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On the ordered branch of tricritical crossover scaling, at zero field and $X\to +\infty$, the minimizing square is $a={\psi }^2=(\sqrt{X^2+4}-X)/2\sim 1/X$. Its stationarity relation $Xa+a^2=1$ gives ${\Phi }_{-}(X,0)=-a/4-a^3/12\sim -1/(4X)$. Thus the minimized free-energy density becomes $-r^2/(4q)$ for fixed $q>0$ as $r\uparrow 0$, recovering the ordinary quartic continuous phase transition and its heat-capacity critical exponent $\alpha =0$. Replacing the scaling function by a nonzero constant in this limit would give an incorrect exponent.

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).