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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?

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  • $\begingroup$ 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. $\endgroup$
    – Mauricio
    Commented Sep 9, 2022 at 14:15
  • $\begingroup$ @Mauricio Ineed, the $\pi^{\pm0}$ can be represented as $|1,1\rangle$, $|1,-1\rangle$, and $|1,0\rangle$. $\endgroup$
    – quantum
    Commented Sep 9, 2022 at 14:30
  • $\begingroup$ Related : physics.stackexchange.com/search?q=user%3A227119+pion $\endgroup$
    – Frobenius
    Commented Sep 9, 2022 at 15:02

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The thesis writer of your reference was demonstrably drunk at that part of his task, and so was his proofreader and, arguably, his certification committee. His formulas are not even wrong--indeed, later on, he does talk about indices 0,1,2 he garbled so spectacularly. Your problem is treated in standard conventions in most decent introductory texts on particle physics.

It is the transition to the adjoint basis of su(2) from the spherical basis, and I review it to coddle your sanity.

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.)

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