At 2:13 that's the position operator in the momentum basis At 8:28 what he did is he put the variable x inside the integration which is with respect to 'p'. And wrote the whole thing more compactly by writing -ihd/dp(e^ipx/h) which is infact equal to xe^ipx/h. Hope this helps!
@@sheikhathar5322 Could you please elaborate? Which part do you say is v and which one is u? And shouldn't you end up with a sum of two terms after integration by parts? Does the boundary term vanish?
So for anyone doing Quantum Mechanics problems late at night - here is a little hint (I did do the integration by parts eventually): remember that the integral of a derivative of a function returns the function itself, so u = d(exp(ipx/hbar)/dp*dp and v = Phi. The boundary term does vanish
Very nicely explained. Great work. Thank you very much. Subbed!
Thank you. I still have to feel my way around Delta functions. This has been helpful. 😎
Thanks a lot!
At 2:13 that's the position operator in the momentum basis
At 8:28 what he did is he put the variable x inside the integration which is with respect to 'p'. And wrote the whole thing more compactly by writing
-ihd/dp(e^ipx/h) which is infact equal to xe^ipx/h.
Hope this helps!
lifesaver :)
How did the full derivative becomes partial derivative after the integration by parts?
Using U. V rule
@@sheikhathar5322 Could you please elaborate? Which part do you say is v and which one is u? And shouldn't you end up with a sum of two terms after integration by parts? Does the boundary term vanish?
@@helenaosedo roughly intg by parts,,you may get it,
@@sheikhathar5322 I didn't, that's why I asked for help 🙂
So for anyone doing Quantum Mechanics problems late at night - here is a little hint (I did do the integration by parts eventually): remember that the integral of a derivative of a function returns the function itself, so u = d(exp(ipx/hbar)/dp*dp and v = Phi. The boundary term does vanish
It's too slow