cannot be satisfied for all fields.

The tree-level superpotential in the magnetic theory is given by: $$\begin{aligned}
  W_{\mbox{tree}}=h\mbox{Tr}\bigl(\phi \Phi \tilde{\phi}\bigr) - h \mu^2
  \mbox{Tr}(\Phi)\end{aligned}$$ where $\Phi$ transforms as $N_f \times \overline{N_f}$, $\phi$: $(\overline{N_f},N)$, $\tilde{\phi}$: $(N_f,\overline{N})$, $N=N_f-N_c$, the number of squark flavours in the magnetic theory and we denote the parts of $\Phi$ that will later obtain expectation values as follows: $\Phi=\left(\begin{array}{cc}
    \Phi_1 & 0 \\ 
   0 & \Phi_0\end{array} \right)$. The Kähler potential is canonical.

Considering the tree-level superpotential in isolation one finds that the lowest energy state is a moduli space parametrised by

$$\Phi=\left( \begin{array}{cc}
      0 & 0 \\
      0 & \Phi_0\end{array} \right), \quad \quad \phi= \left(\begin{array}{c} \phi_0 \\ 0 \end{array} \right), \quad \quad \tilde{\phi}^{T}= \left(\begin{array}{c} \tilde{\phi}_0 \\ 0 \end{array} \right), \quad \quad \phi_0 \tilde{\phi}_0 = \mu^2 \, \mathbb{I}_{N_c \times N_c}.$$

Since SUSY has to be broken, the potential is positive definite and is found to have an expectation value of $V=h^2 N_c \mu^4$. When the one-loop effects are included the moduli space is lifted and, aside from flat directions identified with Goldstone bosons, a unique minimum is found at: $$\Phi=0, \quad \quad\phi_0=\tilde{\phi_0}=\mu \,
  \mathbb{I}_{N_c \times N_c}.$$

In addition one must include the non-perturbative, R-symmetry violating contribution: $$\begin{aligned}
 \label{nnpert}
  W 