Why does the Dirac Lagrangian not already use operators (instead of canonical quantization)? I've learned that in canonical quantization you take a Lagrangian, transform to a Hamiltonian and then "put the hat on" the fields (make them an operator). Then you can derive the equations of motion of the Hamiltonian.
What is the reason that you cannot already put hats in the Dirac Lagrangian? Therefore write the Lagrangian with operators and go straight to Euler-Lagrange equations. Or is there any way this has been tried?
As mentioned in the comments you'd have to make an adjustment to get a real scalar from the operator expression. Can it be done right in the Lagrangian, without artificially putting hats on after a transformation?
 A: In quantization the starting point is a classical theory.

*

*On one hand, in a classical Hamiltonian formulation, there is a (super) Poisson bracket $\{\cdot,\cdot\}$, which we can formally replace with a (super) commutator $[\cdot,\cdot]$, e.g. using deformation quantization, cf. e.g. this related Phys.SE post. In this way the classical Hamiltonian formalism is closely related to the quantum mechanical operator formalism.


*On the other hand, in a classical Lagrangian formulation, it is more indirect how to introduce non-commutativity and operators$^1$.
Of course, the classical Lagrangian and Hamiltonian formulations are formally equivalent, so the two approaches are ultimately related.
E.g. the time evolution operator $\hat{U}_I$ in the interaction picture is
$$\begin{align}
\hat{U}_I~=~& T  \exp\left\{ -\frac{i}{\hbar}\int\!dt~H_{\rm int}(\hat{q},\hat{p})\right\}\cr
~=~&  T_{\rm cov}  \exp\left\{ \frac{i}{\hbar}\int\!dt~ L_{\rm int}(\hat{q},\dot{\hat{q}})\right\}
,\end{align}$$
cf. e.g. my related Phys.SE answer here.
--
$^1$ Note
that inside the path integral the value of the integration variables and  the action itself are (super) numbers, not operators,
A: Note that the Lagrangian density, the Hamiltonian density and the equations of motion all can be deduced from each other via the Legrende transformation for fields and the Euler - Lagrange equation for fields. Regarding the Dirac equation specifically, there is no classical Lagrangian/Hamiltonian density to quantize (since it is dealing with fermions). But for scalars, you deduce the QFT Klein - Gordon equation from a (complex version of a) classical relativistic scalar Lagrangian density (i.e. $\mathcal L= \partial_\lambda \Phi^\dagger \partial^\lambda \Phi-\mu^2 \Phi^\dagger \Phi $ ) and the Euler - Lagrange equation for fields. (What makes something an operator is not determined by whether you start with the Hamiltonian- or Lagrangian density, but by invoking the commutation relations for fields).
