fields. Re-writing this in terms of the tree-level potential gives: $$\begin{aligned}
 V_{\mbox{quad.}} = \frac{\Lambda^2}{16\pi^2} (N_f^2+2 N_f N_c-1) V_{\mbox{tree}} + \frac{\Lambda^2e^K}{16\pi^2} ( ( N_f^2+2 N_f N_c) - 3) W^2 \label{eq:msquared}\end{aligned}$$ and hence $$\begin{aligned}
  V_{\mbox{tree}}+ V_{\mbox{one-loop}} = \left((N_f^2+2 N_f N_c - 1) \frac{\Lambda^2}{16\pi^2} + 1 \right) V_{\mbox{tree}} + \frac{\Lambda^2}{16\pi^2} ( N_f^2+2 N_f N_c - 3) e^K W^2 + V_{\mbox{log}} \label{Eq:One-loop breakdown}\end{aligned}$$

The addition of $V_{\mbox{quad.}}$ to the potential reinforces the tree-level solution, up to additional, gravitationally suppressed contributions from the final term in Eq. (\[eq:msquared\]).

On dimensional grounds, the SUSY parts of the Sugra F-terms will provide the dominant contributions to $V_{\mbox{quad.}}$, except for the moduli fields, which have flat F-terms at the SUSY minimum. This means we expect the minima of $V_{\mbox{quad.}}$ and $V_{\mbox{tree}}$, for the non-moduli fields, to coincide at leading order in $\mu$ (assuming the moduli are taken to be $\sim \mu^2$). However, the gravitational corrections given by $W^2$ and by $K_i W$ come in at the same order of magnitude and hence we do not expect that the same $\Phi_0$ will minimise both $V_{\mbox{quad.}}$ and $V_{\mbox{tree}}$. It is nonetheless clear from Eq. (\[eq:msquared\]) that, if $(N_f^2+2 N_f
N_c)\frac{\Lambda^2}{16\pi^2} \sim 1$, the quadratically divergent corrections will be of equal 