Why usually we don't have a linear term in the potential? [duplicate]

Question

Let us consider scalar field theory. Why usually we do not have a linear term in the potential, like $$V(\phi)=a\phi+\frac{1}{2}m^2\phi^2+\frac{1}{4!}\lambda\phi^4,$$ or equivalently, after a field redefinition, $$V(\phi)=\frac{1}{2}m^2\phi^2+\frac{1}{4!}\lambda(\phi-b)^4?$$

This kind of potential is renormalizable and energy bounded from below. But I have never seen any serious discussions on this type of theory. Are there troubles with this potential?

PS: I have seen the discussions when there is a cubic term $g\phi^3$ in the potential. This happens in, for example, false vacuum decay.

Answer 1

In perturbation theory of QFT, we often (possibly implicitly) Taylor expand the action $$ S[\phi]~=~S[\phi_0] ~+~\underbrace{\left. \frac{\delta S[\phi]}{\delta \phi^{\alpha}}\right|_{\phi_0}}_{=0}\eta^{\alpha}~+~\frac{1}{2}\left. \frac{\delta^2 S[\phi]}{\delta \phi^{\alpha}\delta \phi^{\beta}}\right|_{\phi_0}\eta^{\alpha}\eta^{\beta}~+~{\cal O}(\eta^3) \tag{1}$$ around a stationary field configuration$^1$ $$\phi^{\alpha}~=~\phi_0^{\alpha} ~+~\eta^{\alpha},\tag{2}$$ where $\eta^{\alpha}$ are the quantum fluctuations. Hence the linear term in the action (1) effectively disappears. Trying to perform perturbation theory around a non-stationary field configuration is usually a futile enterprise.

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$^1$ Let us for simplicity assume that the action has a unique stationary field configuration $\phi_0^{\alpha}$.

Answer 2

If you write a general potential as a function of the field and you expand in Taylor series you have a linear term but if you want to have some global minimum of the potential at the field value zero then you have to ensure that there is no linear term. This is the best I can do.