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A normalized blocking kernel assigns probabilities or delta constraints for coarse-grained variables given a microscopic configuration. Multiplying the microscopic Boltzmann weight by this kernel and summing over eliminated variables defines the blocked statistical Hamiltonian. Normalization preserves the partition function exactly when all generated operators and the field-independent constant are retained. A finite-coupling truncation may lose that exactness.

The parameters of a cutoff statistical field theory describe only the retained modes. Changing the ultraviolet cutoff changes which fluctuations have already been integrated out, so its effective mass, interaction coefficients and gradient normalization must change to preserve the same long-distance physics. They are not separately cutoff-independent observables. Because the source's weight is $e^{-H}$, its statistical Hamiltonian is dimensionless, with the physical inverse-temperature factor already absorbed.

For example split the scalar field into slow modes ${\varphi }_{<}$ with $|p|<\Lambda /b$ and shell modes ${\varphi }_{>}$ with $\Lambda /b<|p|<\Lambda$. Define the momentum-shell renormalization group step byThis integrates out short-wavelength fluctuations exactly if all generated terms are retained. The new Hamiltonian can be expanded in local symmetry-allowed operators when the retained external momenta are well below the shell scale. A cumulant expansion of a coarse-grained free energy gives perturbative coefficients. Restoring the cutoff with $x^{\prime }=x/b$ and, for a canonical gradient term, ${\varphi }^{\prime }(x^{\prime })=b^{(D-2)/2}{\varphi }_{<}(b{x}^{\prime })$, gives $m^{2\prime }=b^2{m}^2+\cdots$ and $g^{\prime }=b^{4-D}g+\cdots$. Interaction corrections also change the field normalization. Repeating the step produces the effective theory at successively longer distances.

To obtain the LG theory, assume a short-range scalar theory, slowly varying retained fields, analytic local couplings, positive gradient stiffness and stability, and a regime in which fluctuation corrections at the remaining scales are small. Keeping the leading gradient, quadratic and quartic operators gives a local Landau-Ginzburg theory functional. Its equilibrium in the Landau approximation is a uniform minimum, with a quadratic coefficient proportional to $T-T_c$ after the critical mass has been tuned. The RG explains why this is asymptotically consistent for the ordinary transition above four dimensions: the quartic interaction is irrelevant near the Gaussian fixed point, while it must still be retained to stabilize the ordered phase. Below four dimensions it cannot be dropped in the asymptotic critical region; an interacting Wilson-Fisher fixed point rather than the elementary saddle generally controls the transition. At four dimensions the interaction is marginal and produces logarithmic corrections. Tuning a quartic coefficient through zero requires a sextic stabilizing term and leads to tricriticality.

More concretely, in the Gaussian scaling regime let $u=g/6$ so the local quartic term is $u{\varphi }^4/4$. At a total blocking scale $\ell$, the leading couplings are $r^{\prime }={\ell }^{2}r$, $u^{\prime }={\ell }^{4-D}u$, $h^{\prime }={\ell }^{(D+2)/2}h$. Apply the saddle approximation to the blocked potential and multiply its minimum by ${\ell }^{-D}$ to convert back to original volume units. Rescaling the saddle field by $(|r^{\prime }|/u^{\prime }{)}^{1/2}$ givesAll blocking-scale factors cancel. This explicitly recovers mean-field scalar free-energy scaling with $r\propto t$. Although $u^{\prime }\to 0$ above four dimensions, the saddle free energy is proportional to $1/u^{\prime }$; setting it to zero before minimization would remove the ordered phase. This is the dangerously irrelevant coupling mechanism rather than homogeneous two-variable hyperscaling.

For the perturbative calculation make the source's kinetic convention explicit. Write $K={\alpha }^{-1}>0$. The canonical normalization of a scalar gradient term uses $\psi =\sqrt{K}\varphi$, so the canonical quadratic and quartic coefficients are $r/K$ and $g/K^2$. In what follows $m^2,g$ denote those canonical coefficients; then the reference propagator has denominator $p^2+M^2$. Without this normalization the propagator denominator is $K{p}^2+r$, and the unmodified printed integral would not apply. At one loop the quartic tadpole is momentum independent, so it produces no gradient renormalization at this order.

