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 Answer 1

  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.)

  3. An often-used modified weight assignment don't count sources and external propagators, which works as long as there are no other 1-vertices and no propagators connected to 2 sources.

  4. Finally let us address OP's last question. Since $Y$ is a 1-vertex, it has weight $\frac{i}{\hbar}$.

  • $\begingroup$ Thank you for your answer! It helped me clarify the weight assigning rule. $\endgroup$
    – asdf
    Commented Oct 30, 2020 at 6:14

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