 Solution

Angular momentum and parity. Work in the rho meson rest frame. Since each pion has zero spin, their combined intrinsic spin is $S=0$ and their total angular momentum is entirely relative orbital angular momentum:The parity selection rule for a two-body decay givesThe supplied $P_{\rho }=-1$ requires odd $L$, so both conservation laws agree:Here the statement about two-pion decays is understood as the dominant strong interaction channel; the assertion that it occurs literally always is an idealization, since other interactions can produce other channels.

Exchange symmetry. Pions are spin-zero bosons, so the complete two-particle wavefunction must have bosonic exchange symmetry, including its spatial and isospin factors. The unique spin-zero product is symmetric. Exchanging the particle slots reverses the relative position, so an $L=1$ spatial state has orbital parity $(-1{)}^L=-1$. Its isospin state must therefore be antisymmetric to make the product symmetric.

For two neutral pions, the internal charge state $|{\pi }^0{\pi }^0⟩$ is symmetric under exchange. There is no antisymmetric internal factor to compensate odd $L$. Equivalently, the odd-$L$ projection of two identical spin-zero neutral particles vanishes. ThereforeThis argument does not depend on treating isospin as an exact symmetry of nature.

All nine isospin states. In the following two-pion isospin decomposition, let $|ab⟩=|{\pi }^a{⟩}_1\otimes |{\pi }^b{⟩}_2$ denote ordered particle slots; each of $a,b$ is $+,0,-$. The second label in $|I,q⟩$ is the total electric charge in units of the elementary charge. For this pion system it equals the total isospin projection $I_3$.

Addition of angular momentum for two isospins of one gives $1\otimes 1=2\oplus 1\oplus 0$. Start from the highest-weight quantum state $|2,2⟩=|++⟩$ and apply the sum of the two ladder operators. The ladder recurrence for Clebsch-Gordan coefficients gives the normalized symmetric quintet:The normalized quantum state orthogonal to $|2,1⟩$ in the two-dimensional charge-one subspace is antisymmetric. Lowering it gives the antisymmetric triplet:The remaining charge-zero quantum state must be orthogonal to both displayed charge-zero states and annihilated by the total ladder operators. This gives the symmetric singlet:Together the quintet, triplet and singlet contain $5+3+1=9$ orthonormal states, accounting for the full $3\times 3$ charge-product space. Multiplying an entire isospin multiplet by a common quantum phase changes none of these physical conclusions.

Rho isospin and strong decay modes. Only the $I=1$ two-pion isospin decomposition is antisymmetric under exchange; $I=0,2$ are symmetric. The required $L=1$ spatial state therefore selects $I_{\pi \pi }=1$. Conservation of isospin by the strong interaction, in its usual approximation, impliesThe three charge states have the corresponding strong two-pion channelsThe ordered antisymmetric combinations above describe the internal part; multiplying by the odd spatial state gives the required symmetric complete bosonic state. The neutral $I=1$ state contains no $|00⟩$ component, in agreement with the independent identical-particle argument.

Charge conjugation. Charge conjugation preserves mass and reverses electric charge. A charged rho meson is therefore changed into a distinct state of opposite charge, so neither ${\rho }^{+}$ nor ${\rho }^{-}$ is itself a charge-conjugation eigenstate. The neutral state can be an eigenstate. Acting on its two-pion internal state interchanges the charge labels:The charged-pion quantum phase conventions cancel between the two particles. Thus the two-pion charge-conjugation parity is $C=(-1{)}^L=-1$, and conservation by the strong interaction fixesOnly ${\rho }^0$ among the individual charge states is a charge conjugation eigenstate.