentence in this article "Photon polarization is the quantum mechanical description of the classical polarized sinusoidal plane electromagnetic wave. Individual photons are completely polarized. " - would you call a photon in a superposition of several polarizations (e.g. as the ones used to prove that bell's inequality does not hold) completely polarized? -Lars —Preceding unsigned comment added by Laser Lars (talk • contribs) 09:32, 27 October 2008 (UTC)

three polarization orientations only for massive particles, this is a classic result 2A01:E0A:159:17E0:D088:2EC0:89C6:76C2 (talk) 15:46, 2 October 2023 (UTC)

light has three polarization orientations, not just the two mentioned in the artical

No. Three polarisation base orientations are only possible for waves travelling slower than the universal speed limit. At the speed limit (i.e. for light), relativistic effects remove the longitudinal mode. I do not have a very good explanation, but a clue might be the observation that for anyone travelling at the speed limit, length contraction reduces the universe's dimensionality to two, i.e. the universe has no expansion anymore in the direction of flight.WikiPidi (talk) 14:27, 9 March 2020 (UTC)

Snowflakeuniverse 18:00, 10 January 2007 (UTC)

- light has n polarization orientations, one for each dimension in whatever spatial metric you are operating. Dfruzzetti (talk) 21:24, 31 December 2008 (UTC)

Error: In the section Uncertainty Principle and Mathematical preparation

Error: In the section "Uncertainty Principle" and under the heading "Mathematical preparation" you will find a diagram representing two intersecting vectors v and w and their intersection is incorrectly labeled cos(a). On the diagram, the angle should have measure a rather than the cosine of that measure.

Dfruzzetti (talk) 21:23, 31 December 2008 (UTC)

Photon article says that a photon has two possible polarization states

The Photon article says that a photon has two possible polarization states. This article does not refer to this, but describes the polarization state in terms of a Jones vector, which has three parameters. These two statements seem very inconsistent, and there doesn't seem to be any explanation linking the two. Is there a standard way to represent a polarization state using two parameters? cojoco (talk) 03:14, 26 March 2010 (UTC)

The two possible polarization states the photon article refers to are probably the two basis vectors for the polarization state space, either it be $|x\rangle$ and $|y\rangle$, or $|R\rangle$ and $|L\rangle$ (or any other two orthogonal basis vectors, which two doesn't really matter). To say that the photon only have two possible polarization states clearly seems wrong to me, since it can be in any superposition of the two basis states. On the other hand, upon measurement of which polarization state the photon is in, you will only be able two get two different results (both with different probabilities), which depend on how you perform the measurement (see wave function collapse). --Kri (talk) 14:06, 28 May 2011 (UTC)

article says that a photon can have zero angular momentum

The article says "Photons have only been observed to have spin angular momenta of ±ħ or 0." ±ħ was observed by Beth - Beth, Richard. 1936. “Mechanical Detection and Measurement of the Angular Momentum of Light.” Physical Review 50 (2): 115-125. doi:10.1103/PhysRev.50.115. Photons cannot have zero angular momentum in the direction of propagation. 144.131.117.96 (talk) 00:29, 29 March 2011 (UTC)andrew

Note that the point about saying that a particle is of spin n is that a measurement of its intrinsic angular momentum must be in the set $\{-n\hbar ,-(n-1)\hbar ,...,(n-1)\hbar ,n\hbar \}$. For bosons, zero is a possibility. "Photons have only been observed to have spin angular momenta of ±ħ or 0." is a correct statement, since it does not make any reference to direction of motion. Ricieri (talk) 02:39, 20 August 2014 (UTC)

See my long comment below. 178.38.18.115 (talk) 07:55, 27 June 2015 (UTC)

Repetition

This article repeats, pointlessly, vast amounts of stuff from sub-articles William M. Connolley (talk) 20:52, 24 March 2012 (UTC)

I agree. Further, the article is supposed to be about the polarization of a light quantum, not the collective polarization. Kbk (talk) 02:30, 5 September 2013 (UTC)

Linearly polarized photons?

Isn't this claim false: "Individual photons are completely polarized. Their polarization state can be linear or circular, or it can be elliptical, which is anywhere in between linear and circular polarization." I don't think an individual photon can have any polarization state other than left or right circular. Am I mistaken?

