- #1
kelly0303
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Hello! In several of the derivations I read so far in my QFT books (M. Schawarz, Peskin and Schroeder) they use the fact that "we can safely assume that the fields die off at ##x=\pm \infty##" in order to drop boundary terms. I am not sure I understand this statement in terms of QFT. A field in QFT is an operator which, when applied to vacuum, creates a particle at position ##x## i.e. ##<0|\phi(x)|p>=e^{ipx}## (ignoring the normalizations). But the vacuum is the same everywhere, so how can one assume that the field operator vanishes at infinity. The way I read that is that ##\phi(\pm \infty)|0> = |0>##, or something like this. But from the point of the view of the vacuum all the points are the same so there is no infinity, the operator should behave the same at infinity as it does anywhere else. Can someone explain this to me? Thank you!