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I have some Lagrangian containing a real scalar field $\phi$ with mass $m$. Let $A \in \mathbb{R}$ be some constant. The Lagrangian takes the form:

\begin{equation} \mathcal{L} = -\frac{A}{2} (\partial_\mu \phi)^2 - \frac{1}{2}m^2 \phi^2 + \mathcal{L}_{\phi \phi \phi} + \mathcal{L}_{\phi \phi \phi \phi}, \end{equation}

where the last two terms indicate interaction terms. My question is whether it makes sense to compute the scattering amplitude for the case with $A = 0$?

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  1. On one hand, if $A\equiv 0$, then the field is non-propagating, and one cannot construct a scattering theory. This is OP's case.

  2. On the other hand, if one makes a field-redefinition $$ \phi^{\prime}~=~\sqrt{|A|}\phi, \quad m^{\prime}~=~\frac{m}{\sqrt{|A|}}, \quad g_3^{\prime}~=~\frac{g_3}{|A|^{3/2}}, \quad g_4^{\prime}~=~\frac{g_4}{|A|^2}, $$ and takes the limit $A\to 0$, then the field becomes infinitely massive, and the coupling constants becomes infinitely large.

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  • $\begingroup$ Just to add on to this, in addition to becoming infinitely massive, the couplings become infinitely large, and therefore usual perturbative techniques do not work. $\endgroup$
    – Andrew
    Apr 9, 2022 at 15:49
  • $\begingroup$ $\uparrow$ Right. $\endgroup$
    – Qmechanic
    Apr 9, 2022 at 16:28

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