Vacuum Expectation Value and the Minima of the Potential Often times in quantum field theory, you will hear people using the term "vacuum expectation value" when referring to the minimum of the potential $V(\phi )$ in the Lagrangian (I'm pretty sure every source I've seen that explains the Higgs mechanism uses this terminology).
However, a priori, it would seem that the term "vacuum expectation value" (of a field $\phi$) should refer to $\langle 0|\phi |0\rangle$, where $|0\rangle$ is the physical vacuum of the theory (whatever that means; see my other question).
What is the proof that these two coincide?
 A: They don't excactly coincide. In perturbation theory, the vev $\langle 0|\phi|0\rangle$ is equal to the value of $\phi$ at the minimum of $V(\phi)$ at leading order. The exact value of the vev is equal to this value-at-the-minimum plus perturbative (and at times also nonperturbative) corrections. Saying that they coincide is just a leading-order approximation.
A: We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation:
$$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$
$$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$
Where $\langle\phi\rangle_J$ is the v.e.v  of $\phi$ in presence of external source $J$. That could be considered as a visible "response" of the system on the source and it usually denoted as a new variable, called the "classical field". We would like to find it when there are no external sources: $J=0$.
For that, one then does the Legendre transform trick, arriving at the effective action:
$$\Gamma[\phi_{cl}] = - E - J\phi_{cl}\quad\quad\frac{\delta\,\Gamma}{\delta \phi_{cl}} = - J$$
Remembering our goal to find $\phi_{cl}$ at $J=0$, we arrive at the equation.
$$\frac{\delta\,\Gamma}{\delta \phi_{cl}} = 0$$
Adding an extra assumption that $\phi_{cl}$ is space and time independent: $\phi_{cl}(x) = v$, the effective action functional $\Gamma[\phi_{cl}]$ is then reduced to effective potential $V_{eff}(v)$ and the equation becomes.
$$\frac{dV_{eff}}{dv} = 0$$
Now, as David Vercauteren correctly pointed out, $V_{eff}(v)$ is not the same function as $V(\phi)$. But usually it is a good first approximation, because we usually consider systems where the "real" quantum field  fluctuates weakly around its vacuum: $\phi(x)=v+\eta(x)$ with $\eta$ being small.
