# Photon Spin & Polarization

• I
Homework Helper
Gold Member
2022 Award
TL;DR Summary
Relationship between photon spin and polarisation.
In Griffiths Elementary Particles (2nd, revised edition) there is a footnote on page 241, which states that the photon states with ##m_s = \pm 1## are related to the polarization vector by:
$$\epsilon_+ = \frac 1 {\sqrt 2} (-1, -i, 0) \ \text{and} \ \epsilon_- = \frac 1 {\sqrt 2} (1, -i, 0)$$
But, he doesn't give any justification for this. How do we relate these spatial polarisation vectors to eigenstates of the relevant angular momentum or helicity operator?

Last edited:

Homework Helper
Gold Member
Those are polarization vectors for right and left circular polarized waves.

• StenEdeback
Homework Helper
Gold Member
2022 Award
Those are polarization vectors for right and left circular polarized waves.
Yes, I know. But aren't they also spin eigenstates?

Gold Member
2022 Award
They are helicity eigenstates. Massless fields are different wrt. spin-like degrees of freedom. For spin ##s## there are only 2 helicity-degrees of freedom rather than ##(2s+1)## for nassive particles. The reason is to be found in the analysis of the unitary reps. of the Poincare group. See Weinberg, QT of fields, vol. 1.

Yes, I know. But aren't they also spin eigenstates?

No, because photons have no spin. But they are eigenstates of helicity.

• dextercioby and vanhees71
Homework Helper
Gold Member
2022 Award
No, because photons have no spin. But they are eigenstates of helicity.
How do we show that?

What is the helicity operator for photons?

How do we show that?

What is the helicity operator for photons?
For any massless particle, the helicity operator is the Lorentz generator $J_{12}$ of the stability little group* $G_{p}$ of $p^{\mu} = E (1,0,0,-1)$, with $E>0$.

* $G_{p} = \big\{ \forall \Lambda \in SO(1,3) | \ \Lambda p = p \big\}$

• dextercioby, vanhees71 and PeroK
Homework Helper
Gold Member
2022 Award
For any massless particle, the helicity operator is the Lorentz generator $J_{12}$ of the stability little group* $G_{p}$ of $p^{\mu} = E (1,0,0,-1)$, with $E>0$.

* $G_{p} = \big\{ \forall \Lambda \in SO(1,3) | \ \Lambda p = p \big\}$
That explains why I couldn't figure it out myself. Thanks.

How do we show that?

What is the helicity operator for photons?
Spin is the angular momentum in the rest frame of a particle. As a photon cannot be brought to rest, we can't define its spin.

• 