Why is the time evolution operator not Weyl ordered?

Question

I am reading the book "Structural Aspects of Quantum Field Theory" by Gerhard Grensing. After introducing the Weyl formalism of symmetrizing the coordinates and momenta, he starts with a section on "Generalized Weyl Formalism" and says:

There are operators of fundamental importance that are definitely not Weyl ordered; they also do not belong to one of the known modified ordering schemes. Relevant examples include the time evolution operator $\hat U(t)=\exp(-i\hat H t/\hbar)$, where the Hamiltonian operator is assumed to be Weyl ordered or its imaginary time analogue, $\hat U(\beta)=\exp(-\beta \hat H)$. It is thus of fundamental importance that the Weyl Formalism be generalized.

I don't understand the boldfaced line above. If we have already Weyl ordered the Hamiltonian, won't an expansion of the exponential guarantee that the evolution operator is also Weyl ordered. I guess the problems start when we have to write a time ordered exponential i.e. $[\hat H(t),\hat H(t')]\neq 0$, but I do not understand how.

Any help is appreciated.

Answer 1

1. The defining formulation of the time evolution operator $U(t_2,t_1)$ is indeed time-ordered wrt. the Hamiltonian $H(t)$, cf e.g. this Phys.SE post.

2. If the Hamiltonian $H(t)$ itself has a different operator ordering (say e.g. Weyl-ordering or normal-ordering), then one may in principle$^1$ be able to bring $U(t_2,t_1)$ to the different operator ordering via a nested Wick theorem, cf. e.g. this Phys.SE post.

3. More generally, one may in principle$^1$ rewrite an operator in one operator ordering into another operator ordering via Wick-like theorems. It should be stressed that applying Wick's theorem does not change an operator, it only re-organizes its terms.

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$^1$ The transcription may fail due to divergent terms.