# 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 $$G$$?

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 $${\partial\over\partial t}G(\vec r,t;\vec r',t') =-{i\over\hbar}\left(-{\hbar^2\over 2m}\Delta+V(\vec r)\right) G(\vec r,t,\vec r',t')+\delta(\vec r-\vec r')\delta(t-t')$$