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If I have a simple scalar theory

$$ \mathcal{L}(\phi) = \frac{1}{2} (\partial_\mu \phi)^2 - V(\phi), $$

the effective potential $V_{eff}$, derived from $Z\rightarrow W \rightarrow \Gamma \rightarrow V_{eff}$, has at tree-level precisely the same formula of the potential $V(\phi)$. I guess this makes sense, or they would've chosen a different name.

I can't find an answer to the question: is this always the case, at tree-level? Why or why not? If so, is there a motivation/intuition for this?

Addendum: the question linked in the comment only partially answers my questions. I don't understand where the kinetic term goes, nor how the integral in $S=\int\mathcal{L}$ does not play any role.

Mauro Giliberti
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1 Answers1

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OP's sought-for formula $${\cal V}_\text{eff,tree-level}(\phi_{\rm cl})~=~{\cal V}(\phi_{\rm cl})$$ follows from the following facts:

  1. Ref. 1 defines the effective potential ${\cal V}_{\rm eff}(\phi_{\rm cl})$ for $x$-independent field configurations $\phi_{\rm cl}$ only. Then the kinetic terms don't contribute. Ref. 1 then writes the effective action as $$ \Gamma[\phi_{\rm cl}]~=~-(VT){\cal V}_{\rm eff}(\phi_{\rm cl}), \tag{11.50}$$ where $VT$ is the 4-volume of spacetime.

  2. Similarly, for $x$-independent field configurations $\phi$, we can write the action $$ S[\phi]~=~-(VT){\cal V}(\phi)$$ with the help of the potential ${\cal V}(\phi)$.

  3. Finally use the fact that $$ \Gamma_\text{tree-level}[\phi_{\rm cl}]~=~S[\phi_{\rm cl}], $$ cf. e.g. eq. (12) in my Phys.SE answer here. $\Box$

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eq. (11.50).
Qmechanic
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