In Srednicki's textbook (chpt 5) he has an expression:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int d^3k f(k) \int d^4x (\partial^2 e^{ikx}) \phi(x) [/tex]

and he wants to integrate by parts in order put the Laplacian on the field [tex]\int d^3k f(k) \int d^4x (e^{ikx})\partial^2 \phi(x) [/tex] instead of the the exponential. He says that surface terms vanish because f(k) is chosen so that [tex]\int d^3k f(k)e^{ikx} [/tex] is a wavepacket that will vanish at x=infinity.

However, doesn't the field [itex]\phi(x) [/itex] and its derivatives vanish at infinity anyway, eliminating the need for f(k) and allowing you to use plane-waves? You don't need a wavepacket to vanish at infinity because [itex]\phi(x) [/itex] will.

How else can you justify this manipulation in a path integral:

[tex]e^{i\int d^4x \partial_\mu \phi \partial^\mu \phi}= e^{-i\int d^4x \phi \partial^2 \phi}[/tex]

other than the field vanishes at infinity? In classical field theory, the reason you can write that is because surface terms don't matter as variations don't occur on the surface. But in quantum field theory, what allows you to integrate by parts like that? Don't you have to stipulate that the field and its derivatives vanish at ∞?

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# Do quantum fields vanish at infinity?

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