Interpretation of the quantum field in light of the interpretation of propagators In Page 38 of David Tong's QFT notes and Page 27 (Chapter 2.4 The Klein-Gordon Field in Space-Time under the heading Causality) of Peskin and Schroeder's Introduction to Quantum Field Theory, the propagator $D(x-y)$ is interpreted as 

the amplitude for a particle to propagate from $y$ to $x$ 

and is given by $D(x-y) \ = \ \langle 0 | \phi ( x ) \phi ( y ) | 0 \rangle $, where $\phi (x)$ is the real Klein Gordon field in the Heisenberg Picture. 
This interpretation of the propagator seems to imply that $\phi ( y ) | 0 \rangle$ is a particle that has been prepared at spacetime coordinate $y$ while $\phi ( x ) | 0 \rangle$ is a particle at spacetime coordinate $x$. However, we also know that the mode expansion for $\phi(x)$ is given by the following combination of creation and annihilation operators: 
$$
\phi(x) = \phi ( \mathbf { x } , t ) = \int \frac { d ^ { 3 } p } { ( 2 \pi ) ^ { 3 } } \frac { 1 } { \sqrt { 2 E _ { \mathbf { p } } } } \left( a _ { \mathbf { p } } e ^ { - i p \cdot x } + a _ { \mathbf { p } } ^ { \dagger } e ^ { + i p \cdot x } \right)
$$
which seems nothing like a simple particle at spacetime coordinate $x$. 
It would be a nice interpretation if $\phi ( x ) | 0 \rangle$ is indeed a particle at spacetime coordinate $x$ but this interpretation makes no sense given the mode expansion for $\phi(x)$. 
Hence, is there a self-consistent way to interpret the quantum field given the common interpretation and the form of the propagator? 
 A: Let us define a quantum field as an integral over creation and annihilation operators for each momentum in the Schroedinger picture, i.e. Heisenberg picture at $t = 0$.
$\phi_0(\vec x) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_p}} (a_p e^{i \vec p \cdot \vec x} + a^\dagger_p e^{-i \vec p \cdot \vec x})$
where:
$a_p$ annihilation operator
$a^\dagger_p$ creation operator
$[a_k, a^\dagger_p] = (2 \pi)^3 \delta^3(\vec p - \vec k)$ equal-time commutation relations
$a^\dagger_p |0\rangle = \frac{1}{\sqrt{2 \omega_p}} |\vec p\rangle$ creation of a particle with momentum $\vec p$
$|0\rangle$ vacuum  
To understand what the operator $\phi_0(\vec x)$ does, we can act on the vacuum and project out a momentum.
$\langle \vec p|\phi_0(\vec x)|0\rangle = \langle 0| \sqrt{2 \omega_p} a_p \int \frac{d^3 k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_k}} (a_k e^{i \vec k \cdot \vec x} + a^\dagger_k e^{-i \vec k \cdot \vec x})|0\rangle$
$= \int \frac{d^3 k}{(2 \pi)^3} \sqrt{\frac{\omega_p}{\omega_k}} (e^{i \vec k \cdot \vec x} \langle 0|a_p a_k|0\rangle + e^{-i \vec k \cdot \vec x} \langle 0|a_p a^\dagger_k|0\rangle)$
$= e^{-i \vec p \cdot \vec x}$ 
This is the same as the projection of a position state on a momentum state in one-particle quantum mechanics:
$\langle \vec p| \vec x \rangle = e^{-i \vec p \cdot \vec x}$ 
Therefore $\phi_0(\vec x) |0\rangle = | \vec x \rangle$, that is $\phi_0(\vec x)$ creates a particle at position $\vec x$.
