Potential of the pseudo Nambu-Goldstone boson Pseudo Nambu-Goldstone bosons appear in breaking approximate symmetries, for instance, the QCD axions. Many literatures just write the action of those pseudoscalars without any explanation. 
In general, how is the effective action, especially the potential, of these Pseudo Nambu-Goldstone bosons obtained? 
 A: The easiest way, I think, is by using the spurion analysis. If an operator $\mathcal{O}$ with a small coupling $g$ in the lagrangian breaks a symmetry, you just need to promote $g$ to a field (called spurion) with definite  transformation properties such that $g\mathcal{O}$ is now invariant.  The low-energy effective action now must be written making an expansion in $g$ by writing down all the invariants to any given order, and eventually freeze $g$ to its original value. 
Let me give you the example of chiral $SU(2)$ in QCD. In this theory there is an approximate $G=SU(2)_L\times SU(2)_R$ symmetry of the UV theory for up and down quarks
$$
Q_{L}=\left(\begin{array}{c}u_L\\ d_L \end{array}\right)\rightarrow U_L Q_{L}\,,\quad 
Q_{R}=\left(\begin{array}{c}u_R\\ d_R \end{array}\right)\rightarrow U_R Q_{R}
$$
where $U_{L,R}$ are unitary $2\times 2$ matrices of $SU(2)_{L,R}$. In the IR, the symmetry is broken spontaneously down to a subgroup $H=SU(2)_{L+R}$ isospin so that 3 massless pions $\pi^a$ are present in the low-energy spectrum according to the Goldstone theorem. This pions can be written in a non-linear sigma model fashion as $\Sigma(x)=e^{i\pi^a(x) \sigma^a/f_\pi}$ where $\Sigma\rightarrow U_L \Sigma U^\dagger_R$ non-linearly realize $G$ but linearly realize $H$ (where $U_L=U_R$). The lowest order lagrangian invariant under $G$ is of the schematic form
$$
\mathcal{L}^{(2)}_{eff}=\frac{f_\pi^2}{4}\mathrm{Tr}\left[\partial_\mu\Sigma^\dagger \partial^\mu\Sigma\right].
$$
Now, we add a small explicit breaking of $G$ in the UV, namely a mass term $m_u \bar{u}_L u_{r}+m_d \bar{d}_L d_{r}+h.c.$ which we can write as
$$
\bar{Q}_{L}M Q_{R}+h.c.\,,\qquad M=\left(\begin{array}{cc} m_u & 0 \\ 0 & m_d\end{array}\right)\,.
$$
In fact, this mass term doesn't even respect $H$ (unless $m_u\neq m_d$) and not only the pions will get mass but they will also be split. Anyway, the UV lagrangian can be thought as invariant under $G$ if we promote $M$ to a spurion and transform it as $M\rightarrow U_L M U^\dagger_{R}$. Now, going in the IR and looking at the effective theory of the pions, we can write new invariants in the action for $\pi$. The most important are the ones with the least insertions of $M$ since we assumed that the explicit breaking was small (in this case, compared to $\Lambda_{QCD}$). There is only one operator with just one insertion of $M$, namely
$$
\delta\mathcal{L}=c\mathrm{Tr}\left[\Sigma^\dagger M+\Sigma M\right]\,.
$$
Now one can freeze $M$ to $M=\mathrm{diag}\left(m_u,m_d\right)$, expand $\delta\mathcal{L}$ in powers of the pion fields, and thus get the form of the potential (in particular the mass terms) at leading order in the explicit breaking parameter $M$. 
Just for simplicity, assume now that the explicit breaking of $G$ respects $H$, $m_u=m_d\equiv m$, so that $M=m\mathbf{1}$ is multiple of the identity. In this case, $\delta\mathcal{L}$ can be written in a nice form
$$
\delta\mathcal{L}=c\,m\mathrm{Tr}\left[\Sigma^\dagger+\Sigma\right]=4c\, m\cos\frac{|\pi|}{f_\pi}
$$
where we used that $e^{i\pi^a\sigma^a/f_\pi}=\left(cos\frac{|\pi|}{f_\pi}\mathbf{1}+i\frac{\pi^a\sigma^a}{|\pi|}\sin\frac{|\pi|}{f_\pi}\right)$.
From the very same discussion, but uplifted to chiral $SU(3)$, one can derive the Gellman Okubo mass formula for pions.
