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Under axial transformations, $\sigma$ and $\pi$ are rotated into each other:

$\vec{\pi} \rightarrow \vec{\pi}+ \vec{\theta} \sigma $,

$\sigma \rightarrow \sigma+ \vec{\theta}.\vec{\pi} $.

In arXiv:nucl-th/9706075, page 12, it is the written that if axial symmetry is a symmetry of QCD Hamiltonian, then $\sigma$ and $\pi$ should have the same eigenvalues, i.e., the same masses.

My question is that how this symmetry results in expecting the same masses for $\sigma$ and $\pi$? I know that these mesons have different masses and it is the result of spontaneous symmetry breaking, but if the symmetry was not broken, why should we expect the same masses for these states? How can we prove this?

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  • $\begingroup$ Have you written down the Hamiltonian with different masses and performed the symmetry transformation? $\endgroup$
    – ACuriousMind
    Apr 23, 2015 at 11:28
  • $\begingroup$ It seems that this result is general . For example for vector transformations, pions rotate in isospin space and it is written that if vector transformation is a symmetry of QCD, pion isotriplets should have the same mass. I want to know how these symmetries lead to this result. Before writing any example (Hamiltonian) the author tell this as a general result. Why? $\endgroup$
    – Kheeyal
    Apr 23, 2015 at 13:48

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The mass terms for the $ \sigma $ and $ \vec{ \pi } $ fields are, \begin{equation} m _\sigma \sigma \sigma + m _\pi \vec{ \pi } \cdot \vec{ \pi } \end{equation} You have two terms that are going to turn into each other under a symmetry transformation. Thus they need to have the same coefficient in order to remain invariant under the symmetry (feel free to explicitly stick in the transformations if you are not comfortable with this). If we do have, $ m _\sigma = m _\pi \equiv m $ then the mass terms take the form, \begin{equation} m \Pi \cdot \Pi \end{equation} where $ \Pi \equiv \left( \vec{ \pi } , \sigma \right) $ and \begin{equation} \Pi \rightarrow \left( \begin{array}{cc} 1 & \vec{ \theta } \\ - \vec{ \theta } & 1 \end{array} \right) \Pi \end{equation} (you missed a minus sign above) under the axial symmetry. Since this is a unitary transform, it leaves the mass term invariant as expected. Thus we conclude that for the system to be invariant under axial transformation it requires the masses of the $\sigma $ and $\vec{\pi}$ be degenerate.

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