Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 303 (Statistical Field Theory)

The equilibrium magnetization is a global minimum of the Landau free energy. Its stationary values solve the polynomial equationand a local minimum must satisfyOne compares the value of $f$ at every such local minimum and chooses the smallest. Since $a_6>0$, the polynomial tends to positive infinity as $|m|$ tends to infinity, so a global minimum exists.

Put $t=T-T_c$, so $a_4\sim t$. At zero field the stationary equation factors asFor $t>0$, $m=0$ is the unique minimum. For $t<0$, it is unstable and the two minima areThe order parameter therefore tends continuously to zero, and its order-parameter critical exponent is $\boxed{\beta =1/2}$.

At either ordered minimum the singular free-energy density iswhereas it is zero for $t>0$. Two temperature derivatives give a singular heat capacity proportional to $|t|$ below the transition and zero above it, so the heat-capacity critical exponent is $\boxed{\alpha =-1}$.

The inverse magnetic susceptibility at a stable minimum is the curvature ${\chi }^{-1}={\partial }^{2}f/\partial m^2$. Below $T_c$,and hence ${\chi }_{-}\sim |t{|}^{-2}$ and $\boxed{{\gamma }_{-}=2}$. Above $T_c$, however, the curvature at $m=0$ vanishes for every $t>0$. Indeed, at small field $B=4a_4{m}^3+O(m^5)$, so $m\sim (B/(4a_4){)}^{1/3}$ and the linear susceptibility is already infinite away from the critical point. Consequently the usual magnetic-susceptibility critical exponent ${\gamma }_{+}$ is not defined for this exceptional free energy; assigning it a finite value would incorrectly assume a quadratic term.

At $T=T_c$, the equation of state is $B=6a_6{m}^5$, so $m\sim |B{|}^{1/5}$ and the critical-isotherm exponent is $\boxed{\delta =5}$. Thus the transition is continuous, although its missing quadratic term makes the high-temperature linear response singular throughout that phase.

For every $T<T_c$, the two zero-field minima $m_{+}$ and $m_{-}$ coexist. A positive field selects $m_{+}$ and a negative field selects $m_{-}$, so crossing $B=0$ makes the equilibrium magnetization jump between nonzero values. Hence $B=0$, $T<T_c$, is a line of first-order phase transitions, ending at the continuous critical point $(T_c,0)$.

Write ${\mathbf{\sigma }}_i=\mathbf{m}+\delta {\mathbf{\sigma }}_i$ and neglect products of two fluctuations. ThenEach site has $q$ neighbours and each bond is counted once, so the mean-field approximation givesChoose the $x$ axis along $\mathbf{m}$. For the four-state clock model, the single-site partition function isConsequentlyand therefore

Differentiating the mean-field free energy and imposing stationarity givesThe hyperbolic-function identity $\sinh x/(1+\cosh x)=\tanh (x/2)$ turns this into the self-consistency equation

The Taylor series at $x=0$ isSubstituting $x=Jqm/T$ givesThe quartic coefficient is positive, while the quadratic coefficient changes sign atThis is therefore a continuous mean-field phase transition.

For finite $p=4$, the clock model has a discrete symmetry. Domain walls have finite energy per unit boundary area, so thermal disorder destroys long-range order in one dimension but a finite-temperature ordered phase can exist in two dimensions. Its lower critical dimension is therefore $\boxed{d_{\mathrm{l}}=1}$.

As $p\to \infty$, the permitted angles become continuous and the model becomes the XY model with $O(2)$ symmetry. The Mermin-Wagner theorem forbids spontaneous long-range order at positive temperature in two dimensions, so the lower critical dimension for conventional symmetry breaking is $\boxed{d_{\mathrm{l}}=2}$. The two-dimensional model can nevertheless undergo a Berezinskii–Kosterlitz–Thouless transition between algebraic and exponential correlation decay.

