If I have a quantum system and trace out some degrees of freedom (i.e. in the Green function formalism), I get a subsystem governed by an effective Hamiltonian. Now, if we assume that the full system is in its ground state (let's say a Slater determinant), does that imply that the subsystem will also be in its ground state (a Slater determinant of the new lowest $N$ states of the effective Hamiltonian)? Famously, the tracing out can turn a pure system-environment state into a mixture. But is there any certainty what happens at $T=0$?
I am wondering because what happens when the effective Hamiltonian is Non-Hermitian? It is not clear what is even meant by "ground state" in that case. Would we really need to explicitly trace out the environment degrees of freedom?
For context: I am talking about a topological insulator which is proximity-coupled to a ferromagnet. The form of the TI Hamiltonian is irrelevant, let's say it is given by $H_{TI}$.This is a one-particle Hamiltonian. Now, if we couple the ferromagnet ($H_{full} = H_{TI} + H_{ferro} + H_{coupling}$), we can - by projection onto the surface - get a surface Green function
$g(E) = \frac{1}{E-H_{TI}-\Gamma}$
where $\Gamma$ is non-Hermitian.
Now, we can define the effective Hamiltonian $H_{eff}=H_{TI} + \Gamma$ on the subsystem (the surface).
We have a system of noninteracting fermions, thus the ground state of the full system is a Slater determinant of the lowest energy eingenstates. However, if we are now in the subspace of the surface governed by $H_{eff}$, does the ground state $|0\rangle_{full}$ project onto the ground state of $H_{eff}$ when the non-surface degrees of freedom are traced out?
