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What type of regulation is being employed?

Physics Asked on June 26, 2021

As already mentioned in this post. In the context of QFT, the kernel of integration for the overlap of a field configuration ket, $| Phi rangle$ with the vacuum $|0rangle$ in a free theory is given by (See: S. Weinberg’s Vol. 1 QFT Foundations Ch. 9.2)
$$ mathcal{E}({bf x},{bf y}) = frac{1}{(2pi)^3} int {rm d}^3{bf p}, e^{i{bf p}cdot({bf x}-{bf y})}sqrt{{bf p}^2 + m^2}tag{1}label{eq:kernel}$$
which can be shown to algebraically match the following expression by abusing the Basset Integral for index $nu = -1$.

$$mathcal{E}({bf x},{bf y}) = frac{1}{2pi^2} frac{m}{r} frac{rm d}{{rm d} r} left( frac{1}{r} K_{-1}(m r) right)quad text{for}quad |{bf x – y}| = r tag{2}label{eq:kernel2},$$
where $K_{-1}$ denotes a modified Bessel function of the second kind. It is clear that the integration in Eq.~eqref{eq:kernel} is divergent, while Eq.~eqref{eq:kernel2} is not, so some sort of regularization happened in between these steps. Does anybody know which technique one could use to formalize the relation between the two?

One Answer

If you want to use the standard Fourier transform of functions instead of distributions (see @AccidentalFourierTransform comment), you can regularize as follows, $$mathcal{E}({bf x},{bf y}, t_x,t_y) := mathcal{E}({bf x},{bf y}, t,t)$$ for $$mathcal{E}({bf x},{bf y}, t_x,t_y) = w-lim_{epsilon to 0^+} frac{1}{(2pi)^3} int {rm d}^3{bf p}, e^{i[{bf p}cdot({bf x}-{bf y}) - p^0(t_x-y_y-iepsilon)]}sqrt{{bf p}^2 + m^2}$$ where $p^0:= sqrt{{bf p}^2 + m^2}$ and $w$ denotes the weak limit: first integrate against a smooth compactly support (or Schwartz) function of $({bf x},{bf y}) in mathbb{R}^6$ and next take the limit. Using that procedure one sees that it is equivalent to directly integrate the smooth function against the integral kernel in the right-hand side of (2). Hence (2) is an identity in the sense of distributions in ${cal S}'(mathbb{R}^6)$.

Correct answer by Valter Moretti on June 26, 2021

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