# Vacuum Expectation Value and the Minima of the Potential

Often times in quantum field theory, you will hear people using the term "vacuum expectation value" when referring to the minimum of the potential $V(\phi )$ in the Lagrangian (I'm pretty sure every source I've seen that explains the Higgs mechanism uses this terminology).

However, a priori, it would seem that the term "vacuum expectation value" (of a field $\phi$) should refer to $\langle 0|\phi |0\rangle$, where $|0\rangle$ is the physical vacuum of the theory (whatever that means; see my other question).

What is the proof that these two coincide?

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We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation: $$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi}$$ $$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$ Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be considered as a visible "response" of the system on the source and it usually denoted as a new variable, called the "classical field". We would like to find it when there are no external sources: $J=0$.
For that, one then does the Legendre transform trick, arriving at the effective action: $$\Gamma[\phi_{cl}] = - E - J\phi_{cl}\quad\quad\frac{\delta\,\Gamma}{\delta \phi_{cl}} = - J$$ Remembering our goal to find $\phi_{cl}$ at $J=0$, we arrive at the equation. $$\frac{\delta\,\Gamma}{\delta \phi_{cl}} = 0$$ Adding an extra assumption that $\phi_{cl}$ is space and time independent: $\phi_{cl}(x) = v$, the effective action functional $\Gamma[\phi_{cl}]$ is then reduced to effective potential $V_{eff}(v)$ and the equation becomes. $$\frac{dV_{eff}}{dv} = 0$$ Now, as David Vercauteren correctly pointed out, $V_{eff}(v)$ is not the same function as $V(\phi)$. But usually it is a good first approximation, because we usually consider systems where the "real" quantum field fluctuates weakly around its vacuum: $\phi(x)=v+\eta(x)$ with $\eta$ being small.

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They don't excactly coincide. In perturbation theory, the vev $\langle 0|\phi|0\rangle$ is equal to the value of $\phi$ at the minimum of $V(\phi)$ at leading order. The exact value of the vev is equal to this value-at-the-minimum plus perturbative (and at times also nonperturbative) corrections. Saying that they coincide is just a leading-order approximation.

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So then what is the proof of this leading-order approximation? –  Jonathan Gleason Sep 1 '13 at 22:13
@JonathanGleason How about Peskin Schroeder 11.3 ? –  Kostya Sep 1 '13 at 22:15
You are generally free to take as one of your renormalization conditions that the higher order corrections to the vev vanish. (You don't have to of course, but this is a particularly useful choice for many purposes.) –  Michael Brown Sep 1 '13 at 22:45
@MichaelBrown Indeed, I was under the impression that you had to have this as a re-normalization condition in order to apply LSZ (that is, a hypothesis require for the LSZ Reduction Formula to hold was that $\langle 0|\phi (x)|0\rangle =0$). In fact, I thought this was the entire idea behind the symmetry breaking: you must re-write your Lagrangian in terms of the re-normalized field (with vanishing VEV), and if the bare field had a non-vanishing VEV, this will 'break' the symmetry . . . –  Jonathan Gleason Sep 1 '13 at 23:37
. . . In practice, you do this by writing the Lagrangian in terms of $\phi :=\phi _0-v$, where $v$ is some minimum of the potential and $\phi _0$ is the original field. My question could then be equivalently phrased as "Why does this guarantee that $\langle 0|\phi (x)|0\rangle =0$?". –  Jonathan Gleason Sep 1 '13 at 23:38