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I am learning some basics of Green's functions applied in physics from the article https://arxiv.org/abs/1604.02499

I am struck at equation no (23) which is said to be derived from equation (22) by taking Fourier transform of it. After several attempts, I am unable to find the exact result given in equation (23). It might be pretty elementary to ask at this stage, but being a beginner in this subject I am not able to find any answer.

What am I missing ? What's the trick to achieve equation (23) from (22) ?

\begin{equation} G(\mathbf{r}, t; \mathbf{r}^\prime, t^\prime) = \sum_n \varphi_n(\mathbf{r})\varphi_n^*(\mathbf{r}^\prime)e^{-\frac{i}{\hbar}E_n(t-t^\prime)}\tag{22} \end{equation}

\begin{equation} G\left(\mathbf{r}, \mathbf{r}^\prime; E \right) = \sum_n i \frac{\varphi_n(\mathbf{r})\varphi_n^*(\mathbf{r}^\prime)}{E-E_n}\tag{23} \end{equation}

If I use the following: \begin{align} G\left(\mathbf{r}, \mathbf{r}^\prime; E \right) = \int_0^\infty G(t) e^{{i\over\hbar}Et} dt \end{align} it leads to, \begin{align} G\left(\mathbf{r}, \mathbf{r}^\prime; E \right) = {i\hbar} \sum_n \frac{\varphi_n(\mathbf{r})\varphi_n^*(\mathbf{r}^\prime)}{E-E_n} \end{align} Surely, provided there exists a transformation (might be called a Laplace-Fourier transform) of the following kind, I get the answer. \begin{align} G(E) = {1\over \hbar} \int_0^\infty G(t) e^{{i\over\hbar}Et} dt \end{align}

Thus, my questions remain:

(a) Does such a transformation exist ?

(b) Should not we consider the limit $(-\infty, \infty )$ instead of $(0, \infty )$ ?

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  • $\begingroup$ Isn't it basically a mathematics question? $\endgroup$ – SRS Jun 30 '18 at 13:54
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In the article you quote, have a look at footnote 4. It explains that an ordering should be imposed on $t$ and $t'$, which effectively will lead you to integrate only on $(0,\infty)$.

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