Interpretation of the propagator In quantum mechanics, it is clear that $\langle x|y\rangle = 0$ for $x\ne y$, where $|x\rangle$ is the state with the particle at position $x$. (Notice that this $|x\rangle$ is different from the usual meaning in most textbooks. In this interpretation, we normalized it such that $\langle x | x \rangle=1$.) However, I am a bit confused about this picture in the relativistic quantum field theory. 
The toy example I was playing with is the one particle states in the massive bosonic free field. Assume our Hilbert space at time $0$ to be $\mathcal{H} = \text{span} \{|x\rangle|x^0=0\}$, where $|x\rangle$ represents the state where the  particle is at $x$.
One can find $\langle x |y\rangle=D(x-y)=\int \frac{d^Dk}{(2\pi)^D} \frac{e^{ik(x-y)}}{k^2-m^2+i\epsilon}$ in most textbooks. And after some computation, one can see that $D(0)=\infty$ and $D(x)\ne 0$ even if $x\ne0$ (i.e. when they are spacially seperated). I found it confusing when we put these facts together $\langle x |y\rangle=D(x-y)$:


*

*It is not normalized to $1$ when $x=y$.

*Suppose normalization is not a problem, and indeed $\langle x|y\rangle \ne 0$ for $x\ne y$, how should we understand this difference from quantum mechanics? Does this come from the relativistic effect? 
I am quite interested/surprised/worried about the behavior in 2 since if this were true, it will be somewhat ambiguous to write $|x\rangle$ in the sense that it is also a linear combination of particles at other positions.
edit:
I feel there is a point that is not yet explained in the answers. I will try to show the point using the following question. 
Question: Should $$\int|x,t=0\rangle \langle x,t=0| d^Dx = \text{some const} \cdot I \tag{1}$$  (say we are in D+1 dimension)
This is true in the context of QM and is also true for coherent states $|p,x\rangle$ in a harmonic oscillator.
I believe the above equation is correct because of its symmetry. However, I haven't found a way to prove it and here is a counter-argument. 
$$\begin{align} D(a-b) \sim \langle a|b \rangle &\sim \langle a|\int dx |x\rangle\langle x| b \rangle \\ &\sim \int dx D(a-x)D(x-b), \end{align}$$ which isn't the case for massive field in 1+1.
(Given this, one can try to replace the equation with $\int dx dy A(x,y)|x\rangle\langle y|=I$ for some non diagnal $A$, yet I have no idea why this is necessary and I don't have an explicit construction.)
Despite of that, if we assume the above identity (1) is correct, the constant is determined by computing $$\begin{align}c &=\langle x=0, t=0|c \cdot I|x=0, t=0 \rangle \\ &= \int d^Dx \langle x=0, t=0|x,t=0\rangle \langle x,t=0|x=0, t=0 \rangle \\ &= \int d^Dx D(x) D(-x)\end{align}.$$
Although the identity looks reasonable, it has a nonintuitive consequence:
$$|x=0,t=0\rangle = \int dx |x,t=0\rangle D(x),$$
which is the problem I was worried about.
(One possible bonus of this viewpoint is that the additional linear dependency of the state vectors explains the appearance of Bekenstein bound.)
 A: I think you are confusing the projection of a position quantum state $\vert y \rangle$ against another state $\vert x \rangle$ with the probability amplitude of the transition between different quantum states.  
In the former, the projection is $\langle x \vert y \rangle$ and you have $\langle x \vert y \rangle = 1$ if $y = x$ (we assume the position quantum states are normalized), or $\langle x \vert y \rangle = 0$ if $y \ne x$.  
In the latter, the probability amplitude of the transition between different position quantum states is $\langle x \vert e^{-iHt/\hbar} y \rangle$, in the Schroedinger picture. Even if $y \ne x$ not necessarily it is zero, as it depends on the Hamiltonian $H$ which describes the evolution of the initial state.  
When it comes to the quantum field theory, even if $y \ne x$ the Feynman propagator $\Delta (x - y)$ is not zero either, as the initial state evolves in time.  
Note: A good reference is Srednicki "Quantum field theory", Section 8 -The path integral for free-field theory-.
