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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*}

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  • $\begingroup$ $\psi:=a\phi+b/2a \implies \psi^2=a^2\phi^2+b\ \phi+\mathrm{some}$ $\endgroup$ – Nikolaj-K Dec 10 '13 at 10:26
  • $\begingroup$ Thanks. In hindsight, that was pretty obvious, but I guess that's how it always goes! $\endgroup$ – Danu Dec 10 '13 at 10:39
  • $\begingroup$ However, now that I think about it, this will create new interaction terms! Is that not a problem? $\endgroup$ – Danu Dec 10 '13 at 10:48
  • $\begingroup$ Will depend on your potential. See e.g. here, where they get rid of a cubic term. $\endgroup$ – Nikolaj-K Dec 10 '13 at 10:51
  • $\begingroup$ 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? $\endgroup$ – Danu Dec 10 '13 at 11:04
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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.
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  • $\begingroup$ 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. $\endgroup$ – Jerry Schirmer Dec 10 '13 at 16:42
  • $\begingroup$ @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 ? $\endgroup$ – Trimok Dec 10 '13 at 19:13
  • $\begingroup$ @JerrySchirmer What Trimok is trying to say is that the interpretation in terms of creation and anihiliation of particles is only valid if you have a harmonic potential and you expand around the minimum. Moving to that minimum is exactly what completing the square does. Interpretation as interaction term means that you expand around the minimum, disregarding the shift and taking its effects in only at higher orders in perturbation theory. $\endgroup$ – Neuneck Dec 10 '13 at 20:20

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