Remark: Normalized order costs a numerator order

If a numerator is $N(\hslash )=\hslash N_1+{\hslash }^2{N}_2+\cdots$, then $[\hslash ](\mathrm{i}N/\hslash )=\mathrm{i}{N}_2$. Thus the first positive-$\hslash$ coefficient of a normalized retarded map requires the numerator through order ${\hslash }^2$, not only its leading term CheungIXD.