Physical Interpretation of Diagonal Green's Function In the context of one-particle non-relativstic quantum mechanics, let $H$ be the Hamiltonian for the system and let $z\in\mathbb{C}$ be given.
What is the physical meaning of the following object:
$$ \left|\langle x|\left(H-z\right)^{-1}|x\rangle \right|$$
Apparently it is a transition probability, so it should correspond to some sort of probability of going from one state to another?
Intuitively I feel like it should somehow correspond to the number of times the particle revisits the point $x$, or the likelihood for that to happen?
Any formal answer would be greatly appreciated.
 A: There is one physical problem where this quantity appears directly. Consider the Schrodinger equation with a delta-functional source of a particle at $x=x_0$, $t=t_0$:
$$
\qquad\qquad\qquad\qquad\qquad i\frac{\partial}{\partial t}\psi=H\psi+\delta(x-x_0)\delta(t-t_0).\qquad\qquad\qquad\qquad\qquad(1)
$$
As known, its solution can be expressed through the Green function:
$$
\psi(x,t)=\langle x|G(t-t_0)|x_0\rangle,
$$
where 
$$
G(E)=(E-H)^{-1},\qquad G(t)=\int\frac{dE}{2\pi}\,G(E)e^{-iE t}.
$$
Now introduce the source of wave of frequency $\omega$ at the point $x_0$:
$$
\qquad\qquad\qquad\qquad\qquad i\frac{\partial}{\partial t}\psi=H\psi+\delta(x-x_0)e^{-i\omega t}.\qquad\qquad\qquad\qquad\qquad (3)
$$
Noting that $e^{-i\omega t}=\int dt_0\:e^{-i\omega t_0}\delta(t-t_0)$, we get the superposition of previous solutions:
$$
\psi(x,t)=\int dt_0\:\langle x|G(t-t_0)|x_0\rangle e^{-i\omega t_0}=\{t-t_0=\tau\}=e^{-i\omega t}\int d\tau\:\langle x|G(\tau)|x_0\rangle e^{i\omega\tau}.
$$
Now look at your formula:
$$
\left\langle x\left|\frac1{H-z}\right|x\right\rangle=-G(z)=-\int d\tau\:\langle x|G(\tau)|x\rangle e^{iz\tau}.
$$
Comparing the formulas, we get
$$
\psi(x_0,t)=-e^{-i\omega t}\left\langle x_0\left|\frac1{H-\omega}\right|x_0\right\rangle.
$$
Thus the answer is: if we have a source of quantum wave of a frequency $z$ and unit amplitude at the point $x$ then $|\langle x|(H-z)^{-1}|x\rangle|$ would be the full amplitude of quantum wave at the same point $x$. As concerns complex values of $z$, we need to be cautious with causality of the Green function (I ignored this issue here).
Maybe another formula can be helpful: given stationary states of the Hamiltonian $H|\varphi_i\rangle=E_i|\varphi_i\rangle$, we can relate
$$
\left\langle x\left|\frac1{H-z}\right|x\right\rangle=\sum_i\frac{|\varphi_i(x)|^2}{E_i-z}
$$
to the local density of states
$$
D(E,x)=\langle x|\delta(E-H)|x\rangle=\sum_i\delta(E-E_i)|\varphi_i(x)|^2
$$
in the following way:
$$
\left\langle x\left|\frac1{H-z}\right|x\right\rangle=\int dE\:\frac{D(E,x)}{E-z}.
$$
At every $z=E_i$ the function $\langle x|(H-z)^{-1}|x\rangle$ has the pole of the residue $|\varphi_i(x)|^2$.
UPDATE
Physical interpretation of the Schrodinger equation (1) with a source term $\propto\delta(t-t_0)$ is the following. Consider
$$
\qquad\qquad\qquad\qquad\qquad i\frac{\partial}{\partial t}|\psi\rangle=H|\psi\rangle+\delta(t-t_0)|\psi_0\rangle,\qquad\qquad\qquad\qquad\qquad(2)
$$
where $|\psi\rangle$ is expanded in the eigenvectors of the Hamiltonian:
$$
|\psi\rangle=\sum_ic_i|\varphi_i\rangle,\qquad H|\varphi_i\rangle=E_i|\varphi_i\rangle.
$$
Then for each coefficient $c_i(t)$ we get
$$
i\frac{dc_i}{dt}=E_ic_i+\delta(t-t_0)\langle\varphi_i|\psi_0\rangle.
$$
Using the initial condition $c_i(-\infty)=0$, we have
$$
c_i(t)=\left\{\begin{array}{ll}0,&t<t_0,\\
-i\langle\varphi_i|\psi_0\rangle e^{-iE_i(t-t_0)},&t>t_0.\end{array}\right.
$$
The same solution (except the phase factor) can be obtained from the following equation for $c_i(t)$ without the source term, but with nontrivial initial condition:
$$
\left\{\begin{array}{l}i\frac{dc_i}{dt}=E_ic_i,\\
c_i(t_0)=\langle\varphi_i|\psi_0\rangle.\end{array}\right.\quad\Rightarrow\qquad c_i(t)=\langle\varphi_i|\psi_0\rangle e^{-iE_i(t-t_0)}.
$$
Thus the Schrodinger equation (2) with the source term $\delta(t-t_0)|\psi_0\rangle$ is equivalent to that without source term, but with initial condition $|\psi(t_0)\rangle=-i|\psi_0\rangle$. In particular, Eq. (1) describes wave function propagation beginning from the state $\psi(x_0,t_0)=-i\delta(x-x_0)$, i.e. it is the Green function (up to a phase).
Now about oscillating source term in Eq. (3). Unfortunately, it cannot be interpreted as an initial condition for $\psi$, but we can consider the system if two coupled equations
$$
\left\{\begin{array}{l}i\frac{\partial}{\partial t}\psi_1=H_1\psi_1+\lambda\psi_2,\\i\frac{\partial}{\partial t}\psi_2=H_2\psi_2+\lambda\psi_1,\end{array}\right.
$$
which can describe physically a motion of a particle in two regions of space $R_1$, $R_2$, where
$$
\psi(\mathbf{r},t)=\left\{\begin{array}{ll}\psi_1(\mathbf{r},t),&\mathbf{r}\in R_1,\\
\psi_2(\mathbf{r},t),&\mathbf{r}\in R_2,\end{array}\right.
\qquad H(\mathbf{r})=\left\{\begin{array}{ll}H_1(\mathbf{r}),&\mathbf{r}\in R_1,\\
H_2(\mathbf{r}),&\mathbf{r}\in R_2.\end{array}\right.
$$
The small coupling term $\lambda$ can be responsible for a tunneling between the two regions. So the term $\lambda\psi_2$ acts as a wave function source for a subsystem 1. See the example on the picture: an electron with the energy $\omega$ enters the nanostructure (subsystem 1) from the tip (subsystem 2) at the point $x_0$. It creates the source term $\propto \delta(x-x_0)e^{-i\omega t}$ in the Schrodinger equation for subsystem 1.

