LSZ reduction theorem derivation in Weinberg QFT

Question

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?

Answer 1

Good question. First, I think, we can use Lorentz invariance, i.e, use the properties on how state $\Psi_{\mathbf{q_1} \sigma}$ transforms and how $O_l$ transforms to get to coefficient function, as we did in Chapter 5 when we went down the similar route to obtain coefficient functions from Lorentz transformation of $\Psi_l$ and creation operator (which transforms like a single particle state). Ultimately, we will get again from those conditions that: \begin{align*} (\Psi_0, O_l(0) \Psi_{\mathbf{q_1} \sigma}) = (2 \pi)^{-3/2} N u_l(\mathbf{q_1}, \sigma), \end{align*} where unsurprisingly we will a coefficient function, because in some sense $O_l^\dagger$ is creating a particle (think of a simple case, when $O_l$ is just a scalar field operator); but it could be in general any operator with 'hitherto' unknown normalization, i.e. it might not have the normalization as if it were an operator for a field appearing in the Lagrangian. Basically, we can't say if this general operator will have the same normalization; we are being generic. Next, Weinberg says that if $O_l$ is actually some field operator $\Psi_l$ then at least we should renormalize the fields by $1/N$ so that we can just treat $\Psi_l$ normally in Feynman rules such that: $(\Psi_0, \Psi_l(0) \Psi_{\mathbf{q} \sigma}) = (2 \pi)^{-3/2} u_l(\mathbf{q}, \sigma)$. So, at least in this case N is gone by renormalizing the field operator. This is what we want out of normal field operator (try a scalar field operator for sanity check now after renormalizing in this expression). Hope this helps.