Why does π⁰ have G-parity −1 if its isospin state is |1,0⟩?

[copernicus1]

Since the pi^0 decays to two photons its C-parity is +1, but its isospin ket is $$|1,0\rangle,$$ which makes it a little tricky to understand why its G-parity is -1. Does the rotation around the I_2 axis somehow generate a minus sign? This seems odd since the isospin ket points into the x-y plane, so it seems the rotation by pi wouldn't affect it.

 

[Vanadium 50]

That doesn't look right. It's |1,0>.

 

[copernicus1]

That doesn't look right. It's |1,0>.

You're right that was just a mistake; I fixed it. This is what makes the rotation in isospin space hard to understand. Initially it's pointing into the x-y plane, and after a 180 degree rotation it's still into the x-y plane, but somehow the G-parity is different from the C-parity.

 

[Bill_K]

Isn't this what you'd expect for an isovector? Under a 180° rotation about the I₂ axis, I₁ → - I₁, I₂ → I₂, I₃ → -I₃. And in the spherical basis, π⁺ and π^- are linear combinations of I₁ and I₂, and π⁰ is the I₃ component, so it changes sign.