Expectation values of interacting fields I was motivated to ask this question by the equality claimed in equation 10.3.3 of Weinberg's volume 1 of QFT books. 
My interpretation of that, 
If $O_s$ is a quantum field of spin $s$, $\psi_s$ is the free field of spin $s$, $|p,\sigma\rangle$ is a one-particle state of some interacting theory and $|0\rangle$ is the vacuum state, then there should exist a constant $N$ such that,
$\langle 0|O_s|p,\sigma\rangle =  \frac{N}{(2\pi)^3} \langle 0|\psi_s|p,\sigma\rangle$.
If I understand his equation 10.3.6 then that seems to say that the field $O_s$ will be said to have been "renormalized" if $N$ can be set to $1$. 
So is one saying that all effect of interactions are absorbed into an overall factor at the level of matrix elements? 
That sounds very surprising to me - and it seems that Weinberg claims that it follows merely from the fact that $O_s$ and $\psi_s$ have to transform under the same irreducible representation of the Poincare group. 
I would be glad if someone can elaborate this point. 
Also what happens if one replaces $|p,\sigma\rangle$ by multi-particle states (it's not clear to me as to what the complete set of labels is that will be required to index the continuum of multiparticle states, and clearly total momentum or/and the invariant mass is not enough).
 A: The answer about this equation in Weinberg's book is rather simple. You have to evaluate $\langle 0|{\cal O}_l(0)|q_1,s\rangle$ being $l$ the index due to the corresponding Lorentz transformation. This means that this matrix element should transform as an element of the Lorentz group. Weinberg's assumption is that the corresponding representation is irreducible. If this is true, being this a matrix element between the vacuum and a single particle state, the only factor that has such a property of transformation under Lorentz group is the one of the free particle state and that can be always singled out. What remains is the contribution of the interaction that amounts to a constant. This is just a requirenment of Lorentz invariance. Let us make some examples:
Scalar field:
$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}\sqrt{2E}}N_s$$
Spin-1/2 field:
$$\langle 0|{\cal O}_l(0)|q_1,s\rangle=\frac{1}{(2\pi)^\frac{3}{2}}u_l(q_1,s)N_f$$
and so on.
But this is a well-known quantum field theory theorem: It is named Lehmann-Symanzik-Zimmerman (LSZ)  theorem that Weinberg puts forward in a rather discorsive way in his book at that stage. LSZ theorem permits one to pass from N-point functions to S-matrix elements and to obtain cross sections and decay rates. So, you can write down the partition function of your favourite theory, compute N-point functions for a given reaction and, from it, using LSZ theorem obtain the corresponding observable. Another good presentation of this reduction formula is given on Peskin and Schroeder staring from pag.222.
A: The interpolating field couples to many states.  This is summarized by the Lehaman-Kallen spectral representationfor the interacting propagator. Most field thoery books will have a discussion of this.
