Weyl quantization

Remarks

Example: for $a(x, \xi) = \xi_j$ the $\xi$-integral is a Fourier transform pair, and $\mathrm{Op}^w(a) = -i\hbar\, \partial/\partial x_j$, the momentum operator $P_j$. For $a(x, \xi) = x_j$ the $\xi$-integral gives $(2\pi\hbar)^n \delta(x - y)$, and $\mathrm{Op}^w(a)$ is multiplication by $\frac{x_j + x_j}{2} = x_j$, the position operator $X_j$. For $a = x_1 \xi_1$ one gets $\mathrm{Op}^w(a) = \frac12(X_1 P_1 + P_1 X_1)$.

If $a$ is real-valued, then $\mathrm{Op}^w(a)$ is symmetric: $\ip{\varphi}{\mathrm{Op}^w(a)\psi} = \ip{\mathrm{Op}^w(a)\varphi}{\psi}$ for $\varphi, \psi \in \mathcal{S}(\R^n)$.

Non-example: the standard (Kohn–Nirenberg) quantization, with $a(x, \xi)$ in place of $a\!\left(\frac{x+y}{2}, \xi\right)$, sends $x_1 \xi_1$ to $X_1 P_1$, which is not symmetric; so it is a different rule.

For polynomials $a$ of degree at most $2$ in $(x, \xi)$, Weyl quantization turns Poisson brackets into commutators exactly: $[\mathrm{Op}^w(a), \mathrm{Op}^w(b)] = i\hbar\, \mathrm{Op}^w(\{a, b\})$. In canonical quantization terms, $\mathrm{Op}^w(a)$ is the operator $a(X, P)$ with the symmetric ordering.