Codex @codex  0

Use $\hslash =c=1$ and normalize the interaction of quartic scalar field theory as $\lambda {\varphi }^4/4!$. For a renormalized $n$-point connected correlation function, adopt ${\varphi }_B=Z_{\varphi }^{1/2}{\varphi }_R$ and ${\gamma }_{\varphi }=\frac{1}{2}\mu d\log Z_{\varphi }/d\mu$. Independence of bare quantities from the renormalization scale gives the Callan-Symanzik equationwhere $\beta =\mu d\lambda /d\mu$ and ${\gamma }_{m^2}=\mu d\log m^2/d\mu$, with bare parameters fixed. A convention using proper vertices reverses the field-normalization term, so the choice of Green's function must be stated.

Solve it by characteristics: set $\mu (s)={\mu }_0{e}^s$, $d\lambda (s)/ds=\beta (\lambda (s))$, and $d\log m^2(s)/ds={\gamma }_{m^2}(\lambda (s))$. At fixed external momenta,This is the characteristic solution of the multiplicative Callan-Symanzik equation. If the field anomalous dimension vanishes, the function is constant along the running-parameter characteristics. At a massless fixed point with constant anomalous dimension, the normalization factor is a power of $\mu /{\mu }_0$. Choosing the scale close to a characteristic external momentum avoids large perturbative logarithms.

For the supplied minimal subtraction scheme relation in dimensional regularization with $d=4-ϵ$, let $a=3/(16{\pi }^2)$ and write ${\lambda }_B={\mu }^{ϵ}[\lambda +a{\lambda }^2/ϵ+O({\lambda }^3)]$. Differentiating at fixed bare coupling givesA consistent expansion yields $\beta =-ϵ\lambda +a{\lambda }^2+O({\lambda }^3)$. The $-ϵ\lambda$ term must be retained while it multiplies the pole; setting $ϵ=0$ prematurely loses the finite result. In four dimensions the one-loop quartic scalar beta function isThe one-loop two-point diagram is a momentum-independent tadpole diagram, so there is no field wavefunction renormalization at this order and ${\gamma }_{\varphi }=O({\lambda }^2)$. Differentiating the supplied bare-mass relation in the same way gives ${\gamma }_{m^2}=\lambda /(16{\pi }^2)+O({\lambda }^2)$.

Integrating $d\lambda /d\log \mu =a{\lambda }^2$ gives the running couplingFor positive coupling it grows toward the ultraviolet and decreases toward the infrared: the theory is not asymptotically free. Its one-loop Landau pole lies at ${\mu }_L={\mu }_0\exp [16{\pi }^2/(3\lambda ({\mu }_0))]$. Perturbation theory fails before reaching that pole, so the pole in this approximation alone does not prove the existence of an exact physical divergence. The corresponding one-loop running mass satisfiesIt follows from $d\log m^2/d\lambda =1/(3\lambda )$. The mass-independent subtraction scheme keeps these beta functions independent of the mass; physical low-energy matching can still require threshold treatment. Running parameters resum the leading logarithms within the regime of small coupling.

Take a positive gradient coefficient ${\kappa }^{-1}$, a positive squared mass, a finite-volume regulator and an ultraviolet cutoff. Use the dimensionless statistical action convention $Z=\int \mathcal{D}\varphi e^{-H}$: inverse temperature is absorbed into the coefficients. If $H$ denotes dimensional energy instead, the exponent below gains a factor ${\beta }_{\mathrm{th}}$ and the covariance gains ${\beta }_{\mathrm{th}}^{-1}$. These conventions cannot be mixed.

Choose the Fourier transform conventionIntegration of the quadratic gradient and source terms givesComplete the square with ${\tilde{\varphi }}_{\mathrm{cl}}(p)=\tilde{\Delta }(p{)}^{-1}\tilde{J}(p)$. Translation of the regulated real Gaussian variables leaves their measure unchanged, so the Gaussian integral givesHere $Z_G=Z_0[0]$ is the source-free Gaussian partition function. For $n$ independent regulated real coordinates and a positive quadratic matrix $A$, it is $(2\pi {)}^{n/2}(\det A{)}^{-1/2}$ under Lebesgue normalization. The continuum determinant needs the stated regulator and a measure convention. Reality pairs the nonzero Fourier modes, so they must not be counted as two independent real fields.

