# Massless scalar propagator in Euclidean space and Green's equation

In this paper (Erickson et al, 2000), the authors claim in eq. (46) that the Green's equation corresponding to a bosonic propagator $$\Delta(x)$$ in $$2\omega$$ dimensions is:

$$- \partial^2 \Delta(x) = \delta^{2\omega} (x). \tag{1}$$

But why is there a minus sign in front? If I am not mistaken, the Green's equation in flat Euclidean space usually has no minus. Is it just a matter of convention? I am asking, because propagators would acquire a minus sign if using the "plus convention" compared to the Feynman rules of p.21 in the paper.

EDIT:

Here is a practical example that confuses me. Take the two point function, denoted by:

$$\langle \phi(x_1) \phi(x_2) \rangle. \tag{2}$$

At tree level this is just the propagator, so we would get a plus or a minus sign depending on the convention discussed above. But at leading order, the diagrams involved all contain an even number of propagators (see p.8 in the paper above), so this correction is independent of the convention! Therefore the full result is:

$$\langle \phi(x_1) \phi(x_2) \rangle = \pm f^{(0)}(x_{12}) + f^{(1)}(x_{12}), \tag{3}$$

where the $$\pm$$ depends on the convention. So the relative sign between tree level and leading order seems to be convention-dependent, which is dubious of course. I suspect the vertices must follow the convention that was chosen when defining the Green's equation, but how are those related?

I always put a minus sign ($$-\nabla^2$$) with the Laplace operator because $$-\nabla^2$$ has postive eigenvalues. Putting this sign in the operator definition saves a lot of minus signs in later equations, and so makes things easier to understand and to work with.
• In a Euclidean path integral one always wants $e^{- \int |\nabla \phi|^2 d^dx}\equiv e^{- \int | \phi^* (-\nabla^2 \phi) d^dx}$ for convergence. That fixes all the $n$-point functions. It's just a convention as to whether you write $G(x,x')= \pm [\nabla^{-2}]_{x,x'}$, but choosing the minus sign makes $G(x,x')$ positive. – mike stone Oct 3 '20 at 14:31