0

I found this site to give a nice explanation of isospin. Apparently, the charged pions can be described by a superposition of $\pi^1$ and $\pi^2$.

Let us now consider the $\pi$'s in the context of isospin. There are three different charge states $\pi^+$, $\pi^0$ and $\pi^-$. They fit simply into a $I=1$ multiplet $\pi^a$, corresponding to $I_Z=+1,0,-1$ respectively, with

$\pi^\pm = \frac{1}{\sqrt{2}}(\pi^1\mp i \pi^2)$
$\pi^0=\pi^3$

Where $\pi^\pm$, $\pi^0$ are field operators. However, I fail to understand, what the $\pi^1$, $\pi^2$ and $\pi^3$ are.

Community
  • 1
infinitezero
  • 1,520

1 Answers1

4

Take a look at plain old vectors. Three dimensional space is spanned by ${\bf e}^1$, ${\bf e}^2$, and ${\bf e}^3$, also known as $\hat x$, $\hat y$, and $\hat z$. Of course:

$$ {\bf e}^i \cdot {\bf e}^j = \delta_{ij} $$

and

$$ ({\bf e}^i \times {\bf e}^j)\cdot {\bf e}^k = \epsilon_{ijk} $$

What are the eigenstate of rotations about the arbitrarily chosen $z$-axis? They are (normalized):

$$ {\bf e}^0 \equiv {\bf e}^3 $$

and

$$ {\bf e}^{\pm} = \frac 1{\sqrt 2}({\bf e}^1 \mp i{\bf e}^2)$$

You can show that the ${\bf e}^m$ with $m=(-1, 0, 1)$ are eigenvectors of a $z$-rotation by $\phi$ with eigenvalue:

$$ \lambda = \exp{(im\phi)}$$

This is known as the spherical representation of vectors, in contrast to the more familiar cartesian representation. (It's also called the fundamental representation of SO(3)). They transform exactly like a quantum spin-1 particle, aka, a vector boson.

Change the $\bf e$ to $\pi$, and replace spin with the name isospin, and you have 2 representations of pions.

The "cartesian" version isn't used, because the basis states aren't explicitly eigenstates of rotations about $I_3$--the equivalent $z$-axis in isospin space.

JEB
  • 42,131