# What is photon spin and polarization?

I have literally googled for hours on this subject. I am very curious on what spin really is and how polarization works for photons.

I’ve read that photons have a spin of 1(no idea what that means or how an electron can have a 1/2 spin…).

I’ve read two completely different definitions of photon spin: 1 says that a photon will spin parallel or aniparallel with respect to it’s direction of propagation. 2 says that it spins counter-clockwise or clockwise(left or right) with respect to its propagation direction. Which of these two is true?

I’ve read that linear polarization is horizontal or vertical, but can’t it be at any angle? The electric field could oscillate an any angle can’t it?

Also, I’ve read that linear polarization requires at least two photons with opposite spins, I don’t know if that’s true or not. Is circular polarization even real or is it made of a bunch of out of phase linearly-polarized photons that make it appear to spin on a graph?

In short: I’m not smart and I’m finding so many complicated and conflicting answers online. I’ve also had someone tell me that a photon doesn’t have an electric field and magnetic field, even though my research shows the complete opposite.

Thank you so much.

I’ve read that photons have a spin of 1(no idea what that means or how an electron can have a 1/2 spin…).

There is no easy way to explain this one. The spin expresses several things, but one is a limit on the maximum angular momentum that a plane wave can have. The limit can be zero (no angular momentum, spin-0), or half the energy divided by the wavelength (spin ½), or the energy divided by the wavelength (spin 1), etc., and for subtle reasons, intermediate ratios aren't possible. This property of waves is not quantum-mechanical, though people sometimes say it is. See this answer which is somewhat technical (undergraduate-physics-major level).

When you add quantum mechanics, you get particle-like behavior of the field, and each particle carries a certain amount of the angular momentum.

I’ve read two completely different definitions of photon spin: 1 says that a photon will spin parallel or antiparallel with respect to it’s direction of propagation. 2 says that it spins counter-clockwise or clockwise(left or right) with respect to its propagation direction. Which of these two is true?

The second one is better. The first one is saying that the axis of the rotation is parallel to the direction of propagation of the field. The angular momentum of the light is a vector pointing along the axis, and by an arbitrary convention, one direction of rotation (clockwise or counterclockwise) is assigned to each direction on the axis. So one circular polarization has an angular momentum pointing in the direction of propagation and for the other it points opposite the direction of propagation.

I’ve read that linear polarization is horizontal or vertical, but can’t it be at any angle?

Yes, it can be at any angle. There are also elliptical polarizations, which are intermediate between linear and circular, and can also be tilted at any angle.

What your sources may have been saying is that from sums of two polarization directions, such as horizontal and vertical, you can construct any polarization direction. If you add them in equal amounts with the same phase (so they reach zero at the same time), you get 45° diagonal polarization. If you add them out of phase by the right amount, you get circular polarization:

and if you add them out of phase by a different amount, you get an elliptical polarization.

Also, I’ve read that linear polarization requires at least two photons with opposite spins

That's wrong. A single photon can be linearly or elliptically polarized.

I’ve also had someone tell me that a photon doesn’t have an electric field and magnetic field

Photons are the electromagnetic field. They don't have a field in the way that, say, electrons do. Field lines don't end on photons. The photons just are the field.

• Thanks so much for your long and thorough reply. I’m curious, when you said that the “second one” is better(with spin direction), do you mean that if a photon moves along the Z axis, it spins counter clockwise or clockwise? Or did you mean that it’s only conventional to use that term? Or does the wave literally spin in or against the direction it’s propagating? Also, vertical and horizontal polarization are the only ones ever mentioned..can’t photons have a 45 degree angle or 34 degree angle, etc? I only ever see 90 degree differences mentioned(horizontal or vertical). Thank you again! Sep 16 at 23:36
• @theguineapigking I just mean that when the propagation direction is z, the rotation is in the xy plane. If that's what "spin parallel or antiparallel to its direction of propagation" means, then that description is fine too. Yes, the polarization angle can be 34°. But 34° polarization is a weighted sum of 0° and 90° polarization, so in some sense you only need 0° and 90° to get every polarization. That may be what your sources meant when they suggested there are only two polarizations, but without seeing them I can't be sure. Sep 17 at 19:34
• @theguineapigking Since you've asked another similar question, I'll try to clarify again. If the field propagates in the z direction, then it rotates in the xy plane. That is equivalent to saying that if the field propagates in the z direction, then it rotates around the z axis. That is what your sources were probably saying: the same thing in two different ways. I thought one way of saying it was less confusing than the other, but in hindsight, I probably should have said that both are correct. Sep 19 at 15:31
• Thank you. I posted again because the more I thought about it, the less it made sense. I wanted to know if the photon literally spins left or right when it’s traveling along the z axis. I kept reading (like on Wikipedia too) that it has a helicity which rotates or spins in the z direction or against it, like spinning forwards or backwards. So you’re saying it does indeed spin left or right, in a literal sense? I have Asperger’s so learning stuff is difficult. That’s why I may ask the same question in different wording to truly understand it. Thanks Sep 19 at 16:08
• @theguineapigking Yes, left or right. The rotation axis is in the z direction, but the axis is just a way of representing the rotation, it isn't really the direction of the rotation. In this image, the circle shows the direction of the motion, and the purple line is the vector that, by convention, is used to represent that circular motion. Sep 19 at 16:19

Regarding spin, nothing is spinning at the photon level. (You can have apparent rotating electric and magnetic fields when discussing polarization of classical waves, but we're not doing that).