In the convention fixed by the displayed equation, the truncated two-point function is the one-particle-irreducible two-point vertex, not the connected two-point cumulant itself. If $W[J]=\log Z[J]$ and $\Gamma [\phi ]=\int J\phi -W[J]$ is the Legendre transform, then its second derivative is the inverse of $W^{(2)}=G$. For a translation-invariant background,Decompose the canonically normalized statistical Hamiltonian into a Gaussian part of mass $M^2$, a mass counterterm $\delta m^2=m^2(\Lambda ,T)-M^2$, and the quartic interaction $g{\varphi }^4/4!$. With the Euclidean sign convention in which a positive mass correction increases the inverse propagator, the self-energy expansion isHere $\Sigma$ contains loop corrections from proper two-point diagrams, excluding the separately displayed mass counterterm. The corresponding connected propagator begins $G=G_0-G_0(\delta m^2+\Sigma )G_0+\cdots$. This fixes the sign, which would be reversed if “self-energy” instead denoted the insertion added with a plus sign inside a Dyson series.

The one-loop proper diagram is the tadpole diagram. Attaching two external fields to the quartic vertex gives $4\cdot 3$ contractions, divided by $4!$, so its symmetry factor is $1/2$. With the loop momentum restricted by the cutoff,It is independent of external momentum. Impose the zero-momentum mass condition $M^2=\tilde{\Gamma }(0)=m^2(0,T)$. This gives $\delta m^2+(g/2)I_D(M^2;\Lambda )=0$, henceThis is the one-loop relation using a renormalized mass in the reference propagator, or the corresponding self-consistent tadpole approximation if solved without expanding in $g$. It is not an exact all-orders gap equation. Away from the critical infrared problem, replacing the loop mass by the bare one changes a strict perturbative result only at higher order.

Take $g>0$ smooth and nonzero for an ordinary stable quartic transition. To test the assumed linear thermal mass, work from the disordered side and put $R=m^2(0,T)>0$. For $D>2$, $I_D(0;\Lambda )$ is infrared finite. The critical bare mass is shifted, not generically zero: $m^2(\Lambda ,T_c)=-(g/2)I_D(0;\Lambda )$. Absorb the smooth temperature dependence of couplings into an analytic thermal tuning $\tau =A_t(T-T_c)+\cdots$, with $A_t>0$. Critical subtraction gives the one-loop critical-mass subtractionwhereThe infrared asymptotics of the critical-mass subtraction now distinguish the dimensions. For $D>4$, $J_D(0)=K_D{\Lambda }^{D-4}/(D-4)$ is finite, soFor $D=4$, $J_4(R)=(K_4/2)\log [({\Lambda }^2+R)/R]$ diverges logarithmically. For $2<D<4$, substitution $p=\sqrt{R}q$ givesso the correction to $\tau$ scales as $R^{(D-2)/2}$ and dominates the analytic linear term. Pure mean-field linear mass scaling is therefore consistent only aboveAt the boundary dimension logarithms modify the simple power law. Below it this calculation diagnoses the failure of the Gaussian expansion; the exponent obtained by treating the self-consistent one-loop equation as exact is not automatically the exponent of the interacting scalar theory. For $D\le 2$, the massless subtraction itself has an infrared divergence, so this perturbative argument cannot establish absence of a transition. In particular it does not rule out the two-dimensional Ising critical point.

The same upper dimension follows by engineering dimension counting: a canonical scalar field has dimension $(D-2)/2$, so $[g]=D-4(D-2)/2=4-D$. For a tricritical point tune the renormalized quadratic and quartic terms to zero and retain a positive sextic interaction $v{\varphi }^6$. Its engineering dimension isIt is marginal at $D_c=3$, irrelevant above three, and relevant below three. More generally the upper critical dimension of an even scalar interaction ${\varphi }^{2n}$ is $2n/(n-1)$.

A Ginzburg criterion check gives the same result: at tricritical mean-field scaling $m_0^2\sim |t{|}^{1/2}$ and $\xi \sim |t{|}^{-1/2}$, whereas fluctuations in a correlation volume scale as ${\xi }^{2-D}\sim |t{|}^{(D-2)/2}$. Their ratio to $m_0^2$ is $|t{|}^{(D-3)/2}$, which tends to zero only for $D>3$. ThusThe tricritical tuning concerns renormalized couplings: shell contractions of a sextic term can regenerate quadratic and quartic terms even when their bare coefficients vanish. At three dimensions the marginal sextic coupling produces logarithmic corrections rather than a strictly fluctuation-free mean-field limit.

