In 'Non-abelian Bosonization in Two Dimensions', Witten argues that the Poisson brackets of the currents that generate the $G\times G$ symmetry of the WZW model give rise to a Kac-Moody algebra upon canonical quantization.
The Poisson brackets are calculated on page 465 to be \begin{equation} \begin{aligned} \left[X,Y\right]_{PB}&=-\frac{4\pi}{n}\delta(\sigma-\sigma')\textrm{Tr}\bigg([A,B]\frac{\partial g}{\partial\sigma}g^{-1}\bigg)-\frac{4\pi}{n}\delta'(\sigma-\sigma')\textrm{Tr}~AB. \end{aligned}\tag{29} \end{equation} for $X=\textrm{Tr} A \frac{\partial g}{\partial \sigma}g^{-1}(\sigma)$ and $Y=\textrm{Tr} B \frac{\partial g}{\partial \sigma'}g^{-1}(\sigma')$, where $g$ is a map $g:\mathbb{R}\rightarrow G$, and where $A$ and $B$ are arbitrary generators of $G$.
He then states that the relation between Poisson brackets and quantum mechanical commutation relations implies the commutator
\begin{equation} \begin{aligned} \left[X,Y\right]_{PB}&=i\hbar\frac{4\pi}{n}\delta(\sigma-\sigma')\textrm{Tr}\bigg([A,B]\frac{\partial g}{\partial\sigma}g^{-1}\bigg)+i\hbar\frac{4\pi}{n}\delta'(\sigma-\sigma')\textrm{Tr}~AB, \end{aligned}\tag{30} \end{equation} where I have included Planck's constant explicitly for clarity.
However, it is not clear to me why he does not include terms that are higher order in the Planck constant, which generally occur when going from Poisson brackets to canonical commutators.
One possibility is that dimensional analysis precludes such terms. However, for example, a correction term such as \begin{equation} \hbar^2\textrm{Tr}A \frac{\partial g}{\partial \sigma}g^{-1}(\sigma) B \frac{\partial g}{\partial \sigma'}g^{-1}(\sigma') \end{equation} does seem consistent with dimensional analysis.
How does one show that there are no $O(\hbar^2)$ terms or higher in the commutator (30)?
