Spectrum of sum of positive and negative operators

Question

Let $(\mathscr{H}, \langle \cdot, \cdot \rangle)$ be a separable complex Hilbert space, and let $\mathscr{D}$ be a dense subset of $\mathscr{H}$. Let $P: \mathscr{D} \to \mathscr{H}$ and $N: \mathscr{D} \to \mathscr{H}$ be positive and negative (unbounded) operators, respectively, meaning that $$ \langle Pv,v \rangle \geq 0 \quad \text{and} \quad \langle Nv,v \rangle \leq 0 $$ for all $v \in \mathscr{D}$. Moreover, assume that $P$ and $N$ commute, and $P \neq -N$.

Consider the operator sum $X := P + N$. If $\mathscr{D}$ decomposes as a finite sum of eigenspaces for $X$, is it true that $\mathscr{D}$ also decomposes into eigenspaces for $P$ and $N$, respectively?

Answer 1

No, this isn't working at all.

Since $N,P$ commute (you would have to be more specific what exactly you mean by this since the operators are unbounded, but let's just assume we have the right version), they are functions of a third operator. So passing to a spectral representation, we may assume that $N$ is multiplication by $f(x)\le 0$ and $P$ is multiplication by $g(x)\ge 0$ in $L^2(\mathbb R, \rho)$ (really in a sum of such spaces, but let's ignore multiplicity).

Now the assumption is that $f(x)+g(x)$ takes only finitely many values, and we would like to deduce that $f,g$ are also of this type, but clearly this is completely hopeless.