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A logarithmic divergence grows like a logarithm as a cutoff is removed. For example ${\int }_{ϵ}^{1}dx/x=\log (1/ϵ)\to \infty$ as $ϵ\to 0^{+}$. An integrand $f(x)=C/x+O(1)$ near zero gives $C\log (1/ϵ)+O(1)$. This behavior distinguishes logarithmic sensitivity to an infrared or ultraviolet cutoff from power-law divergence.

Under the bipartite spin rotation, the staggered magnetization becomes the uniform transformed $S^z$. In the Bogoliubov transformation vacuum ${\alpha }_k|0_{\alpha }⟩=0$,The quantum depletion of Néel order therefore gives, with the zero modes regulated,Near each zero of $\sin k$, the integrand behaves as $1/(2|k-k_0|)$, producing a logarithmic divergence. With a finite-size cutoff of order $N^{-1}$, the depletion grows as ${\pi }^{-1}\log N+O(1)$. Thus the large-$S$ expansion about a state with finite Néel order is not self-consistent in the infinite one-dimensional chain. The divergent expression is not a negative physical magnetization; it signals breakdown of that ordered approximation. It does not determine whether the exact excitation spectrum is gapped.

Let $I_b,I_f$ count internal scalar and fermion lines and $L$ count independent loops in a connected amputated Feynman diagram. At large loop momentum, the scalar propagator falls as $k^{-2}$ and the Dirac propagator falls as $k^{-1}$, including its numerator. Each loop integration is $d^{3}k$. The nonderivative interaction vertices add no powers. ThereforeCount endpoints of each kind of line and use the connected-graph loop order identity:Eliminating $L,I_b,I_f$ givesThe cancellation of $n_1,n_3$ reflects their dimensionless couplings; every insertion of the positive-dimensional $g_2$ improves the superficial degree of divergence by one.

The printed exact homogeneity is shorthand for ultraviolet power counting, not an identity for a massive integrated correlation function. For example, even the free amputated scalar two-point vertex ${\Gamma }^{(2)}(p)=p^2-M^2$ obeys ${\Gamma }^{(2)}(\lambda p)={\lambda }^2{p}^2-M^2$, which differs from ${\lambda }^2{\Gamma }^{(2)}(p)$ when $M\ne 0$. Logarithmic divergences, actual numerator cancellations and ultraviolet subdivergences also require separate treatment.

Set the renormalized $g_3$ to zero. For a six-scalar vertex, b gives $\omega =-n_2$, so a new overall logarithmic divergence can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.

At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional toThis vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading $/k$ by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega =-3$. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. ThereforeThe six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in $2+1$ dimensions. In this representation,The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega =0$, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put $\varphi =v+\eta$ with constant $v$. Define the background fermion mass $\mu =m+g_1{v}^2/2$, the background scalar squared mass ${\mathcal{B}}^2=M^2+g_2{v}^2/2$, and the induced three-leg Yukawa coupling $y=g_{1}v$. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, isHere the dots and gamma-matrix traces are Euclidean. With $A=k^2+{\mu }^2$, $B=l^2+{\mu }^2$, $C=(k-l{)}^2+{\mathcal{B}}^2$, its numerator is exactlyThe first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term isFor positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to ${\int }^{\Lambda }dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying $H$ containsThe two-loop sextic counterterm in quartic Yukawa theory follows because its $v^6$ coefficient is $g_1^4\ne 0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. ThusA sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to $g_3=0$ at every scale lacks counterterm closure when $g_1\ne 0$. The decoupled exception $g_1=0$ is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.