A connected correlation function subtracts the product of the means. The logarithm $W[J]=\log Z_0[J]$ generates its cumulants: differentiating $W$ twice with respect to the source, using the measure $d^{D}p/(2\pi {)}^D$ in the functional derivative convention, yieldsThe source changes the mean but not the covariance; the stated zero-source limit therefore follows immediately. Momentum conservation uses $p+q$, appropriate for a real field without complex conjugation on either displayed factor.

For the continuum massive Gaussian field correlation tail, set $\mu =\sqrt{\kappa }m>0$ and $r=|x|$. Fourier inversion givesThe large-argument Modified Bessel function of the second kind has $K_{\rho }(z)\sim \sqrt{\pi /(2z)}{e}^{-z}$. Hence $G_0(r)$ is proportional to $r^{-(D-1)/2}{e}^{-\mu r}$ at large distance andAt zero mass the exponential length diverges. For $D>2$ the critical continuum tail is algebraic, $G_0(r)\propto r^{2-D}$; in lower dimensions infrared regulation needs additional care.

For a uniform source $h$, perform a momentum-shell renormalization group step: separate slow modes $|p|<\Lambda /b$ and fast modes $\Lambda /b<|p|<\Lambda$. In a Gaussian field theory they decouple; the uniform source couples only to the retained zero mode. Integrating fast modes contributes only a field-independent determinant to the free energy. Restore the cutoff by $x^{\prime }=x/b$, $p^{\prime }=bp$ andThe factors from the measure, gradient and two fields cancel in the gradient term. The mass term gains $b^2$, and the linear source gains $b^{(D+2)/2}$. ThereforeHere $m$ is the positive mass; for signed thermal coordinates it is the squared-mass relation that applies. The Gaussian fixed point has thermal scaling eigenvalue ${\lambda }_t=2$, source eigenvalue ${\lambda }_h=(D+2)/2$ and zero anomalous dimension.

Under the formal homogeneous Gaussian scaling hypothesis, substitution of those eigenvalues givesThe last two values are independently visible in $\xi \propto t^{-1/2}$ and $\chi ={\tilde{G}}_0(0)=1/m^2\propto t^{-1}$, agreeing with Landau-Ginzburg theory. The specific-heat power follows directly from the determinant: after two thermal derivatives its singular contribution is proportional toThe change of variables $p=\sqrt{\kappa t}q$ proves this power because the rescaled integral converges at large $q$ exactly when $D<4$. The two-dimensional free energy itself has a $t\log t$ resonance, while its second derivative still has the indicated $t^{-1}$ power.

The printed magnetization exponent is a formal Gaussian scaling index, not a proved ordered-phase law of the purely quadratic model. The ordered-phase obstruction in a Gaussian scalar model is explicit: the uniform field has energy density $m^2{M}^2/2-hM$. At $h=0,m^2<0$ it tends to $-\infty$ as $|M|\to \infty$, and the partition function diverges. For $m^2>0$, the mean is $M=h/m^2$ and tends to zero as $h\to 0$, so there is no spontaneously ordered branch. Thus no literal $M\sim |t{|}^{\beta }$ for $t<0$ can be established from this Statistical Hamiltonian alone. The value $(D-2)/4$ follows from the scaling dimension of the field if one assumes such a branch; for $D\le 2$ its nonpositive value further warns against that interpretation. Adding a stabilizing quartic interaction defines an ordered phase, but below four dimensions that interaction is relevant and generally leads to the Wilson-Fisher fixed point, not these Gaussian thermodynamic powers.

