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Choose $A^{\mu }=(\Phi ,\mathbf{A})$, so that $F_{i0}=E_i$ and $F_{ij}={ϵ}_{ijk}{B}_k$, with ${ϵ}_{123}=1$. The definitions of the polarization-magnetization tensor imply $M^{0i}=-P_i$ and $M^{ij}={ϵ}_{ijk}{M}_k$. Separating temporal and spatial indices,The Levi-Civita symbol contraction equals $2{\delta }_{k\ell }$, givingThus $\mathbf{P}$ is the electric dipole moment per volume, or electric polarization, and $\mathbf{M}$ is the magnetic dipole moment per volume, or magnetization. The signs are those of dipole interactions in a Lagrangian; the associated fixed-dipole potential energies have the opposite signs. Consistently, the equation in part (a) has componentsso ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\mathbf{J}}_b={\partial }_t\mathbf{P}+\nabla \times \mathbf{M}$ are the bound charge and bound current.

The constants specify isotropic, nondispersive linear response: $\kappa$ controls electric polarization, and $\chi$ controls magnetization. Their numerical relation to conventional electric susceptibility and magnetic susceptibility depends on the signs and field normalization in the action. In the rationalized units used here, the printed action gives the constitutive relationsThus its relative permittivity and magnetic permeability areHere conventional ${\chi }_m$ is defined by $\mathbf{M}={\chi }_m\mathbf{H}$, not by $\mathbf{M}={\chi }_m\mathbf{B}$. With the alternative repaired action that yields the PDF's target equation, ${\epsilon }_r$ is unchanged but ${\mu }_r=(1-\chi {)}^{-1}$ and ${\chi }_m=\chi /(1-\chi )$. In particular, a positive conventional electric susceptibility corresponds to negative $\kappa$ in the printed conventions; the problem does not specify signs for these constants.

A Landau free energy is a coarse-grained thermodynamic potential expressed as a function of a proposed macroscopic order parameter. Its coefficients include microscopic and thermal effects already averaged over. At fixed external thermodynamic conditions, equilibrium selects its global minimum; other local minima can represent metastable states. It is not the internal energy of a single microscopic magnetic configuration. Here the real scalar $m$ represents a selected component of magnetization: zero describes the disordered state, while nonzero values distinguish two oppositely ordered states.

Every power is even, so the internal discrete symmetry isFor a magnetic order parameter, this represents reversal of the magnetization, as under time reversal in zero applied magnetic field. A nonzero field would generally add an odd term and break this symmetry explicitly. This scalar model has no continuous internal rotation symmetry; physical rotational symmetry of a vector magnet is not implied by treating one real component. The condition $c>0$ makes the polynomial bounded below at large $|m|$.

For the requested ordered phase on cooling, take the usual implicit assumption $a>0$, and write $r=a(T-T_c)$. If $a<0$ the temperature directions reverse; if $a=0$ this expansion has no temperature-driven transition. The source explicitly states only $c>0$.

For $b<0$ at $T=T_c$, the quartic term lowers the free energy near $m=0$, while the positive sextic term stabilizes it at larger $|m|$. The nonzero stationary points haveThey are global minima. The stationary point $m=0$ has zero quadratic curvature but is a flat local maximum because its leading nonzero term is negative. Thus the system is already ordered at $T_c$, with a finite magnetization.

More explicitly, a continuous onset from $m=0$ could occur only where $f^{\prime \prime }(0)=2r$ vanishes, namely at $r=0$. But the global minima there are the finite values above. For sufficiently large positive $r$, only $m=0$ is stable. Therefore the equilibrium minimum must change at some positive $r$ by crossing between a zero and a finite nonzero order parameter. This is a first-order phase transition. Its actual temperature lies above $T_c$ and is calculated next; $T_c$ itself is the disordered spinodal point, not the coexistence temperature.

Let $\Delta T=T-T_c$, with $a,c>0$ fixed. The equilibrium phase diagram has boundariesFor either branch, larger $\Delta T$ is disordered and smaller $\Delta T$ is ordered. Both signs of the ordered magnetization are symmetry-related global minima throughout the ordered region, rather than separate regions selected by temperature and $b$.

