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?
 A: Hint 1. What happens if you take the vacuum expectation value of equation $(3)$? You should find $\langle\Phi\rangle\neq 0$. Why is this a bad thing?
Hint 2. (pretty much the same thing as Hint 1 actually) What happens under the field redefinition $\Phi\to\Phi+C/m^2$?
A: One obvious change is as well that the usual Klein-Gordon Lagrangian
\begin{equation}
\mathcal{L} = \frac{1}{2}(\partial_{\mu} \Phi^{\dagger} \partial^{\mu} \Phi - m^{2} \Phi^\dagger\Phi)
\end{equation}
has a $U(1)$ symmetry if you consider the field $\Phi$ to be complex.
Meaning that $\mathcal{L}$ is invariant under the transformation
\begin{align}
& \Phi \to \Phi e^{i\phi} \,, \\
& \Phi^{\dagger} \to \Phi^{\dagger} e^{-i\phi} \,,
\end{align}
where $\phi \in \mathbb{R}$ is a real parameter. You could also promote $\phi \to \phi(x)$ making the global $U(1)$ symmetry a local one, hence promoting it to a gauge symmetry.
Adding a linear term explicitly breaks said symmetry! Therefore in principle if you have a initial gauge symmetry adding a linear term is forbidden by the demand for gauge invariance.