From Mathpages.com: "...each absorption of a photon, imparts either +h or -h to the absorbing object, so if the intensity of a linearly polarized beam of light is lowered to the point that only one photon is transmitted at a time, it will appear to be circularly polarized (either left or right) for each photon, which of course is not predicted by classical theory."--Srleffler (talk) 06:11, 21 October 2013 (UTC)

I completely agree! You cannot have linearly polarized photons. They are either completely circularly polarized: right or left. According to the current theory which is referenced by Srleffler, there is no such thing as a linearly polarized photon. --Srodrig (talk) 07:18, 12 December 2013

(1) The state space is a Hilbert space. All state vectors in the Hilbert space are possible, not just the two eigenstates Φ_R and Φ_L of angular momentum. Saying "there are only two states" is just a manner of speaking; in QM one often speaks of the eigenvectors of some observable as if they were "alternatives", but this is not the whole story.

The photon state can also be any superposition of the two chiral states Φ_R and Φ_L. For example, if they are added with equal magnitude, then the photon will be linearly polarized, meaning that it is an eigenstate of an observable that measures linear polarization (for example, a linearly polarizing filter, corresponding to a rank-one projection operator). This is also a pure state. It persists. It is only when we measure angular momentum that we induce the state to jump to Φ_R or Φ_L.

Indeed, if you ask "is the polarization right or left circular", you'll always get one of these two possibilities as an answer. This gives the impression that a photon must always be circularly polarized. Like the guy who was asked "when did you stop beating your wife", the photon appears to be trapped by the excluded middle. But by the same token, if you ask "is the polarization vertical or horizontal", you'll also only ever get one of these two cases. This suggests a photon must always be linearly polarized.

The "paradox" has a particularly strong grip on us because both bases, the chiral Φ_R and Φ_L as well as the linear Φ_x and Φ_y, are intuitively very natural, yet their projection operators don't commute. The bases are aslant to each other. In fact, the bases are maximally dissimilar, somewhat like the position and momentum bases for state vectors of the Schroedinger equation.

Finally one should note that the predicate "is circularly polarized" cannot even be expressed in the quantum-mechanical language (although we may use it in the metalanguage) because it does not correspond to a linear subspace of the state space C^2. In fact, it corresponds to the union of two linear subspaces, just as the predicate "is linearly polarized" corresponds to the union of an S^1 of linear subspaces (it is the cone on the Clifford torus). In both cases, the span is all of C^2 which is why the question "how are you circularly polarized (resp. linearly polarized)" always gets an answer.

(2) If the value of angular momentum can be h, 0, or -h, as stated by 144.131.117.96 and Ricieri above, then the Hilbert space must have complex dimension at least 3. But it is complex dimension 2 in the article (it is the Jones vectors), and the angular momentum operator is given explicitly; its eigenvalues are h and -h.

This point is rather puzzling for me, since I too have heard that 0 is a possible eigenvalue. It seems that for photons of spin h, 0, -h, we really need to be looking at the three-dimensional irreducible representation C^3 = R^3 ⊗ C of Spin(3)=SU(2). It seems that Jones vectors resemble the two-dimensional representation C^2, which is more properly used for fermions with spin -h/2, h/2. This is very confusing.

Perhaps the explanation is that there is an underlying wave-vector k in R^3 which is fixed for the duration of the article, but left implicit; that is the element of R^3 that is acted on by the full representation.

But then what is the full quantum-mechanical state space? Is it a tensor product of R^3 (direction) with C^2 (spin and polarization)? What is the full symmetry group? Is it Spin(3) acting simultaneously on k in R^3 and the Jones vector Φ in the spin space C^2? To fully understand angular momentum of photons, we need to know this. This math is missing from the article (and as far as I can tell, it is not even alluded to), which explains why we can't understand this matter of -h, 0, h properly at this point.

178.38.18.115 (talk) 07:19, 27 June 2015 (UTC)

Here's my take on it:

---

The sinusoidal 'waves' of photons are not actually waves... they're spirals. http://staff.washington.edu/bradleyb/spiralsynth/fig3.1.gif

The image above shows the real (cosine... labeled 'Re' in the image) and imaginary (sine... labeled 'Im' in the image) components of an electromagnetic 'wave'. When viewed in line with its direction of travel, it will appear to be a circle, and when viewed orthogonal to its direction of travel, it will appear to be a sinusoid, when in reality it's a spiral.

This is because a sinusoid is a circular function. https://i.imgur.com/zofvpkI.png

You'll note the peak amplitude of the sinusoid is analogous to the radius of the circle, the peak-to-peak amplitude is analogous to the diameter of the circle, and the frequency of the sinusoid is analogous to the rotational rate of the circle. You'll further note the circumference of the circle is equal to 2 pi radians, and the wavelength of a sinusoid is equal to 2 pi radians, so the wavelength of the sinusoid is analogous to the circumference of the circle.

