# Assigning proper powers of $i$ to vertices of Feynman diagram

I'm reading the chapter 9 "the path integral for interacting field theory" of the Srednicki's QFT book. The lagrangian we are dealing with here is given by $$\begin{gather} \mathcal{L} = \mathcal{L}_0 + \mathcal{L}_1\\ \mathcal{L}_0 = -\frac 12 \partial^\mu\phi\partial_\mu\phi - \frac 12 m^2\phi^2 \,\,\mbox{(free lagrangian)}\tag{9.8}\\ \mathcal{L}_1 = \frac16 Z_g g\phi^3 + Y\phi -\frac 12 (Z_\phi-1)\partial^\mu\phi\partial_\mu\phi - \frac 12 (Z_m-1) m^2\phi^2.\tag{9.9} \end{gather}$$ Considering only the $$\phi^3$$ term, the generating functional is \begin{align} Z_1(J) \propto &~ \exp{\left(\frac i6 Z_g \,g \int d^4x \left(\frac{\delta}{i\delta J(x)}\right)^3 \right)} \,Z_0(J)\\ =& \sum_{V=0}^\infty \frac{1}{V!}\left[\frac i6 Z_g \,g \int d^4x\left(\frac{\delta}{i\delta J(x)}\right)^3 \right]^V \\ &\times \sum_{P=0}^\infty \frac{1}{P!}\left[\frac i2 \int d^4y\,d^4z J(y)\Delta(y-z)J(z) \right]^P \tag{9.11} \end{align} where $$Z_0(J)$$ is the generating functional for $$\mathcal{L}_0$$ and $$\Delta(y-z)$$ is the Feynman propagator. Here $$V$$, $$P$$, and $$E :=2P-3V$$ are the number of the vertices, propagators (edges), and the sources (external lines) of each Feynman diagram, respectively. As a Feynman rule, Srednicki assigns $$iZ_g g \int d^4x$$ for each vertex, $$\frac 1i \Delta(y-z)$$ for each propagator, and $$i\int d^4x J(x)$$ for each source. My question is how the powers of $$i$$ of these assigned values are determined? First I guessed that since each term of $$(1)$$ has $$i^V(\frac 1i)^{3V} i^P = i^{V-P+E}$$ as its prefactor, it is natural to assign $$i$$ for vertex, $$\frac 1i$$ for propagator, and $$i$$ for source as mentioned.

However, my guess turns out to be wrong when considering the next term, $$Y\phi$$. The generating functional is then $$Z_Y (J) \propto \exp{\left(iY \int d^4x \left(\frac{\delta}{i\delta J(x)}\right) \right)} \,Z_1(J). \tag{*}$$ According to my guess, a new kind of vertex introduced by $$Y\phi$$ should stand for $$Y\int d^4y$$ because two $$i's$$ cancel off in $$(*)$$ and a prefactor for each term is still $$i^{V-P+E}$$. But in eq. (9.19) on p. 66 (of the 1st edition), Srednicki assigns $$iY\int d^4y$$ instead and I cannot figure out how $$i$$ appears. Can anyone help me understand this? I appreciate any help.

1. By the way, one can restore the dependence of Planck's constant $$\hbar$$ in eq. (9.11) by replacing $$i\to \frac{i}{\hbar}$$, so the counting of $$i$$s is related to the counting of $$\hbar$$s.
2. OP is discussing Feynman diagrams in the $$J$$-picture, i.e. diagrams of the partition function $$Z[J]$$. (In particular there are no amputated diagrams.) Then the simplest weight assignment goes as follows: All propagators (internal and external) have weight $$\frac{\hbar}{i}$$, and all vertices and sources have weight $$\frac{i}{\hbar}$$. (Sources may be viewed as 1-vertices.)
4. Finally let us address OP's last question. Since $$Y$$ is a 1-vertex, it has weight $$\frac{i}{\hbar}$$.