Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 303 (Statistical Field Theory)

In the Landau theory approximation, choose the global minimum of the Landau free energy as a function of the magnetization. Since ${\mathcal{A}}_6>0$, the polynomial is a coercive function and a global minimum exists. The candidates are stationary points satisfyingCheck their values of the free-energy density; a stationary point or a metastable local minimum alone need not be the equilibrium state. Positive curvature ${\mathcal{A}}_2+3{\mathcal{A}}_4{m}^2+5{\mathcal{A}}_6{m}^4$ ensures local stability; when it vanishes, higher terms must be checked. The selected global minimum is the equilibrium magnetization within this approximation.

At zero magnetic field, equal opposite minima describe distinct phases exhibiting spontaneous symmetry breaking. A spontaneous magnetization is obtained by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A symmetric finite-volume average can instead have zero expectation even when the minima are nonzero. Thus one must specify the phase when interpreting the minimizer as the equilibrium expectation.

Discuss the zero-field phase boundaries, taking $h=0$ and writing $r={\mathcal{A}}_2$, $q={\mathcal{A}}_4$, $s={\mathcal{A}}_6>0$. This qualification matters because a conjugate field generally rounds the zero-field continuous transition. Nonzero stationary points satisfy $r+q{m}^2+s{m}^4=0$.

For $q>0$, the continuous phase transition occurs atOn the $r>0$ side the only stable phase is $m=0$; on the $r<0$ side the equilibrium global minima are $\pm m_0$ with $m_0^2\sim -r/q$. The magnetization therefore grows continuously from zero as $r$ changes sign.

For $q<0$, the first-order phase transition instead occurs while $r$ is positive. At phase coexistence, $m=0$ and $m=\pm m_0$ are equally deep global minimum points. With $a=m_0^2>0$, stationarity and equality of free energies giveEliminating $r$ gives $-q{a}^2/4-s{a}^3/3=0$, henceAt this coexistence value, the sextic even Landau potential factors as $f(m)=s{m}^2(m^2+3q/(4s){)}^2/6\ge 0$. This verifies that the three equal stationary values are global minima. Above this coexistence value of $r$ the disordered phase wins; below it the ordered phase wins, with a finite jump in the phase-selected magnetization. The values $r=0$ and $r=q^2/(4s)$ are the disordered and ordered spinodal points, respectively, not the coexistence boundary: the latter follows from the repeated root of $r+qa+s{a}^2=0$.

The two transition loci meet at the tricritical point,Two independent control parameters such as $T$ and $g$ can tune these two conditions. In a generic local phase diagram their coefficient map has a Jacobian matrix of rank two at the intersection; unspecified functions ${\mathcal{A}}_2(T,g)$ and ${\mathcal{A}}_4(T,g)$ do not by themselves determine the geometric orientation of the boundaries. The sextic term stabilizes this intersection and the negative-quartic first-order side.

These are mean-field critical exponents at a tricritical point. Hold the other control at its tricritical value and take the coefficients to be real analytic functions of temperature along a thermal approach with ${\mathcal{A}}_2=a(T-T_c)+o(T-T_c)$, $a>0$, ${\mathcal{A}}_6\to s>0$, and ${\mathcal{A}}_4=0$ or ${\mathcal{A}}_4=o(|{\mathcal{A}}_2{|}^{1/2})$. At fixed tricritical $g$, a generic real analytic temperature path with nonzero thermal slope has ${\mathcal{A}}_4=O(T-T_c)$ and meets the last condition. Paths tangent to a scaling direction can have different apparent temperature powers.

At zero field on the ordered side with ${\mathcal{A}}_4=0$, stationarity gives $m^4=-{\mathcal{A}}_2/s$; under the more general approach ${\mathcal{A}}_4=o(|{\mathcal{A}}_2{|}^{1/2})$, this relation holds to leading order. Thus the order-parameter critical exponent isFor the zero-quartic representative, putting this minimum into the Landau free energy gives the singular order-parameter contribution (and the leading asymptotic contribution under the stated negligible-quartic approach)Using $C_{\mathrm{s}}=-T{\partial }_T^2{f}_{\mathrm{s}}$ at fixed other control yields $C_{\mathrm{s}}\sim (T_c-T{)}^{-1/2}$, specifically $T{a}^{3/2}/(4\sqrt{s(T_c-T)})$ to leading order. HenceThis heat-capacity critical exponent refers to the nonzero ordered-side singular amplitude; in this mean-field approximation to Landau theory, the disordered global minimum contributes zero singular free energy, while any regular background remains. Fluctuation corrections lie outside the present approximation.

