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?


2 Answers 2


To call a classical extremum of the field a "vacuum state" is a common abuse of terminology. It is not the extremum itself that is a vacuum state, but to every classical extremum there is (to first order) an associated vacuum.

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).

Now, to each different VEV there must belong a different vacuum state - we can't have the same state and the same field giving two different VEVs, and there is nothing else in that expression the VEV could depend on. (Note that whether you say the vacuum state or the representation of the field as an operator is different doesn't matter, you get "different" Hilbert spaces (perturbatively) built on the vacuum state in both cases.) We also should concede that the existence of such vacua associated to each VEV is a postulate of QFT - since we can rarely explicitly construct the space of states of a QFT anyway, it seems infeasible to give any proof of existence of these vacua.


$ϕ(x)$ 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.

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    $\begingroup$ I have trouble deciphering this answer, partially due to its abysmal formatting. Use spaces after punctuation and capitalize beginnings of sentences. The question is why we call the classical minimum the "vacuum state" even though the field itself does not define a state, and a good answer to it would show the correspondence between the classical minimum, the vacuum expectation value, and the vacuum state. $\endgroup$
    – ACuriousMind
    Aug 12, 2016 at 19:41

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