$\rho\to \pi\pi$ parity

Question

I'm having some difficulty understanding why the $\rho^0$$\rightarrow$$\pi^+ \pi^-$ decay conserves parity.

By my understanding, the parity of a meson is given by $P_aP_b(-1)^{L+1}$ and the parity of a two-body state is given by $P_cP_d(-1)^L$

Angular momentum of $\rho=1$ and spin of $\pi=0$ so the final state must have $L=L_{\pi^+}+L_{\pi^-}=1$

Thus total parity for the final $\pi^+\pi^-$ system is given by

$P_{\pi^+}P_{\pi^-}(-1)^L=P_uP_\bar{d}(-1)^{L_{\pi^+}+1}P_{\bar{d}}P_d(-1)^{L_{\pi^-}+1}(-1)^1=1$

since fermions and antifermions have parity $1$ and $-1$ respectively

But the parity of $\rho=-1$ if $L_{\rho}=0,S_{\rho}=1$ so parity isn't conserved

What am I missing?

Answer 1

I am summarizing my comments only to prevent the systematic abuse of a terminology malforming the question; in particular, its erroneous $L=L_{\pi^+}+L_{\pi^-}=1$ part.

Indeed, the (intrinsic) parity of a meson is $(-1)^{L+1}$ where L is the angular momentum of the wavefunction of the quark-antiquark pair. Both the orthoquarkonium and the paraquarkonium, the $\rho$ and the $\pi$, are in s-wave, so, then, the respective L=0 ; and they consequently both have negative parity, $P_\rho=P_\pi=-1$. It is the spin wavefunction that gives the $\rho$ its J and denies it to the pions, but that does not affect parity.

Since the parity of a two-body state is given by $P_cP_d(-1)^L$, and the L of the two-pion wavefunction =1, so as to produce a J=1 particle out of two spinless particles, $$ P_{\pi\pi}=P_\pi P_\pi (-)^{-1}=-1 =P_\rho, $$ as required in a strong decay.

Answer 2

Turns out what I was missing was the presence of a gluon decaying into a $d\bar{d}$ pair, with the gluon adding an extra $(-1)$ term to the total parity equation.