LSZ reduction theorem derivation in Weinberg QFT

When deriving LSZ reduction theorem Weinberg in his QFT book have assumed n-point generalized Green functions,

$$G(q_{1},...,q_{n}) = \int d^{4}x_{1}...d^{4}x_{n}e^{-i\prod_{i =1}^{n}q_{j}x_{j}} \langle |\hat {T}\left( \hat {O}_{l}(x_{1})\hat {A}_{2}(x_{2})...\hat A_{n}(x_{n})\right) |\rangle , \quad (1)$$

where $\hat {O}_{l}(x)$ transforms under the irreducible representation of the Lorentz group as some free field $\hat {\Psi}_{l}(x)$. By insertion between $\hat {O}_{l}(x_{1})$ and $\hat {A}_{2}(x_{2})$ functional unit

$$\sum_{i, \sigma}\int d^{3}\mathbf p | (\mathbf p , \sigma )_{i}\rangle \langle (\mathbf p , \sigma )_{i}|$$

and by allocation of one-particle states from it he have "reduced" (with some hints) $(1)$ to the form

G(q_{1},...,q_{n}) \to f(q)\sum_{\sigma}\langle | \hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle \times $$ $$ \times \int d^{4}x_{2}...e^{-iq_{2}x_{2}-...}\langle (\mathbf q_{1}, \sigma ) |\hat {T}\left( \hat {A}(x_{2})...\right) | \rangle \delta (q_{1} + ... + q_{n}). \qquad (2)

Here $f(q)$ contains the pole of the first order $\frac{1}{q^{2} - m^{2} - i\varepsilon}$ and $q = q_{1} + ... + q_{r}$. After that he says that in $(2)$ there is equality $\hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle = \frac{1}{\sqrt{(2 \pi )^{3}}}Nu^{\sigma}_{l}(\mathbf q_{1})| \rangle$. So I have the question: why was factor $N$ (in comparison with free field-like expression $\hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle = \frac{1}{\sqrt{(2 \pi )^{3}}}u^{\sigma}_{l}(\mathbf q_{1})| \rangle$) appeared? What is its physical sense? Is its appearance connected with the fact that $| \rangle$ doesn't refer to the "usual" vacuum? Can you also comment this statement, if you please?