# Why are constant terms irrelevant in Lagrangians in quantum field theory? [duplicate]

This explanation is solid. I want the same in quantum field theory. For illustration I shall consider two real scalar fields with a mexican hat potential. Using dimensional regularization the action is $$I[\phi_1,\phi_2]=\int d^dx\bigg(\frac{1}{2}\partial_{\mu}\phi_1\partial^{\mu}\phi_1+\frac{1}{2}\partial_{\mu}\phi_2\partial^{\mu}\phi_2+\frac{1}{2}\mu^2(\phi_1^2+\phi_2^2)-\frac{1}{4}\lambda^2(\phi_1^2+\phi_2^2)^2\bigg)$$ where both $\mu$ and $\lambda$ are positive. The potential above is $$V(|\phi|)=-\frac{1}{2}\mu^2|\phi|^2+\frac{1}{4}\lambda^2|\phi|^4$$ has a minimum at $$|\phi|=\frac{\mu}{\lambda}$$ I pick to expand around $$\phi_1=\frac{\mu}{\lambda}\qquad\phi_2=0$$ redefining the parameters $$\eta=\phi_1-\frac{\mu}{\lambda}\qquad\xi=\phi_2$$ and re-expanding the lagrangian $$I[\phi_1=\frac{\mu}{\lambda}+\eta,\phi_2=\xi]\equiv I'[\eta,\xi]= \int d^dx\bigg(\frac{1}{2}\partial_{\mu}\eta\partial^{\mu}\eta-\mu^2\eta^2+\frac{1}{2}\partial_{\mu}\xi\partial^{\mu}\xi-\mu\lambda(\eta^3+\eta\xi^2)-\frac{\lambda^2}{4}(\eta^4+\xi^4+2\eta^2\xi^2)+\frac{\mu^4}{4\lambda^2}\bigg)$$ Now, this action goes in a generating functional. $$\mathcal{Z}[J_{\eta},J_{\xi}]=\int[d\eta][d\xi]\exp(iI'[\eta,\xi]+i\int d^dx\,\bigg[J_{\eta}\eta+J_{\xi}\xi\bigg])$$ I am concerned about the constant piece in the action $$\Delta I=\int d^dx\,\frac{\mu^4}{4\lambda^2}$$ This integral above is infinite even in $d=4-2\epsilon$ dimensions! so, in which sense can we ignore constant terms in the Lagrangian such as above in quantum field theory, and why?