I have the following operator $$\Phi(\chi_A)=\int \text{d}\eta\, \text{d}\zeta\,\chi_A(\eta,\zeta)\,e^{i(\eta \hat{P}+\zeta\hat{Q})}.$$ With $\chi_A$ the indicator function associated to a set $A\subset \mathbb{R}^2$ and $\hat{P},\hat{Q}$ hermitian operators that satisfy $[\hat{P},\hat{Q}]=1$$[\hat{P},\hat{Q}]=-i\hbar$. For which conditions on the set $A$ this operator is positive semi-definite? $$\textbf{Where am I wrong?}$$ One way of proving if $\Phi(\chi_A)$ is positive semi-definite is just by finding an hermitian operator $A$ such that $$\Phi(\chi_A)=A^2.$$ In order to do so I can define the following operator \begin{equation} A=\int\, d\mu\, d\nu\, \mathcal{F}_\sigma(\mathcal{\chi_A})(\mu,\nu)\,e^{i(\mu \hat{P}+\nu\hat{Q})} \end{equation} with $\mathcal{F}_\sigma$ the symplectic Fourier transform, assuming already that $\chi_A(\eta,\zeta)=\chi_A(-\eta,-\zeta)$. Applied twice we find \begin{align} A^2&=\int\, d\mu_1\, d\nu_1\, d\mu_2\, d\nu_2\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_2,\nu_2)\,e^{i(\mu_1 \hat{P}+\nu_1\hat{Q})}\,e^{i(\mu_2 \hat{P}+\nu_2\hat{Q})}\\ &=\int \,d\mu_1\, d\nu_1\, d\mu_2\, d\nu_2\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_2,\nu_2)\,e^{i\hbar(\mu_1\nu_2-\mu_2\nu_1)}\,e^{i((\mu_1+\mu_2) \hat{P}+(\nu_1+\nu_2)\hat{Q})} \end{align} Defining the new variables $\gamma=\mu_1+\mu_2$ and $\delta=\nu_1+\nu_2$ we can write \begin{align} A^2&=\int \,d\mu_1 \,d\nu_1 d\gamma\, d\delta\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\gamma-\mu_1,\delta-\nu_1)\,e^{i\hbar(\mu_1(\delta-\nu_1)-(\gamma-\mu_1)\nu_1)}\,e^{i(\gamma \hat{P}+\delta\hat{Q})}\\ &= \int \,d\mu_1\, d\nu_1\, d\gamma\, d\delta\, e^{i\hbar(\mu_1\delta-\nu_1\gamma)}\,e^{i\hbar(-\mu_1\nu_1+\mu_1\nu_1)} \mathcal{F}_\sigma(\mathcal{\chi_A})(\mu_1,\nu_1)\,\mathcal{F}_\sigma(\mathcal{\chi_A})(\gamma-\mu_1,\delta-\nu_1)\,e^{i(\gamma \hat{P}+\delta\hat{Q})} \end{align} Notice that the integral over the variables $\mu_1$ and $\nu_1$ is the symplectic fourier transform of a convolution, then by the convolution theorem we conclude: \begin{align} A^2&=\int \, d\gamma\, d\delta\, \mathcal{\chi_A}\left(\hbar\gamma,\hbar\delta\right)\,e^{i(\gamma \hat{P}+\delta\hat{Q})} \end{align} This would imply that for any $\chi_A$ there is an operator $A$ verifying the positive semi-definiteness condition, but that is clearly wrong.
