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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?

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    $\begingroup$ intrinsic parities are mostly conventional, unrelated to multipole radiation type $\endgroup$
    – Tanmoy Pati
    Commented Dec 31, 2023 at 9:12
  • $\begingroup$ "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 $\endgroup$
    – anna v
    Commented Dec 31, 2023 at 10:13
  • $\begingroup$ @annav This is only for E1 or electric dipole type radiation. This is not the intrinsic parity of photons I think. $\endgroup$
    – Sagar K. Biswal
    Commented Dec 31, 2023 at 10:57
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    $\begingroup$ 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. $\endgroup$
    – anna v
    Commented Dec 31, 2023 at 13:36

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The parities of E1 and M1 type radiations are the parities of radiation states. These are inferred from various atomic and nuclear transition selection rules.

They have nothing to do with parity of a photon vector field $A^\mu$, which is of course negative.

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    $\begingroup$ The total parity is $(-1)^J$ where $J$ is the total angular momentum of the final state. This includes the photon parity. $\endgroup$
    – my2cts
    Commented Jan 2 at 20:04
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The parity of a the electromagnetic potential is -1, as it is odd under inversion. Therefore the parity of a photon is -1 as well. However an electromagnetic field can also have orbital angular momentum. If this is also odd, the overall parity is +1.

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