Why do we multiply $(1+\delta)$, but just add $\delta$ to construct counterterms for a Lagrangian?

Question

Consider the Lagrangian

$$ L = \frac{1}{2}(\partial^\mu\phi\partial_\mu\phi-m^2\phi^2)+\bar\psi(i\not\partial-m)\psi-g\phi\bar\psi\psi. $$

I was told when we include the counterterms, it becomes

$$ L = \frac{1}{2}(\partial^\mu\phi\partial_\mu\phi)(1+\delta_z)-\frac{1}{2}(m^2+\delta_m)\phi^2+\bar\psi(i\not\partial(1+\delta_z')-(m+\delta_m'))\psi - (g+\delta_g)\phi\bar\psi\psi $$

Where $\delta_z$ and $\delta_m$, etc. are counterterms.

My question is why do we multiply $(1+\delta)$ to some part, while just adding $\delta$ to others?

Answer 1

There are 2 kinds of $Z$-factors in renormalization:

1. A $Z$-factor $Z_{\phi}=1+\delta_{\phi}$ associated with wave function renormalization $\phi_0=Z_{\phi}^{1/2}\phi$.

2. A $Z$-factor $Z_g$ associated with each coupling constant $g$. (The mass $m$ is here viewed as a coupling constant.)

OP's source uses for the 2nd kind an additive prescription for $\delta$. (Be aware that the precise definition of the counterterms $\delta$ may vary from author to author.)