e^{-iHt} |\downarrow\Uparrow\ra \times \la\downarrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Downarrow\ra \cr
       &+& \la\downarrow\Downarrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Downarrow\ra \times \la\uparrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Uparrow\ra \cr
       &+& \la\uparrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Uparrow\ra \times \la\downarrow\Downarrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Downarrow\ra \cr
       &+& \la\downarrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Downarrow\ra \times \la\uparrow\Downarrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Uparrow\ra \cr
       &+& \la\downarrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Uparrow\ra \times \la\uparrow\Downarrow| e^{+iHt} |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Downarrow\ra        \}\end{aligned}$$

$$\begin{aligned}
 \label{a_b_t}
  b(t) = \frac{-1}{2} &\{& \la\uparrow\Downarrow| e^{+iHt} |\downarrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Uparrow\ra \times \la\downarrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\downarrow_N| e^{-iHt} |\downarrow\Downarrow\ra \cr
       &+& \la\downarrow\Downarrow| e^{+iHt} |\downarrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Downarrow\ra \times \la\uparrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\downarrow_N|  e^{-iHt} |\downarrow\Uparrow\ra \cr
       &+& \la\downarrow\Downarrow| e^{+iHt} |\downarrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Uparrow\ra \times \la\uparrow\Uparrow| e^{+iHt} |\uparrow_N\ra \la\downarrow_N|  e^{-iHt} |\uparrow\Downarrow\ra \cr
       &+& \la\downarrow\Uparrow| e^{+iHt} |\downarrow_N\ra \la\uparrow_N|  e^{-iHt} |\uparrow\Uparrow\ra \times \la\uparrow\Downarrow| e^{+iHt} |\uparrow_N\ra \la\downarrow_N|  e^{-iHt} |\downarrow\Downarrow\ra   \}\end{aligned}$$

![(Color online) (a) Entanglement $E(t)$ as a function of $Jt$ for a chain of length $N=40$ (i.e. $N_L=N_R=20$). (b) The maximal entanglement $E_{max}$ versus length $N$.[]{data-label="fig4"}](Fig4.eps){width="9cm" height="4.5cm"}

One can compute the entanglement, quantified by concurrence [@concurrence], between the two qubits which becomes $$\label{concurrence}
    E(t)=max\{0,b(t)-a(t)\}.$$ In Fig. \[fig4\](a) the entanglement $E(t)$ is plotted as a function of time. It is worth mentioning that as entanglement propagates 