Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 303 / 3

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.

When $u_0=0$, the field theory is a Gaussian field theory, so integrating out short-wavelength modes can be done exactly rather than merely at a saddle. Introduce an ultraviolet momentum cutoff $\Lambda$ and a scale factor $b>1$, then split $\varphi ={\varphi }_{<}+{\varphi }_{>}$ into $|q|<\Lambda /b$ and $\Lambda /b<|q|<\Lambda$. Orthogonality of Fourier modes eliminates cross terms in the quadratic Hamiltonian. A constant source couples only to the zero-momentum mode and therefore does not couple to ${\varphi }_{>}$.

Before rescaling, the Gaussian functional integral factorizes:where $H_{<}$ has the original $r_0$ and $h$, restricted to low momenta. Up to a field-independent normalization, the eliminated modes contributeThe determinant affects the additive free energy, but generates neither a mass correction nor a source correction, since no interaction mixes low and high modes. This remains true beyond the Landau approximation: it is an exact integration of a Gaussian field theory, not neglect of the high-mode fluctuations.

To restore the original cutoff, use $x=b{x}^{\prime }$, $q^{\prime }=bq$, and choose the field rescalingSubstitution into the low-mode Hamiltonian gives the Gaussian momentum-shell scalingThe gradient coefficient is unchanged: the volume factor is $b^D$, the two derivatives contribute $b^{-2}$, and the two fields contribute $b^{2-D}$. The mass term gains $b^2$, while the source term gains $b^{(D+2)/2}$. ThusFor a nonconstant source the corresponding formula is $h^{\prime }(x^{\prime })=b^{(D+2)/2}h(b{x}^{\prime })$, after restricting its coupling to retained modes. The constant-source case avoids that extra source filtering.

Writing $b=e^{\ell }$, the exact Gaussian renormalization-group flow hasBoth perturbations are relevant at the massless zero-source Gaussian fixed point, with $y_t=2$ and $y_h=(D+2)/2$. There is no anomalous field rescaling here, so $\eta =0$ and the correlation-length critical exponent is again $\nu =1/y_t=1/2$. The coefficient changes arise from rescaling, not from an interaction-induced shift in $r_0$ during shell integration.

A stable real Gaussian functional integral requires $r_0>0$, or an infrared prescription that treats the zero mode at the massless point. With $u_0=0$ and $r_0<0$, the field energy is unbounded below; the Gaussian field theory alone cannot describe a stable ordered phase. The scaling laws are therefore interpreted about the Gaussian fixed point from the stable side, with finite-volume zero-mode regularization when needed.