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In the convention implied by the printed expansion, the truncated two-point quantity is the inverse connected propagator, or one-particle-irreducible two-point vertex:It is the second functional derivative of the quantum effective action (or statistical Legendre effective action) about a translationally invariant zero-field equilibrium. The relation follows from the inverse Hessian relation for a connected two-point function. The subscript $c$ in $G$ already removes disconnected one-point products; the word “truncated” here is not a request to subtract that product a second time. Nor is a general amputated connected correlation function interchangeable with a one-particle-irreducible correlation function.

Use the zero-momentum renormalized mass $R=m^2(0,T)$ in the free propagator ${\tilde{G}}_0(p)=(p^2+R{)}^{-1}$, and split the quadratic coupling into $R$ plus a mass counterterm $\delta m^2=m^2(\Lambda ,T)-R$. The self-energy $\Sigma (p)$ is the sum of loop 1PI insertions, excluding the separately displayed counterterm. Summing repeated insertions by Dyson resummation givesThe sign convention is that a positive tadpole shifts the inverse propagator upwards. At first order, the connected correlation function correction is $-G_0^2\Sigma$, consistent with this inverse-propagator convention. A different split between the reference mass and counterterm produces the same renormalized result.

For the positive quartic interaction $g{\varphi }^4/4!$, the one-loop tadpole diagram has no external-momentum dependence. Its symmetry factor is $1/2$: assigning the two external legs to four vertex fields gives $4\cdot 3$ contractions, and division by $4!$ gives $1/2$. With the dimensionless statistical-action convention,The physical zero-momentum condition $\tilde{\Gamma }(0)=R$ sets $\delta m^2+{\Sigma }_1(0)=0$, hence the one-loop mass relation isWriting the internal line with $R$ is a renormalized or self-consistent one-loop convention. Away from critical infrared singularities it differs from a bare-mass insertion only at higher perturbative order. This equation does not by itself provide exact critical exponents once loop corrections become large.

For $D>2$, subtract the critical-temperature condition $0=m^2(\Lambda ,T_C)+(g/2)I_D(0)$. Take $g$ and the regular coefficients at their critical values, absorbing smooth changes into a coefficient $A>0$. Sincethe one-loop critical-mass subtraction becomesLet $K_D$ be the area of the unit $(D-1)$-sphere divided by $(2\pi {)}^D$. Radial integration gives $J_D(R)=K_D{\int }_0^{\Lambda }{k}^{D-3}(k^2+R{)}^{-1}dk$.

For $D>4$, $J_D(0)=K_D{\Lambda }^{D-4}/(D-4)$ is infrared finite, so it merely renormalises the coefficient and $R\propto T-T_C$ is consistent. For $2<D<4$, setting $k=\sqrt{R}x$ yieldswith a finite positive dimensionless integral. The correction is singular relative to the term $R$, invalidating the finite-coefficient linear-mass assumption. At $D=4$,so the boundary is logarithmically marginal. For $D\le 2$, even the subtraction using $I_D(0)$ needs an infrared regulator; it cannot be used to restore a finite linear critical expansion. Thus the ordinary upper critical dimension isA fixed-coupling self-consistent one-loop formula is not the full marginal renormalization group analysis, but its logarithm already shows why an uncorrected linear power law is not generic at $D=4$.

At a tricritical point, both the quadratic and quartic scaling directions must be tuned; the leading stabilising interaction is sextic. With canonical scalar-field engineering dimension $(D-2)/2$, the sextic coupling has eigenvalue $D-6(D-2)/2=6-2D$. It becomes marginal at $D=3$, givingLower even couplings generated by coarse-graining must remain tuned. This is why using an untuned quartic tadpole to diagnose a tricritical point would give the wrong boundary. The tricritical sextic beta function supplies marginal logarithmic corrections at three dimensions.

For a general multicritical even Landau potential, assume the lower stabilising even terms have been tuned away and the first remaining one is $A_{2n}{M}^{2n}$, with $n\ge 2$ and $A_{2n}>0$. Minimising the potential on its ordered branch givesIts curvature at the minimum is $4(n-1)a_2|t|$. For a finite positive gradient stiffness $c$, the longitudinal correlation length therefore scales as $\xi \propto |t{|}^{-1/2}$. The Ginzburg criterion compares the order-parameter fluctuation averaged over a correlation volume with this squared mean-field value. Keeping momenta of order ${\xi }^{-1}$ or less,Consequently the multicritical Ginzburg ratio behaves asOnly for a positive exponent do these relative fluctuations vanish on approaching the critical point. Thus the general upper critical dimension isEquivalently the interaction eigenvalue $y_{2n}=2n-(n-1)D$ vanishes there. The cases $n=2$ and $n=3$ reproduce 4 and 3 respectively.

For $D<D_C$, the ratio diverges and the mean-field assumptions lose self-consistency arbitrarily close to the transition. At $D=D_C$ the marginal Ginzburg criterion is scale-independent at this leading estimate, rather than tending to zero; the criterion alone does not prove a divergence or force new power indices. Marginal interactions require a renormalization group calculation and generally give logarithmic corrections, as for the quartic and sextic cases above. The boundary case is marginal, not a strict power-law divergence. This qualifies the printed wording at equality while recovering the requested upper critical dimensions.

The statistical Hamiltonian in this question is already in thermal units, as indicated by $e^{-H}$ without an additional ${\beta }_T$. Define $W[h]=\log Z[h]$ and $F[h]=-W[h]$. The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle $m_h$. It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.

The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, $h$ is chosen to produce $m$, and $\delta {\Gamma }_{\mathrm{L}}/\delta m=h$. Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in $F_{\mathrm{L}}$.

Differentiate the saddle equation with respect to $h(y)$. The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that $G$ is the inverse kernel of ${\delta }^2{\Gamma }_{\mathrm{L}}/\delta m(x)\delta m(y)$. The factor $u_0/2$ follows from twice differentiating the quartic term $u_0{m}^4/4!$; it is not $u_0/6$. This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.

For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so $m_h(x)=m_0$ and $G(x,y)=G(x-y)$. SetThe Fourier transform with the printed positive sign sends $-{\nabla }^2$ to $q^2$, while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is $G(x)=\int d^{D}q(2\pi {)}^{-D}{e}^{-iq\cdot x}\tilde{G}(q)$. For general inhomogeneous $h(x)$, $K$ has nonconstant coefficients and $G$ depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.

At zero source and for $u_0>0$, the homogeneous saddle is $m_0=0$ for $r_0>0$, and $m_0^2=-6r_0/u_0$ on either selected stable ordered branch for $r_0<0$. Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining $m_0=0$ below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning $r_0=a_0(T-T_c)$, $a_0>0$, both branches diverge as $|T-T_c{|}^{-1/2}$, so the correlation-length critical exponent isThe high-temperature amplitude is $\sqrt{2}$ times the low-temperature amplitude for the same $a_0$. This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.