# Why the minima of potential of classical Lagrangian called ''vacuum expectation''?

Why the minima of potential of classical Lagrangian is called the ''vacuum expectation value of Phi(field function)''.Is it really a vacuum expectation value of field operator at the vacuum states(at this state,the potential part of classical Lagrangian equals zero)?

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Well, the answer lies in the fact that you can always define your vacuum as the state with the lowest energy, and since your Lagrangian gives you the energies of your system, the lowest energy state lies in the minimum of the Lagrangian.

It seem to me that the expectation value of operator of field at the vacuum states depends on the properties of the vacuum states.In some case the one point(one leg) correlation function does not equal to zero.Then the expectation value of operator of field does not equal zero.So that the spontaneously breaking of symmetry has the origination from the properties of vacuum states.Is that correct?

Well, the properties of your vacuum state depend on your definition of the Lagrangian. In the case of your vacuum state not being zero, you can simply shift your field by some constant to make it zero. This amounts to a redefinition of your Lagrangian.
Spontaneous symmetry breaking also depends (obviously) on the definition of your Lagrangian. The most prominent example would be the $$\varphi^4$$-potential, which admits a U(1)-symmetry. One has now continuously many possible vacuum states, and by chosing one you break the symmetry of your problem while the Lagrangian itself remains symmetric. The degree of freedom of chosing one of those vacuum states amounts to the existence of the so-called Nambu-Goldstone boson.

Edit: By chosing a vacuum state I mean taking it as a starting point for your perturbative expansion.

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