This is a simplified version of a recent question I asked. My hope is that this simplified version will be easier to tackle. The motivation behind both of these questions is roughly to ask "Given that a Hamiltonian acts simply on each state in some subspace, can it be identified with a single simple operator when acting on the subspace?"
Below, I explain my question and specify what it means to "act simply" on a subspace.
Consider a subspace $W$ of the Hilbert space of an $L$-qubit spin chain.
I have a Hermitian operator $H$ acting on the spin chain with two properties. It preserves the subspace $W$, and, for each vector in $W$, $H$ looks as though it's a sum of single-site operators, though $H$ need not look like a sum of single-site operators on the whole Hilbert space. Specifically:
$$\forall |w\rangle \in W,\,\,\, H|w\rangle \in W$$ $$\forall |w\rangle \in W,\,\,\, H|w\rangle = \sum_{i=1}^L H_i(|w\rangle) |w\rangle$$ where $H_i(|w\rangle)$ is a Hermitian single-site operator acting on the $i$th site. $H_i(|w\rangle)$ may depend on $|w\rangle$.
Can I use these two properties of $H$ to construct a $|w\rangle$-independent sum of single-site operators $\sum_i \tilde{H}_i$ that satisfies
$$\forall |w\rangle \in W,\,\,\, \sum_{i=1}^L \tilde{H}_i |w\rangle = H|w\rangle?$$
