= Solution
At an ordinary zero-field <continuous phase transition>, the positive quartic coefficient remains nonzero while $\mathcal A_2\to0$. 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 $\psi^4+X\psi^2-1=0$, so
$$
\psi^2=\frac{\sqrt{X^2+4}-X}{2}\sim X^{-1}.
$$
At that stationary point, substitution gives $\Phi_-(X,0)=-\psi^2/4-\psi^6/12$. Its <large-quartic asymptotic of tricritical scaling> is consequently $\Phi_-(X,0)\sim-1/(4X)$, and
$$
f_{\rm s}\sim-\frac{|\mathcal A_2|^{3/2}}{4\sqrt{\mathcal A_6}X}=-\frac{\mathcal A_2^2}{4\mathcal A_4}.
$$
If $\mathcal A_2$ crosses zero linearly with <temperature>, this is quadratic in $T-T_c$, giving a finite <heat capacity> jump rather than a power divergence. Therefore
$$
\boxed{\alpha=0.}
$$
For 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 $\Delta C=T_ca^2/(2\mathcal A_4)$. Treating $\Phi$ as a nonzero constant while $X\to\infty$ would incorrectly assign the tricritical exponent to this ordinary transition.
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