In Zworki's book Semiclassical analysis, he gives examples of quantized symbols on page 57 with the following definition: $h > 0$ and $a\in\mathscr{S}(\mathbb{R}^{2n})$ is called a symbol. A semiclassical pseudodifferential operator is defined as
$$\mathrm{Op}_t(a)u(x) := \frac{1}{(2\pi h)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}e^{\frac{i}{h}\left<x - y,\xi\right>}a\left(tx + (1 - t)y,\xi\right)u(y)dyd\xi$$
with $0\leq t \leq 1$. Zworski then claims that if $a(x,\xi) = \xi^\alpha$, with $\alpha = (\alpha_1,\dots,\alpha_n) \in\mathbb{N}^n$ and $\xi^\alpha := \prod_{m=1}^n\xi_m^{\alpha_m}$ then $\mathrm{Op}_t(a)u = (hD)^\alpha u$ with $u\in\mathscr{S}(\mathbb{R}^n)$ and $D$ a differential character with the interpretation
$$(D^\alpha u)(x) = \prod_{m=1}^n\left(\frac{\partial^{\alpha_k}u}{\partial x_k^{\alpha_k}}\right)(x)$$
But since Zworski's Fourier kernel is $e^{\frac{i}{h}\left<x,.\right>}$, shouldn't we multiply $D$ by $\frac{h}{i}$ to cancel the $i$ in partial integration? To be specific, if we consider the case of $n = 1$, then for $u\in\mathscr{S}(\mathbb{R})$:
$$\widehat{u'(x)}(\xi) = \int_{\mathbb{R}}u'(x)e^{\frac{-i}{h}\xi x}dx = -\left(-\frac{i\xi}{h}\right)\int_{\mathbb{R}}u(x)e^{\frac{-i}{h}\xi x}dx \leftrightarrow$$
$$\widehat{u'(x)}(\xi) = \frac{i\xi}{h}\int_{\mathbb{R}}u(x)e^{\frac{-i}{h}\xi x}dx$$
so we get a factor of $\frac{i}{h}$ instead of $h$.