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?

• consider e.g. a harmonic oscillator, with $L=\frac12 m\dot x^2-\frac12 kx^2-cx$. The last term is a constant force that changes the equilibrium position, from $x_0=0$ to $x_0=c/k$. Of course, you can always redefine your coordinate $x$ so that it represents displacement from the equilibrium position, $x\to x-x_0$, which eliminates the linear term. In this sense, the $c$ term does not lead to new physics. The system is the same, in less convenient coordinates. The same thing happens in the KG case, but using $\Phi$ instead of $x$, and $\langle\Phi\rangle$ instead of $x_0$. – AccidentalFourierTransform Aug 26 '17 at 9:28
• @AccidentalFourierTransform Thanks a lot! So the constant is irrelevant as long as we consider a free scalar field $\Phi$, but becomes important as soon as we consider interactions with other fields?! If there are interactions with other fields, the field shift also affects other terms and thus, e.g. leads us to "mass terms" – jak Aug 26 '17 at 11:49
• Possible duplicate of Effect of linear terms on a QFT – tparker Jun 29 at 11:19

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

• thanks for your answer! I would love to know your thoughts on what the correct answers to these questions are. Regarding "Hint 1": A linear term arises sometimes in GUTs, and thus, as you note one gets $\langle\Phi\rangle\neq 0$, which is called an induced VEV. Usually these additional contributions to the symmetry breaking are tiny, because they are surpressed by some large scale. "Hint 2:" I'm not sure about the meaning of such a shift here... – jak Aug 26 '17 at 8:49
• @JakobH recall that for the validity of the LSZ formula we require $\langle\Phi\rangle=0$. In general we use $\Phi$ to calculate $S$-matrix elements and therefore we need its vev to vanish. If you don't want to use $\Phi$ to calculate $S$-matrix elements, then it is perfectly valid for its vev to be different from zero (but bear in mind that that would require one more renormalisation condition). If the vev of $\Phi$ is not forced to vanish, then a linear term in the Lagrangian is valid. But in general we do require $\langle\Phi\rangle$ to vanish, and this explains why we use $C\equiv 0$.(1/2) – AccidentalFourierTransform Aug 26 '17 at 9:19
• (2/2) If you take the vev of your equation (3), you get $m^2\langle\Phi\rangle=C$, and therefore $\langle\Phi\rangle=0$ if and only if $C=0$. Finally, recall that if $\langle\Phi\rangle=v\neq 0$, you can always redefine $\Phi'=\Phi-v$, which has $\langle\Phi'\rangle=0$ by construction. This shift is, by the previous argument, equivalent to $\Phi\to\Phi-C/m^2$. That's why I said that the two hints are actually the same. Again, $v\neq 0$ is not bad per-se. If you insist that it should vanish, then take $C=0$. Otherwise, you may keep $C$ arbitrary and use it to tune $v$ to any value you want. – AccidentalFourierTransform Aug 26 '17 at 9:23

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 thanks for correcting my typo. However, I guess that $\Phi^{2} \equiv \Phi^{\dagger} \Phi$, assuming complex fields. – gothicVI Aug 3 '17 at 12:29
• np. Note that $|\Phi|^2=\Phi^\dagger\Phi$, but $\Phi^2\neq \Phi^\dagger\Phi$ unless $\Phi$ is real. – AccidentalFourierTransform Aug 3 '17 at 12:31
• note also that in the complex case, the Lagrangian does not typically include the $\frac12$ factors. – AccidentalFourierTransform Aug 3 '17 at 14:42