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). 669
I found it confusing when we put these facts together $\langle x |y\rangle=D(x-y)$:
1. It is not normalized to $1$ when $x=y$.
2. 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 <PERSON> bound.) 669