# The isospin of pions

In the thesis(Page:53), the author defined three pion bases, namely,

$$|\pi^{1}\rangle = |1, 1\rangle, \quad |\pi^{0}\rangle=|1,0\rangle, \quad |\pi^{-1}\rangle=|1, -1\rangle.$$ Here, the first number is total isospin and the second number is the third component of isospin. The relation between the charged and the isospin states is given by

$$|\pi^{1}\rangle = \frac{1}{\sqrt{2}}(\pi^{+}+\pi^{-}), \quad |\pi^{0}\rangle=\pi^{0}, \quad |\pi^{-}\rangle=\frac{i}{\sqrt{2}}(\pi^{+}-\pi^{-}). \qquad (*)$$ in isospin SU(3) group, $$\pi^{\pm0}$$ are the third components of isospin 1. We label them as 1, -1, 0, respectively. if we make $$I_{3}$$ act on both sides of the above equations, for instant, we have $$I_{3}|\pi^{1}\rangle = 1 |\pi^{1}\rangle = \frac{1}{\sqrt{2}}(\pi^{+}+\pi^{-}) \neq I_{3} \frac{1}{\sqrt{2}}(\pi^{+}+\pi^{-}) = \frac{1}{\sqrt{2}}(1\pi^{+}+(-1)\pi^{-})$$ From my calculations, I can not understand the relations in eq.(*), Is there a mistake in the thesis or where do I make a mistake for my calculations?

• If I understand your notation, the asterisk equation cannot be right in the sense that on the left-hand side you have a ket and not on the right hand side. Commented Sep 9, 2022 at 14:15
• @Mauricio Ineed, the $\pi^{\pm0}$ can be represented as $|1,1\rangle$, $|1,-1\rangle$, and $|1,0\rangle$. Commented Sep 9, 2022 at 14:30
• Commented Sep 9, 2022 at 15:02

The normalized pion states are usually defined as $$\pi^0\equiv \pi^3 \\ \pi^{\pm}= {1\over \sqrt 2} (\pi^1\pm i \pi^2),$$ where, using the somewhat nonstandard convention of fields creating instead of annihilating the eponymous states when acting on the vacuum, $$|\pi^{+}\rangle = |1, 1\rangle, \quad |\pi^{0}\rangle=|1,0\rangle, \quad |\pi^{-}\rangle=|1, -1\rangle.$$ Consequently, $$I_3 |\pi^{\pm}\rangle = \pm |\pi^{\pm}\rangle, \qquad I_3|\pi^0\rangle =0.$$
If you insist on going to the adjoint basis, where none of the three isospin generators, $$I_{1,2,3}$$, is an eigenoperator of its states, $$\pi^1={1\over \sqrt 2}( \pi^+ + \pi^-),\\ \pi^2={i\over \sqrt 2}( \pi^- - \pi^+),\\ \pi^3= \pi^0.$$
You then see that $$I_{3}|\pi^{1}\rangle = i |\pi^{2}\rangle,$$ etc. (Formally, in angular momentum theory, you rotated around the z- axis by π/2 to get the x axis on the y axis.)