First, recall that on a Riemannian manifold $(M,g)$ the Laplace-Beltrami operator $\Delta_g:C^\infty(M)\to C^\infty(M)$ is defined as
$$
\Delta_g=\mathrm{div}_g\circ\mathrm{grad}_g,
$$
where the gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_g(f)}g=df,$$
$\iota_\bullet$ being the contraction with a vector field, and the divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by
$$
\mathcal{L}_X\,\mathrm{vol}_g=\mathrm{div}_g(X)\,\mathrm{vol}_g,
$$
where $\mathcal{L}_\bullet$ is the Lie derivative and $\mathrm{vol}_g$ the Riemannian volume density on $M$.
On a symplectic manifold $(M,\omega)$ we can proceed by complete analogy and define the symplectic Laplacian
$\Delta_\omega:C^\infty(M)\to C^\infty(M)$ as
$$
\Delta_\omega=\mathrm{div}_\omega\circ\mathrm{grad}_\omega,
$$
where the symplectic gradient of a function $f\in C^\infty(M)$ is defined by $$\iota_{\mathrm{grad}_\omega(f)}\omega=df,$$
and the symplectic divergence of a vector field $X\in \Gamma^\infty(TM)$ is defined by
$$
\mathcal{L}_X\,\mathrm{vol}_\omega=\mathrm{div}_\omega(X)\,\mathrm{vol}_\omega,
$$
where $\mathrm{vol}_\omega=\frac{1}{n!}\omega^n$ is the symplectic volume form on $M$ (here $2n=\mathrm{dim}\,M$).
Now, a fundamental difference between Riemannian manifolds and symplectic manifolds is that on the latter we have the Darboux theorem, and a simple computation shows that in Darboux coordinates $q_1,\ldots,q_n,p_1,\ldots,p_n$, the symplectic Laplacian defined above is given by the formula
$$
\Delta_\omega = \sum_{i=1}^n\Big(\frac{\partial^2}{\partial p_i\partial q_i}-\frac{\partial^2}{\partial q_i\partial p_i}\Big).
$$
By the theorem of Schwarz, this means that
$$
\Delta_\omega=0.
$$
This seems to be a first hint why we don't have a symplectic version of spectral geometry.
I would like to know:
Can we prove or refer to a theorem of the form:
The Darboux theorem is equivalent to the statement that there is no second order differential operator on $M$ that is invariant under all symplectomorphisms ?
A broader question is:
Is there a deeper reason behind the non-existence of a symplectic version of spectral geometry?
Indeed, recall that on a Riemannian manifold the Laplace-Beltrami is not the only natural differential operator that is invariant under isometries, see
Laplace-Beltrami and the isometry group
and the related question
It would be nice to have an argument why in fact there cannot be symplectic analogues of any of the natural differential operators commuting with isometries of a Riemannian manifold.
Some comments on the answers received so far:
I have accepted Ben McKay's answer because it comes closest to what I hoped for. It indeed seems to settle the question for scalar differential operators.
As Igor Rivin pointed out, there is a non-trivial study of operators acting on symplectic spinors. It would be interesting to know whether there are more extra structures apart from symplectic spinor bundles that do not correspond to introducing a Riemannian structure but lead to natural differential operators whose study one could call "spectral symplectic geometry".
I am not sure if I really agree with Dan Fox' point of view that a smooth setting in which there are no local invariants should be regarded as part of differential topology rather than differential geometry. Maybe it boils down to the question if global analysis is to be considered part of differential geometry or rather of differential topology. I would say it is part of both.
Best Answer
The characteristic variety (i.e. vanishing locus of the symbol) of a symplectomorphism invariant scalar differential equation is a real projective hypersurface invariant under the group of projectivized linear symplectic transformations. This group acts transitively on the real points of projective space, so preserves no hypersurface.