The branches meet at the tricritical point $(\Delta T,b)=(0,0)$. Directly on $b=0$, minimizing $r{m}^2+c{m}^6/3$ givesThe transition there remains continuous, with a different mean-field critical exponent. For $b<0$, the metastability boundaries are $\Delta T=0$ for the disordered state and $\Delta T=b^2/(4ac)$ for the ordered state. The first-order phase transition curve lies strictly between them, as shown in the figure. These spinodal curves describe possible hysteretic continuation of local minima, not equilibrium phase boundaries.

At zero field, write the right-hand side of the self-consistency equation as $G_T(S)=g(4\beta JS)$, with $g(x)=2\sinh x/[1+2\cosh x]$. The origin is always a solution, and the slope there isThe critical crossing occurs when this equals one, givingThis is a continuous phase transition in the mean-field approximation. For $x>0$,so the positive branch is a strictly concave function, while $g$ is odd and bounded by one. For $T\ge T_c$, the slope at zero is at most one and strict concavity puts $G_T(S)$ below $S$ for every $S>0$, leaving only the unmagnetized solution. For $T<T_c$, it initially lies above $S$ but has $G_T(1)<1$; hence there is one positive intersection and, by spin inversion symmetry, one negative intersection.

The curvature of the Helmholtz free energy in the mean-field approximation is $4J[1-G_T^{\prime }(S)]$ per site. It is negative at $S=0$ below $T_c$ and positive at the nonzero intersections, identifying the ordered minima and the unstable zero solution.

The critical temperature and branches here belong to the mean-field approximation. In the exact finite system, spin inversion symmetry forces zero average magnetization at zero field; selecting a broken-symmetry phase requires the thermodynamic limit with a vanishing field of specified sign.

The field term explicitly selects the positive $x$ direction. For $B>0$ the uniform minimizing state is ${\mathbf{S}}_j=S\hat{\mathbf{x}}$, with no remaining continuous degeneracy of its orientation. Rotations about the $x$ axis still leave the Hamiltonian operator invariant, and the selected state respects that residual symmetry.

Tilting the magnetization now costs a field energy. Accordingly the transverse spin wave acquires a gapas confirmed directly in d. The factor $S$ is present because the source couples the field as $-BS{\sum }_j{S}_j^x$ rather than $-B{\sum }_j{S}_j^x$. Gapless zero-field Goldstone modes need not survive an explicit symmetry-breaking field.

Choose the positive-$x$ polarized ground state selected by $B>0$, or its symmetry-selected $B\to 0^{+}$ limit. It is exactly the boson vacuum in the Holstein–Primakoff transformation. Thus $n_j|\mathrm{GS}⟩=0$ and $S_j^x=S-n_j$ giveFor the transverse components, $S^y=({\tilde{S}}^{+}+{\tilde{S}}^{-})/2$ and $S^z=({\tilde{S}}^{+}-{\tilde{S}}^{-})/(2i)$. In the vacuum, the only nonzero quadratic ladder-operator expectation iswhich follows using the full ordered square-root operators, or directly from ${\tilde{S}}^{-}|S,S⟩=\sqrt{2S}|S,S-1⟩$. ThereforeAn independent exact spin argument starts from $S_j^x|S,S{⟩}_x=S|S,S{⟩}_x$ and $(S_j^x{)}^2+(S_j^y{)}^2+(S_j^z{)}^2=S(S+1)$. Rotational invariance about $x$ makes the two transverse second moments equal, so each is $[S(S+1)-S^2]/2=S/2$. This agrees exactly with the full boson representation.

The first moments describe the local magnetization. Subtracting their squares from the second moments gives spin component variances: zero longitudinal variance and transverse variance $S/2$. Transverse quantum fluctuations persist in a perfectly polarized state even though its magnon occupation is zero. Their relative scale decreases as $S^{-1/2}$ in the classical limit.

At precisely $B=0$, “the ground state” requires a choice within the ferromagnetic multiplet. For example, when $NS$ is an integer, a maximum-total-spin state with total $x$ projection zero is also a ground state but has $⟨S_j^x⟩=0$. The displayed local moments refer to the explicitly specified polarized member, not to every superposition or rotationally invariant mixture of zero-field ground states.

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.

