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In QCD, due to the work of ‘t Hooft and Vafa-Witten, we know that confinement implies chiral symmetry breaking (David Tong’s gauge theory notes have a clear discussion of this). It is then said that the confined pion $\pi=\bar{\psi}\gamma^5\psi$ is light as it corresponds to the Goldstone mode of chiral symmetry breaking $SU(N_f)_L\times SU(N_f)_L \rightarrow SU(N_f)_V$. How do we see that the pion corresponds to these Goldstone modes?

My eventual goal is to understand which mesons correspond to Goldstone modes in more general symmetry breaking patterns of some chiral gauge theories.

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Due diligence. The article specifies the condition clearly. A minimum competence review provides lots of explicit examples.

The broken generators in your case are the $N^2-1$ generators not in $su(N)_V$, the axial ones, which do not close to a group.

You must first compute the infinitesimal transforms of your $N^2-1$ operators $\bar \psi \gamma^5 T^a\psi$ under these $N^2-1$ axial transforms, and take their QCD vacuum expectation values.

If such a value does not vanish, i.e. if the operator transforms inhomogeneously, the operator interpolates a Goldstone mode. Such non-vanishing v.e.v.s are nontrivial null eigenvectors of the Goldstone mass matrix.

In your case, all those $N^2-1$ do. In fact, the pions and the SSB charges (now gone), have the same quantum numbers, in a 1-to-1 correspondence.

Example Let us illustrate this for N=2, which is to say $SU(2)_L\times SU(2)_R\to SU(2)_V$ easier to picture in the algebraic equivalent language $SO(3)\times SO(3) \to SO(3)_V$. So you know, from your σ model (you wouldn't be asking this question unless you mastered this!), that the SO(4) vector $$ \begin{pmatrix}\pi^1\\ \pi^2\\ \pi^3\\ \sigma \end{pmatrix}$$ has its components scrambled pairwise by 6 generators. The SO(3) vector ones rotate the three pion components among themselves, and the 3 axials (not closing to a Lie algebra, of course) scramble each pion with the scalar σ.

The 0th component of the three vector currents is $$ \vec V^0= \vec \pi \times \partial^0 \vec \pi, $$ which you'll space-integrate to the charge generating the three isospin rotations; while the same for the three axial currents is $$ \vec A^0 = -\partial^0 \sigma ~\vec \pi+ \sigma \partial^0 \vec \pi . $$

Consequently, $$ \delta_A \vec \pi = \vec \theta_A \sigma ,\\ \delta_A \sigma = -\vec \theta_A \cdot \vec \pi, $$ where I have been cavalier with normalization constants.

Now for the crucial part: For vacuum values $\langle \vec \pi\rangle=0$ and $\langle \sigma\rangle=v$, it follows that $\langle \delta \sigma \rangle =0$, and, crucially, $$ \langle \delta \vec \pi \rangle =\vec \theta_A v\neq 0, $$ so each component is a Goldstone boson, a null eigenvector of the Goldstone mass matrix. You also see the direct connection between the three axial parameters with the three pions, i.e. the three SSBroken axial charges.

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