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

Eqs. (\[3\]) and (\[5\]) characterize what may be called exact renormalization group method (ERG) in the sense that the whole spectrum of $H'$ is mapped onto a part (usually the bottom part) of the spectrum of $H$. In practical cases though the exact solution of Eqs. (\[3\]) and (\[5\]) is not possible so that one has to resort to approximations (see later on). Considering Eqs. (\[3\]) and (\[5\]) we can set up the effective Hamiltonian $H'$ as:

$$H' = T^{\dag} H
T                                                   \label{10}$$

This equation does not imply that the eigenvectors of $H'$ are mapped onto eigenvectors of $H$. Notice that Eq.(\[10\]) together with (\[5\]) does not imply Eq. (\[3\]). This happens because the converse of Eq.(\[5\]), namely $TT^{\dag } \neq 1_{\cal
H}$ is not true, since otherwise this equation together with (\[5\]) would imply that the Hilbert spaces ${\cal H}$ and ${\cal H'}$ are isomorphic while on the other hand the truncation inherent to the RG method assumes that $dim {\cal H'} < dim {\cal H}$.

The fact that $T^{\dag } T\neq 1_{{\cal H}}$ reflects nothing but the irreversibility of the RG-transformation. Indeed, 