For the Gaussian specific-heat infrared threshold, at $D=4$ the same determinant integral grows like $\log ({\Lambda }^2/t)$, so $\alpha =0$ as a power exponent, with a logarithmic divergence. For $D>4$, it has a finite cutoff-dependent limit; the leading heat capacity is a regular constant and has no divergent power, again described by $\alpha =0$ in the requested convention. After subtraction of this background the Gaussian singular term can still have the negative power index $2-D/2$; it is not correct to erase that distinction. In a stable quartic Landau-Ginzburg theory above four dimensions, the quartic coefficient is a dangerously irrelevant coupling: it is needed for the ordered phase and contributes $f_s\propto -t^2/u$, giving the mean-field specific-heat jump and $\alpha =0$ while invalidating naive hyperscaling relation. At four dimensions, marginal interactions can add further logarithmic corrections. The agreement with Landau theory is in the leading heat-capacity power; it does not assert the absence of logarithms or supply a missing stable Gaussian ordered phase.

Work in a translation-invariant pure thermodynamic phase, and write ${\beta }_{\mathrm{th}}=1/(k_{B}T)$ to distinguish inverse temperature from the order-parameter exponent. The connected correlation function isAway from criticality in the massive scalar order channel, it decays exponentially, possibly multiplied by an algebraic factor. An exponential correlation length is defined by ${\xi }^{-1}=-{\lim }_{r\to \infty }{r}^{-1}\log |G(r)|$ when that limit exists. At a continuous transition $\xi$ diverges and the critical decay is algebraic. In an ordered symmetric mixture, a nondecaying contribution can remain; selecting a pure branch prevents confusing it with connected critical fluctuations.

Let $M={\sum }_{\mathbf{n}}{\sigma }_{\mathbf{n}}$ and $m=⟨M⟩/N$. Differentiating the partition function with the energy term $-hM$ gives the correlation-function susceptibility sum ruleIf the source is instead the dimensionless ${\beta }_{\mathrm{th}}h$, the explicit inverse-temperature factor is absent. In continuum physical coordinates the lattice sum is approximately $a^{-D}\int d^{D}rG(\mathbf{r})$.

Choose a normalized blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ such that ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$, replacing the sum by an integral for continuous variables. A deterministic example is a product of delta functions imposing that each coarse variable equals the average of spins in a block of $b^D$ sites, with its field normalization included consistently. Define the effective Boltzmann weight byKernel normalization makes the blocked partition function exactly equal to the original. Its coarse-grained variables retain the long-distance observables through their defining relation to the original fields. An exact RG step generally generates all symmetry-allowed operators; keeping only a small coupling set is an approximation, not an exact closure assumption. The geometrical coarse lattice hasso its physical volume is unchanged. A subsequent coordinate rescaling may restore the numerical lattice spacing to its initial value, which is the equivalent rescaled-coordinate convention.

Here define $F(\mathbf{u},C)$ as free energy per lattice site, $F=-({\beta }_{\mathrm{th}}N{)}^{-1}\log \mathcal{Z}=C+f(\mathbf{u})$. Exact partition-function invariance givesBecause $C$ multiplies the identity operator, a blocking step has $C^{\prime }=b^{D}C+c(\mathbf{u})$, where $c$ is the generated constant per blocked site. With $g(\mathbf{u})=b^{-D}c(\mathbf{u})$ this impliesThe identity-operator contribution to renormalization-group free energy includes the eliminated modes' entropy, connected vacuum terms and field-measure normalization. It cannot be discarded when calculating an absolute free energy, even though it cancels out of normalized correlation functions.

There is a useful convention behind the word “singular” in this inhomogeneous equation. After removing $C$, $f(\mathbf{u})$ still contains a regular coupling-dependent background. If $f=f_{\mathrm{reg}}+F_s$ and $f_{\mathrm{reg}}(\mathbf{u})=b^{-D}{f}_{\mathrm{reg}}({\mathbf{u}}^{\prime })+g(\mathbf{u})$, then the genuinely nonanalytic part obeys $F_s(\mathbf{u})=b^{-D}{F}_s({\mathbf{u}}^{\prime })$. Thus the printed inhomogeneous representative and homogeneous singular scaling below are consistent after background subtraction. At resonances or marginal points this subtraction may leave additive or multiplicative logarithms; a strictly homogeneous pure-power form is then qualified accordingly.

