Understanding propagator nature of QM Green's function I'm trying to understand the many-body Green's functions, but first I want to understand Greens functions in QM.
I'm reading this article, but I'm having trouble with eq. 17.
The equation states:
$$
\Psi(\mathbf r, t) = \int d^3\mathbf r' G(\mathbf r, t;\mathbf r', t')\Psi(\mathbf r', t') \tag{1}
$$
(First question: Why do we integrate over $\mathbf r'$, but not $t'$?)
But $(1)$ does not follow from the definition of $G$, since the definition of $G$ is
$$
\left[ i\hbar \partial_t + \frac{\hbar^2}{2m}\nabla^2 \right]
G(\mathbf r, t;\mathbf r', t') = \delta(\mathbf r-\mathbf r')\delta(t-t')
\tag{2}
$$
and thus 
$$
\begin{align*}
\Psi(\mathbf r, t) &= \int d^3\mathbf r'dt' \delta(\mathbf r-\mathbf r')\delta(t-t') \Psi(\mathbf r', t')\\
&= \int d^3\mathbf r'dt'
\left[ i\hbar \partial_t + \frac{\hbar^2}{2m}\nabla^2 \right]
G(\mathbf r, t;\mathbf r', t') \Psi(\mathbf r', t')\\
&= \left[ i\hbar \partial_t + \frac{\hbar^2}{2m}\nabla^2 \right]
\int d^3\mathbf r'dt' G(\mathbf r, t;\mathbf r', t') \Psi(\mathbf r', t').
\end{align*}
$$
Now if I suppose, that $(1)$ is true, then
$$
\begin{align*}
\Psi(\mathbf r, t) &= 
\left[ i\hbar \partial_t + \frac{\hbar^2}{2m}\nabla^2 \right]
\int dt'\Psi(\mathbf r, t)
\end{align*}
$$
which is nonsense, because $\displaystyle{\int dt' = \infty}$.
Also, because by definition,
$$
\left[ i\hbar \partial_t + \frac{\hbar^2}{2m}\nabla^2 \right]\Psi(\mathbf r, t)=V(\mathbf r, t)\Psi(\mathbf r, t).
$$
Question: How can I prove $(1)$ from the definition of $G$?
 A: The reason why there is no integral over $t'$ is that your first equation is actually equivalent to
    $$|{\psi(t)}\rangle=U(t,t')|{\psi(t')}\rangle,\quad\forall t>t'$$
where $U(t,t')$ is the evolution operator. Projecting out on $|{\vec r}\rangle$
$$\langle{\vec r}|{\psi(t)}\rangle=\langle{\vec r}|U(t,t')|{\psi(t')}\rangle
=\int \langle{\vec r}|U(t,t')|{\vec r'}\rangle \langle{\vec r'}|{\psi(t')}\rangle d^3\vec r'$$
To enforce the constrain $t>t'$, one can choose
    $$G(\vec r,t;\vec r',t')= \langle{\vec r}|U(t,t')|{\vec r'}\rangle\theta(t-t')$$
where $\theta$ is the Heaviside function. To proove (2) from (1), take the time-derivative of this relation
    $$\eqalign{
{\partial\over\partial t}G(\vec r,t;\vec r',t')
&=\langle{\vec r}|{\partial\over\partial t}U(t,t')|{\vec r'}\theta(t-t')
+\langle{\vec r}|U(t,t')|{\vec r'}\rangle{\partial\over\partial t}\theta(t-t')\cr
&=-{i\over\hbar}\langle{\vec r}|HU(t,t')|{\vec r'}\rangle\theta(t-t')
+\langle{\vec r}|U(t,t')|{\vec r'}\rangle\delta(t-t')\cr
}$$
using again the fact that $U(t,t')=e^{-iH(t-t')/\hbar}$ and that the derivative of the Heaviside function is the Dirac distribution. Since the last term imposes $t=t'$, the last term can be written
    $$ \langle{\vec r}|U(t,t')|{\vec r'}\rangle\delta(t-t')
=\langle{\vec r}|\mathbb{I}|{\vec r'}\rangle\delta(t-t')
=\delta(\vec r-\vec r')\delta(t-t')$$
while the first becomes for a Hamiltonian $H={p^2\over 2m}+V(\vec r)$
$$\eqalign{
\langle{\vec r}|HU(t,t')|{\vec r'}\rangle
&=\int \langle{\vec r}|H|\vec r''\rangle
\langle{\vec r''}|U(t,t')|{\vec r'}\rangle d^3\vec r''\cr
&=\int \langle{\vec r}|{p^2\over 2m}+V|{\vec r''}\rangle
G(\vec r'',t,\vec r',t')d^3\vec r''\cr
&=\left(-{\hbar^2\over 2m}\Delta+V(\vec r)\right)
G(\vec r,t,\vec r',t')
}$$
Putting everything together gives
    $${\partial\over\partial t}G(\vec r,t;\vec r',t')
=-{i\over\hbar}\left(-{\hbar^2\over 2m}\Delta+V(\vec r)\right)
G(\vec r,t,\vec r',t')+\delta(\vec r-\vec r')\delta(t-t')$$
