quantum mechanics - Hamiltonian in QM/QFT path integral being Wigner transformation (Weyl-symbol)? of Hamiltonian operator?

The question is inspired from the answer of Why path integral approach may suffer from operator ordering problem?. In the answer, it says below equation 5:

where $H(q,p)$ denotes the Weyl-symbol for the Hamiltonian operator $\hat{H}$. Weyl-ordering prescription is better than other operator ordering prescriptions, but it is still an approximation.

I do not understand what this is suppposed to mean. In usual QM, don't we take classical action (and thus Hamiltonian) and use it directly in path integral?

1. Is this saying that Weyl transformation of classical Hamiltonian would be an approximation to actual Hamiltonian operator in QM?


2. Or is it saying that Hamiltonian used in path integral is inverse Weyl transformation (Wigner transformation) and that it is not always true for other operators that classical version would correspond to such inverse Weyl transformation?

For 1, would path integral still turn out to be always correct despite initially using approximations? (both QM/QFT)

Answer

What is most fundamental: The path integral formalism or the operator formalism?

I) Consider first the path integral.

1. 
   

   Firstly, a path integral with a purely classical action is not as commutative and tame as it might naively appear: There is always an implicit underlying non-commutative time slicing procedure assumed which affects all dot variables, i.e. time derivatives. See e.g. this and this Phys.SE answers.

   
   
   


2. 
   

   Secondly, within the path integral formalism, the formula $$ \langle q_f,t_f|q_i,t_i \rangle~\sim~\int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~\exp\left[ \frac{i}{\hbar}S[q,p]\right] \tag{1} $$ becomes merely a postulate rather than something one can prove.

   
   

II) For these reasons, we will instead take the operator formalism as fundamental, and try to derive the path integral formula (1). This is, however, in general easier said than done, as explained in my mentioned Phys.SE answer:

1. 
   

   When we replace the Hamiltonian operator $\hat{H}$ by one of its symbols (via a Wigner-like map), we introduce errors, e.g. since it appears exponentiated, cf. the BCH formula.

   
   



2. 
   

   The Weyl symbol fares slightly better than other symbols (such as, e.g. $\hat{p}\hat{q}$ or $\hat{q}\hat{p}$ symbols), but it is generically still an approximation.