amount of time.

Non-perturbative contributions {#sec:NonPert}
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To estimate the value of the cut-off for which the non-perturbative piece (\[nnpert\]) dominates, and destabilises the potential, we calculate the second derivative of the non-perturbative correction to the potential, evaluated at the minimum of the tree-level + one-loop effective potential. If $\frac{\partial^2
  V_{\mbox{non-pert.}}}{\partial \Phi^2} >\frac{\partial^2\left(
  V_{\mbox{tree}}+V_{\mbox{one-loop}}\right)}{\partial \Phi^2}$, then the non-perturbative piece will likely dominate and the fields will roll to the supersymmetric minimum. This allows us to put rough lower bounds on the cut-off, such that the meta-stable solution is stable. This is in contrast to the global case in which the non-perturbative effects vanish in the tree-level + one-loop minimum. Considering, for simplicity’s sake, $N_f=N_c+1$: $$\begin{aligned}
  \frac{\partial^2V_{\mbox{non-pert.}}}{\partial \Phi_1 \partial (\Phi_0)_i} \sim \sum_j^{N_f-1} 2 h^{N_f/N} \Lambda_m^{-(N_f-3)}  \left( \prod_{k\neq j,i}^{N_f-1} (\Phi_0)_i \right)(-h \mu^2)\end{aligned}$$ where all higher powers of $\Phi$ have been discarded.

Close to the global SUSY minimum the dominant contribution to the effective potential is the globally supersymmetric effective potential derived in [@Intriligator:2006dd]. This means that

$$\begin{aligned}
  \frac{\partial^2 V_{\mbox{log}}}{\partial \Phi_{ii}^2}\sim \frac{\log(4)-1}{4\pi^2}h^4\mu^2\end{aligned}$$

and hence if $\sum_j 2 h^{\frac{N_f+N}{N}} \Lambda_m^{-(N_f-3)} \left(
  \prod_{k\neq j,i}^{N_f-1} (\Phi_0)_i \right)>
\frac{\log(4)-1}{4\pi^2}h^4$ it is clear that the non-perturbative piece will dominate and the fields will evolve into the SUSY minimum. 