DrDu said:
You say that pQFT is a perturbational expansion not only in coupling constant but also in Planck's constant. The latter point is not immediately clear to me.
Here is how to see it:
The explicit ##\hbar##-dependence of the perturbative
S-matrix is
$$
S(g_{sw} L_{int} + j_{sw} A)
=
T \exp\left(
\tfrac{1}{i \hbar}
\left(
g_{sw} L_{int} + j_{sw} A
\right)
\right)
\,,
$$
where ##T(-)## denotes
time-ordered products. The generating function
$$
Z_{g_{sw}L_{int}}(j_{sw} A)
\;:=\;
S(g_{sw}L_{int})^{-1} \star S(g_{sw}L_{int} + j_{sw} A)
$$
involves the
star product of the free theory (the normal-ordered product of the
Wick algebra). This is a
formal deformation quantization of the
Peierls-Poisson bracket, and therefore the commutator in this algebra is a formal power series in ##\hbar## that, however, has no constant term in ##\hbar## (but starts out with ##\hbar## times the Poisson bracket, followed by possibly higher order terms in ##\hbar##):
$$
[L_{int},A] \;=\; \hbar(\cdots)
\,.
$$
Now by
Bogoliubov's formula the quantum observables are the derivatives of the generating function$$
\hat A
:=
\tfrac{1}{i \hbar} \frac{d}{d \epsilon}
Z_{g_{sw}L_{int}}(\epsilon j A)\vert_{\epsilon = 0}
$$
Schematically the derivative of the generating function is of the form
$$
\begin{aligned}
\hat A
& :=
\tfrac{1}{i \hbar} \frac{d}{d \epsilon}
Z_{g_{sw}L_{int}}(\epsilon j A)\vert_{\epsilon = 0}
\\
& =
\exp\left(
\tfrac{1}{i \hbar}[g_{sw}L_{int}, -]
\right)
(j A)
\end{aligned}
\,.
$$
(The precise expression is given by the "
retarded products", see (
Rejzner 16, prop. 6.1).)
By the above, the exponent ##\tfrac{1}{\hbar} [L_{int},-]## here yields a formal power series in ##\hbar##, and hence so does the full exponential.Here is how this relates to
loop order in the Feynman perturbation series:
Each Feynman diagram ##\Gamma## is a finite labeled graph, and the order in ##\hbar## to which this graph contributes is
$$
\hbar^{ E(\Gamma) - V(\Gamma) }
$$
where
- ##V(\Gamma) \in \mathbb{N}## is the number of vertices of the graph
- ##E(\Gamma) \in \mathbb{N}## is the number of edges in the graph.
This comes about (see at
S-matrix -- Feynman diagrams and Renormalization for details) because
1) the explicit ##\hbar##-dependence of the S-matrix is
$$
S\left(\tfrac{g}{\hbar} L_{int} \right)
=
\underset{k \in \mathbb{N}}{\sum} \frac{g^k}{\hbar^k k!} T( \underset{k \, \text{factors}}{\underbrace{L_{int} \cdots L_{int}}} )
$$
2) the further ##\hbar##-dependence of the time-ordered product ##T(\cdots)## is
$$
T(L_{int} L_{int}) = prod \circ \exp\left( \hbar \int \omega_{F}(x,y) \frac{\delta}{\delta \phi(x)} \otimes \frac{\delta}{\delta \phi(y)} \right) ( L_{int} \otimes L_{int} )
\,,
$$
where ##\omega_F## denotes the
Feynman propagator and ##\phi(x)## the (generic) field observable at point ##x## (where we are notationally suppressing the internal degrees of freedom of the fields for simplicity, writing them as scalar fields, because this is all that affects the counting of the ##\hbar## powers).
The resulting terms of the S-matrix series are thus labeled by
1. the number of factors of the interaction ##L_{int}##, these are the vertices of the corresponding Feynman diagram and hence each contibute with ##\hbar^{-1}##
2. the number of integrals over the Feynman propagator ##\omega_F##, which correspond to the edges of the Feynman diagram, and each contribute with ##\hbar^1##.
Now the formula for the
Euler characteristic of planar graphs says that the number of regions in a plane that are encircled by edges, the
faces, here thought of as the number of "loops", is
$$
L(\Gamma) = 1 + E(\Gamma) - V(\Gamma)
\,.
$$
Hence a planar Feynman diagram ##\Gamma## contributes with
$$
\hbar^{L(\Gamma)-1}
\,.
$$
So far this is the discussion for internal edges. An actual scattering matrix element is of the form
$$
\langle \psi_{out} \vert S\left(\tfrac{g}{\hbar} L_{int} \right)
\vert \psi_{in} \rangle
\,,
$$
where
$$
\vert \psi_{in}\rangle
\propto
\tfrac{1}{\sqrt{\hbar^{n_{in}}}}
\phi^\dagger(k_1) \cdots \phi^\dagger(k_{n_{in}}) \vert vac \rangle
$$
is a state of ##n_{in}## free field quanta and similarly
$$
\vert \psi_{out}\rangle
\propto
\tfrac{1}{\sqrt{\hbar^{n_{out}}}}
\phi^\dagger(k_1) \cdots \phi^\dagger(k_{n_{out}}) \vert vac \rangle
$$
is a state of ##n_{out}## field quanta. The normalization of these states, in view of the commutation relation ##[\phi(k), \phi^\dagger(q)] \propto \hbar##, yields the given powers of ##\hbar##.
This means that an actual scattering amplitude given by a Feynman diagram ##\Gamma## with ##E_{ext}(\Gamma)## external vertices scales as
$$
\hbar^{L(\Gamma) - 1 + E_{ext}(\Gamma)/2 }
\,.
$$