quantum field theory - The proof of Goldstone's theorem

On page 352 of Peskin and Shroeder.

The athors show the proof of Goldstone's theorem.

A general continuous symmetry transformation has the form

$$\phi^a \to \phi^a + \alpha\Delta^a (\phi) ,\tag{11.12}$$ where $\alpha$ is an infinitesimal parameter and $\Delta^a$ is some function of all the $\phi$'s. Specialize to constant fields; then the derivative terms in $\cal{L}$ vanish and the potential alone must be invariant under (11.12). This condition can be written $$V(\phi^a)=V(\phi^a+ \alpha\Delta^a (\phi)) \quad \text{or} \quad \Delta^a (\phi)\frac{\partial}{\partial \phi^a}V(\phi)=0.$$ Now differentiate with respect to $\phi^b$, and set $\phi=\phi_0$: $$0=\left( \frac{\partial\Delta^a}{\partial \phi^b} \right)_{\phi_0} \left( \frac{\partial V}{\partial \phi^a} \right)_{\phi_0} + \Delta^a (\phi_0) \left( \frac{\partial^2}{\partial \phi^a \partial \phi^b}V \right)_{\phi_0} .\tag{11 .13}$$ The first term vanishes since $\phi_0$ is a minimum of $V$, so the second term must also vanish. If the transformation leaves $\phi_0$ unchanged (i.e., if the symmetry is respected by the ground state), then $\Delta^a (\phi_0)=0$ and this relation is trivial. A spontaneously broken symmetry is precisely one for which $\Delta^a (\phi_0)\ne 0$; in this case $\Delta^a (\phi_0)$ is our desired vector with eigenvalue zero, so Goldstone's theorem is proved.

Why in spontaneously broken symmetry $\Delta^a (\phi_0)\ne 0$?
And on the other hand when the symmetry is not spontaneously broken, $\Delta^a (\phi_0)= 0$?
Thanks.