In the disordered phase, differentiating the equation of state at $m=h=0$ gives the magnetic susceptibility $\chi =1/{\mathcal{A}}_2$. On the ordered branch its curvature is ${\mathcal{A}}_2+5s{m}^4=4|{\mathcal{A}}_2|$, so $\chi =1/(4|{\mathcal{A}}_2|)$. Both give the magnetic-susceptibility critical exponentAt the tricritical temperature with both lower coefficients zero, $h=s{m}^5$, so $m=\mathrm{sgn}(h)(|h|/s{)}^{1/5}$ and the critical-isotherm exponent isThe Widom scaling relation $\gamma =\beta (\delta -1)$ and Rushbrooke scaling relation $\alpha +2\beta +\gamma =2$ both check these powers. Their validity here is a mean-field conclusion, not a claim that fluctuations in every spatial dimension preserve them.

For ${\mathcal{A}}_2\ne 0$, rescale the order parameter by $m=(|{\mathcal{A}}_2|/{\mathcal{A}}_6{)}^{1/4}\psi$. Write $\sigma =\mathrm{sgn}{\mathcal{A}}_2$ and introduce dimensionless scaling variablesEvery term then has the common energy-density factor $|{\mathcal{A}}_2{|}^{3/2}/{\mathcal{A}}_6^{1/2}$:Minimizing over $\psi$ defines the tricritical crossover scaling functionsFor the minimized potential, following the paper's notation $\mathcal{F}$, this provesThe exponent names $x,y$ here are unrelated to the couplings in question 2. The expression concerns the specified order-parameter polynomial; a general material can also have a smooth, field-independent background free-energy density. Such a background should be separated before writing a homogeneous singular scaling form. A potential at prescribed magnetization, instead of prescribed field, is a different Legendre transform and is not obtained by this minimization.

There is a needed qualification to the printed single $\Phi$ and nonzero-value condition. At $(X,Y)=(0,0)$, the ordered branch has minimizers $\psi =\pm 1$ and ${\Phi }_{-}(0,0)=-1/3$, whereas the disordered branch has minimizer zero and ${\Phi }_{+}(0,0)=0$. ThusFor example, ${\mathcal{A}}_2>0$, ${\mathcal{A}}_4=h=0$ gives an identically zero minimized polynomial, contradicting a nonzero $\Phi (0,0)$ on that side. The valid scaling statement uses separate sign branches, or restricts the asserted nonzero value to approach from the ordered side. At ${\mathcal{A}}_2=0$ the displayed coordinates are singular; the critical isotherm is obtained as a limit or directly from the equation of state.

At an ordinary zero-field continuous phase transition, the positive quartic coefficient remains nonzero while ${\mathcal{A}}_2\to 0$. Approaching from the ordered side therefore sends the variable $X$ in the tricritical crossover scaling form to $+\infty$, rather than to zero. With $Y=0$ the dimensionless minimizer obeys ${\psi }^4+X{\psi }^2-1=0$, soAt that stationary point, substitution gives ${\Phi }_{-}(X,0)=-{\psi }^2/4-{\psi }^6/12$. Its large-quartic asymptotic of tricritical scaling is consequently ${\Phi }_{-}(X,0)\sim -1/(4X)$, andIf ${\mathcal{A}}_2$ crosses zero linearly with temperature, this is quadratic in $T-T_c$, giving a finite heat capacity jump rather than a power divergence. ThereforeFor locally constant ${\mathcal{A}}_4>0$ and slope $a={\partial }_T{\mathcal{A}}_2{|}_{T_c}$, the leading singular heat capacity jump from this term is $\Delta C=T_c{a}^2/(2{\mathcal{A}}_4)$. Treating $\Phi$ as a nonzero constant while $X\to \infty$ would incorrectly assign the tricritical exponent to this ordinary transition.

On a tricritical thermal path, the quartic scaling variable $X$ tends to zero, and at zero field $Y=0$. On the ordered side the tricritical crossover scaling function tends to ${\Phi }_{-}(0,0)=-1/3$. Consequentlyprovided the quadratic coefficient has a nonzero thermal slope and the sextic coefficient has a finite positive limit. Since the heat-capacity critical exponent satisfies $f_{\mathrm{s}}\sim |T-T_c{|}^{2-\alpha }$,Equivalently, twice differentiating this ordered-side free-energy density produces the $(T_c-T{)}^{-1/2}$ singularity. The scaling path and branch qualifications are the same as for the direct Landau theory calculation: a nonzero fixed positive quartic coefficient would instead send $X\to \infty$ and yield the ordinary transition of part (e).

