# Photon Spin & Polarization

• I
• PeroK
In summary, the conversation discusses the polarization vectors of photons with ##m_s = \pm 1## and their relationship to helicity eigenstates, which are different from spin eigenstates due to the massless nature of photons. The helicity operator for photons is the Lorentz generator J_{12} of the stability little group G_{p} of p^{\mu} = E (1,0,0,-1), with E>0. It is explained that spin is the angular momentum in the rest frame of a particle, which cannot be defined for a photon due to its inability to be brought to rest. Helicity, on the other hand, is the component of angular momentum in the direction of the momentum.

#### PeroK

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

StenEdeback
Meir Achuz said:
Those are polarization vectors for right and left circular polarized waves.
Yes, I know. But aren't they also spin eigenstates?

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.

PeroK said:
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
DrDu said:
No, because photons have no spin. But they are eigenstates of helicity.
How do we show that?

What is the helicity operator for photons?

PeroK said:
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
samalkhaiat said:
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.

PeroK said:
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.

"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.

PeroK

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