# Understanding field strength renormalization

When it comes to renormalization in QFT I always considered the renormalization of mass and coupling constant as rather intuitive, whereas I found the field strength renormalization rather counter-intuitive. I do not consider the rescaling of the quantum field $\phi(x)$ as intuitive. I did not like the argument: just rescale the field so that it makes the LSZ-formula work. Neither I like the argument that renormalization, here in particular field strength renormalization works perfectly (predicted results agree with the measured ones) , so it is justified (There is of course the whole formalism of renormalization group which explains "better" the renormalization machinery than the 2 preceding arguments, I guess, but I have not yet thoroughly looked into that).

But recently I found the following argument in the book "Tutorium QFT" (Edelhaeuser, Knochel). If the bare field $\phi_0$ is replaced by $\sqrt{Z}\phi$ where $\phi$ is the renormalized field the lagrange density $L$ changes from

$$L = \frac{1}{2}\partial_\mu \phi_0 \partial^{\mu} \phi_0 - \frac{1}{2}m^2 \phi_0^2-\lambda \frac{\phi^4_0}{4!}$$

to

$$L = \frac{1}{2}Z \partial_\mu \phi \partial^{\mu} \phi - \frac{1}{2}Z m^2 \phi^2-\lambda Z^2 \frac{\phi^4}{4!}$$

then the field equations

$$(\Box +m^2)\phi_0 = -\frac{1}{3!}\lambda \phi^3_0$$

change to

$$(\Box +m^2)\phi = -\frac{1}{3!}\lambda Z \phi^3$$

so actually rescaling the field strength does change the field equations, here in particular the coupling constant. So actually rescaling the field strength must have a sensible effect.

However, is it justified to argue that as the coupling constant $\lambda$ anyway gets renormalized, a rescaled coupling constant $\lambda Z$ does not matter (This is my question) ? The same/similar argument could then also be applied on more realistic QFTs like QED etc., I guess.