# In QFT, is the effective potential at tree-level always the same as the potential? Why?

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.

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. $$\Box$$