# I Why does a photon have parity -1

I am wondering what would be an experiment demonstrating that photon parity is -1. It also occurs to me that one might deduce the parity from Maxwell's equations, though that might be a bit of a stretch since they are classical of course. Also, it occurs to me that parity might be assigned a value to -1 as merely a convention. So, all in all, I am in the dark about this.

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Thanks, Fred. I read the paper you mention. It describes a method of measuring parity, but does not give the result of the measurement, right? (I found this paper rather advanced for my background. Perhaps you can mention some preparatory reading?)

Thanks, that is a very helpful set of notes. It seems to say that photon parity is a matter of convention, not determined by experiment. Also that it is sometimes 1 and sometimes -1 and that it is really not defined since {P,$\vec{p}$} = 0 and the photon always has nonzero momentum. I tried to compute {P,$\vec{p}$} and got stuck on P$\vec{p} e^{i\vec{p}\cdot \vec{x}}$. Couldn't decide whether or not to change the sign of $\vec{p}$.

#### vanhees71

Gold Member
There are aspects of parity for particles other than photons that are a bit more complicated than envisaged in the OP.

First of all parity emerges, as most fundamental things in modern physics, from a symmetry principle, namely that of spatial reflections (usually in the sense of reflections of the spatial components of the position vector at the origin of cartesian coordinate system). Then you describe the particles by relativistic quantum fields, building up a representation of the proper orthochronous Poincare group augmented by space reflections.

Now if particles ought to have a specific parity in the strict sense the theory should have spatial reflections as a symmetry. For the hadrons, which can be described in the low-energy realm by an effective (chiral) theory of strongly interacting hadrons (baryons and mesons). The strong interaction is P (even CP) invariant (which is a puzzle in itself) and thus as long as only strong interactions are taken into account parity is a well-defined quantum number of all hadrons. There's even an approximate spontaneously broken symmetry between even and odd parity states with the pions as the pseudo-Goldstone bosons of this broken symmetry. Now since on the fundamental level of QCD the spontaneous breaking is due to the vacuum expectation value of the scalar order parameter $\langle \bar{\psi} \psi \rangle$ (with $\psi$ describing up and down quarks an their antiparticles), the quark condensate. The Goldstone modes are necessarily the pseudoscalar modes, i.e., $\mathrm{SU}(2)_{\text{L}} \times \mathrm{SU}(2)_{\text{L}}$ is spontaneously broken to the iso-vector symmetry $\mathrm{SU}(2)_{\text{V}}$. Thus the pseudo-Goldstone modes, providing an isotriplet of mesons, must be described by pseudo-scalar fields, and all three pions must have the same parity (-).

Now naively there's also an Abelian axial symmetry, $\mathrm{U}(1)_{\text{A}}$, if you look at the "classical" Lagrangian, but that symmetry is broken by an anomaly, particularly also when taking into account the electromagnetic interaction (which also breaks isospin symmetry by the way). That's why the decay of the neutral pion into two photons is much larger than expected. The decay into 3 photons is highly surpressed if present at all. According to the PDG the partial decay ratio to $3 \gamma$ is $<3.1 \cdot 10^{-8}$. Within the standard model it doesn't decay to 3 photons, because this would violate C (charge-conjugation) symmetry of the electromagnetic interaction.

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#### vanhees71

Gold Member
As long as you consider only strong interactions, Quantum Chromodynamics (QCD), the fundamental theory of the strong interactions within the standard model, tells us that chiral symmetry is a very good approximate symmetry in the light-quark sector (u, and d quarks, to some extent even also including s quarks). Hadron phenomenology tells us that chiral symmetry is not only slightly explicitly broken due to the non-zero light-quark masses but also spontaneously, i.e., there's a mass splitting of chiral partners due to the formation of the "quark condensate", i.e., the fact that the vacuum expectation value $\langle \bar{\psi} \psi \rangle \neq 0$. This leads to the PCAC theory, and it proves to be an excellent description of low-energy QCD in terms of effective hadronic QFT models.

Also the electromagnetic decay of the neutral pion fits very well into the picture, as soon as the chiral anomaly (i.e., the symmetry under $\mathrm{U}(1)_{\text{A}}$ transformations of the classical theory is broken due to quantization). It mostly decays to two photons or a photon and an electron-positron pair, all in very good agreement with the standard model.

Together with the electroweak theory you can systematically map the fundamental currents built by the quark fields (quark-antiquark pairs, three-quark states to get mesons and baryons, repsectively) to the effective hadronic models. Then fitting some phenomenological couplings (like the pion-decay constant) leads to a very good description of low-enery hadron physics. For a very nice introduction, see

https://arxiv.org/abs/nucl-th/9706075

#### PeterDonis

Mentor
Moderator's note: Some posts having to do with personal speculation, and responses to them, have been deleted.

"Why does a photon have parity -1"

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