term in the right-hand side of is the tree level value. The second term comes from a loop diagram as in Figure \[fig:HartreeLoop\], which gives a contribution near $q=q_0$ of the form $$\label{E:HartreeLoopContribution}
\frac{\lambda}{(2\pi)^d} 
\int  \frac{d^d \vec{q}}{M^2+(|\vec{q}|-q_0)^2} \sim 
\frac{\lambda}{(2\pi)^d} S_d q_0^{d-1}
\int_0^\infty 
\frac{dq}{M^2+(q-q_0)^2} = 
\frac{\lambda}{(2\pi)^d} \frac{\pi S_d q_0^{d-1}}{M} \;.$$ There are subleading terms for small $M$, but what is relevant for the analysis below is the $1/M$ pole from the loop. The third term $\lambda a^2$ comes from the quartic coupling combined with the expectation value (\[E:phivev\]) of $\phi$.

![A loop diagram that contributes to the two point correlation function.[]{data-label="fig:HartreeLoop"}](HartreeLoop.jpg){width="7cm"}

The free energy $\mathcal{F}$ for the expectation value can be evaluated by setting [@Brazovskii], $$\label{E:dFda}
\frac{d\mathcal{F}}{da} = 2 M^2 a - \lambda a^3\;.$$ Combining this with $$\frac{da}{dM} = \frac{M}{\lambda a}\left(1+ \frac{s \lambda}{2 M^3}\right)\;,$$ derived from , we find $$\label{E:FreeEnergyExpression}
\begin{split}
\mathcal{F} &= \int \frac{d\mathcal{F}}{da} da = \int \frac{d\mathcal{F}}{da} \frac{da}{dM} dM\\
&=\frac{1}{2\lambda}\left(\frac{M^4} 2 + m^2 M^2 + 3s \lambda M - \frac{s\lambda m^2}{M} -\frac{s^2\lambda^2}{2 M^2}\right)\;.
\end{split}$$

Note that the terms in the parenthesis in depend on $s$ and $\lambda$ only in the combination $s \lambda$. If we rescale $m$ and $M$ by a factor of $\epsilon$ and $s\lambda$ by $\epsilon^3$, the 