Simple Green's function question: Propagator for stationary particle? Suppose the probability a particle transitions into a state of interest at time $t$ having position $x$ is 
$$\omega(x,t).$$
Once a particle enters this state it does not leave it, nor does its position change within it.
I would like to express the probability that a particle is found in this state of interest at position $x$ and time $t$. This involves the propagator or Green's function $g(x,t)$:
$$ p(x,t) = \int_{-\infty}^x dx' \int_{-\infty}^t dt' \omega(x',t') g(x-x',t-t').$$
What Green's function represents the phenomenon that particles in the state of interest stay in the same location for all times? Is it $g(x,t) = \delta(x)\theta(t)$? I believe this implies equal amplitude to be found at $x$ at any time past arrival into the state at $t$, but I'm confused because $\int_{-\infty}^\infty dt' \int_{-\infty}^\infty dx' g(x',t') \neq 1$.  Can anyone clarify this issue for me? 
 A: Let me attempt to answer what I think is the question. You are asking yourself for the probability of finding the particle at some spatial region $(x_0,x_1)$ (points are not really well defined due to uncertainty principle) for $t>t_0$. If you are given the probability density function $\omega(t,x)$, (I am thinking of it as the modulus square of its wave-function in QM), then the integral you are dealing with for a fixed time $t$ would be 
$$P(t,x\in(x_0,x_1)) = \int_{x_0}^{x_1} {\rm d}x\; \omega(t,x) \int_{x_0}^{x_1} {\rm d}x\; |\psi(t,x)|^2$$
however handling time regions will be trickier, are you observing the state somehow at every time? How do you know it has been in this region in the past? In the situations I know the dynamics are not up to us, the particle follows some evolution according to the model at hand. In the simplest case according to some Hamiltonian, this is what determines the propagator, it doesn't correspond to some state that you pick. This information is already encoded in the probability density function (if it is already a function of time). 
The propagator $K(t,t';x,x')$ is nothing more than the Green's function to your equation of motion.  In terms of wave-functions in the non-relativistic case you use them to evolve a known state like this
$$ \psi(t,x) = \int_{-\infty}^\infty {\rm d}x'\; \psi(t',x')K(t,t';x,x')\tag{1}\label{eq:evol}$$
with the interpretation that a particle with wave-function $\psi(t',x')$ at time $t'$, (notice here, $\psi$ for a fixed $t'$ has the information on all $x$'s) will evolve according to your physics into some wave-function $\psi(t,x)$ at time $t$. So $t$ is just parametrizing the states (wave-functions if you want) along which your initial state will be driven by unobserved evolution. Let's say you are now interested in a specific configuration at some time $t_0$ as above, which you want it to be somehow at rest after some $t_0$ so you would say
$$\eta(t,x) = \Theta(t-t_0)\delta(x-x^*)$$
which is constant in space roughly (exaggerating and placing the particle at a point $x^*$) and integrates to 1, (but probably doesn't satisfy its Schrödinger equation, ignoring that...) you can know ask for probability of achieving such configuration by taking the inner product of the wave-function of the system with this special state:
$$P_{\psi\rightarrow\eta}(t) = |\langle \eta|\psi\rangle|^2 = \left(\int_{-\infty}^\infty {\rm d}x\; \eta^*(t,x)\psi(t,x)\right)^2$$
you can then use $\eqref{eq:evol}$ to do the computation. But here I have always fixed the time since one is focused on a state/wave-function as a function of space. If you want several times, perhaps the proper way would be to consider conditional probabilities,
$$P(t\in T | x\in X) = \frac{P(t\in T, x\in X)}{P(x \in X)}$$
where $X$ is the region $(x_0,x_1)$ in our case and $T$ could be $t>t_0$. I hope this clarifies things a bit. Hopefully you can find analogues to relativistic cases if that is what you are after.
