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

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*}

• $\psi:=a\phi+b/2a \implies \psi^2=a^2\phi^2+b\ \phi+\mathrm{some}$ – Nikolaj-K Dec 10 '13 at 10:26
• Thanks. In hindsight, that was pretty obvious, but I guess that's how it always goes! – Danu Dec 10 '13 at 10:39
• However, now that I think about it, this will create new interaction terms! Is that not a problem? – Danu Dec 10 '13 at 10:48
• Will depend on your potential. See e.g. here, where they get rid of a cubic term. – Nikolaj-K Dec 10 '13 at 10:51
• My full Lagrangian is quite messy with a lot of interaction terms involving various powers of the relevant field (and other fields, too). Does this mean I probably cannot get rid of the linear term? And if that's the case, that what do I do with it? – Danu Dec 10 '13 at 11:04

There are two ways to deal with a linear term in $\phi$:
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.
• Couldn't you also use it as a modification to your quantization conditions? $\nabla^{2}\phi + m\phi + constant = 0$ isn't all that hard to solve, and then you're free to define your new $a$ and $a^{\dagger}$ operators in terms of the space of solutions to this new pde. I get that this is almost certainly just equivalent to completing the square and then un-doing the transformation, though. – Jerry Schirmer Dec 10 '13 at 16:42
• @JerrySchirmer : The spatial Fourier transform would give : $\frac{\partial^2\phi_k(t)}{\partial t^2} + (k^2+m^2)\phi_k(t) + constant ~\delta^3(\vec k) = 0$. How would you define the creation and annihilation operators ? – Trimok Dec 10 '13 at 19:13