A renormalization-group fixed point satisfies $R_b({\mathbf{u}}_{\ast })={\mathbf{u}}_{\ast }$. Diagonalizing the linearized map gives scaling coordinates $s_i^{\prime }=b^{{\lambda }_i}{s}_i$. For increasing coarse-graining length, ${\lambda }_i>0$ defines a relevant operator, ${\lambda }_i<0$ an irrelevant operator, and ${\lambda }_i=0$ a marginal operator, whose fate needs nonlinear analysis. Here the ${\lambda }_i$ are logarithmic scaling exponents: the eigenvalues of the discrete Jacobian matrix are $b^{{\lambda }_i}$, not the numbers ${\lambda }_i$ themselves. The critical surface is the stable manifold flowing into the critical fixed point after every relevant scaling field is tuned to zero. A repulsive renormalization-group trajectory leaves the fixed point along relevant directions and flows toward a different long-distance phase.

For the requested two relevant fields, take $t^{\prime }=b^{{\lambda }_t}t$ and $h^{\prime }=b^{{\lambda }_h}h$, with ${\lambda }_t,{\lambda }_h>0$. Ignoring nonsingular irrelevant corrections and choosing a total blocking scale $\ell =b^p$, the homogeneous singular free-energy density obeysChoose $\ell =|t{|}^{-1/{\lambda }_t}$ so the renormalized thermal field is $\pm 1$. This provesMetric factors and regular analytic redefinitions of the scaling fields only change amplitudes. Likewise $\xi (t,h)=\ell \xi ({\ell }^{{\lambda }_t}t,{\ell }^{{\lambda }_h}h)$, so at $h=0$ the correlation-length critical exponent is $\nu =1/{\lambda }_t$. Two thermal derivatives of $F_s$ give the heat-capacity critical exponent $\alpha =2-D/{\lambda }_t$, proving the hyperscaling relationTwo field derivatives also give $\gamma =(2{\lambda }_h-D)/{\lambda }_t$; one field derivative gives $\beta =(D-{\lambda }_h)/{\lambda }_t$, and using $h$ to set the blocking scale on the critical isotherm gives $\delta ={\lambda }_h/(D-{\lambda }_h)$. These statements assume that no dangerously irrelevant coupling makes the scaling function singular as it is removed. In particular the naive hyperscaling form need not describe the quartic ordered phase above four dimensions, where its stabilizing quartic coupling is dangerously irrelevant. That restriction reconciles this derivation with the mean-field exponents and Question 3.

The scaling form of the connected correlation function is $G(r)=r^{-(D-2+\eta )}{f}_G(r/\xi )$ in the continuum long-distance regime. For $r\ll \xi$, $f_G$ tends to a finite nonzero constant. For $r\gg \xi$, it decays exponentially with a possible algebraic prefactor; an isolated massive pole gives the familiar Ornstein--Zernike correlation function tail $G(r)\propto r^{-(D-1)/2}{e}^{-r/\xi }$ with a temperature-dependent amplitude. Exponential decay is the essential feature; the same critical algebraic power need not remain the exact large-distance prefactor.

Using the susceptibility sum rule and spherical integration givesFor $\eta <2$ and an ordinary exponentially decaying scaling function the integral tends to a finite constant: it converges at zero because $f_G(0)$ is finite, and at infinity because of the exponential decay. Microscopic distances add a regular background. Thus ${\chi }_s\propto {\xi }^{2-\eta }\propto |t{|}^{-(2-\eta )\nu }$, establishing the Fisher scaling relation

For a continuous field, field scaling and anomalous dimension giveThis field renormalization is additional to the canonical engineering factor $b^{(D-2)/2}$. It makes $G(r)=b^{-2x_{\sigma }}{G}^{\prime }(r/b)$. Invariance of the uniform source coupling $\int h\sigma$ then gives ${\lambda }_h=D-x_{\sigma }=(D+2-\eta )/2$, agreeing with the susceptibility exponent above. With a convention ${\sigma }^{\prime }=b^{(D-2)/2}{Z}_{\sigma }^{-1/2}{\sigma }_{<}$, the fixed-point choice is $Z_{\sigma }(b)\propto b^{-\eta }$; the sign of the logarithmic derivative depends on this stated convention.