Take the couplings in the Hamiltonian to include inverse temperature, so the Boltzmann factor is $e^{-H}$. If physical energies are used instead, first replace $J,L,D,K$ by their products with $1/(k_{B}T)$. Split the on-site term equally between its two bonds. For spin states $s,t\in \{1,0,-1\}$ the spin-chain transfer matrix has entriesUsing the specified coupling coordinates, in the order $(1,0,-1)$ this isFor example, $e^{D/2}=x/z$ and $e^{-J+L+D}=y/z$. This is the nearest-neighbour Blume–Emery–Griffiths model with an additive constant and the paper's sign convention for $D$.

Summing the periodic spin chain gives the partition function $Z_N=\mathrm{tr}{W}^N={\sum }_{a=1}^3{\lambda }_a^N$. For finite real couplings all entries of $W$ are strictly positive; the Perron–Frobenius theorem gives a unique positive dominant eigenvalue with ${\lambda }_1>|{\lambda }_a|$ for $a\ne 1$. Since $W$ is a symmetric matrix, all its eigenvalues are real. Thus in the thermodynamic limit,The positivity condition is stronger and more useful than mere ordering by signed value: subdominant eigenvalues can be negative. Also, the printed strict ordering between the other two is not guaranteed for all couplings. For instance, $J=L=D=0$ makes $W$ the positive constant matrix $e^{K}11^T$ with two equal zero eigenvalues. Degeneracy there does not affect the largest-eigenvalue limit; no explicit generic eigenvalues are needed.

For spin magnetization, introduce a dimensionless field $h$ through $-h{\sum }_i{\sigma }_i$, or insert $S=\mathrm{diag}(1,0,-1)$ into the trace. Thenwhere $v_1$ is a normalized eigenvector of the largest eigenvalue and $W_{st}(h)=e^{h(s+t)/2}{W}_{st}(0)$. Spin inversion symmetry gives $v_{1,+}=v_{1,-}$, since the positive eigenvector of the largest eigenvalue is unique. Thereforeat zero field, both at finite $N$ and in the finite-coupling thermodynamic limit. The finite-$N$ result follows directly by pairing each configuration with its spin-reversed partner. Eigenvalues at a single fixed field do not determine a general observable: a field derivative of the largest eigenvalue, or its eigenvector, is required. Positivity and the real analytic dependence on the couplings exclude a finite-temperature spontaneous symmetry breaking transition in this one-dimensional finite-range chain; singular zero-temperature coupling limits require separate treatment.

For even $N$, spin decimation on alternate sites sums the middle spin of each two-bond segment, hence the coarse spin-chain transfer matrix is $W^{\prime }=W^2$. DefineDirect multiplication gives $W^2=e^{2K}{z}^{-2}\left(\begin{array}{ccc}A& B& C\\ B& E& B\\ C& B& A\end{array}\right)$. Matching its entry ratios to the original parameterization yields the spin-1 chain decimation recursionThe remaining overall positive factor is absorbed into $K^{\prime }$. Keeping that factor preserves the free energy as well as normalized spin probabilities. Indeed $\mathrm{tr}(W^2{)}^{N/2}=\mathrm{tr}{W}^N$, exactly; on an odd ring an unmatched boundary segment needs separate handling rather than assuming a uniform two-site block decomposition.

The fixed-point equation for $u$ factors asThe cubic derivative is $3u^2+2u+3>0$ for every real $u$; its values at zero and one have opposite signs. Thus the possible first coordinates in the stated domain are exactly $u=0,u_{\ast },1$.

For finite $v$, the second fixed-point equation is $v[v^3-(1+u^2{)}^2]=0$. The renormalization-group fixed points are consequentlyThere are six finite points and three boundary points at infinity. Infinity is interpreted in a compactified domain, not as an ordinary real number. The reciprocal coordinate at infinite coupling $w=1/v$ transforms as $w^{\prime }=(1+u^2{)}^2{w}^4$, making $w=0$ a well-defined boundary fixed point.

