[proofplan] We first regularize the Weyl oscillatory integral by inserting a symmetric compact cutoff in the variables $(x,y,\xi)$, so every expression becomes an absolutely convergent Lebesgue integral. For the regularized operators, the adjoint identity follows from taking the complex conjugate in the $L^2$ inner product and interchanging the variables $x$ and $y$; the Weyl midpoint $(x+y)/2$ is exactly what makes the symbol transform into $\overline{a}((x+y)/2,\xi)$. Finally we pass to the oscillatory integral limit, using the defining continuity of Weyl quantization on $\mathcal{S}(\mathbb{R}^n)$ and $\mathcal{S}'(\mathbb{R}^n)$. [/proofplan]

[step:Regularize the Weyl pairing by a symmetric cutoff] Choose a function \begin{align*} \chi:\mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R} \end{align*} with $\chi \in C_c^\infty(\mathbb{R}^{3n})$, $\chi(x,y,\xi)=\chi(y,x,\xi)$ for all $(x,y,\xi)$, and $\chi=1$ on a neighbourhood of $(0,0,0)$. For $\varepsilon \in (0,1]$, define \begin{align*} \chi_\varepsilon:\mathbb{R}^n \times \mathbb{R}^n \times \mathbb{R}^n &\to \mathbb{R} \end{align*} \begin{align*} (x,y,\xi) &\mapsto \chi(\varepsilon x,\varepsilon y,\varepsilon \xi). \end{align*} For each $\varepsilon \in (0,1]$, define the regularized pairing as the map \begin{align*} I_\varepsilon(a;\cdot,\cdot):\mathcal{S}(\mathbb{R}^n) \times \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}. \end{align*} For $u,v \in \mathcal{S}(\mathbb{R}^n)$, its value is \begin{align*} I_\varepsilon(a;u,v) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) e^{i(x-y)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) u(y)\overline{v(x)} \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} Because $\chi_\varepsilon$ has compact support and the remaining factors are smooth on that compact set, this is an absolutely convergent Lebesgue integral. We use the standard cutoff-independence theorem for oscillatory integrals: if an amplitude is a symbol in the oscillatory variables and is paired with Schwartz functions in the remaining variables, then inserting any cutoff $\chi_\varepsilon(x,y,\xi)=\chi(\varepsilon x,\varepsilon y,\varepsilon \xi)$ with $\chi=1$ near the origin gives the same oscillatory integral in the limit $\varepsilon \to 0^+$. Its hypotheses apply here because $a \in S^m(T^*\mathbb{R}^n)$ has polynomial growth with symbol estimates, while $u,v \in \mathcal{S}(\mathbb{R}^n)$ provide rapid decay in the spatial variables; after the usual integration-by-parts operator in $\xi$ and in $x-y$, the regularized integrands are dominated by integrable seminorm bounds independent of $\varepsilon$ on compact parameter ranges. Therefore the cutoff limit represents the distributional pairing of the Weyl operator with $v$: \begin{align*} (\operatorname{Op}_h^w(a)u,v)_{L^2(\mathbb{R}^n)} = \lim_{\varepsilon \to 0^+} I_\varepsilon(a;u,v). \end{align*} [/step]

[step:Expand the regularized right-hand pairing and exchange the two spatial variables]For each $\varepsilon \in (0,1]$, define the regularized right-hand pairing as the map \begin{align*} J_\varepsilon(\overline a;\cdot,\cdot):\mathcal{S}(\mathbb{R}^n) \times \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}. \end{align*} For $u,v \in \mathcal{S}(\mathbb{R}^n)$, its value at $(v,u)$ is \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) u(x) \overline{ e^{i(x-y)\cdot \xi/h} \overline a\!\left(\frac{x+y}{2},\xi\right) v(y) } \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} This integral is absolutely convergent because $\chi_\varepsilon$ has compact support. Since $\chi_\varepsilon$ is real-valued, taking the complex conjugate inside the integrand gives \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) e^{i(y-x)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) u(x)\overline{v(y)} \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} Now apply the linear change of variables $(x,y,\xi) \mapsto (y,x,\xi)$. Its absolute Jacobian determinant is $1$, so the product measure $\mathcal{L}^n \otimes \mathcal{L}^n \otimes \mathcal{L}^n$ is preserved. The cutoff is unchanged because $\chi_\varepsilon(y,x,\xi)=\chi_\varepsilon(x,y,\xi)$, and the Weyl midpoint is unchanged because \begin{align*} \frac{y+x}{2}=\frac{x+y}{2}. \end{align*} Therefore \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) e^{i(x-y)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) u(y)\overline{v(x)} \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} The right-hand side is exactly $I_\varepsilon(a;u,v)$.[/step]

