In deriving the expression for the position operator in momentum space

Question

Consider this: $$\langle \mathbf{r} | \hat{\mathbf{P}} | \psi \rangle = \displaystyle\int d^3\mathbf{r}'\displaystyle\int d^3\mathbf{r}''\langle \mathbf{r}|\mathbf{r'}\rangle\langle\mathbf{r}'|\hat{\mathbf{P}}|\mathbf{r}''\rangle\langle\mathbf{r''}|\psi\rangle = \\ = \displaystyle\int d^3\mathbf{r}'\displaystyle\int d^3\mathbf{r}''\langle \mathbf{r}|\mathbf{r'}\rangle\Big(-i\hbar\nabla_{\mathbf{r}'}\delta^3(\mathbf{r'}-\mathbf{r''})\Big)\psi(\mathbf{r''}) $$ Where $\hat{\mathbf{P}}$ is the momentum operator in three dimensions and $\langle\mathbf{r}|$ is the position bra.

Can I move the gradient to the outer integral? I appreciate any tips on this.

Answer 1

OK, if indeed calculus is your problem, I'll just remind you of the elementary calculus move, with one-dimensional integrals and $\delta(-x)=\delta(x)$ s, $$ \int \!\! dy~dz~\delta(x-y) ~\partial_y \delta (y-z) ~\psi(z)\\ =\int \!\! dz~ \partial_x \delta (x-z) ~\psi(z)= \partial_x \int \!\! dz~\delta (x-z) ~\psi(z)= \partial_x \psi(x). $$

In bracket notation, this is self-evident by inspection from $$ \hat{\mathbf{P} } = \int\! d^3 \mathbf{r} ~~| \mathbf{r}\rangle ( - i \hbar \nabla) \langle \mathbf{r}| ~, $$ so, in words, "enter the coordinate representation, take the gradient, and exit the coordinate representation". It's up to you, then, what else to do with the answer.