Integration by parts of derivative of expectation value problem

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Homework Help Overview

The discussion revolves around the integration by parts of a derivative related to expectation values in quantum mechanics. Participants are trying to clarify the steps involved in the integration process as presented in a textbook.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the integration steps taken in the textbook, particularly regarding the presence of a negative sign in the integral. There is confusion about whether the integration has been completed and how it relates to the original expressions.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about the integration process and the implications of the negative sign. Some have acknowledged misunderstandings, while others are attempting to clarify the steps taken in the textbook.

Contextual Notes

There is a mention of a specific equation from a photo that is referenced but not provided in the discussion. Participants are also navigating their own assumptions about the integration process and the notation used in the textbook.

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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Welcome to PF!

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 -[itex]\int[/itex][itex]\frac{\partial\Psi^*}{\partial\ x}[/itex][itex]\Psi[/itex]dx-[itex]\int[/itex][itex]\frac{\partial\Psi}{\partial\ x}[/itex]×-[itex]\Psi^*[/itex]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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