I consider an operator $T: L^2(\mathbb R^2) \to L^2(\mathbb R^2)$ which acts as $$ \langle \psi, T \phi \rangle = \int dp dq~ a^2(p,q) \Big( \overline{\widetilde{f \psi}(p,q)} \widetilde{f \phi}(p,q) - \overline{\widetilde{f \psi}(p,q)} \widetilde{f \phi}(q,p) \Big), $$ where $a$ is a real-valued function and $f(x,y) = f(y,x)$ is symmetric.
Q: Is it a positive operator or has a spectrum bounded from below?
The operator can be written as $T_1 - T_2$ for the first and second term in the brakets. $T_1$ is obviously positive as it is of the form $T_1 = L^\star L$. $T_2$ is nearly of this form, but has an additional variable flip $T_2 = L^* L F$ if $(F \phi)(x,y) = \phi(y,x)$.
Is there anything known about such a flip operator and the positivity properties of ($1 - F$), that could be used to show positivity (or boundedness from below) for the above operator?
Thinking finite dimensional, a matrix of this form $A = \begin{pmatrix} ~~~1 & -1 \\ -1 & ~~~1 \end{pmatrix}$ would be positive, since $\langle \psi, A \psi \rangle = |\psi_1 - \psi_2|^2 \geq 0$.
$T$ is in general not a positive operator!
This is because for some function $g(p,q)$, $$ |g(p,q)|^2 - \overline{g(p,q)} g(q,p) \not\geq 0. $$ So for any $a(p,q)$, we can find a $g(p,q)$ such that the integral $$ \int dp dq ~ a^2(p,q) \Big( |g(p,q)|^2 - \overline{g(p,q)} g(q,p) \Big) < 0. $$ However, there is a special case (which is relevant in my case):
If $a(p,q) = a(q,p)$, then $T$ is a positive operator!
In this case, one can symmetrize the brakets and $p,q$ and obtain $$ |g(p,q)|^2 + |g(q,p)|^2 - \overline{g(p,q)} g(q,p) - \overline{g(q,p)} g(p,q) = |g(p,q) - g(q,p)|^2 \geq 0. $$ Then, the integral is positive because the integrand is positive.