divergence of the axial vector current is $$\partial_{\mu}{\bf A}^{\mu} \,=\, -V_{SB}'(\sigma)
\mbox{\boldmath $\pi$} \, ,$$ where the prime denotes differentiation with respect to $\sigma$. It is clear that PCAC is an operator identity only if $V_{SB}$ is linear in $\sigma$. The ground state of this theory ought to occur at $\sigma(x) = \sigma_{\rm gs} > 0$ and $\mbox{\boldmath $\pi$}(x) = \bf 0$. We must immediately impose three conditions; that the minimum of the full potential occur at $\sigma_{\rm gs}$, that PCAC hold for small fluctuations about $\sigma_{\rm gs}$, and that the pion has its physical mass there. Therefore, $$V_{SB}'(\sigma_{\rm gs}) \,=\, -f_{\pi}m_{\pi}^2 \, ,$$ $$\sigma_{\rm gs}(\lambda \sigma_{\rm gs}^2 - c^2) + V_{SB}'
(\sigma_{\rm gs}) \,=\, 0 \, ,$$ $$\lambda \sigma_{\rm gs}^2 - c^2 \,=\, m_{\pi}^2 \, ,$$ and so $\sigma_{\rm gs} = f_{\pi}$.

One must still ensure that the global minimum is really obtained when $\mbox{\boldmath $\pi$}(x) = \bf 0$. To investigate this problem, let us expand the fields about an arbitrary point as follows. $$\begin{aligned}
\sigma(x) &=& v\,\cos\theta + \sigma'(x) \, ,\\
\mbox{\boldmath $\pi$}(x) &=& {\bf v}\,\sin\theta +
\mbox{\boldmath $\pi$}'(x) \, .\end{aligned}$$ The primes denote fluctuations about the given point. The full potential is $$V(v,\theta) \,=\, \frac{\lambda}{4} \left(v^2 -c^2/\lambda\right)^2
