TL;DR: The $n$'th Born approximation of a particle of mass $m$
m----V--------V--------V--------V----m
in scattering perturbation theory of QM (cf. Fig. 11.12 in Griffiths) corresponds to a tree-level Feynman diagram
M-----M M-----M M-----M M-----M
| | | |
| | | |
m------------------------------------m
in QFT, where $m$ interacts with (usually a much heavier) particle of mass $M$ via a mediating/force-carrying particle. There are many other tree-level diagrams in QFT.
It is not clear how to model a generic potential in QM via a force-carrying field in QFT, except for a few cases, such as e.g. the Coulomb and the Yukawa
potential.
In the rest of this answer, we will carefully compare the $\hbar$-dependence in QM and QFT.
Example: Rutherford scattering $m+M\to m+M$.
According to the $\hbar$/loop-expansion$^1$ the amputated connected Feynman diagram in QFT contains a factor $\hbar^{L-1}$, where $L$ is the number loops. There is one $\hbar$ for each internal propagator, and one $\hbar^{-1}$ for each vertex.
However for this to be true, there are a couple of assumptions:
4-momenta $p^{\mu}=\hbar k^{\mu}$ are replaced with 4-wavevectors $k^{\mu}$.
The action $S$ has no explicit $\hbar$-dependence. The 2nd-quantized QED action contains a mass parameter $m_2=\frac{mc}{\hbar}$ and a coupling constant $e_2=\frac{e}{\hbar c}$.
We identify the scattering amplitude
$$\langle k_1\alpha_1, k_2\alpha_2|{\cal M}|k_3\alpha_3, k_4\alpha_4\rangle \quad=\quad\left(\underbrace{\text{ext. half-leg}}_{=\sqrt{\hbar}}\right)^4\quad\times\quad\underbrace{\begin{matrix}\text{amputated connected}\cr\text{Feynman diagram}\end{matrix}}_{\text{tree-level with factor } \hbar^{-1}}.$$
Alternatively, we can completely eliminate all explicit $\hbar$-dependence in the path integral $Z[J]=\overline{Z}[\overline{J}]$ by normalizing$^2$
$$\overline{S}~=~\frac{S}{\hbar}, \quad
\overline{\phi}~=~\frac{\phi}{\sqrt{\hbar}}, \quad
\overline{J}~=~\frac{J}{\sqrt{\hbar}}, \quad
\overline{m}~=~\frac{mc}{\hbar}, \quad
\overline{e}~=~\frac{e}{\sqrt{\hbar c}}, \quad\ldots$$
Then the scattering amplitude
$\langle k_1\alpha_1, k_2\alpha_2|{\cal M}|k_3\alpha_3, k_4\alpha_4\rangle$ has no explicit $\hbar$ dependence at all.
The differential cross-section of $2\to 2$ scattering in a $d$-dimensional spacetime is
$$\begin{align} \left(\frac{d\sigma}{d\Omega}\right)_{\rm CM}~=~&\frac{\theta\left(\frac{E_{\rm CM}}{c^2}-m_3-m_4\right)}{64\pi^2\omega_{\rm CM}^2}\frac{|{\bf k}_f|^{d-3}}{|{\bf k}_i|}
\sum_{\rm spin}|\langle k_1\alpha_1, k_2\alpha_2|{\cal M}|k_3\alpha_3, k_4\alpha_4\rangle|^2, \cr
E_{\rm CM}~=~&\hbar\omega_{\rm CM}, \qquad {\bf p}_f~=~\hbar{\bf k}_f, \qquad {\bf p}_i~=~\hbar{\bf k}_i.\end{align}\tag{5.32} $$
The differential cross-section of $2\to 2$ elastic scattering in 3+1D QFT is
$$\begin{align} \left(\frac{d\sigma}{d\Omega}\right)_{\rm CM}~=~&\frac{1}{64\pi^2\omega_{\rm CM}^2}\sum_{\rm spin}|\langle k_1\alpha_1, k_2\alpha_2|{\cal M}|k_3\alpha_3, k_4\alpha_4\rangle|^2, \cr
E_{\rm CM}~=~&\hbar\omega_{\rm CM}.\end{align}\tag{5.33} $$
On the other hand, we may completely eliminate all explicit $\hbar$-dependence in the Schrödinger equation and the scattering theory of QM by normalizing$^2$
$$\overline{H}~=~\frac{H}{\hbar c}, \quad
\overline{E}~=~\frac{E}{\hbar c}, \quad
\overline{\bf p}~=~\frac{\bf p}{\hbar}, \quad
\overline{m}~=~\frac{mc}{\hbar}, \quad
\overline{e}~=~\frac{e}{\sqrt{\hbar c}}, \quad\ldots$$
The 1st Born approximation in 3+1D QM is
$$\begin{align}\left(\frac{d\sigma}{d\Omega}\right)_{\rm CM}~=~&\left(\frac{1}{4\pi}\frac{2m}{\hbar^2}\right)^2|\widetilde{V}(k)|^2\cr
~=~&\left(\frac{me^2}{2\pi\hbar^2k^2}\right)^2\cr
~=~&\left(\frac{m_2\hbar c e_2^2}{2\pi k^2}\right)^2\cr
~=~&\left(\frac{\overline{m}~\overline{e}^2}{2\pi k^2}\right)^2, \end{align}\tag{5.34+5.36}$$
where we use the Coulomb potential
$$V(r)~=~\frac{e^2}{4\pi r} \quad\Leftrightarrow\quad \widetilde{V}(k)~=~\frac{e^2}{k^2}.$$
In particular we see that the powers of $\hbar$ match in the various approaches, cf. OP's question.
References:
- M.D. Schwartz, QFT & the standard model,
2014; Sections 5.1 + 13.4. Be aware that Ref. 1. uses units where $c=1=\hbar$, cf. App. A. Also $\epsilon_0=1$.
$^1$ For a proof, see e.g. my Phys.SE answer here.
$^2$ Any physical quantity $Q$ can be normalized to a barred quantity $\overline{Q}=Q\hbar^rc^s$ with dimension $[\overline{Q}]=\text{Length}^p$. A purist can easily modify this normalization to avoid introducing the speed of light $c$ into a non-relativistic problem.
