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I am reading Greiner's "Quantum Electrodynamics". In example 1.5 he derives the Green's function for diffusion. I am stuck on a step in the derivation.

He has the defining differential equation as $$ {\bf \nabla}^\prime G-a^2{\partial G \over \partial t^\prime}=-4\pi\delta^3({\bf x}^\prime-{\bf x})\delta(t^\prime-t) \label{eq4} \tag{1} $$ where $a$ is a constant. He then defines \begin{align} \tau&=t^\prime-t\\ {\bf R}&={\bf x}^\prime-{\bf x}. \tag{2} \label{eq5} \end{align} The Fourier transform of the Green's function is given as $$ G({\bf x}^\prime,t^\prime,x,t)={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot({\bf x}^\prime-{\bf x})} g({\bf p},\tau)\, . \tag{3} \label{eq6} $$ We insert Eq. (\ref{eq6}) into the left hand side of Eq. (\ref{eq4}) to get $$ {\bf \nabla}^\prime G-a^2{\partial G \over \partial \tau}={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}} \left(-p^2 g-a^2{\partial g \over \partial \tau}\right)\, . \tag{4} \label{eq8} $$ Using, $$ \delta^3({\bf R})={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}} \tag{5} \label{eq7} $$ this results in a differential equation for $g$: $$ a^2{\partial g\over \partial \tau}+p^2 g=4\pi\delta(\tau)\,. \label{eq9} \tag{6} $$ Greiner then claims that this equation has the following solution: $$ g={4\pi \over a^2}e^{-(p^2\tau/a^2)}\Theta(\tau) \label{eq10} \tag{7} $$ To prove this he substitutes Eq. (\ref{eq10}) into Eq. (\ref{eq9}) and uses the relation $$ { {\rm d}\Theta \over {\rm d} \tau}=\delta(\tau) \label{eq11} \tag{8} $$ to get $$ 4\pi e^{-p^2\tau/a^2}{{\rm d} \Theta \over {\rm d} \tau}=4\pi\delta(\tau) \,. \label{eq12} \tag{9} $$

As far as I can see Eqs. (\ref{eq12}) and (\ref{eq11}) imply that $e^{-p^2\tau/a^2}=1$. But I don't think this is correct. My question is what am I missing or is this an error in the textbook?

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    $\begingroup$ According to the definition of the delta function both sides of eq. (9) are zero for $\tau\neq 0$. In case that $\tau=0$ the expression $e^{-p^2\tau/a^2}=1$. So there is no contradiction. $\endgroup$ – Frederic Thomas Jan 22 at 10:49
  • $\begingroup$ @FredericThomas Thanks, I think that is the correct answer. $\endgroup$ – Virgo Jan 22 at 17:20
  • $\begingroup$ @FredericThomas I think your proof is incorrect since is based on 1$\cdot$singularity$\boldsymbol{=}$singularity. The equality of two singularities has no sense. $\endgroup$ – Frobenius Jan 23 at 1:06
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Let a real function $\;f(x)\;$ of the real variable $\;x\in\mathbb{R}\;$ for which \begin{align} f(x)\boldsymbol{=}0 \quad & \text{for any} \quad x\boldsymbol{\ne} x_{0} \quad \textbf{and} \tag{01a}\label{01a}\\ \int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{01b}\label{01b} \end{align} Under these conditions it seems that this function is not well-defined at $\;x_{0}$, may be because of a singularity at this point. But we have good reasons to $^{\prime\prime}$believe$^{\prime\prime}$ that
\begin{equation} f(x)\boldsymbol{\equiv}\delta\left(x\boldsymbol{-}x_{0}\right) \tag{02}\label{02} \end{equation} since equations \eqref{01a},\eqref{01b} remind us the defining properties of Dirac delta function on the real axis $\mathbb{R}$.

Now let the function \begin{equation} f(\tau)\boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\,, \quad b\in\mathbb{R} \tag{03}\label{03} \end{equation} We have \begin{align} f(\tau)\boldsymbol{=}e^{\boldsymbol{-}b^2 \boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\boldsymbol{=}0 \quad & \text{for any} \quad \tau\boldsymbol{\ne} 0 \quad \textbf{and} \tag{04a}\label{04a}\\ \int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}f(\tau)\mathrm d\tau\boldsymbol{=}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}e^{\boldsymbol{-}b^2 \boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\mathrm d\tau\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{04b}\label{04b} \end{align} Equation \eqref{04b} is proved by integrating by parts \begin{align} \int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\mathrm d \Theta & \boldsymbol{=}\left[e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\Theta\right]_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\boldsymbol{-}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\Theta\,\mathrm d \left(e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right) \nonumber\\ & \boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\varepsilon}}\boldsymbol{-}\int\limits_{0}^{\boldsymbol{+\varepsilon}}\mathrm d \left(e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right) \boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\varepsilon}}\boldsymbol{-}\left[e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right]_{0}^{\boldsymbol{\varepsilon}} \boldsymbol{=}1 \tag{05}\label{05} \end{align} So \begin{equation} \boxed{\:\: e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\boldsymbol{=}\delta\left(\tau\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \quad b\in\mathbb{R} \tag{06}\label{06} \end{equation}

$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

$\textbf{NOTE :}$

We treat the 3-dimensional case on the same reasoning :

So, let a real function $\;F(\mathbf{r})\;$ of the vector variable $\;\mathbf{r}\in\mathbb{R}^{\bf 3}\;$ for which \begin{align} F(\mathbf{r})\boldsymbol{=}0 \quad & \text{for any} \quad \mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0} \quad \textbf{and} \tag{n-01a}\label{n-01a}\\ \iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}F(\mathbf{r})\mathrm d^{\bf 3}\mathbf{r}\boldsymbol{=}1\quad & \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0 \tag{n-01b}\label{n-01b} \end{align} where $\;\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$ a ball with center at $\;\mathbf{r}_{0}\;$ and radius $\;\boldsymbol{\varepsilon}$.

Under these conditions it seems that this function is not well-defined at $\;\mathbf{r}_{0}$, may be because of a singularity at this point. But we have good reasons to $^{\prime\prime}$believe$^{\prime\prime}$ that
\begin{equation} F(\mathbf{r})\boldsymbol{\equiv}\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right) \tag{n-02}\label{n-02} \end{equation} since equations \eqref{n-01a},\eqref{n-01b} remind us the defining properties of Dirac delta function in the real space $\;\mathbb{R}^{\bf 3}$.

An example is the representation of the Laplacian of $\;1/\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert\;$ by Dirac delta function \begin{equation} \boxed{\:\: \nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}\boldsymbol{-}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tag{n-03}\label{n-03} \end{equation} useful in Electrostatics.

For a proof of \eqref{n-03} see my answer therein : Divergence of Electric Field Due to a Point Charge.

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