Is $\phi^\dagger \phi (\phi + \phi^\dagger)$ a possible interaction for scalar fields in QFT? In my QFT course, we are looking at writing the most general lorentz invariant, renormalizable lagrangians with hermitian interactions.
However I have never seen the interaction in the title mentioned, where $\phi$ is a complex scalar field. To me it seems this would fulfill all the criteria: in 4 dimensions, it has mass dimension 3, so the theory would be super renormalizable. Is there a reason this is not allowed?
 A: it is odd under $\phi \to -\phi$ and therefore the energy will be unbounded from below, which is a requirement
Edit following discussion in comments:
We demand that our theories will have an energy spectrum bounded from below. Taking the proposed term to be the only term in the potential energy will violate this, as it is odd under $\phi\to -\phi$. If we combine this with the knowledge that $\langle E \rangle \geq E_{\rm{gs}}$ for any state, where $E_{\rm{gs}}$ is the ground state energy, we see that if this is the highest-power term in our Lagrangian will lead to a configuration that will allow $\langle E \rangle \to -\infty$ (we will just push the VEV of $\phi$ to $\infty$ or $-\infty$).
However, if we have higher-power term with a finite positive coefficient, let's say $(\phi^{\dagger}\phi)^2$, then the energy can be bounded again from below.
One cal also consider such a term in the context of perturbation theory and calculate its effects and contributions, while keeping in mind that the underlying theory could be problematic without a necessary term that will prevent these divergences.
From symmetries point of views, as it was commented, this term violates the $U(1)$ symmetry for a complex field, which is something we would like to keep. But this in itself doesn't make such a term invalid. If you don't mind violating this symmetry, one can add other renormalizable terms (from dimension counting perspective) such as $\phi^2+\rm{h.c.}$.
A: It seems like it's allowed to me. But this term precludes a straightforward coupling to a $U(1)$ gauge field, and we usually want to do that: under $\phi\to \exp(i\alpha)\phi$, this term is not invariant. 
