Similar to how we can construct a density matrix $\rho_A$ that represents states in a subsystem $A$ by performing a partial trace on $\rho$ (the full density matrix of the whole subsystem (say $AB$)), is there a similar operation we can do on the hamiltonian $H$?
Consider $$ i\frac{d}{dt} \rho_B = i\frac{d}{dt} \text{Tr}_A (\rho) = \text{Tr}_A ([H,\rho]), $$ where the last step is true by linearity. If for simplicity we are in a bipartite system such that $H=H_A \otimes H_B$ then we can further massage above expression \begin{align} i\frac{d}{dt} \rho_B & =\text{Tr}_A ([H,\rho]) \\ & = \sum_i (\langle \phi_i|_A\otimes 1_B)[H_A \otimes H_B,\rho](| \phi_i \rangle _A\otimes 1_B) \\ & =\sum_i E_i^A\underbrace{( 1_A \otimes H_B)}_\text{abuse of notation}(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B) \\ & \qquad -E_i^A(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B)\underbrace{( 1_A \otimes H_B)}_\text{abuse of notation} \\ & =\left[H_B, \sum_iE_i^A(\langle \phi_i|_A\otimes 1_B)\rho(| \phi_i \rangle _A\otimes 1_B)\right] \\ & = [H_B,\rho_B], \end{align} where the abuse of notation is that the identity now $1_A:\mathcal{H}^* \rightarrow \mathcal{H}^*$ and not $1_A:\mathcal{H} \rightarrow \mathcal{H}$, $|\phi_i\rangle_A$ is the eigenket of $H_A$ with eigenenergy $E_i^A$ and in the last line I redefined the basis of the partial trace.
This calculation, if correct implies that $H_B$ is the "reduced" Hamiltonian on $B$, is this correct?
To me this is not trivial as there might be interactions between $A$ and $B$ (nor that it matters why is not trivial to me).