If an $SU (2)$ isospin transformation converts a proton to a neutron, how does a pion transform under the same transformation?

Question

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 $SU(2)$?

Answer 1

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.