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

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Discussion Overview

The discussion revolves around the concept of why the minima of the potential in classical Lagrangian mechanics is referred to as the "vacuum expectation value" of a field function. It explores the relationship between the vacuum state, energy levels, and the properties of the Lagrangian, touching on themes of symmetry breaking and field operators.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asks why the minima of the potential in classical Lagrangian is termed the "vacuum expectation value" and questions if it truly represents the vacuum expectation value of the field operator at vacuum states.
  • Another participant suggests that the vacuum can be defined as the state with the lowest energy, linking this to the minima of the Lagrangian.
  • A participant notes that the expectation value of the field operator at vacuum states may depend on the properties of those vacuum states, indicating that the one-point correlation function can be non-zero, which relates to spontaneous symmetry breaking.
  • Another response emphasizes that the properties of the vacuum state are contingent on the definition of the Lagrangian, mentioning that one can shift the field to redefine the vacuum state and discussing the implications for spontaneous symmetry breaking in the context of the \varphi^4-potential.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the vacuum state and the Lagrangian, particularly regarding the implications of vacuum state properties and symmetry breaking. There is no consensus on the interpretations presented.

Contextual Notes

The discussion includes assumptions about the definitions of vacuum states and Lagrangians, as well as the implications of symmetry breaking, which remain unresolved. The dependence on specific models and definitions is acknowledged but not fully explored.

ndung200790
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Please teach me this:
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)?
Thankyu very much in advanced.
 
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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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