0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Cool with this? Of course the gradient is acting on the delta. $\endgroup$ – Cosmas Zachos Jun 12 '20 at 0:26
  • $\begingroup$ Ok, that was a stupid question. But sending me the wikipedia page on braket notation isn't helpful. $\endgroup$ – Rafael Mancini Jun 12 '20 at 0:45
  • $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Jun 12 '20 at 1:42
  • $\begingroup$ 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. $\endgroup$ – Rafael Mancini Jun 12 '20 at 1:53
  • $\begingroup$ @CosmasZachos: I am not sure the gradient acts on the delta. $\endgroup$ – fra_pero Jun 12 '20 at 5:55
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.