What are exactly the loop correction to the potential? [duplicate]

I am struggling with the concept of quantum effective action, but first recall the definition : given a Wilsonian effective action $$W[J]$$ of our theory, the quantum effective action is just $$\Gamma[\phi_{J}]\equiv W[J]-\phi_{J}\cdot J\tag{1}$$ where the dot notation is used to indicate an integral over $$d^{4}x$$, and I have denoted with

$$\phi_{J}(x)\equiv<\phi>_{J}\equiv Z[J]^{-1}∫ \mathcal{D}\phi e^{-\frac{S[\phi]+J\cdot\phi}{\hbar}}\phi=\frac{δ W[J]}{δ J(x)}\tag{2}$$

the quantum average in presence of the external source $$J$$. If I understand correctly, $$\Gamma[\phi_{J}]$$ is just an object that, if somehow known, would let us to compute the partition function $$\mathcal{Z}$$ simply solving its equation of motion:

$$\frac{δ\Gamma}{δ\phi_{J}}\bigg|_{\phi_{J}=\phi_{c}}=0\iff\phi_{c}=\phi_{J=0}=<\phi>.\tag{3}$$

So plugging this "classical" solution into the quantum action gives

$$\mathcal{Z}[J]=e^{-\frac{1}{\hbar}W[J]}=e^{-\frac{1}{\hbar}(\Gamma[\phi_{J}]+J\cdot\phi_{J})}\Rightarrow \mathcal{Z}=∫ \mathcal{D}\phi e^{-\frac{S[\phi]}{\hbar}}=e^{-\frac{1}{\hbar}\Gamma[\phi_{c}]}.\tag{4}$$

which encodes the information about transition amplitudes. This shows how the quantum effective action "contains" all the quantum effects associated to the measure $$\mathcal{D}\phi$$. Here my questions:

• Why is this relevant if we are not able to compute fully $$\Gamma$$? I mean, it's usually said that it helps because it enable us to correct the potential with loop contributions $$V_{eff}=V+V_{1-loop}+...,\tag{5}$$ but then what do we do with that? To me this seems not easier than just do the loop calculations in perturbation theory using just $$S[\phi]$$, could you please explain how to put this things down to Earth? I don't think I really understand what does it mean to correct a potential with loops.
• Possible duplicates: physics.stackexchange.com/q/337898/2451, physics.stackexchange.com/q/414617/2451 and links therein. Commented Apr 12 at 7:14
• In eq. (4) is $\mathcal{Z}=\mathcal{Z}[J=0]$? Commented Apr 12 at 8:49
• yes sorry but I see your are already fixing Commented Apr 12 at 8:51
• How does $\mathcal{Z}=\mathcal{Z}[J=0]$ (which is essentially just a number; more precisely: a function of some coupling constants) depend on $\phi_c$ in eq. (4)? Commented Apr 12 at 9:01
• Well I just put $J=0$ from the previous equation, I am not really sure about it I just wrote what I thought was possible... if this is not correct please let me know. Commented Apr 12 at 9:09