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Let me answer to your second query and make $h=1$. You have $$ Op_1(a(x) \xi)= a(x) D_x,\quad \text{with $D_x=-i\partial_x$}, $$ $$ Op_0(a(x) \xi)= D_x a(x), $$ $$ Op_{1/2}(a(x) \xi)= \frac 12D_x a(x) …

Let me answer in terms of operator theory. The uncertainty principle can be interpreted as some particular inequality (as you say) such as $$ \frac{\hbar}{2}\Vert{u}\Vert^2\le \Vert{D_xu}\Vert\Vert{xu …

Well, you have $$ [\frac{1}{idx},x]=1/i $$ although $\frac{1}{idx}$ is far from the Identity.

With $z=x+iy$, we use the Fourier transformation in $(x,y)$ to see that $H$ is unitarily equivalent to $$ \frac12\begin{pmatrix}2m&\xi-i\eta\\ \xi+i\eta&-2m\end{pmatrix}, \text{whose eigenvalues are } …

Let $\mathbb H$ be a Hilbert space and let $\mathcal B(\mathbb H)$ be the bounded operators on $\mathbb H$. Let $J,K\in \mathcal B(\mathbb H)$ such that $ J=J^*, K=-K^*. $ Then the commutator $[J,K]$ …

Consider the most classical example, $D=\frac{d}{dx}-x$, which is the creation operator. Note that $D$ is injective since, with $L^2$ norms and dot-products and say $u$ in the Schwartz class, $$ \Vert …

With $D_x=\frac{d}{i dx}$, the Heisenberg uncertainty principle in its most classical form can be deduced from the equality $$ 2\Re \langle \hbar D_x u, ix u \rangle_{L^2(\mathbb R)}= \langle \bigl[\h …

Too long for an additional comment. I guess that you can keep the assumption $\hat P, \hat Q$ Hermitian and require $$ [\hat P, \hat Q]=1/(2πi), $$ as it is the case with the prototypical example $ \h …