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

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?

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.)