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One of the first assumptions, when introducing the Lagrangian and Hamiltonian in an undergraduate course on QFT is $$ \phi(x)=0\,\text{on the boundary} $$ and this is widely used in many situations (e.g when calculating a surface integral on the boundary). Is there a physical reason for such assumption?

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  • $\begingroup$ If the field is radiated during a finite time, it is absent at large enough distances, so the integral of its energy density is finite. Thus, it is convenient to use simple harmonics that vanish on boundaries and carry elementary quanta of energy. $\endgroup$ – Vladimir Kalitvianski Sep 30 '12 at 15:49
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    $\begingroup$ Also, in the limit as the boundary goes to infinity, a non-zero boundary value will correspond to an infinite total energy. $\endgroup$ – Benjamin Horowitz Sep 30 '12 at 17:43
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In general, boundary conditons must be adapted to the real situation.

Zero boundary conditions are just for the sake of simplicity. But they are realistic only when the field is really zero for some definite reason.

If the boundary is at infinity, zero boundary conditions means that everything of interest happens in a finite domain and cannot be noticed from far away. If this is really the case, these boundary conditions are appropriate. In particular, one needs zero boundary conditions for square integrable states.

But scattering requires different boundary conditions as something moves across space and time from infinity to infinity. The corresponding scattering states are not square integrable.

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  • $\begingroup$ In my case, I'm using field theory to study fluctuations on the sound wave propagation speed (phonons) inside a disordered solid. It yields the SCBA approximation, as depicted by W. Schirmacher. Following the calculations, it seems clear that in some places he exploits the $\varphi=0$ assumption, and I think that it's exactly one of the "finite domain of interest" case. Thx for your answer. Anyway, as this could be a quite interesting topic, I'll leave it open a little more, and set your answer as correct if there won't be other answers :) $\endgroup$ – Juan Sebastian Totero Oct 1 '12 at 8:43

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