= Solution
For candidate (ii), $|\mathbf p|=p_0$ everywhere and each component has wavevector magnitude $q_0$. The quadratic density is therefore $\widehat a p_0^2/2$, while the quartic density is $bp_0^4/4$ without trigonometric averaging:
$$
\frac{F_{(ii)}}V=\frac{\widehat a}{2}p_0^2+\frac b4p_0^4.
$$
For $\widehat a<0$,
$$
p_0^2=-\frac{\widehat a}{b},
\qquad
\boxed{\frac{F_{(ii)}}V=-\frac{\widehat a^2}{4b}.}
$$
Since $(-1/4)/(-1/6)=3/2$, the helical structure's free energy is $50\%$ more negative than that of candidate (i).
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