# Why do we ignore surface terms in the action in Quantum Field theory ?

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I have trouble understanding why do we ignore surface terms. To get the on-shell fields we fix the fields at infinity so these terms do not matter. However , In the path integral , We integrate over all field configurations. So we must integrate over all values of the fields at infinity. So , one shouldn't set terms like $e^{i\oint dn^{\mu}\psi_{\mu}}$ equal to one and they should contribute to the partition function. Is this correct ?

In the field theory textbook I use , The path integral is written as $\int D\phi e^{iS}$ So I'm confused. The total derivative terms are evaluated at the boundary of spacetime by gauss theorem. Right ? do we keep the fields at spatial & temporal infinity fixed ? I'm confused by this point.
If you remember, the path integral can be introduced in calculation of the amplitude $\langle x(t_1)|x(t_2)\rangle$, $t_2 > t_1$. So the initial and the final "positions" (states) are considered fixed, known. It is written in many textbooks.