If an $SU (2)$ isospin transformation converts a proton to a neutron, how does a pion transform under the same transformation? I read in a particle physics note that if an $SU(2)$ isospin transformation makes $p\rightarrow n$ then under the same transformation pions go like $\pi^+ \rightarrow\pi^-$. I'm assuming that this means $\pi^0$ remains unchanged under this transformation.
Now if I represent $p$ to be $\begin{pmatrix} 1\\
0 \end{pmatrix}$
And $n$ to be $\begin{pmatrix} 0\\
1\end{pmatrix}$
The $SU (2)$ element with a two dimensional representation $\begin{pmatrix} 0&1\\
1&0\end{pmatrix}$ is the required transformation.
But pions are isospin triplet hence we need the representations
$\pi^+ = \begin{pmatrix} 1\\0\\
0\end{pmatrix}$
$\pi^0 = \begin{pmatrix} 0\\1\\
0\end{pmatrix}$
$\pi^- = \begin{pmatrix} 0\\0\\
1\end{pmatrix}$
Now I (vaguely) understand that $SU(2)$ can have a three-dimensional representation. But the matrix representation of the element that makes these transformations is as follows
$\begin{pmatrix} 0&0&1\\
0&1&0\\
1&0&0 \end{pmatrix}$
This is indeed unitary, but how do I know that this is a representation of an $SU(2)$ element?
How can I construct the above matrix from the three dimensional generators of $SU(2)$?
 A: There is a geometrical interpretation of this fact. You need to look at the adjoint action of the group $SU(2)$ on its Lie algebra representing isospin. This can be identified by the rotation group $SO(3)$.
You want to send $p\to n$, which amounts to flipping the sign of $I_3$. You cannot do this by leaving the other components unchanged, you need to change them as well. Geometrically, you are looking for a rotation that flips the sign of the $z$ axis. One possible candidate os the $\pi$ rotation about the $y$ axis. Any other rotation can obtained by adding a rotation about the $z$ axis.
This immediately gives you the result, if the rotation switches $p\to n$ then it switches $\pi^+\to\pi^-$ leaving $\pi^0$ unchanged up to possible phase changes.
Explicitly, in matrix notation this is tabulated by the Wigner $d$ matrices (hence the choice of the $y$ axis). This gives in the fundamental representation $p,n$:
$$
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
$$
up to a multiplication of:
$$
\begin{pmatrix}
e^{i\phi/2} & 0\\
0 & e^{-i\phi/2}
\end{pmatrix}
$$
Btw your matrix was not even in $SU(2)$.
And in the adjoint representation $\pi^{+,0,-}$:
$$
\begin{pmatrix}
0 & 0 & 1 \\
0 & -1 & 0 \\
1 & 0 & 0
\end{pmatrix}
$$
up to a multiplication by:
$$
\begin{pmatrix}
e^{i\phi} & 0 & 0\\
0 & 1 & 0 \\
0 & 0 & e^{-i\phi}
\end{pmatrix}
$$
Btw the result is more natural in the $x,y,z$ basis in which case you get:
$$
\begin{pmatrix}
0 & 0 & -1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
$$
Up to a multiplication by:
$$
\begin{pmatrix}
\cos\phi & -\sin\phi& 0\\
\sin\phi & \cos\phi & 0 \\
0 & 0 & 1
\end{pmatrix}
$$
Hope this helps.
