Below, I will give a brief introduction to the difference between the interacting fields and the free field. This is discussed in the first few chapters of Srednicki.
The point is
- A vacuum expectation value of field operator gives one particle state.
- In general, one particle state of interacting system is different from that of free theory, but its difference is only the existence of wave-function (= field strength) renormalization constant $Z$ (see below). This is an effect of interaction.
- In particular, condensed matter people calls such a one-particle state in an interacting system as a quasiparticle excitation.
Let me elaborate a bit more.
1st point is simple. If we consider the free theory, one particle state can be obtained by the following relation
$$\phi_0(x)|0\rangle\sim a^\dagger|0\rangle.$$
Samely, in the interacting system, we can define the one particle state as a state that is created by field operator (more rigorous definition of one particle state is found in Srednicki or Duncan)
$$\phi(x)|0\rangle=|\mathrm{one}\ \mathrm{particle}\ \mathrm{state}\rangle,$$
where both $\phi$ and $|0\rangle$ are interacting ones.
However, this interacting one particle state is not the same to the one particle state of free theory. We can easily understand this fact.
Firstly, in general the difference between interacting field and free field will take the following form:
$$\phi(x)=\phi_0(x)+g\delta \phi(x,g),$$
where $g$ is a coupling constant and $\delta \phi(x)$ is a function of $g$, but is regular in the limit of $g\rightarrow 0$.
Notice that not only the first term but the second term have a contribution to the one particle state.
Let’s consider one example: $\phi^3$ theory. In this case, EoM is given by
$$(\Box+m_0^2)\phi(x)=-\frac{g}{3!}\phi(x)^3.$$
For simplicity, we ignore the mass renormalization and assume that renormalized mass is equal to that of free theory $m_0^2=m^2$. In real world, this relation is modified to $m^0=m^2+g\delta m^2$, but this term does not contribute to the wave-function renormalization.
By substituting third eq. to forth eq., at the first order of $g$, we will obtain
$$(\Box+m_0^2)\delta\phi(x,g=0)=-\frac{1}{3!}\phi_0(x)^3.$$
The general solution of this equation is given by
$$\delta\phi(x,g=0)=\alpha \phi_0(x) -\frac{1}{3!}\int_y iG(x-y)\phi_0^3(y),$$
where $\alpha$ is arbitrary constant. A generalization of this equation is called as the Yang-Feldman equation.
From the above, it is clear that $\delta\phi$ can have a contribution to free one particle state.
Thus, the general interacting field is written as the following form:
$$\phi(x)=(1+\alpha g+\cdots)\phi_0(x)+g\delta \phi(x,g)_{n\geq 2},$$
where $\delta \phi(x,g)_{n\geq 2} $ denotes the quantum correction from more than 2 particle states.
The coefficient of the first term is exactly the wave-function renormalization: $\sqrt{Z}:= (1+\alpha g+\cdots)$.
From these discussion, we can understand the following formula:
$\phi(x)|0\rangle=\sqrt{Z}\phi_0(x)|0 \rangle.$
That is, the change of normalization about one-particle state due to the interaction is $Z$. Moreover, because one particle can interact with infinite number of particles around him, this coefficient $Z$ is usually divergent in QFT.
I give a few final comments.
These results are correct only if the interaction results in essentially no change in the particle depiction. Sometimes, the definition of “particle” is strongly modified by the interaction. (e.g. Tomonaga-Luttinger fluid)
In this case, we cannot find the simple formula like
$$\phi(x)|0\rangle=\sqrt{Z}\phi_0(x)|0 \rangle$$
between the interacting field and the free field.
In this sense, the discussion above is applied only to the case that the effect of interaction essentially does not modify the definition of particle. This type of discussion is found in the topics around Landau-quasiparticle in condensed matter textbook.
