(i) The pions form an isospin triplet with $\pi^{+}=|1,1\rangle, \pi^{0}=|1,0\rangle$ and $\pi^{-}=|1,-1\rangle$, whilst the nucleons form an isospin doublet with $p=\left|\frac{1}{2}, \frac{1}{2}\right\rangle$ and $n=\left|\frac{1}{2},-\frac{1}{2}\right\rangle$. Consider the isospin representation of two-particle states spanned by the basis

$$T=\left\{\left|\pi^{+} p\right\rangle,\left|\pi^{+} n\right\rangle,\left|\pi^{0} p\right\rangle,\left|\pi^{0} n\right\rangle,\left|\pi^{-} p\right\rangle,\left|\pi^{-} n\right\rangle\right\}$$

State which irreducible representations are contained in this representation and explain why $\left|\pi^{+} p\right\rangle$ is an isospin eigenstate.

Using

$$I_{-}|j, m\rangle=\sqrt{(j-m+1)(j+m)}|j, m-1\rangle,$$

where $I_{-}$is the isospin ladder operator, write the isospin eigenstates in terms of the basis, $T$.

(ii) The Lie algebra $s u(2)$ of generators of $S U(2)$ is spanned by the operators $\left\{J_{+}, J_{-}, J_{3}\right\}$ satisfying the commutator algebra $\left[J_{+}, J_{-}\right]=2 J_{3}$ and $\left[J_{3}, J_{\pm}\right]=\pm J_{\pm}$. Let $\Psi_{j}$ be an eigenvector of $J_{3}: J_{3}\left(\Psi_{j}\right)=j \Psi_{j}$ such that $J_{+} \Psi_{j}=0$. The vector space $V_{j}=\operatorname{span}\left\{J_{-}^{n} \Psi_{j}: n \in \mathbb{N}_{0}\right\}$ together with the action of an arbitrary su(2) operator $A$ on $V_{j}$ defined by

$$J_{-}\left(J_{-}^{n} \Psi_{j}\right)=J_{-}^{n+1} \Psi_{j}, \quad A\left(J_{-}^{n} \Psi_{j}\right)=\left[A, J_{-}\right]\left(J_{-}^{n-1} \Psi_{j}\right)+J_{-}\left(A\left(J_{-}^{n-1} \Psi_{j}\right)\right)$$

forms a representation (not necessarily reducible) of $s u(2)$. Show that if $J_{-}^{n} \Psi_{j}$ is nontrivial then it is an eigenvector of $J_{3}$ and find its eigenvalue. Given that $\left[J_{+}, J_{-}^{n}\right]=$ $\alpha_{n} J_{-}^{n-1} J_{3}+\beta_{n} J_{-}^{n-1}$ show that $\alpha_{n}$ and $\beta_{n}$ satisfy

$$\alpha_{n}=\alpha_{n-1}+2, \quad \beta_{n}=\beta_{n-1}-\alpha_{n-1}$$

By solving these equations evaluate $\left[J_{+}, J_{-}^{n}\right]$. Show that $J_{+} J_{-}^{2 j+1} \Psi_{j}=0$. Hence show that $J_{-}^{2 j+1} \Psi_{j}$ is contained in a proper sub-representation of $V_{j}$.