My answer is based on the book "Quantum field theory in a nutshell, by A. Zee, Chapter IV.3 : Effective Potential"

Take a scalar field theory $(\phi)$, when you have a classical potential $V(\phi)$, you can find the state V.E.V. (vacuum expectation value) $\phi_0$ by minimizing this potential $V(\phi)$.
But quantum fluctuations change this potential, so to calculate the modifyed state V.E.V. $\phi_{eff}$, you have first to calculate the modifyed potential $V_{eff}(\phi)$.

Define $W(J) = i log Z(J)$, You have V.E.V. $\phi(x) = \frac{\delta W(J)}{\delta J(x)}$ (a particular example of Green function obtained by diffentiating $W$ by the source $J$ repeatedly).
By using a Legendre transformation, define $\Gamma(\phi) = W(J) - \int d^4x J(x) \phi(x)$.
Note that there is an implicite dependence of $J$ on $\phi$, and you have $-J(x) = \frac{\delta \Gamma(\phi)}{\delta \phi(x)}$.

You may define $V_{eff}$ by : $\Gamma(\phi) = \int d^4x (- V_{eff}(\phi) + Z(\phi) (\partial\phi)^2) + ...$, which is justifyed because you can show that $V^\prime_{eff}(\phi) = J$ (rigourously when $\phi$ and $J$ do not depend on x). This means, that without source, $V^\prime_{eff}(\phi) = 0$, so $V_{eff}$ is the correct potential which, minimized, gives the ground state.

With step descent approximation applyied to $Z$, around the step descent point field $\phi_s$, where $V^{\prime}(\phi_s) = 0$ you will find, with explicit $\hbar$:
$$Z(J) = \int D\phi\;e^{(i/\hbar)[S(\phi)+ J\phi]}$$
$$Z(J) \approx e^{(i/\hbar)[S(\phi_s)+ J\phi_s]} \int D\phi\;e^{(i/\hbar)\int d^4x\,\frac{1}{2}
[(\partial\phi)^2) - V^{\prime\prime}(\phi_s)\phi^2]}$$
In the previous expression, the new integration variable is $\phi= \phi-\phi_s$.
$$Z(J) \approx e^{(i/\hbar)[S(\phi_s)+ J\phi_s] - \frac{1}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi_s)) - Tr\, log (\partial^2))}$$
Note that the appearance of the term $Tr\, log (\partial^2))$ may be seen as a renormalization factor.

$$W(J) = S(\phi_s) + J\phi_s + i\frac{\hbar}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi_s)) - Tr\, log (\partial^2)) +O(\hbar^2)$$
$$\Gamma(\phi) = S(\phi) + i\frac{\hbar}{2} (Tr\, log (\partial^2 +V^{\prime\prime}(\phi)) - Tr\, log (\partial^2)) +O(\hbar^2)$$

With $\phi$ independant of x, you can write : $$Tr\, log (\partial^2 +V^{\prime\prime}(\phi)) = \frac{1}{(2\pi)^4}\int d^4x \int d^4k \,\,log \frac{(-k^2 + V^{\prime\prime}(\phi))}{-k^2}$$

So, finally, by the definition of $V_{eff}(\phi)$ from $\Gamma(\phi)$ $$V_{eff}(\phi) = V(\phi) - i\frac{\hbar}{2} \frac{1}{(2\pi)^4} \int d^4k \,\,log \frac{(k^2 - V^{\prime\prime}(\phi))}{k^2} + O(\hbar^2)$$

In practice, this integral is quadratically divergent, so you have to put a cut-off $\Lambda$, and you get : $$V_{eff}(\phi) = V(\phi) + \frac{\Lambda^2}{32\pi^2}V^{\prime\prime}(\phi) - \frac{1}{64\pi^2}[V^{\prime\prime}(\phi)]^2 log\frac{e^{1/2}\Lambda^2}{V^{\prime\prime}(\phi)}$$
So, by minimizing this effective potential $V_{eff}$, you can find the new state V.E.V. $\phi_{eff}$.

Now, in 3+1 dimensions, a scalar like $\phi_{eff}$ has the dimension of a mass (or energy), so by some Higgs mechanism, it can give mass to spin 1 gauge bosons, but the minimum of the effective potential $V_{eff}$ and the effective V.E.V. field $\phi_{eff}$ are 2 different things.

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