Given some function $f \in L^2(\mathbb{R}^2)$, I'm interested in finding a positive semi-definite differential operator $\mathcal P: L^2(\mathbb{R}^2) \rightarrow L^2(\mathbb{R}^2)$ that is *quadratic* in $f$ and invariant under the the action of $\textrm{SL}{(2, \mathbb{R})},$ such that $\forall A \in \textrm{SL}(2, \mathbb{R})$ and $\forall {\bf x} \in \mathbb{R}^2,$
$$ {\mathcal P} f(A {\bf x}) = [{\mathcal P} f] (A {\bf x} ).$$

After thinking for some time, I've come up with two operators that are invariant and P.S.D, but not quadratic.

For example, suppose we consider the operator $${\mathcal P} = \left( \frac{\partial^2}{\partial x^2} \frac{\partial^2}{\partial y^2} - \left[\frac{\partial^2}{\partial x y}\right] \right)^2,$$ which is the squared determinant of the Hessian. It's clear that it is P.S.D. and invariant under transformations in $\textrm{SL}(2, \mathbb{R})$, though it is quartic in $f$.

Furthermore, letting $H$ denote the Hessian and $J \in \textrm{SO}(2)$ be a rotation by $90^\circ$, the operator $$ {\mathcal P} = \left(\nabla^T J^T H \ J \ \nabla\right)^2,$$ is also invariant and P.S.D., but is not quadratic.

I'm asking this question in the hope that someone might know of such a quadratic P.S.D. differential operator that is invariant under $\textrm{SL}(2, \mathbb{R})$ (if it even exists) or be able to point me toward a with a few other ideas I could try.

Some possibly related question(s):

Projective invariant differential operator

Classification of diffeomorphisms by association of differentials with Lie groups