Understanding the Relationship Between Photon Spin and Polarization Vectors

[PeroK]

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?

 

[Meir Achuz]

Those are polarization vectors for right and left circular polarized waves.

 

[PeroK]

Those are polarization vectors for right and left circular polarized waves.

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

 

[vanhees71]

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.

 

[DrDu]

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

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

 

[PeroK]

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

How do we show that?

What is the helicity operator for photons?

 

[samalkhaiat]

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 $E>0$.

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

 

[PeroK]

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.

 

[DrDu]

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.

 

[Meir Achuz]

"Spin is the angular momentum in the rest frame of a particle "
That is a property of spin for a massive particle, not a definition.
Helicity is the component of angular momentum in the direction of the momentum.
It's just a quibble over words.