When learning angular momentum in quantum mechanics, a spin 1 particle have 3 states. Then I saw from sakurai's modern quantum mechanics that photon's two polarization are just like spins, but with several differences. I looked up on the web, and find that: first, the zero mass of photon is the reason that when calculating spins we can not choose a rest frame, which will induce the spin of photon have only 2 values instread of 3. Second, there is a paragraph says that actually photon does not have spin at all, it only have helicity, which is the projection of spin on the propagation direction. And there is this concept "chirality" which is connected with circularly polirization of light. I got confused with relations of this 4 concept of photon: polarization, spin ,helicity, chirality. Can anybody tell me about it?


Consider a one-particle state of a relativistic quantum field theory, and let this state be an eigenstate of the 4-momentum operator, $\hat P^\mu |p^\mu\rangle = p^\mu|p^\mu\rangle$. Other than the 4-momentum, what other quantum numbers can the state have, and how should they transform? That is, if there are any quantum numbers collectively labeled by $s$, then for a Lorentz transformation $\Lambda$, $$\Lambda |p^\mu,s\rangle = \sum_{s'} D_{ss'}|\Lambda^\mu{}_\nu p^\nu, s'\rangle $$ what can we say about the quantities $D_{ss'}$?

There are two distinct classes of physical 4-momenta: $p^2 = m^2 > 0$ and $p^2 = 0$, both with $p^0 > 0$. We can pick standard 4-momenta for these classes, say $p^\mu =(m, 0,0,0)$ and $p^\mu = (\omega, 0,0,\omega)$. Any 4-momentum in a class can be boosted to this standard momentum. Physically, we can go to a restframe of a massive particle, or align the propagation of a massless particle with the $z$-axis.

The conditions described above do not fix the frames completely. In the massive case, there is still $SO(3)$ freedom. A rotated rest frame is still a rest frame. In the massless case you can rotate around the $z$-axis, but you can also do null-rotations around $(\omega, 0,0,\omega)$. The group that preserves the standard momentum is called the little group.

With some algebra, which you can find in full in [Weinberg], Chapter 2, one can work out that the quantities $D_{ss'}$, up to a normalization factor, have to form a representation of the little group. The representations of $SO(3)$ (actually, of its double cover $SU(2)$) are familiar: for each representation you can use the eigenvalue of $J_z$ as a quantum number; this is the spin.

The massless little group has three generators, but two of those have a continuous spectrum. Because no continuous quantum number other than 4-momentum is observed, we ignore them (discard states not in their nullspace). The third generator is $J_z$. An irreducible representation of a one-dimensional group is characterized by the eigenvalue of the single generator; this is the helicity. This is what people mean when they say that because the photon is massless you can only measure $J$ along its momentum. It is also what is meant by that photons have helicity, not spin.

It seems like I have argued that the photon field has only one polarization. But note: helicity is the component of $J$ along the momentum, $J$ is a pseudovector, the momentum is a vector, hence helicity changes sign under parity. Therefore, in a parity-invariant theory like QED you have to include both the $+1$ and $-1$ helicity photon fields. The representations do not have to be irreducible, so no one can stop us from thinking about a photon field with two polarizations.


  • [Weinberg] Weinberg, S. The Quantum Theory of Fields, Volume 1 (Cambridge University Press, 2005)
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  • $\begingroup$ I was looking for a quick summary of the mathematical formulation of photons helicity, and this excellent quick summary of Weinberg by @RobinEkman was just what I was looking for -- thanks! I will note one quick point that I fear sometimes gets overlooked in many physics formulations, which is that sometimes the deepest answer is not the math, but rather "because that is what we see experimentally." So, why really do photons have only two discrete helicity states?: "Because no continuous quantum number other than 4-momentum is observed, we ignore them (discard states not in their nullspace)." $\endgroup$ – Terry Bollinger May 28 '16 at 12:34

Indeed, the photon doesn't have spin, what it has is POLARIZATION. For a photon emitted from an atom the polarization is CIRCULAR. That means, if we look in a fixed plane perpendicular on its motion, we see that the electric vector does not have a fixed position, it rotates.

If so, if the electric vector rotates counter-clockwise the circular polarization is RIGHT, and if the electric vector rotates counter-clockwise the circular polarization is LEFT. You can see this in Wikipedia


Now, about what you saw, when the photon is right-polarized, it behaves as if carrying an angular momentum of ħ ALONG the direction of flight. When left-polarized it behaves as if it carries an angular momentum of -ħ along the flight direction.

Thus, when in an atom the electron jumps from a level with, say ℓ = 1 to a level with ℓ = 0, a photon is emitted with right-polarization, i.e. it carries the difference of ħ in linear momentum.

So, the circular polarization of the photon is another phenomenon than, e.g. the spin of an electron. You see, the circularly polarized photon has a rotating electric field, while the electron has no such thing.

Now, chirality is the property of an object to be non-identical with its image in a mirror. If we send a photon to a mirror, its image will look as having opposite helicity.

I hope it helps.

Good luck.

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  • $\begingroup$ Thanks a lot. So when we talk about helicity of photon, we are talking about the projection of its polarization on the propagating direction, right? $\endgroup$ – BNHSX Dec 3 '14 at 21:33
  • $\begingroup$ No, you'll understand better if you look at the animated picture in Wikipedia, address: en.wikipedia.org/wiki/… The electric vector rotates in a plane PERPENDICULAR on the direction of motion. When the photon comes TOWARD YOU, and the electric vector appears to rotate against the clock, this is right polarization. Well, the right polarization is EQUIVALENT to an angular momentum ALONG the direction of motion. In short, polarization is PERPENDICULAR to the direction of motion, the angular momentum is ALONG the direction of motion. $\endgroup$ – Sofia Dec 3 '14 at 21:56

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