1
$\begingroup$

By using Feynman rules of the interacting theory, one obtains the scattering amplitude

$$\mathcal{M} = \mathcal{M}_0 + \mathcal{M}_1 + \cdots = \sum^{\infty}_{i = 0}\mathcal{M}_i\tag{1}$$

Where $\mathcal{M}_i$ denotes $i$-level loop scattering amplitude. If $i = 0$, then obtains tree-level contribution to the process. However, it seems that we can power-series expand in coupling constant $g$ or Planck constant $\hbar$:

$$\mathcal{M}(g) = \sum^{\infty}_{i = 0}g^i\mathcal{M}_i\tag{2}$$

$$\mathcal{M}(\hbar) = \sum^{\infty}_{i = 0}\hbar^i\mathcal{M}_i\tag{3}$$

What exactly is the difference between $(2)$ and $(3)$, if there is?

From I understand, power-series expanding around a small parameter $\epsilon$ makes higher-order contributions (generally) less impactful:

$${\displaystyle \mathcal{M}(\varepsilon) = \mathcal{M}_{0}+\varepsilon^{1}\mathcal{M}_{1}+\varepsilon ^{2}\mathcal{M}_{2}+\varepsilon^{3}\mathcal{M}_{3}+\cdots}\approx \mathcal{M}_0+\epsilon\mathcal{M}_1\quad (\epsilon\ll 1)$$

$\endgroup$

1 Answer 1

2
$\begingroup$

Feynman diagram techniques are inherently perturbative.

Diagramatically, in a theory like scalar $\phi^3$ or QED with a photon and an electron where you have one vertex, an expansion in the coupling constant is an expansion in the number of vertices in the diagram.

Meanwhile, an expansion in $\hbar$ is an expansion in terms of the number of loops in the diagram.

This is covered in more detail in standard QFT sources.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.