 Solution

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.