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Quick correction from my previous derivation: Sign error by me, such that ##\omega = \frac{\hat{H}}{\hbar}##, which for natural unts, ##\omega = \hat{H}##.I will go through the derivation.Equation 17.21:
\begin{equation}
\begin{split}
| \psi_t \rangle &= e^{-\frac{i}{\hbar} \hat{H} (t-\acute{t})} | \psi_{\acute{t}} \rangle
\end{split}
\end{equation}Given:
\begin{equation}
\begin{split}
\omega &= \frac{\hat{H}}{\hbar}
\end{split}
\end{equation}\begin{equation}
\begin{split}
| \psi_t \rangle &= e^{-i \omega (t-\acute{t})} | \psi_{\acute{t}} \rangle
\end{split}
\end{equation}
Which is my version of the time evolution operator, ##\hat{U}(t,\acute{t})##.Equation 17.26:
\begin{equation}
\begin{split}
\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\delta(t-\acute{t})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
G_j^{+}(t,\acute{t})&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) \left(-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)
\\
&=\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)
\end{split}
\end{equation}And:
\begin{equation}
\begin{split}
\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=\delta(t-\acute{t})
\\
-i\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\\
\left(-i \frac{\partial}{\partial t}- \frac{E_j}{\hbar} \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\end{split}
\end{equation}
Given ##\omega_j=\frac{E_j}{\hbar}## and ##\partial_t = \frac{\partial}{\partial t}##:
\begin{equation}
\begin{split}
\left(-i \partial_t- \omega_j \right ) \left( e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t})&=G_j^{+}(t,\acute{t})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\left(-i \partial_t- \omega_j \right ) G_j^{+}(t,\acute{t})&=-i \delta(t-\acute{t})
\end{split}
\end{equation}Again, this is the same as my derivation.Equation 17.28:
\begin{equation}
\begin{split}
G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j e^{-\frac{i}{\hbar}E_j (t-\acute{t})}\theta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\end{split}
\end{equation}Equation 17.31:
\begin{equation}
\begin{split}
\left(i\hbar \frac{\partial}{\partial t} -\hat{H} \right) G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j \left(i\hbar \frac{\partial}{\partial t} -E_j \right) G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\sum_j \delta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\delta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=\delta^4(\vec{x}-\vec{\acute{x}})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
\sum_j u_j(\vec{r}) u_j^*(\vec{\acute{r}})&=\delta^3(\vec{r}-\vec{\acute{r}})
\end{split}
\end{equation}And:
\begin{equation}
\begin{split}
G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=\sum_j G_j^+(\vec{r},t;\vec{\acute{r}},\acute{t})
\\
&=\sum_j G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\sum_j G_j^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=G^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=-\frac{i}{\hbar} e^{-i \omega (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\end{split}
\end{equation}Which is the SAME expression for the Green function that I derived (assuming some algebra with ##i## and ##\hbar##, which I demonstrate above). In other words, the result is the same. Trying to distinguish my derivation from the one in the paper by asserting one involves position and the other energy in order to accept one and not the other is simply not tenable.
\begin{equation}
\begin{split}
| \psi_t \rangle &= e^{-\frac{i}{\hbar} \hat{H} (t-\acute{t})} | \psi_{\acute{t}} \rangle
\end{split}
\end{equation}Given:
\begin{equation}
\begin{split}
\omega &= \frac{\hat{H}}{\hbar}
\end{split}
\end{equation}\begin{equation}
\begin{split}
| \psi_t \rangle &= e^{-i \omega (t-\acute{t})} | \psi_{\acute{t}} \rangle
\end{split}
\end{equation}
Which is my version of the time evolution operator, ##\hat{U}(t,\acute{t})##.Equation 17.26:
\begin{equation}
\begin{split}
\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\delta(t-\acute{t})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
G_j^{+}(t,\acute{t})&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) G_j^{+}(t,\acute{t}) &=\left( i\hbar \frac{\partial}{\partial t}-E_j \right ) \left(-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)
\\
&=\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)
\end{split}
\end{equation}And:
\begin{equation}
\begin{split}
\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=\delta(t-\acute{t})
\\
-i\left( \frac{\partial}{\partial t}+ \frac{i}{\hbar}E_j \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\\
\left(-i \frac{\partial}{\partial t}- \frac{E_j}{\hbar} \right ) \left( e^{-\frac{i}{\hbar} E_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\end{split}
\end{equation}
Given ##\omega_j=\frac{E_j}{\hbar}## and ##\partial_t = \frac{\partial}{\partial t}##:
\begin{equation}
\begin{split}
\left(-i \partial_t- \omega_j \right ) \left( e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t}) \right)&=-i \delta(t-\acute{t})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
e^{-i \omega_j (t-\acute{t})} \theta(t-\acute{t})&=G_j^{+}(t,\acute{t})
\end{split}
\end{equation}Such that:
\begin{equation}
\begin{split}
\left(-i \partial_t- \omega_j \right ) G_j^{+}(t,\acute{t})&=-i \delta(t-\acute{t})
\end{split}
\end{equation}Again, this is the same as my derivation.Equation 17.28:
\begin{equation}
\begin{split}
G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j e^{-\frac{i}{\hbar}E_j (t-\acute{t})}\theta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\end{split}
\end{equation}Equation 17.31:
\begin{equation}
\begin{split}
\left(i\hbar \frac{\partial}{\partial t} -\hat{H} \right) G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=-\frac{i}{\hbar}\sum_j \left(i\hbar \frac{\partial}{\partial t} -E_j \right) G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\sum_j \delta(t-\acute{t})u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\delta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=\delta^4(\vec{x}-\vec{\acute{x}})
\end{split}
\end{equation}Where:
\begin{equation}
\begin{split}
\sum_j u_j(\vec{r}) u_j^*(\vec{\acute{r}})&=\delta^3(\vec{r}-\vec{\acute{r}})
\end{split}
\end{equation}And:
\begin{equation}
\begin{split}
G^+(\vec{r},t;\vec{\acute{r}},\acute{t})&=\sum_j G_j^+(\vec{r},t;\vec{\acute{r}},\acute{t})
\\
&=\sum_j G_j^+(t-\acute{t}) u_j(\vec{r}) u_j^*(\vec{\acute{r}})
\\
&=\sum_j G_j^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=G^+(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=-\frac{i}{\hbar} e^{-\frac{i}{\hbar} E (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\\
&=-\frac{i}{\hbar} e^{-i \omega (t-\acute{t})} \theta(t-\acute{t})\delta^3(\vec{r}-\vec{\acute{r}})
\end{split}
\end{equation}Which is the SAME expression for the Green function that I derived (assuming some algebra with ##i## and ##\hbar##, which I demonstrate above). In other words, the result is the same. Trying to distinguish my derivation from the one in the paper by asserting one involves position and the other energy in order to accept one and not the other is simply not tenable.