Integration by parts of derivative of expectation value problem

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The discussion revolves around confusion regarding the integration by parts of a derivative related to expectation values in quantum mechanics. A participant questions how a book derived a specific equation, particularly the presence of a negative sign. Clarifications reveal that the integration has not been completed in the initial statement, and the negative sign is standard in integration by parts. Participants confirm that the omitted terms in the equation are zero, leading to the conclusion that only the integral term remains with a negative sign. Overall, the conversation highlights the nuances of integration by parts in the context of quantum mechanics.
Normalization
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Homework Statement


I don't know how the writer of the book took integral of the first statement and got the second statement? Can anybody clarify on this?
Problem.png

Homework Equations


Given in the photo

The Attempt at a Solution


When I took the integral I just ended up with the exact same statement but without the negative sign behind the (ihbar/2m)
 
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Hi Normalization! Welcome to PF! :smile:
Normalization said:
When I took the integral I just ended up with the exact same statement but without the negative sign …

But that negative sign is always there in integration by parts. :confused:

(see the last equation in your photo)
 
Oh w8 (facepalm) they haven't actually integrated yet? Wow... I feel so stupid right now. I thought they already integrated both products inside the brackets in which case -\int\frac{\partial\Psi^*}{\partial\ x}\Psidx-\int\frac{\partial\Psi}{\partial\ x}×-\Psi^*dx Which would be the exact same thing. Alright thanks I guess...
 
I'm not convinced you've got this. :confused:

They have integrated that big bracket, and so they also differentiate the x (to get 1) …

the […] term is 0 (so they haven't written it), and that only leaves the ∫ term, which always has a minus in front of it. :smile:
 
tiny-tim said:
I'm not convinced you've got this. :confused:

They have integrated that big bracket, and so they also differentiate the x (to get 1) …

the […] term is 0 (so they haven't written it), and that only leaves the ∫ term, which always has a minus in front of it. :smile:

Yes,yes,yes I know but I thought they had already taken the integral of -int(df/dx g dx)
 
Last edited:
Which by nature should be equal to the left side of the equation. So it's fine I was just being daft
 

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