Finding the effective Hamiltonian in a certain subspace In order to find the effective Hamiltonian in a subspace which is energetically well separated from the rest of the Hilbert space people try to find a unitary transformation which makes the Hamiltonian block-diagonal in that subspace. Usually this procedure is done perturbatively and the corresponding formulae -usually to second order- is available. But I saw somewhere that the effective Hamiltonian satisfies the compact relation :
$$ \frac{1}{E-H_{eff}}=P_s \frac{1}{E-H} P_s$$
Where $P_s$ is the projection operator into the subspace which we want its effective Hamiltonian. 
So where does the above relation come from? Also it will be very helpful if you mention some references about about different ways of obtaining effective Hamiltonian systematically. 
I saw the above formula in the book "Interacting Electrons and Quantum Magnetism" by Auerbach.
 A: Everett explained the compact relation nicely. Regarding references, 
https://arxiv.org/abs/1005.2495
and
https://arxiv.org/abs/1105.0675 (more mathematical)
are nice.
Also, the effective Hamiltonian is calculated to 6th order in appendix B of
https://arxiv.org/abs/1704.03870
using the Schrieffer-Wolff formulation of degenerate perturbation theory.
(I am an author of this one.)
A: This approach is straightforward to understand if you realize that $1/(E-H)$ is nothing but the propagator (the Green's function) $G(E)=(E-H)^{-1}$. So this approach simply means that the effective propagator $G_\text{eff}(E)=(E-H_\text{eff})^{-1}$ is obtained by restricting the full propagator to the subspace of interest $G_\text{eff}(E)=P_sG(E)P_s$. One may wonder why not projecting the Hamiltonian directly to the subspace but projecting the propagator. The reason is that all physical observables are measured with respect to the density matrix $\rho(E)=-2\Im G(E+i0_+)$, which is the imaginary part of the propagator. For example, the expectation value of an operator $A$ evaluated on an eigenstate of the energy $E$ is given by
$$\bar{A}(E)=\text{Tr}\hat{A}\rho(E)=-2\text{Tr}\hat{A}\Im G(E+i0_+).$$
Now suppose we are only interested in the physical observables in the Hilbert subspace $\mathcal{H}_s$, then the information of the propagator $G(E)$ in this subspace will be sufficient to reproduce all measurement result, and hence an "effective" description of the subsystem. The Hamiltonian that will produce the effective propagator is therefore considered as the effective Hamiltonian for the subsystem. Of course, the effective Hamiltonian is typically only calculated perturbatively to some order, so approximations are introduced. But imagine if we could find the effective Hamiltonian to all orders, then it would agree with the full Hamiltonian on any physical measurements that take place in the subsystem (or the subspace).
Take a simple quantum mechanical problem for example. Consider a two-level system described by the Hamiltonian
$$H=\left[\begin{matrix}0&t\\t&U\end{matrix}\right],$$
where $t\ll U$ is treated as a perturbation. In the limit of $t\to 0$, we get two levels of the energies 0 and $U$ respectively. Now we are interested in the energy correction to the low-energy level (the level around energy 0). So we first calculate the propagator of the system
$$G=\frac{1}{E-H}=\left[\begin{matrix}\frac{E-U}{E^2-E U-t^2} & \frac{t}{E^2-E U-t^2} \\
 \frac{t}{E^2-E U-t^2} & \frac{E}{E^2-E U-t^2} \end{matrix}\right].$$
The effective propagator for the low-energy level is obtained by restricting the propagator to the low-energy subspace, i.e. by taking the $\mathcal{P}_1 G(E) \mathcal{P}_1=G(E)_{11}$ component (at the first line and first column),
$$G_\text{eff}(E)=\frac{E-U}{E^2-E U-t^2}.$$
Now we wish to construct an effective Hamiltonian $H_\text{eff}$ such that the effective propagator can be produced by $G_\text{eff}(E)=1/(E-H_\text{eff})$. We find
$$H_\text{eff}=\frac{t^2}{E-U}.$$
We note that $H_\text{eff}$ is also a function of $E$, because the physics can change with respect to the energy scale. To find the eigen energy, one may solve the Schrodinger equation $H_\text{eff}(E)|\psi\rangle=E|\psi\rangle$. Because the subspace only contains a single state, in this case, the eigenstate is simply fixed with respect to the basis of the subspace (but actually implicitly varies with $E$ in the original basis of the full space), and the eigenenergy is given by $t^2/(E-U)=E$, whose solution is
$$E=\frac{1}{2}\big(U\pm\sqrt{U^2+4t^2}\big).$$
One can see the effective Hamiltonian, if calculated exactly to all ordered, still contains the spectrum of the full system. But in general, we can only compute the effectively Hamiltonian perturbatively. In that case, it will only make sense to evaluate the effective Hamiltonian around the unperturbed energy level. So we evaluate $H_\text{eff}(E)$ at $E=0$ and find the second-order perturbation result $H_\text{eff}(E=0)=-t^2/U$. To obtain higher-order corrections, we feed back the second-order energy to the effective Hamiltonian and find the result that is accurate to the fourth-order in $t/U$, i.e. $H_\text{eff}(E=-t^2/U)=-t^2/U+t^4/U^3+\mathcal{O}[t^6]$. In this way, we can obtain the perturbative corrections order by order recursively.
