# Why do we need the coupling small when doing perturbative QFT calculation?

I don't really understand why, when we calculate say the 2-point Greens function in a scalar QFT with interaction $$\lambda \phi^4$$, we need the coupling constant $$\lambda$$ to be small?

Everywhere I look it seems to be a result of having factors of the form $$e^{\int \mathcal{L}_{\text{INT}}}$$ and then requiring that $$\mathcal{L}_{\text{INT}} = \lambda \phi^4$$ be sufficiently small such that we can expand the exponential into its power series, and so we take $$\lambda$$ to be sufficiently small.

I don't understand why we need this assumption given the exponential is analytic everywhere and thus its Taylor series is well defined and equal to the exponential even when $$\mathcal{L}_{\text{INT}}$$ is large. We could therefore expand without this assumption, no? At what point are we using $$\lambda \ll 1$$?

Take for example the explanation at the beginning of the following notes on page 26 https://www.southampton.ac.uk/~doug/ft1/ft17.pdf

• Do you have a source that it should be analytic in lambda? The usual argument in QED that it is not analyric should apply he as well (negative lambda shd lead to divergent integrals, so function cannot be analytic at 0) – lalala Oct 9 at 10:26
• Hmm okay, maybe this is where I'm misunderstanding things. In my mind we have $e^{\lambda k}$ where for this example $k = \int d^4x\phi^4$ and so we can just Taylor expand like normal (don't quite see how it couldn't be analytic unless the exponent was infinite in which case having a small $\lambda$ won't change things), and so we should get $e^{\lambda k}= \sum \frac{(\lambda k)^n}{n!}$ irrespective of the value of $\lambda$? – jojo Oct 9 at 10:38

Consider the toy integral $$Z(\lambda)= \int_{-\infty}^{\infty} e^{-x^2-\lambda x^4}dx$$ Lets expand in powers of $$\lambda$$. We find $$Z(\lambda)= \int_{-\infty}^{\infty}dx e^{-x^2} \sum_{n=1}^\infty (-1)^n\frac 1{n!} \lambda^n x^{4n}\\ = \sum_{n=1}^\infty (-1)^n\frac 1{n!} \lambda^n\int_{-\infty}^{\infty}dx e^{-x^2} x^{4n}\\ = \sum_{n=1}^\infty (-1)^n\frac 1{n!}\lambda^n \Gamma(2n+1/2).$$ Now $$\Gamma(2n+1/2)\sim (2n)!\sim (n!)^2$$, so the sum has radius of convergence zero. (I'll leave you to figure out where the calculus went wrong.) We have found only an asymptotic expansion that is useful only near $$\lambda=0$$. QFT perturbation theory works the same way.