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.

  • 1
    $\begingroup$ At every order in perturbation theory, you use the same renormalized coupling and fix counter-terms to enforce your favourite scheme. It just so happens that $MS$ type schemes prevent the counter-terms from contributing at tree level. $\endgroup$ Commented Jul 22, 2022 at 23:01
  • $\begingroup$ Hi Ryder Rude. Consider to provide references for context and conventions. Which pages? $\endgroup$
    – Qmechanic
    Commented Jul 24, 2022 at 14:47
  • $\begingroup$ @Qmechanic It's about the 2- vertex loop diagram of $\phi ^4$ self interaction. It starts around page 453 of Jacob's No nonsense Quantum Field Theory. He replaced the coupling constant of this diagram with a coupling dependent on $\Lambda$. But didn't make the same replacement for the tree level coupling constant $\endgroup$
    – Ryder Rude
    Commented Jul 24, 2022 at 15:02
  • $\begingroup$ @ConnorBehan I don't know what counter terms means. Please explain this as an intro to renormalisation. The scheme used in the book was "momentum cut-off" scheme I think $\endgroup$
    – Ryder Rude
    Commented Jul 24, 2022 at 15:05
  • $\begingroup$ Are you sure he didn't do the opposite? When the $\lambda$ in the tree level term is expressed up to $O(\lambda_R^2)$, this is what will cancel divergences at one loop. Taking $\lambda^2$ from one loop and making the same substitution will just produce $\lambda_R^3$ and $\lambda_R^4$ terms which can be dropped. $\endgroup$ Commented Jul 24, 2022 at 15:07

1 Answer 1


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.

  • $\begingroup$ I get that the final perturbation series that we obtain is in terms of $\lambda (s, s_0)$, where $s$ is the energy of our experiment and $s_0$ is our chosen reference energy. But still, the initial perturbation series that we derived, using Dyson series, was in terms of the Langrangian parameter $\lambda$. I'm not finding the transition from the series in terms of $\lambda$ to the series in terms of $\lambda (s,s_0)$ logically consistent $\endgroup$
    – Ryder Rude
    Commented Jul 28, 2022 at 4:08
  • $\begingroup$ The first two terms of $\phi ^4$ interaction, in terms of the Langrangian parameter, turned out to be $-iV\lambda$ and $-\lambda ^2 C(\ln s -\ln \Lambda ^2)$. Now we make the transition from the series in terms of $\lambda$, to in terms of $\lambda (s,s_0)$. We substitute $\lambda = \lambda _R +$ ($\Lambda$ dependent terms). This substitution makes the loop term finite as it cancels out $\Lambda$. But, to be consistent, we must make the same substitution in the first term $-iV\lambda$.This term will diverge if we make that substitution because there's nothing to cancel $\Lambda$ $\endgroup$
    – Ryder Rude
    Commented Jul 28, 2022 at 4:19
  • $\begingroup$ Sorry I mis-read his proof. The substitution didn't cancel the divergence in $-\lambda ^2C(\ln s -\ln \Lambda ^2)$. It instead cancelled the divergence in $-\lambda -\lambda ^2Cln (\frac{s}{\Lambda^2})+.....$. So it pretty much cancelled the divergences in all terms at once, instead of just the second term. $\endgroup$
    – Ryder Rude
    Commented Jul 28, 2022 at 4:47

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