[guided]The desired identity compares two different pairings, not $I_\varepsilon(a;u,v)$ with its own conjugate. Because the $L^2(\mathbb{R}^n)$ inner product is linear in the first entry and conjugate-linear in the second, the regularized version of $(u,\operatorname{Op}_h^w(\overline a)v)_{L^2(\mathbb{R}^n)}$ is \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) u(x) \overline{ e^{i(x-y)\cdot \xi/h} \overline a\!\left(\frac{x+y}{2},\xi\right) v(y) } \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} Now we take the conjugate of the operator output. The phase changes sign, the conjugated symbol $\overline a$ becomes $a$, and $v(y)$ becomes $\overline{v(y)}$: \begin{align*} \overline{ e^{i(x-y)\cdot \xi/h} \overline a\!\left(\frac{x+y}{2},\xi\right) v(y) } = e^{i(y-x)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) \overline{v(y)}. \end{align*} Thus \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) e^{i(y-x)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) u(x)\overline{v(y)} \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} We now exchange the spatial variables. The map $T:\mathbb{R}^{3n}\to\mathbb{R}^{3n}$ defined by $T(x,y,\xi)=(y,x,\xi)$ is linear and has absolute Jacobian determinant $1$, so the measure $d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x)$ is preserved. The cutoff is symmetric, so $\chi_\varepsilon(y,x,\xi)=\chi_\varepsilon(x,y,\xi)$. The midpoint also survives the exchange: \begin{align*} \frac{y+x}{2}=\frac{x+y}{2}. \end{align*} After this change of variables, the phase $e^{i(y-x)\cdot \xi/h}$ becomes $e^{i(x-y)\cdot \xi/h}$, the factor $u(x)$ becomes $u(y)$, and $\overline{v(y)}$ becomes $\overline{v(x)}$. Hence \begin{align*} J_\varepsilon(\overline a;v,u) = (2\pi h)^{-n} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} \chi_\varepsilon(x,y,\xi) e^{i(x-y)\cdot \xi/h} a\!\left(\frac{x+y}{2},\xi\right) u(y)\overline{v(x)} \, d\mathcal{L}^n(\xi)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(x). \end{align*} This is exactly $I_\varepsilon(a;u,v)$. The midpoint in the Weyl quantization is the structural reason the argument works: exchanging $x$ and $y$ leaves $(x+y)/2$ fixed, so the same quantization convention reappears after the adjoint operation.[/guided]

[step:Pass from the regularized identity to the oscillatory integral identity] From the preceding step, for every $\varepsilon \in (0,1]$, \begin{align*} I_\varepsilon(a;u,v)=J_\varepsilon(\overline a;v,u). \end{align*} Taking the limit as $\varepsilon \to 0^+$ on both sides gives \begin{align*} (\operatorname{Op}_h^w(a)u,v)_{L^2(\mathbb{R}^n)} = (u,\operatorname{Op}_h^w(\overline a)v)_{L^2(\mathbb{R}^n)}. \end{align*} The left-hand side is the cutoff-independent oscillatory limit for the pairing of $\operatorname{Op}_h^w(a)u$ with $v$. The right-hand side is the corresponding cutoff-independent oscillatory limit for the pairing of $u$ with $\operatorname{Op}_h^w(\overline a)v$; the same oscillatory-integral theorem applies because $\overline a \in S^m(T^*\mathbb{R}^n)$ whenever $a \in S^m(T^*\mathbb{R}^n)$, and $u,v \in \mathcal{S}(\mathbb{R}^n)$. [/step]

[step:Identify the formal adjoint and the self-adjoint real-symbol case] The formal adjoint $(\operatorname{Op}_h^w(a))^*$ is defined by the identity \begin{align*} ((\operatorname{Op}_h^w(a))u,v)_{L^2(\mathbb{R}^n)} = (u,(\operatorname{Op}_h^w(a))^*v)_{L^2(\mathbb{R}^n)} \end{align*} for all $u,v \in \mathcal{S}(\mathbb{R}^n)$, interpreted as an identity of distributions. The identity proved above therefore gives \begin{align*} (\operatorname{Op}_h^w(a))^*=\operatorname{Op}_h^w(\overline a) \end{align*} as continuous maps $\mathcal{S}(\mathbb{R}^n)\to \mathcal{S}'(\mathbb{R}^n)$ after taking formal adjoints. If $a$ is real-valued, then $\overline a=a$, and hence \begin{align*} (\operatorname{Op}_h^w(a))^*=\operatorname{Op}_h^w(a). \end{align*} Thus $\operatorname{Op}_h^w(a)$ is formally self-adjoint. [/step]