= Solution
For candidate (i), $\mathbf p=p_0\widehat{\mathbf p}\cos q_0x$. Since $G(q_0)=\widehat a:=a-a_c$, $\langle\cos^2q_0x\rangle=1/2$, and $\langle\cos^4q_0x\rangle=3/8$, its mean-field free-energy density is
$$
\boxed{\frac FV=\frac{\widehat a}{4}p_0^2+\frac{3b}{32}p_0^4.}
$$
For $\widehat a<0$, stationarity gives
$$
p_0^2=-\frac{4\widehat a}{3b},
$$
and substitution yields
$$
\boxed{\frac{F_{(i)}}V=-\frac{\widehat a^2}{6b}.}
$$
For $\widehat a\geq0$, the minimum is $p_0=0$. The amplitude therefore vanishes continuously as $p_0\propto(a_c-a)^{1/2}$ on approaching $a_c$ from below, which is a continuous <mean-field approximation>[mean-field transition].
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