# Parity of Photons

In nuclear physics, while studying gamma decay (Nuclear physics, Roy and Nigam, 1st ed, pp 450) I have read that the parity of photons depends on the type of multipole radiation they represent. Means for electric type parity is $$(-1)^L$$. For magnetic type parity is $$(-1)^{L+1}$$. So, for E1 type radiation, parity is negative, and for M1 type radiation, parity is positive.

But in particle physics, I am reading (Intro. to elementary particles, Griffiths, revised 2nd ed, pp 141) that photons are vector particles with intrinsic parity -1. So, are we considering only E1 type radiation here? Or these parities are completely unrelated?

• intrinsic parities are mostly conventional, unrelated to multipole radiation type Commented Dec 31, 2023 at 9:12
• "Photons are emitted via atomic dipole transitions where $Δ L = ±1$ . Hence the atomic parity changes by (-1) during these transitions and for the overall parity of the system (atom + photon) to be conserved (electromagnetic interaction conserves parity) we must have that the photon has negative intrinsic parity." from lecture notes, looking up "photon" alpha.physics.uoi.gr/foudas_public/APP/Lecture8-Pion-Exp.pdf Commented Dec 31, 2023 at 10:13
• @annav This is only for E1 or electric dipole type radiation. This is not the intrinsic parity of photons I think. Commented Dec 31, 2023 at 10:57
• It is the reason the intrinsic parity assigned to standard model photons is -1. Photons in the standard model have only a four vector with zero mass, spin +/- 1, they do not radiate, they are point particles. Dipoles etc come from confluence of many photons and then it is a parity of a system of particles, not one photon. Commented Dec 31, 2023 at 13:36

They have nothing to do with parity of a photon vector field $$A^\mu$$, which is of course negative.
• The total parity is $(-1)^J$ where $J$ is the total angular momentum of the final state. This includes the photon parity. Commented Jan 2 at 20:04