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Motivation: The neutral pion decays to 2 photons ($\pi^0\to\gamma\gamma$) most of the time. For the decay of the neutral to 3 photons ($\pi^0\to 3\gamma$) we have an upper limit on the branching ratio of $3.1 \cdot 10^{-8}$ in the Particle Data Book (2012). The explanation is that this decay would violate charge conjugation.

I haven't found anything about angular momentum conservation in this decay: The pion has spin zero, the photon is a spin-one particle, but, being massless, a (free) photon cannot have 0 for the projection of the spin.

Now, my question is: Can I combine three photons to give a spin-zero state such that angular momentum in the $\pi^0\to 3\gamma$ decay is conserved?

Thoughts: For "ordinary" spin-zero particles the spin addition is described by the Clebsch-Gordan coefficients. From the tabulated CG coefficients I see that I can combine $1\times1$ to give a $J=1$ state with zero contribution from $m_1=m_2=0$. So I could add two photons, giving $J=1$ and $M=0$ (the important point being that all other possibilities are ruled out by the requirement of the photon not having $m=0$??). I could then add the third photon on top, argueing along the same lines.

Does that make sense? Is the application of the formalism correct?

(Sorry for writing so much text about a simple question.)

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The decay of a neutral $\pi^0$ to three photons would indeed violate charge conjugation.

The charge conjugation argument goes as follows: The reaction $$\pi^0 \to 3 \gamma$$ is mediated by electromagnetism. QED has a charge conjugation symmetry, so you should be able to apply a charge conjugation to both sides of the equation. Under charge conjugation, $\pi^0 \to \pi^0$ while $\gamma \to - \gamma$ (this follows from the gauge principle). Therefore, the initial state has an even C transformation while the final state has an uneven C transformation. In other words: Applying these transformation properties to both sides of the above equation gives you a minus sign (this may seem like a bogus calculation, but you can work it out with the actual fields if you like).

The angular momentum thing is more difficult. You have to take into account that the system of three photons does not only have spin, but can also have an non-zero 'regular' angular momentum (e.g. the photons are emitted in a p-wave and not an s-wave).

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  • $\begingroup$ I know the C violation argument, the question is really more about the spin addition of photons (not taking into account orbital angular momentum because I do not see how that would help here). $\endgroup$ – fuenfundachtzig May 16 '13 at 11:11
  • $\begingroup$ Your calculations would be correct, if the photon had a $S=1, M=0$ state. However, the photon only allows $S = 1, M = \pm 1$ states. If you add the spin of two photons the possibilities are $ J = 0, 1, 2$ with $ M = 0 = -1 +1 $ or $M = 2 = +1 +1$. If you add another photon your options are $ J = 0, 1, 2, 3$ and $M = 1 = 0 + 1 = 2 - 1$ or $M = 3 = 2 + 1$. (and of course, for every M also -M). So there is no possibility to reach $M = 0$ with three photons, as the photon does not have a $ S = 1, M = 0$ component. $\endgroup$ – Neuneck May 16 '13 at 11:30
  • $\begingroup$ I agree with the first part. I'm not sure about the second -- I thought I had been taught that I can do the addition of spins stepwise (i.e. it's associative)? $\endgroup$ – fuenfundachtzig May 16 '13 at 14:51
  • $\begingroup$ Yes, it is associative. Nevertheless adding +1 or -1 to 0 or 2 will never give you 0. And 0 or 2 are the only possibilities for the addition of two photon spins, irregardless of order. $\endgroup$ – Neuneck May 17 '13 at 6:02
  • $\begingroup$ I guess my wording in the heading is misleading, I tried to improve on that: I'm mostly interested in the conservation of angular momentum in the $3\gamma$ decay. I don't need a spin-0 state for that, only a state which allows the projection of the spin to be zero. $\endgroup$ – fuenfundachtzig May 17 '13 at 8:11
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I had the same doubt as I was figuring out if a certain coupling of angular momenta in the 3$\gamma$ could be found such that parity would be conserved.

Turns out (if I am correct) you cannot have three photons with coupled total angular momentum $J=0$ in the first place.

For the coupled photon angular momentum $L(3\gamma)$ and spin $S(3\gamma)$ to be able to couple to $J(3\gamma)=0$, we must have

$$L(3\gamma)=S(3\gamma)\hspace{2cm}(1)$$

This poses no big limitation, as photon are spin 1 particles, so coupling spins (independently of orbital angular momenta) can give $S(3\gamma)$=0,1,2,3, and coupling orbital angular momenta (independent of spins) can give any positive integer $L(3\gamma)$.

In order to preserve the Bose Symmetry of the photons though, the spin and space Wave Functions must be either $\textit{simultaneously}$ symmetrical, or $\textit{simultaneously}$ antisymmetrical. This is expressed by the condition: $$\text{$L$ even (symmetrical space WF) $\iff$ $S$ odd (symmetrical spin WF)}$$ $$\text{$L$ odd (antisymmetrical space WF) $\iff$ $S$ even (antisymmetrical spin WF)}$$ or combined $$L(3\gamma)+S(3\gamma)+1\hspace{0.5cm}\text{even.}\hspace{2cm}(2)$$ Now (1) and (2) are clearly incompatible, hence we cannot have $J(3\gamma)=0$.

So the decay $\pi^0\rightarrow3\gamma$ would be ruled out by simple angular momentum conservation.

This seems a little weird to me, since the impossibility of the decay is usually accounted to $C$ violation, but again I have seen the formula $C(n\gamma)=(-1)^n$ for the charge conjugation eigenvalue of $n$ photons, which would only be universable from the formula for bosons $$C(n\hspace{1mm}\text{bosons})=P(\text{boson})^n\cdot(-1)^L\cdot(-1)^{S+1}$$ if $L+S+1$ were always even for photons ($P(\gamma)=-1)$.

So if I did do a mistake, it is probably in the coupling and hence the condition (1).

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