There are on this site a few questions about photons and circular polarization, but none of them give satisfactory answers:

Connection between spin angular momentum of a photon and circular polarization of light

I have read this question:

What does a circularly polarized electromagnetic plane wave look like in a co-rotating reference frame?

Where annav says:

This illustration explains how the photons, which can only have spin +1 or -1 to their direction of momentum, build up a polarized beam

enter image description here

where Sean E. Lake says:

"As there exist no circularly polarized photons" That seems incorrect, since circularly polarized photons are photons of definite helicity, which is just spin measured along the direction or propagation. The different polarization states would correspond, in principle, to measuring spin along different axes than the propagation one, I think.

And from wiki:

In the quantum mechanical view, light is composed of photons. Polarization is a manifestation of the spin angular momentum of light. More specifically, in quantum mechanics the direction of spin of a photon is tied to the handedness of the circularly polarized light and the spin of a beam of photons is similar to the spin of a beam of particles, such as electrons.[12]


An individual photon can be described as having right or left circular polarization, or a superposition of the two.


Now this one specifically states that single photons do have polarization and can have circular polarization:

Does polarization happen with single Photon?

Now these are two different explanations, as the former describes photons as QM entities, that on their own can only have spin of 1 or -1, that's it. In this description, only the confluent classical EM wave, being built up by a large number of photons, can have circular polarization.

But the latter describes photons as QM entities, and even single photons as having circular polarization of their own.

The closest to this topic, I have found in another question on this site, which describes orbital angular momentum:

What is the orbital angular momentum (OAM) of individual photons?

This describes single photons as having OAM (in addition to spin or helicity), but one of the answers says it does exist for single photons, it is just hardly measurable. The other answer states the opposite, saying that since photons are in potential wells, and not orbits, there is no OAM for single photons.

So there are two completely opposing views, these are:

  1. single photons are QM entities, but all they can have is simply a spin of 1 or -1, that's it, single photons cannot have circular polarization on their own, only the classical EM wave they build up can have it

  2. single photons are QM entities, and even so they can possess superposition of polarization (circular polarization is a superposition of linear)


  1. Can a single photon have circular polarization?

4 Answers 4


I do not understand why you suggest there is a contradiction. The possible polarization outcomes are eigenvalues of operators, and this does not depend on the choice of basis, so if only the eigenvalues $\pm 1$ and $0$ are possible in one basis then only those will be possible in any other basis.

In this perspective polarization is basically spin measured in a non-Cartesian basis. The fact that it’s a complex combination is no more strange than measuring spin along some arbitrary direction, with a spin wave function given by some complex combination of the $\vert \pm z\rangle$ spin states.

Back to polarization: if you take a linearly polarized photon and pass it through a circular filter, then it will emerge as a circularly-polarized photon or it will not emerge at all.

  • 1
    $\begingroup$ There's actually a relativistic argument that eigenstate $\vec\sigma\cdot\vec p =0$ is forbidden for massless particles like photons. This makes photon spin behave like a two-state system. The linear polarization states are the two pairs of orthogonal linear combinations of the two circular-polarized spin states with $\vec\sigma\cdot\vec p = \pm 1$. That, and the selection rule that photons can't be emitted from transition between spinless states $(0\to0)$, suggest a model where all photons are at least born with circular polarization. $\endgroup$
    – rob
    Sep 23, 2020 at 21:12

For light travelling in some direction a photon can have spin either clockwise (+1) or anti clockwise (-1) around a ray in said direction. After passing a circular polarization filter any photon will have such defined spin. Any photon which didn’t pass had the other spin. In general a photon could be in a mixture of the two possible states. An equal mixture will give linear polarization (the transverse E field direction depending on the phase relation between the 2 circular states). An uneven mixture can give an elliptical polarised wave. A mixed photon will have a certain probability to be either one or the other and therefore may or may not pass a circular Polaroid with said probabilities. Note a spin +1 photon on passing a linear Polaroid will only pass with 50 % probability. And the emerging photon now has mixed spin +1 and also spin -1 with well defined phase relation. The apparent jump of spin from +1 to -1 for some of the photons is typical quantum measurement uncertainty phenomena. Wierd but unavoidable.


If one agrees that EM radiation is produced exclusively by excited subatomic particles, then one must also agree that EM radiation always consists of photons.

Each photon of the EM radiation has exactly the properties that are generally attributed to all EM radiation according to Maxwell. In fact, the two components of EM-radiation are measurable at radio waves.
The acceleration of the electrons at the antenna rod generates an electric field directed along the antenna rod. The magnetic field induced thereby is perpendicular to it.

If one uses the second and third finger to represent the two field components and the thumb to indicate the forward motion of the photon, then this directional arrangement is always the same for all photons emitted by electrons.
But there is a second chirality, corresponding to the right and left hand or a right and left oriented coordinate system. These two orientations correspond to photon spin.

It is possible to set a photon in rotation. This can happen, for example, during the transition between two media.A coordinate system - where Z describes the direction of the photon - rotates around Z in the case of circular polarization, i.e. the E and B fields rotate together around Z.

Compact answer: A single photon can be made to rotate around its axis of motion. The photon is then circularly polarized.


Try out this metaphor.

Quantum voter theory says that voters are either Republicans or Democrats. The way you test whether a voter is a Republican or a Democrat, is by having an election and see which way he votes. If he doesn't vote for either one then he is not a voter. So there are only two states.

The very act of having an election can cause voters to change state. If their party wins they are likely to be dissatisfied with their government afterward and switch state. It is theorized that the opposite could occur. The other party wins and they like the outcome so much that they change state. Experiments are underway to find examples of this, using multi-billion-dollar propaganda campaigns. This voter may be statistically proven to exist any year now.

It could be theorized that voters have a voting spin state, they could be for example 72% likely to vote Republican and 28% likely to vote Democrat. This interpretation is compatible with the available evidence.

Which theory to use? Both of them fit the data. So use the one that better meets your needs. If voters are in fractional states, then it may make sense to try to move their state a little to get more voters into the state you prefer. If voters are mostly fixed in one of two states, then try to get more voters of one type to vote, and persuade more voters of the other type not to vote. If you are more interested in state change use one theory, and if you are more interested in turnout use the other.

Just don't use them both at the same time because they don't fit together.



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