At any finite fixed point $(\tilde{u},\tilde{v})$, let $\delta u=u-\tilde{u}$, $\delta v=v-\tilde{v}$. The Jacobian matrix linearization isThe only point with both coordinates strictly interior is $\tilde{u}=u_{\ast }$, $\tilde{v}=(1+u_{\ast }^2{)}^{2/3}$. Using its fixed-point relations, the matrix becomesIts eigenvalues are $\rho$ and $4$. In particular, $u_{\ast }<1/\sqrt{3}$ because the increasing cubic is already positive there, so $\rho >1$ (numerically about $1.68$). The corresponding eigenvectors can be chosen as $(1,c/(\rho -4){)}^T$ and $(0,1{)}^T$. Both discrete multipliers exceed one, henceBoth perturbations are relevant directions of a fixed point under repeated coarse-graining, rather than one stable and one unstable direction. A length-rescaling factor was not specified, so these multipliers should not be assigned numerical critical scaling exponents without additional information. At $a=0,1$, the $u$ multiplier is zero; at $v=0$ or $w=0$ the other multiplier is also zero. Thus the four corner points attract locally within the domain, the two positive finite boundary points at $a=0,1$ are saddles, and the points $(u_{\ast },0),(u_{\ast },\infty )$ have one repulsive and one attractive direction.

Use a dimensionless Hamiltonian in the functional weight $e^{-H}$, assume $\alpha >0$, and let $\varphi (\mathbf{x})={\int }_{|\mathbf{p}|\le \Lambda }{d}^{D}p(2\pi {)}^{-D}{e}^{i\mathbf{p}\cdot \mathbf{x}}\tilde{\varphi }(\mathbf{p})$. Reality means $\tilde{\varphi }(-\mathbf{p})=\tilde{\varphi }(\mathbf{p}{)}^{\ast }$. At zero field, the quadratic form isIn the momentum-shell renormalization group, split into ${\varphi }_{<}+{\varphi }_{>}$ below and above $\Lambda /b$. Disjoint Fourier supports make the quadratic form split into $H_0[{\varphi }_{<}]+H_0[{\varphi }_{>}]$. Integration over the shell gives a source-independent Gaussian functional integral multiplying the partition function, or an additive constant in the effective free energy; it leaves the slow-mode quadratic coefficients unchanged before rescaling. A uniform field has support only at zero momentum, so it does not change this shell integration.

Restore the ultraviolet cutoff by ${\mathbf{p}}^{\prime }=b\mathbf{p}$, ${\mathbf{x}}^{\prime }=\mathbf{x}/b$, and chooseThe measure, two gradients and two fields have scale factors $b^D$, $b^{-2}$ and $b^{-(D-2)}$, whose product is one. The mass term has factor $b^D{b}^{-(D-2)}=b^2$, while the uniform source term has factor $b^D{b}^{-(D-2)/2}=b^{(D+2)/2}$. Hence the Gaussian momentum-shell scaling isFor a slowly varying nonuniform source the corresponding formula is $h^{\prime }({\mathbf{x}}^{\prime })=b^{(D+2)/2}h(b{\mathbf{x}}^{\prime })$. Here the nonuniform source on the right is its projection onto the retained Fourier modes; an eliminated source component contributes only to the field-independent Gaussian normalization. The field engineering dimension is $(D-2)/2$ and its Gaussian anomalous dimension is zero. The mass and uniform source are relevant perturbations of the Gaussian fixed point. Positivity of $\alpha$ is necessary for a stable kinetic term near this point. One can use a finite volume and positive mass as infrared regulators and then take the critical limit; at exactly zero mass the integral over the zero Fourier mode alone is not a normalized finite-volume Gaussian measure.

Compute the coupling flow at zero magnetic field, or assume the source has support only in retained Fourier modes, so that the shell Gaussian measure is centered. First justify the shell propagator. For a positive quadratic kernel on the shell, the Gaussian functional integral gives covariance equal to the inverse kernel,with both momenta restricted to $\Lambda /b<|\mathbf{p}|\le \Lambda$. Inserting the Fourier transforms and using the Dirac delta function yields the Gaussian shell covarianceChanging $\mathbf{p}$ to $-\mathbf{p}$ gives the negative-exponent convention as well. The dependence is on the separation, as required by translation invariance. The printed numerator uses $\mathbf{x}$ alone: it is correct only if that symbol means the separation or if $\mathbf{y}=0$. For example at $\mathbf{x}=\mathbf{y}\ne 0$, the covariance must equal the constant coincident value $G_{>}(0)$, while the literal printed integral generally depends on $\mathbf{x}$. Thus the missing separation is a real source qualification, not a change of Fourier-sign convention.

