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In my assignment, I am asked to show whether $\rho^0\to3\pi^0$ is allowed. From the particle data group, I cannot find the decay mode; hence, I am guessing it is not allowed. Based on the knowledge that $\rho^0\to2\pi^0$ is not allowed, I focus on parity. However, the action contains an angular momentum of 3 bodies. Then I get stuck. Can someone tell me whether this action is allowed? Or at least give me some hint. (I have not learned weak interaction and strong interaction yet.)

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  • $\begingroup$ Well, the decay certainly violates G-parity, but so does the suppressed but allowed decay $\pi^0\pi^-\pi^+$, since isospin is an imperfect symmetry, hence also G. Parity violation would make the decay weak, so enormously suppressed. Note your 3-body wf must be completely symmetric, but, can it be? $\endgroup$ – Cosmas Zachos Feb 7 '18 at 2:28
  • $\begingroup$ @CosmasZachos Yes, my question now becomes for many bodies (or 3 bodies) problem, will the orbital angular momentum still have the same spherical harmonic function? If so, then the $Y_1^m$ will change the sign under parity; hence, 3 $\pi^0$ cannot be in the same orbit because they are indistinguishable. $\endgroup$ – Hamio Jiang Feb 7 '18 at 2:38
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I confess I might be overthinking this, having missed something more obvious (C? -- violated by the weak interactions).

The G-parity of the ρ is +, so it prefers to decay to 2π s, since the π has G parity -. (It requires an antisymmetric state because of angular momentum, as you already see.)

But G-parity relies on isospin, which is broken somewhat in the strong decay, so there is also the 3π mode at the $10^{-4}$ level, where all 3π s are different, instead of your 3π0 which has to be fully symmetric.

You cannot first combine the two π0s to a vector by an L=1 as you notice, by their unavoidable symmetry. So you have to combine them to a symmetric S-wave spinless neutral dipion, and then combine that antisymmetrically with an L=1 to the remaining neutral π0 to net a vector ρ0.

However, this is impossible; ignoring normalizations, $$ (12+21)3-3(12+21)= 123+213-312-321= (123-321)+(-312+213), $$
but the terms in the first and second final parentheses are 1-3 antisymmetric, and 3-2 antisymmetric respectively, which is impossible for identical neutral pions.

This argument fails, of course, (phew!), for the extant decay mode π0π+π- which is allowed, although ferociously suppressed.

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