# Is the scalar-field Feynman propagator at the origin ($x=0$) equal to 1?

I was reading about Feynman rules for scalar field in $$\phi^4$$ theory in section 4.6, pages 113-114 of Peskin & Schroeder, and, calculating amplitudes for processes, the authors show that Feynman propagators $$\Delta_{(0)}$$ appear as multiplicative factors. Until that moment I thought that Feynman propagators $$\Delta_{(x-y)}$$ were only defined for $$x\neq y$$, and, in fact, if the formula

$$\Delta_{(x-y)}=\int^{+\infty}_{-\infty}\frac{d^4p}{(2\pi)^4}\frac{e^{-ip(x-y)}}{p^2-m^2+i\epsilon}\tag{1}$$ is valid for $$x=y$$, then $$\Delta_{(0)}=\int^{+\infty}_{-\infty}\frac{d^4p}{(2\pi)^4}\frac{1}{p^2-m^2+i\epsilon}\tag{2}$$ would diverge (?).

Also, it is written at some point in Peskin, discussing feynman diagrams, that a diagram, obtained from the expansion of the Green function for the process considered, needs to be amputated before it is used to calculate amplitudes. For "amputated" the authors mean that every external line should be removed together with "bulb" which represent, precisely, propagators which starts and end at the same vertex, i.e multiplicative terms built like $$\Delta_{(0)}$$. My question is: does it makes sense to define $$\Delta_{(0)}=1$$, so that even this Feynman rule makes sense (the authors don't give convincing reasons to drop $$\Delta_{(0)}$$)? It could be the case, being the the propagator interpreted as the amplitude for a particle to move from a point to another, so that $$\Delta_{(0)}=1$$ would mean the particle has a probability of 1 to move from a point to the same point.

Edit: Peskin (and many other authors) explicitly defines the propagator $$(1)$$ only for different times (formula $$2.60$$ in Peskin & Schroeder), i.e. $$x^0 or $$x^0>y^0$$. It is true that, for $$x\rightarrow y$$, $$\Delta_{(x-y)}\rightarrow \infty$$, but still $$\Delta_{(0)}$$ is not defined.

2. The propagator $$\Delta(x-y=0)$$ at coinciding spacetime points $$x=y$$ is (without regularization, such as e.g. dimensional regularization) $$\infty$$; not $$1$$.
3. $$\Delta(x-y=0)$$ is an amputated self-loop diagram in momentum space, cf. OP's eq. (2). The contributions of self-loop diagrams are often cancelled via renormalization conditions, cf. e.g. this related Phys.SE post.
• I saw that post you are referring. 1. You said that we need renormalization (this means that we need to renormalize at first order in scalar $\phi ^4$ field theory? 2. That self loop appear at first order, if I'm not wrong) but the only other response is that, precisely, we could define $\Delta _{0} =1$. 3. I also read somewhere that the propagator is properly a distribution, this means that we could simply redefine it in a discrete number of points (is this correct?). 4. Could you please give references to renormalization of self-loop in $\phi ^4$ theory? Commented Nov 15, 2023 at 11:17
• 5. Furthermore Peskin explicitly define the propagator (formula 2.60) only for different time points ($x^0>y^0$ or $x^0<y^0$) so, it remains undefined for $x=y$. In the limit $x\rightarrow y$ would be $\Delta_{(x-y)}\rightarrow \infty$, but explicitly $\Delta_{(0)}$ is undefined, right? Commented Nov 15, 2023 at 12:05