Effective potential in scalar-vector interaction I am reading the thesis of Erick James Weinberg from arXiv: https://arxiv.org/abs/hep-th/0507214. On page 22 he finds the effective potential in scalar-vector interaction in one-loop. I am having a hard time to understand and to calculate the numerical factor $3$ in front of the relation. Also, what exactly is $G_{ij}(\phi_c)$?
 A: The factor 3 arises while calculating the functional determinant for the gauge fields which in turn gives the contribution of all the one loop diagrams in the effective potential. Let me illustrate this with the example of Scalar QED given on page 26 and 27 of the same thesis.
In scalar QED, the functional determinant with respect to the gauge field $A_{\mu}$ in the Landau gauge turns out to be:
\begin{equation}
\tag{1}
i{\Delta}^{-1}_{\mu\nu}(x-y) = -\delta^{4}(x-y)\bigg[ g_{\mu\nu}\Box - \partial_{\mu}\partial_{\nu} \bigg] - e^{2}\hat{\phi}^{2}g_{\mu\nu}\delta^{4}(x-y) 
\end{equation}
In the momentum space, the determiant can be written as:
\begin{equation}
\tag{2}
i{\Delta}^{-1}_{ab}(\hat{\phi},p) = \bigg[g_{\mu\nu}p^{2} - p_{\mu}p_{\nu}\bigg] - e^{2}\hat{\phi}^{2}g_{\mu\nu}
\end{equation}
On evaluating the determinant of this 4$\times$4 matrix (I did the same using Mathematica) you get a pretty simplified result
\begin{equation}
\tag{3}
e^{2}\hat{\phi}^{2}[p^{2}-e^{2}\hat{\phi}^{2}]^{3}
\end{equation}
Since the logarithm of the functional determinant goes into the effective potential, the factor 3 in the exponential comes at the front of the expression.
The point to note is that the calculation followed above as well as in the thesis are done in the Landau gauge which simplies the summing up of all the diagrams to an extent as there is no kinetic mixing between the gauge and the scalar field i.e a term like $\text{e}A^{\mu}\epsilon_{ab}\phi_{\text{a}}(x)\partial_{\mu}\phi_{\text{b}}(x)$ where $a,b \in \{1,2\}$ and $\epsilon_{ab} = \begin{bmatrix} 1 & 0 \\ 0 & -1  \end{bmatrix} $. These terms are set to $0$ in the Landau gauge where $\partial_{\mu}A^{\mu}=0$  by using the derivative in the expression to perform integration by parts. 
But even for an arbitrary choice of gauge, when the kinetic mixing terms are $\not=$0, it can be shown that a similar factor of 3 arises on the evaluation of the functional determinant.
