Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential? Why is there a specific negative sign in front of the $m_{12}$ term of the 2HDM Higgs potential?
(but not for the $m_{11}$ and $m_{22}$)
See for example: https://arxiv.org/abs/1106.0034
Eq. (2) Page 6:
$$
V = m_{11}^2\Phi_1^\dagger\Phi_1 + m_{22}^2\Phi_2^\dagger\Phi_2 -m_{12}^2(\Phi_1^\dagger\Phi_2+\Phi_2^\dagger\Phi_1) + \frac{\lambda_1}2(\Phi_1^\dagger\Phi_1)^2+\frac{\lambda_2}2(\Phi_2^\dagger\Phi_2)^2+\lambda_3\Phi_1^\dagger\Phi_1\Phi_2^\dagger\Phi_2+\lambda_4\Phi_1^\dagger\Phi_2\Phi_2^\dagger\Phi_1+\frac{\lambda_5}2\left[(\Phi_1^\dagger\Phi_2)^2+(\Phi_2^\dagger\Phi_1)^2\right].\tag{2}
$$
 A: The authors are writing down the most general potential consistent with the symmetries of the problem. The constants $m_{ij}^2$ are free parameters, and the sign is purely conventional. For example, if I were to write the most general linear function of $x$, I could write $f(x)=\alpha+\beta x$ for some parameters, or $f(x)=-\alpha+\beta x$, or $f(x)=\alpha-\beta x$, etc. All these parametrizations are equivalent, as $\alpha,\beta$ are free parameters and thus I am free to define their sign however I want.
In the case at hand this freedom in choosing the parametrization becomes even more clear due to the fact that we can change the sign in front of $m_{12}^2$ by the field redefinition $\Phi_1\mapsto-\Phi_1$ (or same with $\Phi_2$), which flips the sign of $(\Phi^\dagger_1\Phi_2+\Phi^\dagger_2\Phi_1)$, but leaves the rest of the Lagrangian invariant. The sign of $m^2_{12}$ is irrelevant, as it depends on our conventions. (Note that the signs of the eigenvalues of the mass matrix don't care about the sign of $m_{12}^2$, so this sign has nothing to do with symmetry breaking!) The authors are just choosing a particular sign that they found convenient for some reason. But there is no physics behind it.
A: Usually, arbitrary signs and phases are chosen to minimize the number of explicit signs and phases that appear later, for convenience. In the paper you linked, the sign chosen for $m_{12}^2$ ensures that it appears with a positive sign in the mass terms for the charged scalars in equation (5), and with a positive sign on the diagonal elements for the mass matrix for uncharged scalars in equation (7).