Thus the magnetic field and electric field (oscillating in quadrature) of a photon is a circle geometrically transformed into a spiral by the photon's movement through space-time. This is why all singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless and hence having no rest frame, which precludes their exhibiting the third state expected of a spin-1 particle (for a spin-1 particle at rest, it has three spin eigenstates: +1, -1, 0, along the z axis... no rest frame means no 0-spin eigenstate). A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly or elliptically polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

---

— Preceding unsigned comment added by 71.135.44.22 (talk) 19:43, 29 September 2019 (UTC)

Source for uncertainty relation in "Application to angular momentum"

It would be great to have a citation or reference for the uncertainty relation in the section "Application to angular momentum": $\Delta {\hat {l}}_{z}\,\Delta {\theta }\geq {\frac {\hbar }{2}}$.

I just briefly searched the web and some books on quantum optics. It seems that this kind of uncertainty relation, relating the polarization angle and angular momentum, is not very common. So, it would be nice to elaborate a bit on that or add a reference for better understanding. I would be interested in some explanations or derivations myself. How is the general relation in 3D?

--Schiefesfragezeichen (talk) 08:53, 24 June 2014 (UTC)

Dear Schief, I would also like to know this, thanks for asking. 178.38.18.115 (talk) 07:20, 27 June 2015 (UTC)

I do not have a particular source for this relation, but you need to realise that you can write uncertainty relationsships like this for any couple of canonically conjugated variables. The one with x and p is only one (and not even a special) case, in quantum mechanics you derive p=d/dx (this should be a partial derivative), i.e. the momentum is proportional to the derivative with respect to position, which makes them conjugated. Similarly, E=d/dt (time evolution, Schrödinger equation), so energy and time build another uncertainty relation, and in the same way L=d/dAngle. WikiPidi (talk) 14:18, 9 March 2020 (UTC)

zero polarization possible?

See http://www.dailymail.co.uk/sciencetech/article-3644747/Does-Nasa-s-fuel-free-thruster-invisible-exhaust-New-theory-explain-EmDrive-man-Mars-10-weeks.html — Preceding unsigned comment added by 90.154.68.99 (talk) 18:33, 23 June 2016 (UTC)

Dirac dubious tags

The long quotation from Dirac's book was tagged dubious in two places in this diff: [1]. The reasons for the tags are given in the tags, where it is unlikely to be seen by readers (they are visible in the diff). It seems to me that this is WP:OR. If Dirac needs correcting it should be in the article and cited from a published reliable source. If we feel the Dirac quote is too misleading and we don't have a way to include a clarification, we can delete the quote. Either way, the tags should go.--agr (talk) 15:04, 27 December 2017 (UTC)

To clarify this: It is correct that Dirac wrote that. However it is commonly believed that this is one of the (very few) mistakes Dirac made. There is a theory which is supported by experiments that two different photons may interact. Moreover the property of "begin different" is not such a clear dichotomic property as we might want to believe. — Preceding unsigned comment added by 217.95.166.99 (talk) 22:48, 24 March 2019 (UTC)

Is this right?

I am new at this, so I'm not confident.

If unit vectors are defined such that

$|\mathrm {R} \rangle \ {\stackrel {\mathrm {def} }{=}}\ {1 \over {\sqrt {2}}}{\begin{pmatrix}1\\i\end{pmatrix}}$

and

$|\mathrm {L} \rangle \ {\stackrel {\mathrm {def} }{=}}\ {1 \over {\sqrt {2}}}{\begin{pmatrix}1\\-i\end{pmatrix}}$

then an arbitrary polarization state can be written in the "R-L basis" as

$|\psi \rangle =\psi _{\rm {R}}|\mathrm {R} \rangle +\psi _{\rm {L}}|\mathrm {L} \rangle$

where

$\psi _{\rm {R}}=\langle \mathrm {R} |\psi \rangle ={\frac {1}{\sqrt {2}}}(\cos \theta \exp(i\alpha _{x})-i\sin \theta \exp(i\alpha _{y}))$

and

$\psi _{\rm {L}}=\langle \mathrm {L} |\psi \rangle ={\frac {1}{\sqrt {2}}}(\cos \theta \exp(i\alpha _{x})+i\sin \theta \exp(i\alpha _{y})).$

Are the signs reversed here?

Jethomas5 (talk) 08:56, 30 August 2022 (UTC)