For the Gaussian critical exponents, use dimensionless statistical-action conventions, with the Boltzmann factor $e^{-H}$. If $H$ is an energy, apply the following to $H/(k_{B}T)$ and absorb that smooth factor into the couplings. Choose the Fourier transform conventionReality gives $\tilde{\varphi }(-p)=\tilde{\varphi }(p{)}^{\ast }$. The Fourier-space statistical Hamiltonian isThe source term is not multiplied by one half. Completing the Gaussian functional integral gives $⟨\tilde{\varphi }(p)⟩=\tilde{J}(p)/({\kappa }^{-1}{p}^2+m^2)$ and the connected Fourier-space two-point correlation functionAssume $m^2>0$ for the stable massive Gaussian field theory. In the continuum long-distance theory, the real-space correlation function is the Green kernel of $-{\kappa }^{-1}{\nabla }^2+m^2$. Away from its source, a radial ansatz $G(R)\sim R^{-(D-1)/2}{e}^{-R/\xi }$ gives, at large $R$, $m^2-{\kappa }^{-1}{\xi }^{-2}=0$. Equivalently the propagator pole is at $p^2=-\kappa m^2$. Thus the exponential correlation length isAt $m=0$ the correlation length diverges. A sharp momentum regulator itself introduces artificial oscillatory long-distance tails; this length describes the physical continuum pole or a local/smoothly regulated model on distances large compared with $a$.

For a momentum-shell renormalization group step with $b>1$, split the Fourier modes into retained modes $|p|<\Lambda /b$ and eliminated modes $\Lambda /b<|p|<\Lambda$. Define ${\tilde{\varphi }}_B(p)$ as the retained field. Integrating the eliminated modes is exact for a Gaussian field theory; a uniform source acts only on the zero mode, so this integration produces a field-independent free-energy term. Rescale $q=bp$ and set ${\tilde{\varphi }}_B(p)=\tilde{Z}{\tilde{\varphi }}^{\prime }(q)$. The retained quadratic action becomesKeeping the gradient coefficient fixed requires $b^{-D-2}{\tilde{Z}}^2=1$. For $J=h$, the source term is $-h{\tilde{\varphi }}_B(0)$, so the complete Gaussian rescaling isThe corresponding real-space field obeys ${\varphi }_B(x)=b^{(2-D)/2}{\varphi }^{\prime }(x/b)$; confusing that real-space factor with $\tilde{Z}$ would change the source exponent incorrectly.

Let $F$ be free energy per original volume, with $t$ proportional to $m^2$. Rescaled volume is $V^{\prime }=V/b^D$, so the Gaussian free-energy scaling relation isA field-normalisation Jacobian can be included in the field-independent shell term. For any finite step on positive-mass modes, that background is analytic near $t=0$. Subtract regular backgrounds and choose $b=t^{-1/2}$ on the massive side. The usual homogeneous singular scaling then givesThese are Gaussian critical exponents for the stipulated Gaussian approximation. A useful check is its source-dependent contribution $-h^2/(2m^2)$: $D/2-2\Delta =-1$ gives precisely that power of $t$.

There is a qualification to a strictly pure-power singular free energy. The Gaussian free-energy logarithm at effective dimension two in $D=2$ contains $t\log t$ after subtracting analytic terms. Thus the displayed exponent assignments remain the formal power indices, but at $D=2$ the advertised homogeneous expression needs this additive logarithmic term. A purely quadratic theory also cannot stabilise the ordered $t<0$ phase or the zero-mass zero mode at nonzero $h$; the scaling calculation is taken from $t>0$, and an ordered-phase continuation requires stabilising interactions.

At a uniaxial Lifshitz point, the quadratic kernel instead isThe vanishing coefficient of $p_{\parallel }^2$ is an additional tuning that distinguishes this Lifshitz point from an ordinary critical point. Use a factorised cutoff, retain $|p_{\perp }|<{\Lambda }_{\perp }/b$ and $|p_{\parallel }|<{\Lambda }_{\parallel }/c$, and integrate its complement. A rectangular or cylindrical retained region is suitable; its precise boundary is not an exponent. This is a uniaxial Lifshitz Gaussian renormalization step.

