Definition of propagator in QFT Srednicki defines the exact propagator as $\langle 0 \mid \text{T} \varphi(x) \varphi(y) \mid 0 \rangle$, where T is the time ordering symbol and $x,y$ are four-vectors. What I am wondering is whether $\mid 0 \rangle$ refers to the ground state of the free Hamiltonian or the perturbed one. I tried Googling this but found no clarification. If it is the perturbed vacuum then to what extent is the physical interpretation of $\varphi(x) \mid 0 \rangle$ as a point particle at $x$ still valid?
 A: The answer is negative. It arises from my discussion below which is not completely rigorous because I will completely disregard several issues concerning domains of operators (which could be fixed using the universal enveloping algebra, the Garding domain and some machinery essentially introduced by Nelson or, alternatively, the theory of induced representations by Mackay).
First of all, $|0\rangle$ is the Poincaré invariant state of the complete theory, the ground state of the complete Hamiltonan and $\phi(f) = \int \phi(x) f(x) d^4x$ is the field operator of the interacting theory formally smeared against the smooth compactly supported function $f$.
The question can be rephrased like this, 
whether or not $\phi(f)|0\rangle$ can be considered a one-particle vector state of an elementary particle.
Elementary particles in QFT are defined as irreducible representations of Poincaré group. Therefore, more precisely, the question is
whether or not the closed subspace $\cal H_0$ spanned by all vectors $\phi(f)|0\rangle$ - varying $f$ in the set of complex-valued compactly supported smooth functions - is a unitary irreducible representation
 of Poincaré group with respect to the restriction of the Poincaré unitary representation of the complete interacting theory.
$\cal H_0$ is invariant under the action of the group since $|0\rangle$ is invariant under this representation and  the field operator satisfies
$$U^*_L \phi(f) U_L = \phi(f_L)\quad\mbox{where } \quad f_L(x) := f(L^{-1}x)$$
for every element $L$ of Poincaré group.
In particular, the action of the translation subgroup generated by the four-momentum $P_\mu$ is the standard one $$U^*_a \phi(f) U_a = \phi(f_a)\quad\mbox{where } \quad U_a = e^{-ia^\mu P_\mu}$$
and $$f_a(x) := f(x-a)\:.$$
As a consequence of this relation, taking into account the invariance of $|0\rangle$ under the Poincaré group we easily find
$$P_\mu P^\mu \phi(f) |0\rangle = \phi(\partial_\mu\partial^\mu f)|0\rangle=
(\partial_\mu\partial^\mu \phi)(f) |0\rangle\:.$$
Here is the crucial point. If $\phi$ were the free field it would satisfy the free equation of motion 
$$\partial_\mu\partial^\mu \phi = -m^2 \phi\tag{1}$$
giving rise to 
$$P_\mu P^\mu \phi(f) |0\rangle = -m^2 \phi(f) |0\rangle$$
so that $P_\mu P^\mu$ would be a constant in the subspace $\cal H_0$. 
$$P_\mu P^\mu |_{\cal H_0} = -m^2 I \tag{2}$$
Eq. (2) is a necessary condition for irreducible representations because  $P_\mu P^\mu$ commutes with all generators of Poincaré group and (a certain version of) Schur's lemma holds.
In the considered case however (1) is false, just because $\phi$ is the interacting field and satisfies an equation encompassing the (self)interaction 
$$\partial_\mu\partial^\mu \phi = -m^2 \phi + F(\phi) \tag{1'}$$
 Therefore we cannot conclude that $\cal H_0$ supports and induced irreducible representation of Poincaré group and thus the interpretation of that space as the one-particle subspace of an elementary particle is untenable.
With very concrete words, the considered states cannot be considered states of an elementary particle because that particle would not have a definite Poincare' invariant mass. If introducing vector or spinor fields, similar problems would appear for the spin.