If it's not spinning, what's going on? Well, it has angular momentum. A spin $$s$$ particle has total angular momentum:

$$S = \sqrt{s(s+1)}\hbar$$

which is a vector, and we can talk about which way it points, not which way something spins.

Because position and momentum don't commute: $$[x_i, p_j] = i\hbar \delta_{ij}\ \ \ \ \ \ (i,j\in (1,2,3))$$

we can't know them both at the same time. That affects:

$$\vec L = \vec X \times \vec P$$

(here $$\vec L$$ is orbital angular momentum, but it holds for spin too).

The commutation relations are:

$$\vec S \times \vec S = i\hbar\vec S$$

That's right, the cross product of a vector with itself is not zero.

That means we're only able to know the projection of the angular momentum onto a single axis. For general spin-1 particles or orbital states, we choose to call that axis $$z$$ so that the possible states look like:

The magnitude of the angular momentum is

$$||S|| = \sqrt 2 \hbar$$

and the projection onto $$z$$ is

$$S_z = s_z\hbar \ \ \ \ (s_z \in (-1, 0, 1))$$

In a lot of discussions, where we say "spin-1", it's important to remember that that is the projection onto an axis that is quantized, and the total spin itself if $$\sqrt 2$$, relative to $$\hbar$$.

A general spin eigenstate is written in terms of the spin quantum number, $$s \in (\frac 1 2, 1, \frac 3 2, 2, \ldots)$$, and the $$z$$-projection quantum number, $$s_z$$ as:

$$\chi = |s, s_z\rangle$$

For the photon, $$s=1$$, and $$s_z$$ cannot be zero (because of gauge invariance), so you can substitute:

$$|1, \pm 1\rangle \rightarrow |\pm 1\rangle$$

These two states are positive (negative) helicity eigenstates. With a propagating (in all frames) photon, the $$z$$-axis is not arbitrary, rather, it is defined by the photon's momentum $$\vec p$$. Thus the two positive (negative) helicity states are defined unambiguously by the projection of the angular momentum onto the momentum (with nothing spinning):

$$H = \frac{\vec S \cdot \hat p}{\hbar}$$

The positive (negative) helicity state is also called right (left) circularly polarized. It is common to write these two states as $$|R\rangle$$ and $$|L\rangle$$, for right and left circularly polarized.

Now once you have a two state system, all possible pure (mixed) states can be described as by point on (in) the Bloch Sphere. It could be a two level atomic system, or a spin 1/2 particle, or photon polarization. They all have the same allowed mathematical mixtures, it's just that the physical interpretation depends on the underlying system.

For example, for the spin 1/2 particle, and pure superposition of up and down lies on the Block sphere, and represents a spin eigenstate along a different axis.

Meanwhile, transverse polarizations of the photon are not possible, and the superpositions of $$|R\rangle$$ and $$|L\rangle$$ do not allow a spin pointing in any direction. Rather, see the left figure:

The poles are the helicity eigenstates. Note that the states on the equator, such as:

$$|H\rangle = \frac 1 {\sqrt 2}\big(|R\rangle +|L\rangle \big)$$ $$|V\rangle = \frac 1 {\sqrt 2}\big(|R\rangle -|L\rangle \big)$$

are linearly polarized (Here, horizontal and vertical...which are orthogonal).

Now the astute reader will note that $$|H\rangle$$ and $$|V\rangle$$ are 180 degree apart, while (anti)diagonal polarizations, are at 90 degrees relative to $$|H\rangle$$ and $$|V\rangle$$. Aren't we missing 180 degrees of support?

No.

Linear polarization is a tensor polarization. It does not define an axis, it defines an alignment, which is a line without a preferred direction, and that requires 180 degrees of support to cover all directions, err, alignments.

Now you don't need two photons to make linear polarization, but it is the superposition of two circular basis states. Nevertheless, the basis is arbitrary, as long as it is made of 2 antipodal points on the Bloch sphere, it's orthonormal and complete.

Finally, all the allowed complex linear combinations of the basis states make up circular, linear and elliptical polarizations, lying on the surface of the sphere.

Inside the sphere are mixed states (e.g., density matrices), with the center being totally unpolarized.

Finally: note that the Bloch sphere for light polarization is called the Poincare sphere.

The previous answers do not make clear how the concept of spin arose at the quantum mechanical level of particles molecules and lattices.

When experiments in the small dimensions showed that the atoms of Democritus were not following the rules of classical mechanics and classical electrodynamics the theory of quantum mechanics was developed to fit and be predictive of new data.For the smooth continuity of the mathematics between classical and quantum theories it was necessary that the conservation laws holding classically would be imposed on the mathematics of quantum theory .

This meant that when particle-particle interactions were studied with an experiment, energy, momentum and angular momentum conservation should hold as laws too. There was a problem for angular momentum conservation: in order to be true, the quantum particles had to be assigned a spin, that turns out to be the value of spins we see at the elementary particle table. These spins made conservation of angular momentum true at the quantum level.

So spins are the outcome of experiments, they came up quantized and the value of 1 or 1/2 to the particles are imposed in the theory in order for angular momentum conservation to hold at the quantum level.

See this wiki article for light, to see the complexity.