Effect of linear terms on a QFT

Question

I was told when first learning QFT that linear terms in the Lagrangian are harmless and we can essentially just ignore them. However, I've recently seen in the linear sigma model, \begin{equation} {\cal L} = \frac{1}{2} \partial _\mu \phi _i \partial ^\mu \phi _i - \frac{m ^2 }{2} \phi _i \phi _i - \frac{ \lambda }{ 4} ( \phi _i \phi _i ) ^2 \end{equation} with $m ^2 =-\mu^2 > 0$, adding a linear term in one of the fields $\phi_N$, does change the final results as you no longer have Goldstone bosons (since the $O(N)$ symmetry is broken to begin with).

Are there any other effects of linear terms that we should keep in mind or is this the only exception of the "forget about linear terms" rule?

Answer 1

Linear terms are important. But in a Poincare covariant QFT, one can always remove them by shifting the field by a constant computed as a stationary point of the Lagrangian.

If there is only one stationary point, it must be a minimizer (to have the energy bounded below), then this gives a unique normal form without linear terms.

If there are multiple stationary points, not all of them are minimizers, but to get physical results, one must shift by a minimizer (usually the global minimizer). This is not always unique.

Thus linear terms can be ignored ''without loss of generality'' in the theory. But in any particular model they cannot be ignored but must be properly removed by shifting the field, and this field shifting affects all other constants in the Lagrangian. Simply ''forgetting'' the linear terms would give wrong results.

Answer 2

Linear terms can be thought as source terms. They are important to define the effective potential (which is the Legendre transform of the (log of) the partition function with respect to the source).

I'm not sure why one would say that one can forget about them, since, for instance, they imply a non zero value of $\langle \phi\rangle$ even in the symmetric phase. This implies, for example, that the tadpole diagrams are non-zero, that you have effectively $\phi^3$ vertices, etc. Maybe the reason is that if you shift the field $\phi\to\phi-\langle\phi\rangle$ the source disappears from the Lagrangian...

Answer 3

Adam's answer from a slightly different perspective. Linear terms are source terms, which are essentially equivalent to boundary conditions. Allowing nontrivial boundary conditions considerably enriches the mathematical behavior these models exhibit. In particular, you shouldn't be surprised that boundary conditions can lead to interesting phase structures. Even in the Ising model, something has to choose +1 or -1 magnetization.