Set $q_{\perp }=b{p}_{\perp }$, $q_{\parallel }=c{p}_{\parallel }$, and ${\tilde{\varphi }}_B=\tilde{Z}{\tilde{\varphi }}^{\prime }$. The momentum measure acquires $b^{-(D-1)}{c}^{-1}$. Keeping both kinetic coefficients fixed requiresTherefore the anisotropic blocking and source factors areThe anisotropic effective dimension is $d_{\mathrm{eff}}=D-1+1/2=D-1/2$: one parallel coordinate contributes half the scaling weight of a perpendicular coordinate. The rescaled-volume factor is $b^{-d_{\mathrm{eff}}}$, soChoosing $b=t^{-1/2}$ gives the requested Lifshitz Gaussian exponents:Again $(D-1/2)/2-2\Delta =-1$, checking the uniform-source susceptibility. The associated length powers are ${\xi }_{\perp }\sim t^{-1/2}$ and ${\xi }_{\parallel }\sim t^{-1/4}$. A Gaussian determinant has the analogous logarithmic exception if its effective dimension equals 2. Interactions can alter these exponents; dimensional rescaling here solves the specified Gaussian model.

Use inverse temperature ${\beta }_T=(k_{B}T{)}^{-1}$, reserving $\beta$ without a subscript for the critical exponent. Write $c=qJ$; each undirected bond occurs once, so the number of bonds is $Nq/2$. In the mean-field theory of the Ising model, write each Ising spin as ${\sigma }_i=m+\delta {\sigma }_i$ and neglect the product of fluctuations:The resulting independent-spin statistical Hamiltonian isThe constant corrects the double counting of interaction energy. The Ising spin sums now factorize and can all be evaluated:This is the approximate partition function at an assumed mean field; equilibrium fixes $m$ self-consistently. The corresponding Ising auxiliary mean-field free energy isIts equilibrium value gives the mean-field Helmholtz free energy in the imposed field. Away from a stationary point its parameter $m$ is an assumed field variable, not necessarily the actual mean Ising spin of that independent-spin distribution.

For the full small-$m$ expansion at fixed $h$, set $t_h=\tanh ({\beta }_{T}h)$ and $s_h={\mathrm{sech}}^2({\beta }_{T}h)$. Differentiating $\log \cosh z$ gives successive derivatives $t_h$, $s_h$, $-2t_h{s}_h$, and $2s_h(2-3s_h)$ at $z={\beta }_{T}h$. ThereforeAt zero field, spin inversion symmetry eliminates odd powers and this simplifies toThe quadratic coefficient changes sign and the quartic coefficient is positive atThus the zero-field mean-field prediction is a continuous, continuous phase transition: the stable zero spin magnetization develops two symmetry-related nonzero minima continuously below $T_c$.

Both requested routes give the same mean-field self-consistency equation. First, the one-spin expectation in the effective field isSecond, differentiating the Ising auxiliary mean-field free energy giveswhose stationary condition is precisely that mean-field self-consistency equation. Choose its stable, lowest-free-energy branch rather than every algebraic solution.

An equally useful mean-field approximation parametrizes the trial distribution by its actual mean Ising spin. Its probabilities are $(1\pm m)/2$, giving the Bragg-Williams free energy of the Ising modelThis energy-minus-entropy function has expansionIts stationary equation is $h=k_{B}T\mathrm{arctanh}m-cm$, again equivalent to the mean-field self-consistency equation. The two functions differ away from equilibrium, but agree on stationary branches: for $z={\beta }_T(h+cm)$ and $m=\tanh z$, the entropy bracket equals $mz-\log (2\cosh z)$. Their small-$m$ coefficients therefore need not agree at arbitrary temperature; at $T_c$ they have the same leading critical quartic coefficient $c/12$. This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.

For the order-parameter critical exponent, expand the equation of state at $h=0$:On a nonzero stable branch,Hence $\beta =1/2$. The spontaneous magnetization is understood by selecting a branch with an infinitesimal field after the thermodynamic limit; a finite symmetric sample has zero exact zero-field mean Ising spin.

For the magnetic susceptibility per site, define $\chi =\partial m/\partial h$ with $h$ in energy units. Implicit differentiation givesAbove $T_c$, $m=0$ and $\chi =1/[k_B(T-T_c)]$. Below $T_c$, evaluated on a selected ordered branch, the expansion of $m^2$ gives $\chi \sim 1/[2k_B(T_c-T)]$. Thus the magnetic-susceptibility critical exponent is $\gamma =1$ on both sides, with different amplitudes. At $T=T_c$, the equation of state becomesso the critical-isotherm exponent is $\delta =3$. These are mean-field critical exponents, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.

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.