# 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.

• So it's a convention? It does not affect unitarity whether there is a minus or a plus? Then the vertices should follow that sign in some way I guess, right? Otherwise you would get different signs.
– Pxx
Oct 3, 2020 at 13:06
• I realize my last comment might be confusing, so I added a practical example in the question.
– Pxx
Oct 3, 2020 at 14:07
• 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. Oct 3, 2020 at 14:31
• Mm okay. And the relative sign between the leading order and the tree level that I discuss in the edit is then also irrelevant? That sounds strange to me, because I encountered a more complicated example in which the anomalous dimensions change sign depending on the choice of convention. I don't think that this is normal behavior, so I assumed the vertices have to adapt to that convention, but maybe I am still not seeing something.
– Pxx
Oct 3, 2020 at 14:42
• Yes. Changing the convention can't change an output of the calculations such as the anomalous dimension. I always expand the exponential to get the perturbation expansion. I never rely on a set of "Feynman rules," as such rules are convention dependent. Oct 3, 2020 at 14:55