At $B=0$ a global reversal ${\sigma }_j\mapsto -{\sigma }_j$ leaves the energy and statistical weight unchanged but changes the sign of the one-spin observable. Pairing reversed configurations givesEquivalently, $⟨\pm |{\sigma }^z|\pm ⟩=0$ in the transfer eigenbasis, so $\mathrm{Tr}({\sigma }^z{T}^N)=0$. This holds for every finite even ring and all finite temperatures, and for the zero-field Gibbs limit. A symmetry-selected zero-temperature alternating state can have nonzero staggered magnetization; its uniform magnetization is still zero. The classical spin variable in this model has no separate quantum spin-operator dynamics despite the conventional matrix notation ${\sigma }^z$ used for the insertion.

The Hamiltonian has the discrete symmetry described in solution, whereas either selected easy-axis ground state has a nonzero order parameter $⟨S^x⟩$ and is not invariant under it. This is spontaneous symmetry breaking: the equations preserve the symmetry while the selected macroscopic quantum state does not.

An infinitesimal longitudinal magnetic field selects one orientation. The defining order of limits isin the ordered zero-temperature phase, with the opposite sign for $h\to 0^{-}$. A symmetric finite-volume quantum state can instead have zero magnetization. Quantum ordered states are generally dressed by fluctuations rather than exact fully polarized product states, because unequal transverse exchange can create pairs of spin waves. Discrete easy-axis symmetry does not imply a gapless Goldstone mode; indeed solution finds a nonzero spin wave gap. Nor does this argument assert positive-temperature order in a one-dimensional short-range chain.^[ref]

Insert the decorating-spin observable into the one bond that contains it. The decorated-spin occupation insertion uses the matrix:The partition function still contains $L$ decorating-spin sums; one of them has been replaced by this weighted sum. HenceTaking the dominant eigenvector givesThis is the probability that the decorating site is in either nonzero state. It tends to $2/3$ at high temperature, when its three states are equally likely, and to one at low temperature, when the nonzero aligned states gain bond energy. Unlike the magnetization, this even observable is not forced to vanish by spin reversal.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

Write $a=a_0(T-T_c)$, $a_0>0$, at fixed $b\ge 0$, and define the critical exponent by $|m|\sim (T_c-T{)}^{\beta }$ on the ordered side. For fixed $b>0$ and small negative $a$, the stationary equation gives $m^2=-a/b+O(a^2)$. ConsequentlyAt the tricritical point approached along the tuned path $b=0$, the stationary equation instead gives $m^4=-a/c$, soFor fixed $b<0$, the transition occurs before $a$ reaches zero and the magnetization jumps; it is not a second-order transition and has no vanishing-order-parameter exponent of this form. A path that does not keep the quartic coefficient tuned to zero generally crosses over to the ordinary result.

The continuous internal symmetry is a common rotation around $z$:Both the transverse exchange and the Dzyaloshinskii-Moriya interaction contain one raising and one lowering spin operator, so the phases cancel. Equivalently, $[H,{\sum }_j{S}_j^z]=0$ and the continuous group is U(1).

The discrete internal symmetry is time reversal, $\mathcal{T}{\mathbf{S}}_j{\mathcal{T}}^{-1}=-{\mathbf{S}}_j$, together with $\mathcal{T}i{\mathcal{T}}^{-1}=-i$. Each term in $H$ contains two spin operators, so it is invariant. The operation is antiunitary; treating the reversal of all three spin components as a unitary rotation would violate the spin commutation relations. It reverses the magnetization and interchanges the two easy-axis ordered orientations. A convenient equivalent representative, composed with a $z$ rotation, is $R_z(\pi )\mathcal{T}$, which reverses $S^z$ while preserving $S^x,S^y$ as operators.

The statement about one continuous and one discrete internal symmetry describes generic parameters. At $D=0$, additional unitary $\pi$ rotations about transverse axes are allowed; at $D=u=0$, the continuous internal symmetry enlarges to full spin rotation symmetry. Spatial translations of the uniform chain are separate symmetries of the sites.

