Past exam of the theoretical physics major option course of the University of Oxford / 2015 / 3

For an analytic, local Landau theory with independent global rotations of the magnetization vector, a uniform scalar potential can depend only on $\rho =\mathbf{m}\cdot \mathbf{m}$. Keeping terms through fourth order in the field, dropping an additive constant, and retaining the leading isotropic positive gradient term gives the displayed quadratic-plus-quartic functional. The stiffness has been absorbed into the normalization of length and field. Higher powers and higher derivatives are neglected in this long-wavelength approximation.

The quartic truncation is stable at large $|\mathbf{m}|$ when $u>0$. The coefficient $t$ changes sign at the mean-field critical point; close to it one commonly writes $t\propto T-T_c$. If $u<0$, higher stabilizing powers are essential, so the displayed truncation alone cannot determine a stable first-order phase diagram. For $u=0$ and $t<0$ it is likewise unbounded.

For $u>0$, spatially uniform stationary points obey $(t+4u\rho )\mathbf{m}=0$. The global minima areThe corresponding minimum free-energy density is zero and $-t^2/(16u)$, respectively. The disordered phase gives way continuously to a ferromagnet with an arbitrary direction of magnetization. At this mean-field transition the amplitude vanishes as $(-t{)}^{1/2}$; long-wavelength critical fluctuations can modify mean-field exponents in three dimensions.^[ref]

Vary each field component independently and integrate the gradient term by parts, taking fixed boundary values or vanishing surface variations. With $\rho =m_1^2+m_2^2+m_3^2$, the Euler-Lagrange equations for the free-energy saddle areThe anisotropy shifts the first two inverse susceptibilities in opposite directions. The derivative of $u{\rho }^2$ is $4u\rho m_j$, which fixes the factor needed for both the magnetization and the free energies in c. Under natural free-boundary conditions the integration by parts instead supplies ${\partial }_n{m}_j=0$ at the surface.

Assume $u>0$, as required for the stable quartic model, and retain the stated domain $t>0$. For the opposite-axis quadratic anisotropy of a vector order parameter, fix the magnitude $\rho$. Then the anisotropy lowers the energy most by putting all magnetization along component 1 if $r<0$ and component 2 if $r>0$. Component 3 has positive quadratic coefficient $t$ and cannot lower the uniform minimum. Minimizing the remaining amplitude givesThe component-1 and component-2 ordered regions each break a discrete sign symmetry, giving two easy-axis orientations rather than a sphere of equivalent directions. To check stability, in the component-1 minimum the longitudinal curvature is $-2(t+r)>0$, while the two transverse curvatures are $-2r>0$ and $-r>0$. The component-2 case similarly has longitudinal curvature $-2(t-r)$ and transverse curvatures $2r,r$, all positive in its region. The disordered state is stable exactly when $t\pm r\ge 0$.

Both boundaries $r=-t$ and $r=t$ are continuous, second-order mean-field transitions: the selected component vanishes with a square root, and the first derivatives of the minimum free energy agree across the boundary. In the stipulated upper half-plane $t>0$, the two ordered phases are separated by a disordered wedge. There is no direct first-order boundary between them in that domain. Extrapolating a phase boundary through $t<0$ would answer a different, larger phase-diagram question.

Specify the Fourier transform before performing the Gaussian integral:At $u=0$ the functional is diagonal in both component and momentum, with inverse kernels $q^2+t+r$, $q^2+t-r$, and $q^2+t$. The condition $0<r<t$ makes all three positive. The covariance of a Gaussian distribution with statistical weight $e^{-F/(k_{B}T)}$ is $k_{B}T$ times the inverse kernel, henceIf $F$ is already the dimensionless free energy divided by $k_{B}T$, the numerators are one instead. Translational invariance forces opposite momenta to pair. There is no complex conjugate on the second field, so the delta function is at $\mathbf{p}=-\mathbf{q}$, not at $\mathbf{p}=\mathbf{q}$. A correlation with $m_j(\mathbf{p}{)}^{\ast }$ would have the latter convention. Rotational symmetry in real space leaves dependence on $q^2$ alone once the momentum-conservation delta function is factored out. The correlation lengths are ${\xi }_1=(t+r{)}^{-1/2}$ and ${\xi }_2=(t-r{)}^{-1/2}$; the second component becomes soft as $r\uparrow t$. The skipped letter in the source is preserved: this solution is part (e), with no inferred part (d).