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We all have been told when we were kids that constant terms in Lagrangians are irrelevant because when doing classical dynamics the Euler Lagrange equations involve derivatives of the Lagrangians so the constant terms just vanish.

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?


marked as duplicate by Qmechanic Jun 28 '18 at 19:03

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