A momentum-shell renormalization group step has three parts. First split the Fourier transform of the field into slow modes ${\varphi }_{<}$ with $|k|<\Lambda /\zeta$ and fast modes ${\varphi }_{>}$ with $\Lambda /\zeta <|k|<\Lambda$, then perform the functional integral over ${\varphi }_{>}$. Second rescale momenta by $k^{\prime }=\zeta k$, equivalently coordinates by $x^{\prime }=x/\zeta$, to restore the cutoff from $\Lambda /\zeta$ to $\Lambda$. Third rescale the field so that the coefficient of $(\nabla \varphi {)}^2/2$ again has its chosen normalization. The effective free energy contains every operator allowed by the symmetries, with transformed coefficients. Repeating the step composes these coefficient maps and produces a renormalization-group flow.

The free energy is dimensionless in units with $k_{B}T=1$, so the integrand has momentum dimension $d$. Since a derivative has dimension one, the kinetic term givesThe mass term then givesThese are engineering dimensions.

The engineering value follows from the Gaussian kinetic term. At an interacting renormalization-group fixed point, momentum-dependent self-energy diagrams change the kinetic coefficient, and restoring its normalization requires wave-function renormalization. The resulting anomalous dimension $\eta$ changes the full scaling dimension to

The operator ${\nabla }^{2m}{\varphi }^{2n}$ has engineering dimensionThe action integral is dimensionless, soThus $\alpha$ is a relevant coupling, marginal coupling, or irrelevant coupling according asrespectively.

At order $g_0^2$, use the two-point sunset diagram: two quartic vertices are joined by three internal propagators, with one external line attached to each vertex. Unlike the one-vertex tadpole diagram, its self-energy $\Sigma (p)$ depends nontrivially on the external momentum $p$. The coefficient of $p^2$ in the expansion of $\Sigma (p)$ changes the kinetic term, so normalizing that term requires wave-function renormalization and gives a nonzero field anomalous dimension.

No. With only ${\lambda }_0{\varphi }^6$, the free energy has an exact Z2 symmetry $\varphi \mapsto -\varphi$. Integrating out fast modes and rescaling preserve that symmetry, whereas ${\varphi }^5$ is odd. Therefore no five-point vertex and no correction to $\gamma (\zeta )$ can be generated at any order in ${\lambda }_0$.

Choose four of the six fields at one sextic vertex to be slow and contract the remaining two fast fields into a tadpole diagram. There are $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ choices. Ifthen the first term of the cumulant expansion contributesAfter the canonical coordinate and field rescaling, its contribution to the quartic coupling is

In the displayed connected Feynman diagram, each sextic vertex carries two slow external legs and the four remaining legs at each vertex are paired across the vertices. The second cumulant expansion has a factor $-1/2$, the slow legs can be selected in ${\left(\genfrac{}{}{0ex}{}{6}{2}\right)}^2$ ways, and the four cross-contractions can be paired in $4!$ ways. The coefficient is thereforeWriting ${\mathcal{S}}_{\zeta }=\{q:\Lambda /\zeta <|q|<\Lambda \}$, the local zero-external-momentum contribution iswith the integral restricted further so that $q_1+q_2+q_3\in {\mathcal{S}}_{\zeta }$. The three independent internal momenta agree with the diagram's loop order $L=4-2+1=3$.

Classify the connected contractions by the numbers $(r,4-r)$ of external slow legs on the two sextic vertices and by the number $c$ of fast propagators joining them. Besides the displayed $(2,2;c=4)$ graph, the distinct topologies are:

Exchanging the two vertices gives no new topology. All four are connected Feynman diagrams selected by the logarithm in the cumulant expansion. With an ideal sharp momentum shell and a projection at exactly zero external momentum, the single joining line in the last topology cannot carry shell momentum, so that topology gives zero to the local quartic coupling; it is still the remaining formal connected contraction.

