rotation group is broken down to the rotation around the z-axis. The states $| \frac{3}{2},  \frac{1}{2} \rangle$ and $| \frac{1}{2},
\frac{1}{2} \rangle_0 $ are mixed in the new ground state which is given by:

$$| + \frac{1}{2} \rangle = \frac{1}{\sqrt{1+2 x^2}}  (
2 | \frac{1}{2},  \frac{1}{2} \rangle_1 + \sqrt{2} x| \frac{3}{2},
\frac{1}{2} \rangle ) \label{11xh13a}$$

where

$$x = \frac{ 2 (\Delta -1)}{8 + \Delta + 3
\sqrt{\Delta^2 + 8}}              \label{11xh13b}$$

and its energy is,

$$E_B = -\frac{J}{4} [\Delta + \sqrt{\Delta ^2 +
8}]                                       \label{11xh13c}$$

along with its $| - \frac{1}{2} \rangle$ partner. This are now the two states retained in the RG method. To be more explicit, we have

$$| + \frac{1}{2} \rangle = \frac{1}{\sqrt{6(1+2 x^2)}}
[(2 x + 2) |\uparrow \downarrow \uparrow \rangle +  (2 x - 1)
|\uparrow \uparrow \downarrow \rangle +  (2 x - 1) |\downarrow
\uparrow \uparrow \rangle ] \label{11xh14a}$$

$$| - \frac{1}{2} \rangle = -\frac{1}{\sqrt{6(1+2
x^2)}}  [(2 x + 2) |\downarrow \uparrow \downarrow \rangle +  (2 x -
1) |\downarrow \downarrow \uparrow \rangle +  (2 x - 1) |\uparrow
\downarrow \downarrow \rangle ]
\label{11xh14b}$$

The intertwiner operator $T_0$ reads then,

$$T_0 = | + \frac{1}{2}\rangle \langle  \uparrow |'
+     | - \frac{1}{2}\rangle \langle  \downarrow
|'                                   