Consider the vacuum expectation value of a (for simplicity scalar) field $\phi$, we know that its vacuum expectation value can be expressed as
$$\langle\phi\rangle=\frac{1}{\mathcal{Z}}∫\mathcal{D}\phi e^{\frac{i}{\hbar}S[\phi]}\phi.$$
If we introduce a source $J(x)$ then we can also define
$$\phi_{J}\equiv\langle\phi\rangle_{J}\equiv\frac{1}{\mathcal{Z}}∫ \mathcal{D}\phi e^{\frac{i}{\hbar}(S[\phi]+J\cdot\phi)}\phi.$$
Recall also that the quantum effective action is $$\Gamma[\phi_{J}]=W[J]-\phi_{J}\cdot J$$ with $W[J]$ being the Wilsonian effective action of our theory. My question is about the classical solution of $\Gamma$, in particular since
$$\frac{\partial\Gamma[\phi_{J}]}{\partial\phi_{J}(x)}=-J(x)\implies \frac{\partial\Gamma[\phi_{J}]}{\partial\phi_{J}(x)}|_{J=0}=0.$$
Now, my question is rather simple: Does it mean that the minima of the quantum effective action is just $\langle\phi\rangle$ without any source? Because if the answer is positive, then why we do say that the minima of the potential is not really $\langle\phi\rangle$ but can change due to quantum corrections?
For example, for the Higgs we have that the minima $v=\langle\phi\rangle$ is obtained by simply minimizing the tree-level potential but could be corrected using quantum correction (for example using the Coleman-Weinberg potential?), but this seems in contrast with the fact that $\langle\phi\rangle$ should minimize the fully quantum corrected $\Gamma[\phi_{J}]$. So, is possible that the minima of $$V_{\mathrm{Higgs}}(\phi)=\frac{1}{2}\mu^{2}\phi^{2}+\frac{1}{4}\lambda\phi^{4}$$ is not really $\langle\phi\rangle$ but rather something close to it (so in this sense we should write $\langle\phi\rangle=v+\mathrm{corrections}$)?
