What to do with a $\phi$ term in a Lagrangian?

Question

I am considering a Lagrangian that is of the following form: $$\mathcal{L}=-{1\over 2}\partial_\mu\phi\partial^\mu\phi+2\mu^2\phi^2+2\sqrt{6}{\mu^3\over \lambda}\phi + {9\mu^4\over 2\lambda} + \text{interactions}$$ Now, I am unsure what to do with a the term that is proportional to $\phi$: I think I heard at some point that terms like these can be absorbed in $\phi$ by a suitable redefinition, but I can't really see how that could happen here. Does anyone know what exactly to do with the term? Any help would be much appreciated

EDIT: Lagrangian fully written out is the following: There are $N-1$ fields labeled as $\chi_a$ and one field labeled as $\sigma$. \begin{align*} \mathcal{L}&=-{1\over 2} \sum_{a=1}^{N-1} \Biggl(\partial_\mu\chi_a\partial^\mu\chi_a+\partial_\mu\sigma\partial^\mu\sigma\Biggr) +\mu^2\Biggl(\sum_{a=1}^{N-1}\chi_a^2+2\sigma^2\Biggr)+2\mu^3\sqrt{{6\over \lambda}}\sigma+{9\mu^4\over 2\lambda}\\ &\quad\ -{\lambda\over 12}\Biggl({1\over 2}\sum_{a=1}^{N-1}\chi_a^4 +\sum_{a=1}^{N-1}\sum_{b\neq a}\chi_a^2\chi_b^2 +{1\over 2}\sigma^4 +\sum_{a=1}^{N-1}\chi_a^2\sigma^2\Biggr) -\mu\sqrt{{\lambda\over 6}}\Biggl(\sum_{a=1}^{N-1}\chi_a^2\sigma+\sigma^3\Biggr) \end{align*}

Answer 1

There are two ways to deal with a linear term in $\phi$:

1. Complete the square, as was suggested in the comments. This is very often possible, but sometimes you do not want to do that.
2. Interpret it as an interaction term with a $\phi$ particle popping out of the vacuum or vanishing. This will lead to non-zero tadpoles in your Feynman diagrams, so additional care is needed when performing calculations.