One consequence of the Ward identity (cf. Di Francesco et al) is that it means variation of correlators under infinitesimal transformation is zero. This can be seen by integrating the ward identity, and using the Gauss divergence theorem to integrate over the surface at infinity, and setting the surface term to zero.

$$ \delta_{\omega} \langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle =\int \frac{\partial}{\partial x^{\mu}}\langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle = \int_{\Sigma} ds_{\mu} \langle j^{\mu}_a \Phi(x_1) \ldots \Phi(x_n) \rangle \\ =0$$.

DiFrancesco says the integrand goes to zero at infinity because the divergence of the correlator vanishes at away from the points $x_1 \ldots x_n$. How does the vanishing of divergence mean vanishing of the correlator? **In general when can we assume a correlator to vanish at infinity "sufficiently fast"?**

In this particular case, from familiar examples of the free boson and fermion, I know that $j$ is proportional to the gradient of the fields, so does this mean we are considering field solutions which attain a constant value? **How is it different from solitonic solutions?**

I have seen surface term to be vanishing sufficiently fast in many places in QFT, cannot seem to remember them now ,but this seems to be quite an adhoc assumption, is it because you want the drama of the theory to happen only in a finite range, and you cannot consider particles at inifinity?

This post imported from StackExchange Physics at 2014-06-21 09:00 (UCT), posted by SE-user ramanujan_dirac