$n_s$ either analytically or numerically.

Eq. (\[21\]) suggests that we should search for solutions of the intertwiner equation (\[3\]) in the form of a perturbative expansion in the interblock coupling constant parameter $\lambda $, namely,

$$T = T_0 + \lambda T_1 +  \lambda^2  T_2 +
\ldots                    \label{22a}$$

$$H' =H'_0 + \lambda H'_1 +  \lambda^2 H'_2 +
\ldots                    \label{22b}$$

To zeroth order in $\lambda $ Eq. (\[3\]) becomes

$$H_B T_0 = T_0
H_0'                                                 \label{23}$$

Since $H_B$ is a sum of disconnected block Hamiltonians $h_{j'}^{(B)}$, $j'=1,\ldots ,N'$ implicitly defined through the relation

$$H_B = \sum
_{j'=1}^{N'}h_{j'}^{(B)}
\label{24}$$

one can search for a solution of $T_0$ in a factorized form

$$T_0 = \prod _{j'=1}^{N'}
T_{0,j'}                                                \label{25}$$

and an effective Hamiltonian $H_0'$ which acts only at the site $j'$ of the new chain,

$$H_0' =   \sum _{j'=1}^{N'}h_{j'}^{(s')} =
H'_{s'}                                                \label{26}$$

Observe that $ H'_{s'}$ is nothing but a site-Hamiltonian for the new chain. Eq. (\[23\]) becomes for each block:

$$h_{j'}^{(B)}  T_{0,j'}   =
T_{0,j'}h_{j'}^{(s')}
\label{27}$$

The diagonalization of $h_{j'}^{(B)}  $ for $j'=1,\ldots
,N'$ will allow us to write

$$h_{j'}^{(B)} =  \sum _{i=1}^{k'} |i\rangle_{j'} \
\epsilon _i \ _{j'} \hspace*{-1pt} \langle i| +  \sum _{\alpha
=1}^{k^{n_s}-k'} |\alpha \rangle_{j'}  \ \epsilon _{\alpha }  \ _{j'}
\hspace*{-1pt} \langle 