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I am a bit confused by what is defined to be a vacuum in field theory.

Classically a vaccum state is defined to be the state where the field sits at some minima of the potential $\frac{\partial V}{\partial \phi}=0$.

My main confusion is that when we go to quantum framework the vacuum state is defined to be a unique state in the Fock space with no-particles $|0\rangle$? I would be grateful if anyone can clarify the correspondence between these various definitions.

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  • $\begingroup$ Well, it's when the potential term in the Lagrangian $V(\phi)$ is at a minimum, i.e., when $\partial V/\partial\phi = 0$... $\endgroup$ – Alex Nelson Feb 12 '14 at 17:15
  • $\begingroup$ How does it correspond to the notion in QFT? $\endgroup$ – user40469 Feb 12 '14 at 17:19
  • $\begingroup$ IIRC, the vacuum expectation value $\langle\phi\rangle$ relates to the classical notion when taking the classical limit; but it has been too long since I have examined this in detail, so I don't want to mislead you posting an answer... $\endgroup$ – Alex Nelson Feb 12 '14 at 17:32
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    $\begingroup$ @user40469 The classical limit can be dealt with in the functional integral formulation using the stationary phase approximation. Try reading a bit about it; it won't answer all of your questions, but it does address at least some of them. $\endgroup$ – joshphysics Feb 12 '14 at 18:21
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In classical field theory, the system will indeed be in the minima of the potential, i.e., the point at which $\partial V/\partial \phi_i=0$ for all fields $\phi_i$ for all fields $\phi_i$ (more precisely one should include the fermionic fields here as well but fermions don't exist in a classical world).

In Quantum Field Theory the same applies as above except that particles are as excited states of the field. With no particles (the vacuum) the field is its ground state, the classical minima. Having one particle will correspond to one field configuration and having two particles will correspond to another. Now instead of remaining at the minima, the field can also move around the minima.

There is also a beautiful analogy provided in this post for how to picture these excitations (I prefer David Z's answer, but that's just me).

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  • $\begingroup$ Are you saying $<\phi> \to \text{solution of }\frac{\partial V}{\partial \phi}$ in the $\bar h \to 0$ limit $\endgroup$ – user40469 Feb 12 '14 at 17:45
  • $\begingroup$ I think that is correct, though I never thought much about the $\hbar \rightarrow 0$ in QFT. $\endgroup$ – JeffDror Feb 12 '14 at 17:53
  • $\begingroup$ The vacuum is not the classical minimum, since there are fluctuations that change the vacuum state. The vacuum is the minimum of the effective action $\Gamma[\varphi]$, where $\varphi=\langle \phi\rangle$. $\endgroup$ – Adam Feb 12 '14 at 18:13

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