# Isospin Rotation

1. May 29, 2007

1. The problem statement, all variables and given/known data

Show that
$$R_2 \left| I \; 0 \right> = (-1)^I \left| I \; 0 \right>$$
where
$$R_2 = e^{i \pi I_2}$$

N.B. the isospin states are represented by the usual notation of
$$\left| \textrm{(total isospin}) \qquad \textrm{(isospin component in the z direction)} \right>$$

3. The attempt at a solution

I guess I may be using the angular momentum analogy a bit too strongly here but my understanding is that we are asked for the eigenvalues of the operator representing a $$\pi$$ rotation about the $$I_2$$ axis.

The states we are working with have no component of isospin in the z (or if you rather $$I_3$$ ) direction, so it's all in the 1-2 plane. My main conceptual problem is that I thought (after knowing total and z component of isospin) we can't tell anything else about the orientation. I guess this could be where I'm taking the angular momentum analogy too far but I'll continue... So if we rotate it about the 2 axis how are we to know it's not all in the 2 direction anyway (giving us a 1 eigenvalue) or all in the 1 direction (a -1 eigenstate) or neither?
Clearly the correct answer doesn't bode well with my thinking.

I have tried treating this purely mathematically by the way but haven't got anywhere and probably isn't worth showing. I'd appreciate any help with this - be it conceptual or mathematical.

Last edited: May 29, 2007