In their 2011 paper, D. Poulin and coauthors show that the size of "physically" accessible states in Hilbert space for local Hamiltonians is much, much smaller than the total Hilbert space. To put it simply, there is an enormous space of states in Hilbert space which take an exponentially long time to prepare, and thus are not physically accessible by a quantum computer or any real-life "analog" quantum mechanical system.
However, the result by Poulin et al only holds true for local Hamiltonians, so I am interested in what happens when you relax that assumption. Locality here means that each qubit is only coupled to at most $k$ other qubits, where $k$ is fixed and does not scale with $N$. As we know from solid-state physics, most Hamiltonians in real-life systems are non-local because electromagnetism is non-local. This is most obvious in the Coulomb interaction between electrons, which is non-local and goes as $V(r)\sim \frac{\rho_1 \rho_2}{r}$. Similar effects occur with dipole-dipole interactions, etc. which generally have the form $V(r) \sim r^{-\alpha}$ for different exponents $\alpha$. In all of these cases, each qubit is coupled to all $N$ other qubits.
Intuitively, I would expect the result of Poulin et al to still hold to some extent depending on the exponent $\alpha$. So my question is has a non-local extension to Poulin et al's proof been made? Or, equivalently, how does the size of physically accessible states in Hilbert space grow with the non-locality of interactions $\alpha$.
I would expect that if we could have a completely non-local Hamiltonian with all-to-all coupling ($\alpha=0$), then all states in Hilbert space are physically accessible. But, as we localize couplings ($\alpha \rightarrow \infty)$), this space should decrease. The mostly physically relevant non-local coupling would be the Coulomb interaction, which I am interested most in.
Edit: Since there is some confusion, the Coulomb interaction I am referring to takes the following form $$\mathcal{H}_C = \frac{1}{2}\sum_{i,j} \frac{\hat{\rho}_i \hat{\rho}_j}{\vert r_i - r_j\vert}$$ where $\hat{\rho}_i$ would be an operator acting on the the $i$-th qubit and $r_i$ is its spatial position. This interaction is non-local both physically and computationally because it both couples qubits which are infinitely far apart in space, and couples all qubits to all other qubits (i.e., an all-to-all interaction). Note that the $r^{-1}$ is a very slow drop off, so there are logarithmic corrections to wavefunctions even as $r\rightarrow \infty$, as described in Sakurai's Quantum Mechanics textbook. So rigorously speaking, one camnot argue that this falls under $k$-locality. Moreover, one can choose a smaller decay of $r^{-\alpha}, \alpha<1$ if $\alpha=1$ is not "slow" enough to break out of the Poulin et al argument.
