Pions and $SU(2)$ representation

Question

I was reading about the pion ($\pi$) SU(2) representation and stumbled upon an expression for the isospin operator, $$ I_i=\epsilon_{ijk}\int d^3x\phi_j \dot\phi_k=-i\int d^3x \dot{\phi}^T t_i \phi ,$$ where $\phi$ stands for the isospin-vector field that can be decomposed into three real components $\phi_1, \phi_2,\phi_3$, or one complex field $\Phi=\frac{1}{\sqrt{2}}(\phi_1-i\phi_2)$ plus one real field $\phi_3$.

And then it's said that the $t_i$ for which $$ [t_i,t_j]=i\epsilon_{ijk} t_k \qquad \hbox{and} \qquad (t_i)_{jk}=-i\epsilon_{ijk}$$ form the adjoint representation of the isospin group.

And here I am a bit confused: if the $t_i$ form the adjoint representation of the isospin group, what are the $I_i$?

I also read in one book that there exist two representations of the isospin group the canonical one (written in the $\pi^+,\pi^0,\pi^- $ basis):

$T_1=\sqrt{2}\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix} $ $, \quad T_2=\sqrt{2}\begin{bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{bmatrix} $

$T_3=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{bmatrix} $

and the above-mentioned adjoint one. Are the $I_i$ maybe the canonical representation?

If the $t_i$ are written as: $(t_i)_{jk}=-i\epsilon_{ijk}$, and I calculate a matrix for the $t_i$ operator, which basis is meant? the $\pi^+,\pi^0,\pi^-$ basis or the $\pi_1,\pi_2,\pi_3$ basis?

Answer 1

I think I've managed to clear up my confusion and would like to summarize the above comments in an answer. Pions are described in QFT using three different fields, all satisfying the K-G equation. Due to their (almost) equal mass the three fields correspond to the same isospin number $T=1$ and form a basis $\{\pi^+,\pi^0,\pi^-\}$ for the $SU(2)$ representation, when stacked in matrix form:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\phi=$$ \begin{pmatrix} \pi^+ \\ \pi^0 \\ \pi^- \\ \end{pmatrix} $.

The isospin generators of $SU(2)$ in the 3-dimensional isospin space spanned by $\{\pi^+,\pi^0,\pi^-\}$, then take the "canonical" form:

$\qquad\qquad\qquad T_1=\sqrt{2}\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix} $ $, \quad T_2=\sqrt{2}\begin{pmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{pmatrix}, $ $\quad T_3=\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{pmatrix} $.

We can also choose a different basis, for example the "adjoint" basis $\{\pi^1,\pi^2,\pi^3\}$. Now the generators take the "adjoint" form:

$\qquad\qquad\qquad t_1=i\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} $ $, \quad t_2=i\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{pmatrix}, $ $\quad t_3=i\begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} $,

the transformation $\{\pi^+,\pi^0,\pi^-\}\rightarrow \{\pi^1,\pi^2,\pi^3\}$ is given in terms of an unitary operator $t_i=U^{\dagger}T_iU$, as can be read here (see answers). The $t_i$ form the adjoint represntation, so that their elements are given as structure constants of the $SU(2)$ group, meaning $(t_i)_{jk}=-i\epsilon_{ijk}$, where $\epsilon_{ijk}$ are the structure constants: $[t_i,t_j]=i\epsilon_{ijk} t_k$.

The $ I_i=-i\int d^3x \dot{\phi}^T t_i \phi$ on the other hand represent the QFT analogy of $T_i$ and $t_i$ and are the isospin operators in QFT, for example the $I_3$ gives $+1,0,-1$ if the field $\phi$ is in the $\pi^+,\pi^0,\pi^-$ state respectively. This is realized in terms of the Jordan map, but in a QFT sense, where each field is treated as an infinite sum of oscillators.