$\pi -\pi$ Systems

Let us consider the isospin structure of the simplest hadronic system composed of two objects, two pions. The spin $J$and parity $\pi$, which are combined into the usual notation $J^\pi$, of the pion is $J^\pi~=~0^-$. It is spinless, and has intrinsically odd-parity. In a system comprised of two pions, the intrinsic parity is $(-1)(-1)=+1$, intrinsically even. Therefore the parity of a two-pion system is given entirely by the parity of the relative orbital angular momentum state they are in $(-1)^L$.

Looking at the isospin structure of a two pion system. Each pion has $I=1$, and therefore the two pion system can have total isospin $I_{\rm total}~=~0$, $1$, $2$. The $I_{\rm total}~=~0$ and $2$ configurations are symmetric under the interchange of the pions, while the $I_{\rm total}~=~1$ is odd under such an interchange. In terms of tensors, we are able to simply construct the irreducible representations of SU(2) from ${\bf 3}\otimes {\bf 3}~=~{\bf 5}\oplus {\bf 3}\oplus {\bf 1}$. The objects that transforms as a ${\bf 5}$ of SU(2)($I~=~2$, a symmetric two-index tensor), as a ${\bf 3}$ of SU(2)($I~=~1$, a one-index tensor), and as a ${\bf 1}$ of SU(2)($I~=~0$, a zero-index tensor), are

 

$$T^{ab}\sim \pi^a\pi^b-{1\over 3}\delta^{ab} \pi^\alpha\pi^\alpha$$

$$T_a \sim \varepsilon_{abc}\pi^a\pi^b$$

$$T\sim \pi^\alpha\pi^\alpha \ \ \ ,$$ (34)

where $\varepsilon_{abc}$ is the totally antisymmetric tensor. As each of these objects is viewed as a pair of annihilation operators that will act on the appropriate wavefunction and produce the vacuum state, it is easy to write down the isospin wavefunction for two pions in states of pure isospin. In contrast, when acting on a wavefunction of different isospin, the state is destroyed. It is easy to verify that

$$\vert(\pi\pi); I=0, I_z=0\rangle \ =\ {1\over\sqrt{3}} \left[ \vert\pi^0\pi^0\rangle\ +\ \vert\pi^+\pi^-\rangle\ +\ \vert\pi^-\pi^+\rangle \right]$$

$$\vert(\pi\pi); I=1, I_z=0\rangle \ =\ {1\over\sqrt{2}} \left[ \vert\pi^+\pi^-\rangle\ -\ \vert\pi^-\pi^+\rangle \right]$$

$$\vert(\pi\pi); I=2, I_z=0\rangle \ =\ {1\over\sqrt{6}} \left[ 2 \... ...^0\rangle\ -\ \vert\pi^+\pi^-\rangle\ -\ \vert\pi^-\pi^+\rangle \right] \ \ \ .$$ (35)

For two pions in a relative S-wave ($L~=~0$, parity $+$) or any other even orbital angular momentum state, the requirement of a totally symmetric wavefunction means that they can only be in an $I~=~0$ or $I~=~2$ state. On the other hand, if the pions are in a relative P-wave ($L~=~1$, parity $-$) or any other odd orbital angular momentum state, the requirement of a totally symmetric wavefunction means that they can only be in an $I~=~1$ state.