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jak
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What would change if we add a linear term in the Klein-Gordon Lagrangian?

The usual Klein-Gordon Lagrangian reads

\begin{equation} \mathscr{L}= \frac{1}{2}( \partial _{\mu} \Phi \partial ^{\mu} \Phi -m^2 \Phi^2) \, . \tag1\end{equation}

Without additional symmetry beyond Lorentz symmetry, nothing forbids an additional linear term:

\begin{equation} \mathscr{L}= \frac{1}{2}( \partial _{\mu} \Phi \partial ^{\mu} \Phi -m^2 \Phi^2) - C \Phi \, , \tag2\end{equation} where $C$ is some constant.

This modified Lagrangian leads to a modified Klein-Gordon equation

$$( \partial _{\mu} \partial ^{\mu}+m^2)\Phi =C \, .\tag3$$

What would be the interpretation of this modified Klein-Gordon equation? Why do we usually neglect the linear term and hence the possible constant in the Klein-Gordon equation?

jak
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