= Solution
\b[<Order parameter> and the free-energy landscape.] In <Landau theory>, take a scalar <magnetization> component at fixed <volume (thermodynamics)> and composition. It is an <order parameter>: its nonzero spontaneous value distinguishes an ordered phase with <ferromagnetism> from the symmetric high-temperature phase. One first constructs a <constrained order-parameter free energy> at specified <magnetization> by summing over microscopic states compatible with it. An external <magnetic field> is an independent control, so it can also appear in that constrained function. For a total <magnetic moment> the conjugate term is $-BM$; if $M$ is a <magnetization> density, the total-energy term is $-VBM$, or $-BM$ in <free-energy density>. Minimizing over $M$ then yields the equilibrium thermodynamic potential as a function of the imposed field.
Thus treating both $M$ and $B$ as arguments of a Landau function does not treat them as independent natural variables of the already-minimized equilibrium <free energy>. The first is the fluctuating or constrained <order parameter>, and the second its <conjugate field>.
At zero field, reversal symmetry requires $F(M)=F(-M)$, excluding odd powers. Near a continuous transition, assume the homogeneous constrained <free energy> is analytic for small $M$, that the quadratic coefficient crosses zero linearly, and that the quartic coefficient is positive. Then
$$
F(T,M)=F_0(T)+b(T-T_c)M^2+dM^4,
\qquad b>0,\quad d>0.
$$
The <temperature> $T_c$ is where the zero-field quadratic coefficient vanishes and spontaneous ordering begins in this approximation. Higher powers are negligible sufficiently close to it; spatial fluctuations are neglected in this mean-field treatment.
Stationarity gives $2M[b(T-T_c)+2dM^2]=0$. Above $T_c$, the only minimum is at zero. Below it, zero becomes a maximum and two minima appear:
$$
\boxed{M_{\mathrm{eq}}=0\quad(T>T_c),\qquad
M_{\mathrm{eq}}=\pm\sqrt{\frac{b(T_c-T)}{2d}}\quad(T<T_c).}
$$
The two signs describe <spontaneous symmetry breaking>: they are the average magnetizations in the symmetry-broken pure phases. A strictly finite system at exactly zero field samples the two wells symmetrically and has <ensemble average> $\langle M\rangle=0$. A spontaneous pure-phase value is defined by selecting a phase, for example taking the <thermodynamic limit> before removing a small field. This qualification distinguishes a minimum's location from the average over a symmetric two-well distribution.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2018-b4-landau-free-energy.png]
{title=Quartic Landau free energy above and below the transition, showing the disordered and symmetry-broken minima}
\b[Minimum <free energy> and the heat-capacity jump.] At a nonzero minimum, $M^2=-b(T-T_c)/(2d)$. Substituting this value gives
$$
b(T-T_c)M^2+dM^4
=-\frac{b^2(T-T_c)^2}{2d}+\frac{b^2(T-T_c)^2}{4d}.
$$
Hence
$$
\boxed{F_{\min}(T)=F_0(T)-\frac{b^2(T-T_c)^2}{4d}\quad(T<T_c),}
$$
while $F_{\min}=F_0$ above the transition. Take the <temperature> <derivatives> only after minimizing, so the equilibrium <order parameter> is allowed to change with <temperature>. With $C_0=-TF_0''(T)$,
$$
C_V=\begin{cases}
C_0+Tb^2/(2d),&T<T_c,\\
C_0,&T>T_c.
\end{cases}
$$
Therefore
$$
\boxed{\Delta C_V\equiv C_V(T_c^-)-C_V(T_c^+)=\frac{T_cb^2}{2d}.}
$$
This is the <heat-capacity jump in quartic Landau theory>. The <entropy> is continuous at $T_c$, so there is no latent <heat>. If $F$ denotes <free-energy density> rather than total <free energy>, these expressions denote <heat capacity> per unit <volume (thermodynamics)> with the same normalization.
\b[Stabilizing a tricritical expansion.] At a <tricritical point>, the quartic coefficient is tuned to zero at the same point where the quadratic coefficient changes sign. Reversal symmetry still excludes odd powers. The next available stabilizing term is therefore
$$
F=F_0+b(T-T_c)M^2+d'M^6,
\qquad\boxed{d'>0.}
$$
A negative or vanishing leading sextic coefficient would not stabilize this truncation for large $|M|$; a higher positive term would then be needed and the present tricritical form would be insufficient. The ordinary quartic theory has a continuous transition for $d>0$, while a stabilized theory with $d<0$ can have a first-order transition. Their meeting at zero field is what makes the simultaneous tuning tricritical.
\b[The tricritical minimum.] At zero field and $T<T_c$, stationarity gives
$$
\frac{\partial F}{\partial M}=2M[b(T-T_c)+3d'M^4]=0,
\qquad
\boxed{M_{\mathrm{eq}}^4=\frac{b(T_c-T)}{3d'}.}
$$
The nonzero solutions are minima: their curvature is $2b(T-T_c)+30d'M^4=8b(T_c-T)>0$. Put $t=T_c-T>0$. Since $d'M^6=(bt/3)M^2$ at stationarity,
$$
F_{\min}-F_0=-btM^2+\frac{bt}{3}M^2
=-\frac23bt\sqrt{\frac{bt}{3d'}}.
$$
Thus
$$
\boxed{F_{\min}(T)=F_0(T)-\frac{2b^{3/2}(T_c-T)^{3/2}}{\sqrt{27d'}}.}
$$
The <order parameter> now vanishes as $(T_c-T)^{1/4}$ rather than the quartic theory's square root.
\b[Heat-capacity singularity.] Define $K=2b^{3/2}/\sqrt{27d'}>0$, so the ordered-side singular <free energy> is $F_{\mathrm{sing}}=-K(T_c-T)^{3/2}$. Its <derivatives> are
$$
\frac{dF_{\mathrm{sing}}}{dT}=\frac32K(T_c-T)^{1/2},\qquad
\frac{d^2F_{\mathrm{sing}}}{dT^2}=-\frac34K(T_c-T)^{-1/2}.
$$
Consequently
$$
\boxed{C_V^{\mathrm{sing}}=-T\frac{d^2F_{\mathrm{sing}}}{dT^2}
=\frac{Tb^{3/2}}{2\sqrt{3d'}}(T_c-T)^{-1/2},
\qquad\alpha=\frac12.}
$$
This is the <tricritical heat-capacity amplitude in a sextic Landau expansion>. It is positive on the ordered side and diverges as $T\to T_c^-$, rather than remaining a finite jump. This value of the <heat-capacity critical exponent> is the mean-field result along the path keeping the quartic term zero; a detuning to $d>0$ ultimately restores ordinary quartic behavior. Fluctuation corrections to the <mean-field theory> are not included in the stipulated <polynomial> minimization.
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