= Solution
Take the ordered phase selected by a small positive real field, so $\phi$ is real and positive before sending $B\to0^+$. Stationarity gives $2a\phi+2\phi^3=B$. Differentiation yields $\chi=(2a+6\phi^2)^{-1}$. Therefore
$$
\boxed{\chi=\frac1{2a}\quad(a>0),\qquad
\chi_{\parallel}=\frac1{4|a|}\quad(a<0).}
$$
Both have mean-field <critical exponent> one, with a factor-of-two amplitude difference. The below-transition result is the longitudinal response of a selected pure phase. The transverse phase direction is soft, so a transverse zero-field response is singular; an unselected phase average is not the same susceptibility. At $a=0$, $\phi=(B/2)^{1/3}$ and the linear susceptibility diverges.
Back to article page
