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?


2 Answers 2


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.


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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    $\begingroup$ So then what is the proof of this leading-order approximation? $\endgroup$ Sep 1, 2013 at 22:13
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    $\begingroup$ @JonathanGleason How about Peskin Schroeder 11.3 ? $\endgroup$
    – Kostya
    Sep 1, 2013 at 22:15
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    $\begingroup$ 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.) $\endgroup$
    – Michael
    Sep 1, 2013 at 22:45
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    $\begingroup$ @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 . . . $\endgroup$ Sep 1, 2013 at 23:37
  • $\begingroup$ . . . 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 $\langle 0|\phi (x)|0\rangle =0$?". $\endgroup$ Sep 1, 2013 at 23:38

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