= Solution
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 are
$$
m_h(x)=\frac{\delta W}{\delta h(x)}=-\frac{\delta F}{\delta h(x)},\qquad
G(x,y)=\frac{\delta m_h(x)}{\delta h(y)}
=\langle\phi(x)\phi(y)\rangle-\langle\phi(x)\rangle\langle\phi(y)\rangle.
$$
These 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 equation>
$$
\boxed{-\nabla^2m_h(x)+r_0m_h(x)+\frac{u_0}{6}m_h(x)^3=h(x).}
$$
This 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, are
$$
\boxed{F_{\mathrm L}[h]=H[m_h,h],\qquad
\Gamma_{\mathrm L}[m]=F_{\mathrm L}[h]+\int d^Dx\,h(x)m(x)
=\int d^Dx\left[\frac12(\nabla m)^2+\frac12r_0m^2+\frac{u_0}{4!}m^4\right].}
$$
In 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 obeys
$$
\boxed{\left[-\nabla_x^2+r_0+\frac{u_0}{2}m_h(x)^2\right]G(x,y)
=\delta^{(D)}(x-y).}
$$
Therefore
$$
\boxed{K(x)=-\nabla_x^2+r_0+\frac{u_0}{2}m_h(x)^2.}
$$
Equivalently, 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 $u_0m^4/4!$; 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 $G(x,y)=G(x-y)$. Set
$$
M^2=r_0+\frac{u_0}{2}m_0^2>0.
$$
The <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> has
$$
\boxed{\widetilde G(q)=\frac{1}{q^2+M^2}
=\frac{1}{q^2+\xi^{-2}},\qquad
\xi=\left(r_0+\frac{u_0}{2}m_0^2\right)^{-1/2}.}
$$
The inverse convention is $G(x)=\int d^Dq\,(2\pi)^{-D}e^{-iq\cdot x}\widetilde G(q)$. 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> is
$$
\boxed{\xi=\begin{cases}r_0^{-1/2},&r_0>0,\\(-2r_0)^{-1/2},&r_0<0.\end{cases}}
$$
In 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> is
$$
\boxed{\nu_{\mathrm{Landau}}=\frac12.}
$$
The high-temperature amplitude is $\sqrt2$ 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.
Back to article page
