In Chapter 9 in Quantum Field Theory written by Srednicki.

This chapter discusses why $Z_{i}=1+O(g^{2})$ and $Y=O(g)$, given specific values of $m$, $g$, and normalization conditions $$\langle k|\phi(x)|0\rangle =e^{-ikx}\qquad\text{and}\qquad \langle 0|\phi(x)|0\rangle =0,\tag{9.2}$$ when considering an interacting quantum field with Lagrange density: $${\cal L}=-\frac{1}{2}Z_{\phi}\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}Z_{m}m^{2}\phi^{2}+\frac{1}{6}Z_{g}g\phi^{3}+Y\phi.\tag{9.1}$$

Denote $$Z(J)\equiv\langle 0|0\rangle_{J}=\int {\cal D}\phi\, e^{i\int d^{4}x\,({\cal L}_{0}+{\cal L}_{1}+J\phi)}\tag{9.5}$$

where $J$ is an external field source, and

$${\cal L}_{0}=-\frac{1}{2}\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}\tag{9.8}$$

$${\cal L}_{1}=-\frac{1}{2}(Z_{\phi}-1)\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}(Z_{m}-1)m^{2}\phi^{2}+\frac{1}{6}Z_{g}g\phi^{3}+Y\phi.\tag{9.9}$$

$$\begin{align}Z_{J} =~&e^{i\int d^{4}x\, {\cal L}_{1}\left[\frac{1}{i}\frac{\delta}{\delta J(x)}\right]}\int {\cal D}\phi\, e^{i \int d^{4}x\, ({\cal L}_{0}+J\phi)}\cr \propto~& e^{i\int d^{4}x\, {\cal L}_{1}\left[\frac{1}{i}\frac{\delta}{\delta J(x)}\right]}Z_{0}(J)\end{align}\tag{9.6}$$

where $Z_{0}(J)$ is (9.5) in the free field theory.

  1. Here is my first question: Why is the equal sign in (9.6) is replaced $\propto$? In my understanding, since $\phi$ is already renormalized, it should be the same as the field operator in free-field theory; thus it should still be $=$ rather than $\propto$ .

  2. The second problem is I cannot find the part in which the writer proves $Z_{i}=1+O(g^{2})$, nor could I understand why this is required.

By assuming $Z_{g}=1+O(g^{2})$ and taking $Z_{g}=1$, the writer if without $$-\frac{1}{2}(Z_{\phi}-1)\partial^{\mu}\phi\partial_{\mu}\phi-\frac{1}{2}(Z_{m}-1)m^{2}\phi^{2}+Y\phi,$$ $$\langle 0|\phi(x)|0\rangle =\frac{1}{2}ig\int d^{4}y\, \left[\frac{1}{i}\Delta(x-y)\frac{1}{i}\Delta(y-y)\right]+O(g^{3})\tag{9.18}$$ where $\Delta(x-y)$ is the Feynman propagator.

Clearly (9.18) violates our assumption that $\langle 0|\phi(x)|0\rangle=0$, so $Y\phi$ must be incorporated and $Y=O(g)$, but I cannot find the contents describing $Z_{i}=1+O(g^{2})$.

Could anyone help me with these two questions?


1 Answer 1

  1. The infinite-dimensional integration measure in the path integral may include an undetermined normalization/proportionality factor, cf. eqs. (9.6) & (9.7).

  2. $$Z_{\phi}=1+O(g^{2})\qquad\text{and}\qquad Z_m~=~1+O(g^{2})\tag{Z}$$ are only explained later in chapter 14 when we impose renormalization conditions (whose explicit forms depend on which renormalization scheme we choose), but they are always tied to which Feynman diagrams contribute to the self-energy. Eq. (Z) follows from the fact that such diagrams always have an even number of vertices in $\phi^3$-theory.


  1. M. Srednicki, QFT, 2007; chapter 9 + 14. A prepublication draft PDF file is available here.

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