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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?

Qmechanic
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jak
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2 Answers2

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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 $\Phi\to\Phi+C/m^2$?

AccidentalFourierTransform
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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.

AccidentalFourierTransform
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gothicVI
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