I'm trying to do a homework problem where they have asked me to recover the usual coordinate space momentum operator from its Hilbert space equivalent. It requires you to show (which I have done):
$$\langle \vec{r}|\hat{p}|\vec{r}'\rangle=i\hbar\nabla_{r'}\delta(r-r')$$
The problem is that now I have done this my professor has listed that to get the proper coordinate space operator I need to put it into an equation which is
$$\psi_b(r)=\int_{-\infty}^\infty i\hbar\nabla_{r'}\delta(\vec{r}-\vec{r'})\psi_a(\vec{r'})~d^3r'$$
I'm up to the last line in the solutions my professor has given me but I don't know how he has gone from what you see previously that last equation there to
$$\psi_b(r)=-i\hbar\nabla_r\psi_a(\vec{r})$$ notice there is no primes in this one.
I know that the derivative of the Dirac delta would give me the negative in front. But it just seems like my teacher has just ignored that the del has been used on the Dirac delta, and doesn't seem like it has been used properly. Why didn't he use the product rule? Why didn't he differentiate both the $\psi(r)$ and the Dirac delta?
Any help is appreciated.
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\bra}[1]{\left\langle #1\right\vert} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\eofm}[3]{\left\langle #1\left\vert #2\right\vert #3\right\rangle} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align}\color{#c00000}{\eofm{\vec{r}}{\hat{p}}{\vec{r}\,'}}& =\bra{\vec{r}}\hat{p} \int\dd^{3}\vec{p}\,\ket{\vec{p}}\bra{\vec{p}}\left.\vec{r}\,'\right\rangle =\int\dd^{3}\vec{p}\,\eofm{\vec{r}}{\hat{p}} {\vec{p}}\bra{\vec{p}}\left.\vec{r}\,'\right\rangle =\int\dd^{3}\vec{p}\,\vec{p}\bra{\vec{r}} \left.\vec{p}\right\rangle\bra{\vec{p}}\left.\vec{r}\,'\right\rangle \\[3mm]&=\int\dd^{3}\vec{p}\,\vec{p}\, {\expo{\ic\vec{p}\cdot\vec{r}/\hbar} \over \pars{2\pi\hbar}^{3/2}} {\expo{-\ic\vec{p}\cdot\vec{r}\,'/\hbar} \over \pars{2\pi\hbar}^{3/2}} =\int{\dd^{3}\vec{p} \over \pars{2\pi\hbar}^{3}}\,\vec{p}\, \expo{\ic\vec{p}\cdot\pars{\vec{r} - \vec{r}\,'}/\hbar} \\[3mm]&=\hbar\int{\dd^{3}\vec{p} \over \pars{2\pi}^{3}}\,\vec{p}\, \expo{\ic\vec{p}\cdot\pars{\vec{r} - \vec{r}\,'}} =-\ic\hbar\nabla_{\vec{r}}\ \overbrace{\int{\dd^{3}\vec{p} \over \pars{2\pi}^{3}}\, \expo{\ic\vec{p}\cdot\pars{\vec{r} - \vec{r}\,'}}} ^{\ds{=\ \delta\pars{\vec{r} - \vec{r}\,'}}} \end{align}
$$ \mbox{Note that}\quad \nabla_{\vec{r}}\delta\pars{\vec{r} - \vec{r}\,'} =-\nabla_{\vec{r}\,'}\delta\pars{\vec{r} - \vec{r}\,'} $$ such that \begin{align} \color{#c00000}{\psi_{b}\pars{\vec{r}}}& =\int_{-\infty}^{\infty}\ic\hbar\bracks{\nabla_{\vec{r}\,'} \delta(\vec{r} - \vec{r}\,')}\psi_{a}\pars{\vec{r}\,'}\,\dd^{3}\vec{r}\,' =\int_{-\infty}^{\infty}\ic\hbar\bracks{-\nabla_{\vec{r}} \delta(\vec{r} - \vec{r}\,')}\psi_{a}\pars{\vec{r}\,'}\,\dd^{3}\vec{r}\,' \\[3mm]&=-\ic\hbar\int_{-\infty}^{\infty}\nabla_{\vec{r}}\bracks{% \delta\pars{\vec{r} - \vec{r}\,'}\psi_{a}\pars{\vec{r}\,'}}\,\dd^{3}\vec{r}\,' \\[3mm]&=-\ic\hbar\nabla_{\vec{r}}\ \overbrace{\int_{-\infty}^{\infty} \delta\pars{\vec{r} - \vec{r}\,'}\psi_{a}\pars{\vec{r}\,'}\,\dd^{3}\vec{r}\,'} ^{\ds{=\ \psi_{a}\pars{\vec{r}}}} \end{align}