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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 $G$?

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

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