# Lagrangian density of an interacting real scalar field theory

Srednicki writes the Lagrangian density of an interacting scalar field theory as $$\mathcal{L} = -\frac{1}{2} Z_\phi \partial^\mu \phi \partial_\mu \phi -\frac{1}{2} Z_m m^2 \phi^2 + \frac{1}{6} Z_g g \phi^3 + Y\phi$$

He claims that $Y=O(g)$ and $Z_i=1+O(g^2)$. What is the basis of this claim? Why do we ignore the $Y_0$ in the expansion of $Y(g)$: $$Y(g) = Y_0 + Y_1 g + Y_2 g^2 + \cdots \cdots$$ and again why do we take the $Z_0 = 1$ in the expansion of $Z_i(g)$: $$Z(g) = Z_0 + Z_1 g + Z_2 g^2 + \cdots \cdots$$

Why $Z_1 = 0$ ?

Are there any references where $\phi^3$ diagrammatics is explained clearly?

## 1 Answer

The free field, non-interacting Lagrangian is given by

$$\mathcal{L}=-\frac12\partial^\mu\phi\partial_\mu\phi-\frac12 m^2\phi^2.$$

In the limit $g\rightarrow0$, the interacting Lagrangian you wrote down is required to reduce to the free field version. This means that the counterterm part has to vanish, which is only possible if this limit yields $Y\rightarrow0$ and $Z_i\rightarrow 1$. This would not be possible if there was a constant $Y_0$ in the expansion.

The reason that there is no $O(g)$ term in the expansion for $Z_i$ can be found in figure 9.4 of Srednicki. The diagrams contributing to the vacuum expectation value of the field $\phi(x)$ all contain three vertices. Having an $O(g)$ term would lead to counterterms with two vertices, which is a mismatch.