$$\frac{d\langle p\rangle}{dt}=-i\hbar \int_{-\infty}^{\infty}\frac{d\psi^*}{dt} \frac{d\psi}{dx}+\psi^*\frac{d}{dt}\Bigr(\frac{d\psi}{dx}\Bigr)$$ 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
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2$\begingroup$ That integral needs parentheses and a differential. $\endgroup$– G. SmithJun 27, 2020 at 19:51
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$\begingroup$ Are you asking where the second term on the right comes from? $\endgroup$– G. SmithJun 27, 2020 at 19:51
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$\begingroup$ \partial which goes to $\partial$ $\endgroup$– DanielCJun 27, 2020 at 20:40
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1 Answer
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$$\langle p \rangle = -i\hbar\int\psi^*\frac{d \psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\frac{d}{dt}\int\psi^*\frac{d\psi}{dx}dx$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int\frac{d}{dt}\left(\psi^*\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By Leibniz's Rule}$$ $$\implies\frac{d\langle p \rangle}{dt} = -i\hbar\int \left(\frac{d\psi^*}{dt}\frac{d\psi}{dx} + \psi^*\frac{d}{dt}\frac{d\psi}{dx}\right)dx \ \ \ \ \text{By the Product Rule}$$