Integrating out the shell gives $H_{\mathrm{eff}}[{\varphi }_{<}]=H_0[{\varphi }_{<}]-\log ⟨e^{-V[{\varphi }_{<}+{\varphi }_{>}]}{⟩}_0^{\mathrm{shell}}$, up to a field-independent constant. At first order in $u_0$, the cumulant expansion of a coarse-grained free energy retains $⟨V⟩$. Odd moments of the centered Gaussian measure vanish and Wick theorem gives $⟨{\varphi }_{>}^4⟩=3G_{>}(0{)}^2$. Writing $I_1=G_{>}(0)$,The last term affects only the constant; the second is a momentum-independent tadpole diagram correction. Before rescaling it changes $r_0$ to $r_0+u_0{I}_1/2$, leaves the quartic coefficient at $u_0$, and produces no gradient correction. Combining with part (a) gives the leading coupling recursionThis is one-loop shell mass renormalization in scalar quartic theory. Here $I_1={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}\alpha /(p^2+\alpha r_0)$. The shell Gaussian is well defined if $p^2+\alpha r_0>0$ there; near $r_0=0$ this follows from $\alpha >0$. A negative slow-mode mass may be stabilized by the retained quartic term and does not require pretending that the full unconstrained quadratic measure at negative mass is normalizable.

Linearizing about $(r_0,u_0)=(0,0)$ givesIts mass and quartic eigenvalues are $b^2$ and $b^{4-D}$. The off-diagonal term is an additive critical-mass shift: setting the bare $r_0=0$ is not generally the critical tuning when $u_0>0$. The mass remains relevant for every $D$; at $D=2$ the two linear eigenvalues coincide and the matrix can have a Jordan block, but both perturbations are still relevant.

For $D>4$, the quartic interaction is an irrelevant operator, and the Gaussian fixed point attracts weak quartic perturbations after the mass and magnetic field have been tuned. The quartic can nevertheless be a dangerously irrelevant coupling: its positive value stabilizes the ordered phase. For $D<4$, it is a relevant operator, so the Gaussian description is unstable toward interactions. At $D=4$ it is a marginal operator; first-order perturbation theory alone does not decide its fate.

To settle that borderline case, retain the leading quartic contribution from the second cumulant, at zero external momentum in the local expansion. The two-fast-field part of $V$ is $(u_0/4)\int {\varphi }_{<}^2{\varphi }_{>}^2$. Since $⟨{\varphi }_{>}^2(\mathbf{x}){\varphi }_{>}^2(\mathbf{y}){⟩}_c=2G_{>}(\mathbf{x}-\mathbf{y}{)}^2$, its connected second cumulant contributesKeeping the local quartic term and using Parseval identity gives $\int d^{D}y{G}_{>}(\mathbf{x}-\mathbf{y}{)}^2=I_2$, where $I_2={\int }_{\mathrm{shell}}{d}^{D}p(2\pi {)}^{-D}[\alpha /(p^2+\alpha r_0){]}^2$. Thus the one-loop shell quartic renormalization isThe coefficient follows from multiplying the quartic density correction $-u_0^2{I}_2/16$ by $4!=24$. This local, zero-external-momentum coupling extraction is not a claim that the exact finite-shell effective action retains only its original polynomial; further operators are generated.

More explicitly, for a thin shell $b=e^{d\ell }$, let $c_D=\mathrm{area}(S^{D-1})/(2\pi {)}^D$, $R=\alpha r_0/{\Lambda }^2$ and $g=c_D{\alpha }^2{\Lambda }^{D-4}{u}_0$. With increasing length scale $\ell$, the leading flow isThis convention has the opposite scale direction to a renormalization-group beta function defined using increasing momentum. On the tuned critical surface at $D=4$, $R=O(g)$ and $dg/d\ell =-3g^2/2+O(g^3)$, so weak positive $g$ is marginally irrelevant, tending to zero with logarithmic corrections rather than remaining an exactly marginal parameter. Just below four dimensions, $D=4-\epsilon$ with $0<\epsilon \ll 1$, the same calculation yields the Wilson-Fisher fixed point $g_{\ast }=2\epsilon /3+O({\epsilon }^2)$, $R_{\ast }=-\epsilon /6+O({\epsilon }^2)$. It does not justify extrapolating a small-$\epsilon$ expansion to every lower dimension. The ordinary scalar quartic upper critical dimension is therefore