The one loop Coleman-Weinberg contribution of a scalar field to the effective potential (in $\overline{MS}$-scheme) is: \begin{equation} \mathrm{const.} \times m^4(\phi_c) \left( \log \left( \frac{m^2(\phi_c)}{\mu^2}\right) -\frac{3}{2} \right) \end{equation}
Now, I have a problem with this formula. In theories with spontaneous symmetry breaking, like the standard model, the background field dependent mass will actually be negative. For the SM effective potential (usually calculated in Landau gauge) we have the Higgs field and the Goldstone fields, with:
\begin{eqnarray} m_H(\phi_c) = 3 \lambda \phi_c^2 - m^2 \\ m_G(\phi_c) = \lambda \phi_c^2 - m^2 \end{eqnarray} At the Higgs VEV $m_H(v)=2m^2$ and $m_G(v)=0$. Where $m$ and $\lambda$ are the renormalized parameters of the tree level Higgs potential.
My problem is that this implies, for example, for $\phi_c < v$ the Goldstone boson $\phi_c$ dependent mass is negative, and the logarithm is complex, also as $\phi_c \to v$ the real part of the log goes to $-\infty$.
Am I doing something silly or does the formula really break down? Or is there a way of making sense of the imaginary potential?