Landau theory expands a thermodynamic potential in an order parameter, constrained by symmetry. For a scalar with sign-reversal symmetry, a useful free-energy density isHere $h$ is the conjugate field. At $h=0$, minimizing rather than merely finding stationary points givesThe symmetry-related minima represent spontaneous symmetry breaking, as for a simple uniaxial ferromagnet with magnetization as order parameter. An infinitesimal field selects one sign. Above $T_c$, the susceptibility is $1/[a(T-T_c)]$; below it the stable-branch susceptibility is $1/[2a(T_c-T)]$. Minimization lowers $f$ by $-a^2(T-T_c{)}^2/(4b)$ below $T_c$, producing a heat-capacity density jump $T_c{a}^2/(2b)$ with continuous entropy.

A negative quartic coefficient requires a stabilizing $+c{\eta }^6/6$ with $c>0$. It can produce a first-order transition before the quadratic coefficient changes sign: coexistence occurs at $a(T-T_c)=3b^2/(16c)$ and ${\eta }^2=-3b/(4c)$. Odd terms are allowed when the assumed symmetry is absent. Landau theory predicts mean-field exponents; spatial fluctuations can change critical behaviour, especially close to a critical point in low dimensions. A gradient term extends the model to nonuniform order parameters but does not by itself supply dynamical equations.

For two localized energy levels $0,\delta$ with degeneracies $d_0,d_1$, let $x=\delta /(k_{B}T)$. Canonical equilibrium givesHere $p_1$ is the total upper-level occupation probability and $u,s,c$ are per unit. For nondegenerate levels the entropy rises from zero to $k_B\ln 2$, while the heat capacity vanishes at both low and high temperatures and has a Schottky maximum near $k_{B}T=0.417\delta$. A spin one-half in a field, or a localized defect with two configurations, supplies an example. Higher levels must be thermally inaccessible; unequal degeneracies change the populations and peak. Distinguishable localized units have total partition function $z^N$, unlike a gas of indistinguishable particles sharing just two orbitals.

For the magnetic atom, choose level labels with $E_m=-mg{\mu }_{B}B$, $m=-J,\dots ,J$. Put $x=T_B/T=g{\mu }_{B}B/(k_{B}T)$. Its partition function and exact entropy of a finite magnetic multiplet areThe ratio form has removable limit $Z_J(0)=2J+1$. Since $s_J$ is a function of $x$ only, it depends on field and temperature solely through $B/T$. For fixed $J$ at small $x$, symmetry removes the odd powers and the supplied sum of $m^2$ givesFor large $J$, the expansion requires $J|x|\ll 1$, not merely $|x|\ll 1$.

For demagnetization with a linear-heat-capacity load, integrate its specified $C_L/T$:Isothermal magnetization at $T_i$ removes spin entropy; the bath receives the corresponding heat. During the following reversible, thermally isolated demagnetization, it is the combined entropy of spins and load that stays constant. At the minimum-field endpoint $B_f=0$, the ideal independent spins reach $s_{\max }=k_B\ln (2J+1)$. ThereforeThe bracket is the initial magnetic entropy deficit. The PDF's intermediate equality uses the initial entropy itself instead; with its preceding definition of $s$, that equality omits the maximum-entropy term. The corrected balance recovers its final temperature formula:Without the small-field expansion, the exact independent-spin prediction is $T_i-(N\theta /M)[\ln (2J+1)-s_J(T_i,B_i)/k_B]$. As a check, zero initial field gives zero entropy deficit and no cooling. The incorrectly printed intermediate balance would fail this check.

The result requires the load law throughout the temperature interval, hence $T_i\ll \theta$; a small initial Zeeman splitting, quantitatively $J{T}_{B,i}/T_i\ll 1$; positive final temperature; negligible additional heat capacities and heat leaks; thermal equilibration between subsystems; and a reversible sweep slow enough to maintain equilibrium. A negative extrapolated temperature is a breakdown of this approximation, not a cooling prediction. Spins alone would keep $B/T$ fixed, but the attached load changes that adiabat by transferring entropy to them as it cools.

Interactions invalidate the product of independent one-site partition functions. The entropy must be computed from the many-body energy spectrum or a justified interacting approximation. Correlations and possible magnetic ordering introduce an interaction scale and can remove the independent-spin entropy available for cooling near zero field. Internal fields and ordering commonly limit practical demagnetization. The detailed correction and any residual entropy depend on the interaction Hamiltonian, dimensionality and frustration; neither a universal ordering temperature nor a numerical final-temperature correction is specified here.