# Why can a neutral pion decay into 2 photons but not one? [duplicate]

Hi Sorry if this is a simple question I just can't seen to get my head at what law is being violated due to a neutral pion decaying into a photon (I have an inclining its to do with 4-vector momentum however that further makes the neutral pion decaying into 2 photons less sense to me.) If anyone could help that would be great, thanks!

• Possible duplicate, and links therein.
– rob
May 24 '21 at 18:49
• @rob: I must agree that's a dupe. I considered typing a worse version of the top answer to that question before checking the link. May 25 '21 at 4:14

The pion has a rest frame where it has 0 momentum. Photons do not have rest frames, so if the pion decays into one photon, that photon will have momentum in every frame. So momentum cannot be conserved. With two photons, the net momentum between them can be zero, and so momentum conservation in the pion's rest frame is possible.

• Thank you very much that makes so much sense now! May 24 '21 at 19:15
• Incidentally "Photons do not have rest frames" isn't a prerequesite for this. Even if you can assign a rest frame to a photon it doesn't work because you can't balance both energy and momentum at the same time. May 25 '21 at 4:14
• Yes, @Joshua. The question asked for a reason, so I thought this one was particularly simple. As other answers are adding now, there are a variety of reasons this particular decay is forbidden. May 25 '21 at 12:36

Why can a neutral pion decay into 2 photons but not one?

It is due to the law of conservation of energy and momentum. A particle with a given mass cannot "decay" to only one smaller mass particle because the decay to one particle at the center of mass would not conserve energy, unless it was in motion, which would not conserve momentum. It is more true for this case, as the photon has zero mass and a single photon does not have a center of mass as it always travels at energy c.

Besides energy-momentum conservation, there are other conservation laws that a $$\pi^0\to\gamma$$ transition would violate: the conserved discrete quantum numbers $$J^{PC}$$ of the neutral pion are $$0^{-+}$$ (it is a pseudoscalar of positive charge parity, the signs are shorthand for $$\pm1$$) whereas those of the photon are $$1^{--}$$ (it is a vector of negative charge parity). While the weak interaction violates $$PC$$-conservation, the intrinsic angular momentum $$J$$ is conserved by all known interactions, and thus the transition is strictly forbidden by conservation laws even taking the weak interaction into account.

I think it makes sense to briefly point out how the decay into two photons is possible and how these conservation laws limit the possible configurations in the final state. It is an electromagnetic decay so all of the quantum numbers $$J^{PC}$$ are conserved separately, meaning that the two-photon system must be in a configuration where its total $$J^{PC}$$ equal those of the pion. The key is that the two-photon-system can have different quantum numbers depending on the polarization of the photons and the orbital angular momentum between them.

• Charge parities $$C$$ multiply, so $$C$$ for the two-photon system is $$(-1)\times(-1)=+1$$ which conveniently is the charge parity of the pion. Besides not preventing the decay, we don't learn much from it.

• one needs to find a configuration of the spins (technically, "helicities") of the photons and the orbital angular momentum between them $$L$$ which gives total angular momentum zero. This allows us to exclude $$L\gt2$$ because the photons don't have enough spin to bring the total angular momentum down to zero again ($$1+1 < 3$$). Hence the decay has to proceed with $$L=1$$ or $$L=2$$.

• parity $$P$$ is also a multiplicative quantum number, but for a two-body system it not only depends on the spins of the particles but also on $$L$$ which adds an additional factor $$(-1)^L$$. For the two-photon system we thus have the combined parity $$P_{2\gamma}=(-1)\times(-1)\times(-1)^L=(-1)^L$$. Since this has to match the negative parity of the pion it means that the decay can only happen for odd $$L=1,3,5,\dots$$ and since $$L$$ cannot be greater than two, we are left with $$L=1$$.

• there is one more symmetry because the two photons are identical particles of even spin: Bose symmetry. The wave function of the final state must be symmetric under exchange of the two photons. $$L=1$$ implies that the orbital part of the wave function changes sign under exchange of the two photons, and so the spin part of the wave function must do so as well. Symmetry arguments (Landau-Yang) or a stupid search through Clebsch-Gordan coefficients (tobi_s) then leads to the conclusion that the photons must be polarized oppositely, as is borne out by experiment, which confirms the assignment of negative parity to the $$\pi^0$$.

Taking all this together, symmetries completely constrain the parameters of the decay. In the rest frame of the $$\pi^0$$ the only unknown is the direction of one of the photons and its polarization. Once this is measured, everything else is determined.