When all masses are equal, the free energy depends on the real vector $\mathbf{\varphi }$ only through inner products, so its symmetry is the orthogonal group $O(N)$. The mean-field potential isFor ${\mu }_0^2>0$, its unique minimum is $\mathbf{\varphi }=0$, which preserves $O(N)$. For ${\mu }_0^2<0$, the minima form the sphereChoosing one minimum leaves the subgroup $O(N-1)$ that fixes its direction, so spontaneous symmetry breaking gives $\boxed{O(N)\to O(N-1)}$.

The leading mass corrections are one-vertex tadpole diagrams. In index notation, one contraction closes a freely summed component loop and is proportional to $N$; the two exchange contractions force the internal component to equal the external one and are not proportional to $N$. Including their multiplicities givesThus the $N$ in $N+2$ comes from the closed index loop, while the $2$ comes from the two same-component contractions.

The leading quartic corrections contain two quartic vertices joined by two internal propagators: the three familiar exchange channels distribute the four external legs in the $s$, $t$, and $u$ pairings. One index contraction contains a freely summed closed component loop and is proportional to $N$; the remaining contractions have indices fixed by the external legs. Their sum givesThe $N$ term is the closed-index-loop topology and the eight is the combined multiplicity of the other contractions.

Under $x^{\prime }=x/\zeta$ and the Gaussian field rescaling ${\varphi }^{\prime }(x^{\prime })={\zeta }^{(d-2)/2}\varphi (x)$, a mass squared has engineering dimension two and a quartic coupling has dimension $4-d$. Therefore

LetFor a radial integrand $h(q)$, differentiating the thin shell at $\zeta =e^s$ givesReplacing the bare parameters by running ones after each infinitesimal step yields the beta functions

For $d=4-ϵ$, define the dimensionless couplings $r={\mu }^2/{\Lambda }^2$ and $u=\tilde{g}={\Lambda }^{-ϵ}g$. Since $K_4=1/(8{\pi }^2)$, the leading epsilon expansion of the flow isThere is a Gaussian fixed point $(r_{\ast },u_{\ast })=(0,0)$. Its thermal eigenvalue is $y_t=2$, so its correlation-length critical exponent is $\boxed{{\nu }_{\mathrm{G}}=1/2}$.

The interacting Wilson-Fisher fixed point isLinearizing the renormalization-group flow gives the thermal eigenvalueTaking its reciprocal gives

Write $r_a={\mu }_{0,a}^2$. The stationary equations forareFor $r_1,r_2>0$, the disordered minimum is $(0,0)$. If $r_1<0$ and $r_1<r_2$, the first component orders with ${\varphi }_1^2=-r_1/(4g_0)$ and ${\varphi }_2=0$. If $r_2<0$ and $r_2<r_1$, the second component orders analogously.

The positive $r_2$ half of $r_1=0$ and the positive $r_1$ half of $r_2=0$ are continuous-transition lines. Along the negative diagonal $r_1=r_2<0$, the potential has an enhanced $O(2)$ symmetry and a circle of minima. Crossing that diagonal exchanges the two ordered axes and makes derivatives of the minimum free energy jump, so it is a first-order phase transition line. The origin, where the two continuous lines meet the first-order line, is a bicritical point.

Define the two shell integralsFor an external ${\varphi }_1$, the ${\varphi }_1^4$ part of the interaction gives the scalar tadpole coefficient $12g_0{I}_1$, while $2g_0{\varphi }_1^2{\varphi }_2^2$ gives $4g_0{I}_2$. Interchanging the components givesWhen the masses agree, $I_1=I_2$ and both formulas reduce to $4(N+2)g_0{I}_1$ with $N=2$.

No. Even when the two masses differ, the free energy retains an independent Z2 symmetry for each component: ${\varphi }_1\mapsto -{\varphi }_1$ and ${\varphi }_2\mapsto -{\varphi }_2$. The bilinear ${\varphi }_1{\varphi }_2$ is odd under either transformation, so integrating out modes cannot generate it.