= Solution
With $Z[J]=e^{iW[J]}$, the logarithm $W$ is the <connected generating functional>. Its derivative $\varphi=\delta W/\delta J$ is the mean field. The specified <Legendre transform> satisfies $\delta\Gamma/\delta\varphi=J$, and the <quantum effective action> generates amputated <one-particle-irreducible vertices> by further <functional derivatives>. In this question $\Gamma=-W+\int J\varphi$ is the negative of the commonly used Minkowski effective-action convention; consequently its tree term is $-S[\varphi]$. Its diagrammatic content is unchanged if the convention is used consistently.
Expand the action around a stationary background in the presence of the source. For $\phi=\varphi+f$, <integration by parts> gives
$$
S[\varphi+f]+\int J(\varphi+f)=S[\varphi]+\int J\varphi+\int\left(\frac{\delta S}{\delta\varphi}+J\right)f-\frac12\int f\Delta f+O(f^3),\qquad\Delta=-\partial^2+V''(\varphi).
$$
The linear term vanishes at the tree-level saddle, $\delta S/\delta\varphi+J=0$. Keeping the <Gaussian fluctuation approximation>, and fixing the field-independent normalization, gives
$$
Z[J]\simeq e^{iS[\varphi]+i\int J\varphi}(\det\Delta_F)^{-1/2},\qquad W[J]=S[\varphi]+\int J\varphi+\frac i2\log\det\Delta_F
$$
to one-loop order. The shift from the tree saddle to the exact mean field cancels from the tree Legendre transform at this order. Hence the <one-loop scalar effective action> in the question's convention is
$$
\boxed{\Gamma[\varphi]=-S[\varphi]+\frac1{2i}\log\det\Delta_F.}
$$
The <functional determinant> has a <Feynman i-epsilon prescription> before <Wick rotation>. Higher loops come from the omitted cubic and higher fluctuation vertices.
To see the one-loop <one-particle-irreducible Feynman diagrams>, write $\Delta=\Delta_0+U(\varphi)$. Its determinant expansion is
$$
\operatorname{Tr}\log\Delta=\operatorname{Tr}\log\Delta_0+\sum_{r\ge1}\frac{(-1)^{r+1}}r\operatorname{Tr}\bigl[(\Delta_0^{-1}U)^r\bigr].
$$
Each trace closes a single scalar loop, with propagators $\Delta_0^{-1}$ and background insertions $U$. No internal edge is a bridge, so these graphs are <one-particle irreducible>. Expanding the background factors produces their external legs.
For a constant $\varphi$, set $A=V''(\varphi)>0$. The <quantum effective potential> is defined here by $\Gamma[\varphi]=\int d^dx\,V_{\rm eff}(\varphi)$. A plane-wave basis diagonalizes $\Delta$, and the supplied <Wick rotation> converts its correction per unit volume to
$$
V_1(A)=\frac12\mu^\epsilon\int\frac{d^{4-\epsilon}k_E}{(2\pi)^{4-\epsilon}}\log(k_E^2+A)=-\frac{\mu^\epsilon\Gamma(-2+\epsilon/2)}{2(4\pi)^{2-\epsilon/2}}A^{2-\epsilon/2}.
$$
Here the power of the <renormalization scale> keeps the continued integral in four-dimensional units. The <Gamma function recurrence> and expansion give
$$
\Gamma(-2+\epsilon/2)=\frac1\epsilon+\frac34-\frac\gamma2+O(\epsilon),\qquad A^{2-\epsilon/2}=A^2\left(1-\frac\epsilon2\log A+O(\epsilon^2)\right).
$$
Consequently
$$
V_1(A)=-\frac{A^2}{32\pi^2\epsilon}+\frac{A^2}{64\pi^2}\left(\log\frac A{\mu^2}-\frac32+\gamma-\log4\pi\right)+O(\epsilon).
$$
The divergence is local, so subtract it with a potential <counterterm>. In the <modified minimal subtraction scheme>, remove its associated $\gamma-\log4\pi$ constants as well. The <one-loop scalar effective potential> becomes
$$
\boxed{V_{\rm eff}(\varphi)=V(\varphi)+\frac{[V''(\varphi)]^2}{64\pi^2}\left(\log\frac{V''(\varphi)}{\mu^2}-\frac32\right),}
$$
up to an arbitrary field-independent vacuum-energy constant. Another allowed finite subtraction replaces the nonlogarithmic constant. Restoring a loop-counting parameter would multiply the displayed correction by $\hbar$. The logarithm assumes positive curvature $A$; a background with negative curvature requires an analytic continuation and can have an imaginary part signalling instability.
The quartic case is special because of <quartic closure of scalar effective-potential counterterms>. If $V$ has degree at most four, then $V''$ has degree at most two and $(V'')^2$ has degree at most four: its pole is removable by the original polynomial couplings and a constant. For example, $V=m^2\varphi^2/2+\lambda\varphi^4/4!$ gives
$$
\frac{(V'')^2}{32\pi^2\epsilon}=\frac{m^4}{32\pi^2\epsilon}+\frac{\lambda m^2\varphi^2}{32\pi^2\epsilon}+\frac{\lambda^2\varphi^4}{128\pi^2\epsilon}.
$$
The last two terms reproduce the <one-loop massive phi-fourth counterterms>, and the first is a vacuum-energy subtraction. A degree-$r$ potential with $r>4$ instead generates degree $2r-4>r$, forcing new operators. This is the potential-level manifestation of <power counting in quantum field theory> and four-dimensional <renormalizability>, even though the finite effective potential itself need not remain polynomial.
Back to article page