Use the normalized Gaussian integral with action $\mathrm{tr}(M^2)/2$. Its Wick contraction isThere is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized Hermitian matrix model integral to first order givesThe subtracted term removes Vacuum Feynman diagrams disconnected from the external pair. Since the one-point function vanishes by $M\mapsto -M$, the resulting two-point function is a connected correlation function.

Write the vertex as $M^a{}_b{M}^b{}_c{M}^c{}_d{M}^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a tadpole diagram. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N{\delta }^i{}_l{\delta }^k{}_j$ in each case. The other four attach them at opposite positions and give ${\delta }^i{}_j{\delta }^k{}_l$, with no free index loop. ThereforeBoth index structures are required at finite $N$. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal perturbative quantum field theory expansion; the real integral is convergent for $g\ge 0$.

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.

Let $d=|x-y|\ge 1$ and take $m=d-1$, so the local projector cone in a frustration-free chain for $A_{x(m)}$ does not contain $y$. It follows that $[A_{x(m)},B_y]=0$. The exact bra identity from (ii) and this commutation giveSince $P_0{A}_x|{\psi }_0⟩=⟨A_x⟩|{\psi }_0⟩$, subtracting the product of expectations yieldsThus the exponential clustering from layered projectors bound isThe connected correlation function therefore has correlation length at most $1/\alpha$ in lattice-spacing units. In particular $\alpha =\Delta /6+O({\Delta }^2)$ for a small spectral gap. The same-site case is covered by the elementary bound $2\Vert A_x\Vert \Vert B_y\Vert$.

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.

For a centered Gaussian random field, Wick theorem takes the classical form of the Isserlis theorem: every even correlator is a sum over pairings of two-point correlators, and every odd correlator vanishes. For example, with $G_i={\zeta }_G({\mathbf{x}}_i)$,A $2n$-point correlator has $(2n-1)!!$ pairings. Equivalently, all connected correlation functions beyond order two vanish. The covariance completely determines the centered Gaussian statistics; higher ordinary correlators can be nonzero, but contain no independent connected information. For a field with nonzero mean, apply these statements to the centered field and then restore its mean.

With Euclidean source convention $+\int d^{d}x{J}_a{\varphi }_a$, the generating functional isHere $J_a(x)$ is a nondynamical source. Its functional derivatives insert fields:This is why $Z$ “generates” correlation functions.

With the notation of the question, $W[J]=-\log Z[J]$ is the Euclidean connected generating functional; it is distinct from a Wilsonian effective action. Define the classical fieldThe quantum effective action is the Legendre transformwhere $J$ is eliminated in favor of $\Phi$. Its first derivative isso at zero source the quantum expectation value is a stationary point of $\Gamma$. Moreover,where $G_{ab}=⟨{\varphi }_a{\varphi }_b{⟩}_{J,\mathrm{conn}}$ is the exact connected correlation function. Thus the second derivative of $\Gamma$ is the inverse exact propagator.

Perturbatively, $Z[J]$ sums all Feynman diagrams, including disconnected ones. The exponential formula for combinatorial structures says that its logarithm selects connected Feynman diagrams, so $W$ sums connected diagrams with the sign dictated by the convention above. The Legendre transform removes diagrams that disconnect when one internal line is cut; consequently $\Gamma$ sums one-particle-irreducible Feynman diagrams. Equivalently, every connected diagram is a tree assembled from one-particle-irreducible vertices and full propagators, and the Legendre transform inverts that tree construction.

Now let $J_a^{\prime }=J_b{U}_{ba}$. ThenChanging variables to ${\varphi }_b^{\prime }=U_{ba}{\varphi }_a$ and using invariance of both the action and functional measure gives $Z[J^{\prime }]=Z[J]$, hence $W[J^{\prime }]=W[J]$. In the Legendre transform, the pairing obeysChanging the source variable and using the invariance of $W$ therefore givesThe symmetry of the classical action and measure is inherited by the full quantum effective action when it has no quantum anomaly.