For tree-level amplitude, we're using the finite constant determine using experiments ($\lambda _R$)
For diverging amplitudes, we're using a different constant: $\lambda _R+ C\ln \frac{\Lambda ^2}{s_0} \lambda_R^2+….$. This constant depends on $\Lambda$ so that it gets cancelled out with the $\Lambda$ in the integration cut-off, in the $\Lambda \rightarrow \infty$ limit.
My question is, how can we justify using different coupling constants for different terms in the Dyson series? The derivation of the Dyson series has the same coupling constant for every term.
Best Answer
I assume/hope the following post should clarify the issue: I'm missing the point of renormalization in QFT.
We don't use two different coupling constants, we use the same constant.
The perturbative expansion is always in terms of $$ \lambda:=G_4(p)|_{p^2=\mu^2} $$ where $G_4=\langle\phi\phi\phi\phi\rangle$ is the four-point function and $\mu$ is some energy scale of interest. You can choose whatever $\mu$ you want, typical choices are $\mu=0$, $\mu=m$ (the pole mass), or $\mu=\sqrt s$, where $s$ is the energy of whatever process you are measuring.
Perturbative predictions always take the form $$ A=a_0+a_1\lambda+a_2\lambda^2+\cdots $$ where $a_i$ are numerical coefficients. This is the expression that you share with your experimental friend in order to compare to their measurements.
Note that $\lambda$ is not a parameter that appears in your Lagrangian. If you let $\lambda_0$ be the coefficient of $\frac{1}{4!}\phi^4$ in your Lagrangian, then you can write $\lambda=\lambda_0+c_1\lambda_0^2+c_2\lambda_0^3+\cdots$ for some coefficients $c_i$. Of course, given that $\lambda=\lambda_0+\cdots$, you can also express the perturbative expansion of $A$ as a series in $\lambda_0$. But this is not useful, since $\lambda_0$ is unmeasurable by itself. On the other hand, $\lambda$ can be measured in the lab, and hence it is much more convenient to express predictions as a series in $\lambda$.
If you only care about tree-level calculations, you can take $\lambda=\lambda_0$, and forget about the fact that $\lambda$ and $\lambda_0$ are two different objects. But if you want to look at higher order contributions, then you have two different objects, and the distinction becomes important.