# In deriving the expression for the position operator in momentum space

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.

• Cool with this? Of course the gradient is acting on the delta. Jun 12, 2020 at 0:26
• Ok, that was a stupid question. But sending me the wikipedia page on braket notation isn't helpful. Jun 12, 2020 at 0:45
• Write out the δ function in r'-r, and collapse the r' integral . Pull out the gradient, now in r, and collapse the r'' δ function to get the gradient acting on ψ. The reason I sent you to that section of WP is because you see the answer by inspection there: that's where it basically came from. This looks like homework. Jun 12, 2020 at 1:42
• It ir homework, haha. Thank you for the answer. I'm having some trouble lately with these integrals that come from working in the x and p space. So my doubt, I think, was more about calculus than braket notation. Jun 12, 2020 at 1:53
• @CosmasZachos: I am not sure the gradient acts on the delta. Jun 12, 2020 at 5:55

## 1 Answer

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.