One more time about LSZ-theorem This question is the continuation of this one.
For simplicity, let's use $(1)$ from the linked question (it is called n-point Green function and in particle case coincides with internal diagram),
$$
G(p_{1},...,p_{n}) = \int d^{4}x_{1}...d^{4}x_{n}e^{-i\prod_{i =1}^{n}p_{j}x_{j}} \langle |\hat {T}\left( \hat {O}_{l}(x_{1})\hat {A}_{2}(x_{2})...\hat A_{n}(x_{n})\right) |\rangle , \quad (1)
$$
(in accordance with the linked question $\hat {O}_{l}(x)$ transforms as some free field $\hat {\Psi}_{l}(x)$ under the irreducible rep of the Lorentz group, and the other operators are local functions of the free fields) 
for the case when $\hat {A}_{2}(x) = \hat {O}_{l'}^{\dagger}(x)$ and the other operators $\hat {A}_{i}(x_{i})$ is equal to the unity. Then it's not hard to show that 
$$
G (p_{1}, p_{2}) \to \frac{-2i|N|^{2}\sqrt{\mathbf p_{1}^{2} + m^{2}} }{p_{1}^{2} - m^{2} - i\varepsilon}\sum_{\sigma}u^{\sigma}_{l}(\mathbf p_{1})(u_{l'}^{\sigma}(\mathbf p_{1}))^{*}\delta (p_{1} + p_{2})(2 \pi )^{4}, \qquad (2)
$$
which coincide with propagator of the free field $\langle |\hat {T}\left(\hat {\Psi}_{l}(x) \hat {\Psi}_{l'}(y)\right)|\rangle$ in the momentum representation excepting $|N|^{2}$ factor. 
For the simplicity lets assume that $\hat {O}_{l}(x) = \hat {O}(x)$, i.e. we working with the scalar operator. Then it's not hard to show that $|N|^{2} = (2 \pi )^{3}\left|\langle \mathbf p| \hat {O}(0) |0\rangle \right|^{2} = \frac{Z}{2\sqrt{\mathbf p^{2} + m^{2}}}$ (this result may be easily compared with the paragraph about renormalization of the field strength of Peskin and Schroeder QFT; the result above coincides with the result from Peskin QFT up to the normalization condition). 
So, the question: why in the LSZ formalism $|N|^{2} \neq \frac{1}{2\sqrt{\mathbf p^{2} + m^{2}}}$ as for case of free theory? I.e., why $Z \neq 1$?
 A: If $Z=1$, then $\phi^{\dagger} (0)|0\rangle$ is a one-particle state (given the right normalization of $u (\ell , \sigma)$, as defined in Weinberg's book, chapter 5) because 
$$
\sum_{p,\sigma}| \langle p,\sigma|\phi^{\dagger}(0)|0\rangle|^2 =1
$$
and is assumed that $ \phi^{\dagger}(0)|0\rangle$ is a normalized state, i.e. the field is defined in such a way that this is a normalized state. In Weinberg, chapter 10, equation 10.7.17 says precisely this, with some extra manipulations.
If $Z\neq 1$ then $ \phi^{\dagger}(0)|0\rangle$ is a multi-particle state and $Z$ should be small than $1$ and greater than $0$.
In the case of $Z=1$, for any n-point Green's function, if you put all the $n-1$ legs on-shell  (loking at the residue of the on-shell poles of this legs) you are going to find the projection of the state $ \phi^{\dagger}(0)|0\rangle$ with n-particle state, but this is zero since $Z=1$. So, there is no scattering in this theory, since the S-matrix is obtained by put all legs of a Green's function on-shell.
