our numerical studies demonstrate that if $W_0\sim \mu^2$ then $\Phi_0$ gets expectation values of order 1, but the expectation value shrinks as $W_0 \rightarrow 0$, going to zero in that limit.

One loop potential {#sec:One Loop}
==================

Since the interplay between the supergravity and one-loop effects is so important to our results it is worth discussing the details of the one-loop calculation, highlighting the approximations we have made. First of all we note that the mass matrices, $\mathcal{M}$, in the well-known formula: $$\begin{aligned}
  V_{\mbox{one-loop}} = V_{\mbox{quad.}}+ V_{\mbox{log}} =
  \frac{1}{32\pi^2}\mbox{STr}\mathcal{M}^2 \Lambda^2 + \frac{1}{64
    \pi^2}\mbox{STr}\mathcal{M}^4\log\frac{\mathcal{M}^2}{\Lambda^2}
  \label{eq:V_one-loop}\end{aligned}$$ are given by the supergravity corrected masses [@Chung:2003fi; @Cremmer:1982en]. This modifies the mass squared matrices at the level of $\mu^4$ (i.e. a rather small shift, but calculable and necessary for the computation of $V_{\mbox{quad.}}$). The term quadratic in $\mathcal{M}$ is generically $\sim m_{3/2}^2 \Lambda^2$ whereas the contribution from $V_{\mbox{tree}}$ is $-3 m_{3/2}^2$, hence the $\mathcal{M}^2$ term can be disregarded if $\Lambda \ll
M_{P}$, but not otherwise. Unfortunately Eq. (\[eq:V\_one-loop\]) is modified when $V\neq 0$ (for discussions of this point, see [@deAlwis:2006nm] and [@Gaillard:1993es]). Even though we are expanding about $V=0$, there will be corrections to this expression since $V=0$ is only true at that point and 