$\cal H'$ (effective Hamiltonian).

-   $T  $ : embedding operator : ${\cal H }'\longrightarrow {\cal H}$

-   $T^{\dagger }  $ :truncation operator : ${\cal H}
    \longrightarrow {\cal H}'$

The problem now is to relate $H$, $H'$ and $T$. The criterium to accomplish this task is that $H$ and $H'$ have in common their low lying spectrum. An exact implementation of this is given by the following equation:

$$H T = T
H'                                                   \label{3}$$

which imply that if $\Psi '_{E'}$ is an eigenstate of $H'$ then $T\Psi '_{E'}$ is an eigenstate of $H$ with the same eigenvalue (unless it belongs to the kernel of $T$: $T\Psi '_{E'}=0$), indeed,

$$HT\Psi '_{E'} =  T H'\Psi '_{E'} = E' T\Psi
'_{E'}                                                  \label{4}$$

To avoid the possibility that $T\Psi '=0$ with $\Psi '\neq 0$, we shall impose on $T$ the condition,

$$T^{\dagger } T = 1_{\cal
H'}                                                   \label{5}$$

such that

$$\Psi = T\Psi '  \Rightarrow \Psi ' = T^{\dag}
\Psi                                                  \label{6}$$

Condition (\[5\]) thus stablishes a one to one relation between $\cal H'$ and Im($T$) in $\cal H$.

Observe that Eq. (\[3\]) is nothing but the commutativity of the following diagram:

$$\begin{array}{llcll}
  & {\cal H'} & \stackrel{T}{\longrightarrow} & {\cal H}  &  \\
 H' & \downarrow 