Eigenstates of total isospin, I In the book from which I'm studying particle physics (by Mark Thompson) it is stated that states of two quarks of third component of isospin = 0 (like ud or du) are not eigenstates of the total isospin. And it does not elaborate. I fail to see why. Probably this can be shown from the eigenvalue equation for the isospin operator but I'm looking for an intuition in terms of a geometrical interpretation in the isospin space rather than algebraic. Can anyone help?
 A: Instead of answering your question directly let me detour  into a few details about isospin. 
Isospin is a global $SU(2)$ symmetry rotating between the ``up-down'' quark content. Three important $SU(2)$ multiplets are the singlet, doublet (fundamental) and the triplet representations given by. There are two orthogonal doublets given by,
\begin{align} 
& u \equiv \left| \frac{1}{2} ,\frac{1}{2} \right\rangle =  \left( \begin{array}{c} 
1 \\  
0 
\end{array} \right) \quad d \equiv \left| \frac{1}{2} ,-\frac{1}{2} \right\rangle = \left( \begin{array}{c} 
0 \\  
1 
\end{array} \right) 
\end{align}
From the studies of spin angular momentum, which is also an $ SU(2) $ symmetry, we know that two doublets combine to form a triplet and a singlet. In particular,
\begin{equation} 
\left|  u \right\rangle  \left|  d \right\rangle  = \frac{1}{\sqrt{2}} ( \left|  1 ,0 \right\rangle  + \left|  0 ,0 \right\rangle ) 
\end{equation} 
(I'm not sure about the sign of the expression above)
Since the product of these states form a linear combination of two states with different total isospin it isn't an eigenstate of the total isospin operator, even though it is an eigenstate of the $ I _z $, the isospin along the $ z $ axis. 
The intuition here is identical to the reason a spin up and spin down particle aren't an eigenstate of the total spin operator.
