2
$\begingroup$

$$\frac{d\langle p\rangle}{dt}=-i\hbar \int_{-\infty}^{\infty}\frac{d\psi^*}{dt} \frac{d\psi}{dx}+\psi^*\frac{d}{dt}\Bigr(\frac{d\psi}{dx}\Bigr)$$ I didn't know the coding of partial derivative which are inside the integral There is a question about this on site, but I don't have much experience with bra -ket notation. Ehrenfest's theorem derivation. The first integrand makes sense but I'm completely at odds with the second one

$\endgroup$
  • 2
    $\begingroup$ That integral needs parentheses and a differential. $\endgroup$ – G. Smith Jun 27 at 19:51
  • $\begingroup$ Are you asking where the second term on the right comes from? $\endgroup$ – G. Smith Jun 27 at 19:51
  • $\begingroup$ \partial which goes to $\partial$ $\endgroup$ – DanielC Jun 27 at 20:40
3
$\begingroup$

$$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int\frac{d}{dt}\left(\psi^*\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By Leibniz's Rule}$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int \left(\frac{d\psi^*}{dt}\frac{d\psi}{dx} + \psi^*\frac{d}{dt}\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By the Product Rule}$$

| cite | improve this 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.