# Why is a minimum field configuration called a vacuum state in SSB?

Most explanations of Spontaneous Symmetry Breaking (SSB) go like this: They take a scalar field Lagrangian with the "Mexican hat" potential $V(\phi)=−10\phi^2+\phi^4$ and argue that since the potential has an infinite number of possible minima, the QFT has an infinite number of vacuum states (see for example Wikipedia).

Why do we call the minimal fields (the $\phi$'s that minimize the Lagrangian) vacuum states? If the fields are operators, not states, then how can a field be a vacuum state?

Since the LSZ formalism and the Feynman rules need vanishing vacuum expectation value (VEV) to work, we need to rewrite fields with non-zero VEV in terms of perturbations about their VEV, i.e. $\tilde{\phi} := \phi - \langle 0 \vert \phi \vert 0\rangle$ are our dynamical fields. To first order, the VEV is well-approximated by the classical minimum, cf. this question and its answers and this answer by Prahar. At weak couplings, perturbative renormalization can ensure that the VEV of the perturbed field is zero order by order in perturbation theory (this should be shown in e.g. Coleman's Aspects of Symmetry).
$ϕ(x)$ is the operator which acts on vacuum creates a particle at x. The Vacuum is the state in which field is at rest with no particles there. So always the field which minimizes the action is called vacuum states. Another way of doing quantum field theory is doing Euclidean path integral,and perturbation theory just correspond to studying the small oscillation around a minimum of the euclidean action. In normal case $\phi$=0 minimizes the action so we have vacuum(at $\phi$=0 ) means no particles. With the change in sign of $\mu^2$ ,$\phi$=0 is not minimum, it is maximum.
After Spontaneous Symmetry Breaking (SSB) at different field configuration action is minimized at $\phi$=$\nu$.So this is called vacuum state. The field is said to have acquired a vacuum expectation value.