When calculating the momentum expectation value

Click For Summary

Homework Help Overview

The discussion revolves around calculating the momentum expectation value in quantum mechanics, specifically focusing on the application of the momentum operator and its interaction with wave functions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to clarify the differentiation process involving the complex and normal wave functions in the context of the momentum operator. Some participants provide hints regarding the momentum operator's definition and its implications for the calculation.

Discussion Status

Participants are exploring the correct application of the momentum operator and its sign. There are hints and confirmations regarding the differentiation process, but some uncertainty remains about the implications of the negative sign in the operator.

Contextual Notes

There is a mention of a potential misunderstanding regarding the sign in the momentum operator, which may affect the calculation. The discussion reflects a focus on ensuring clarity in the mathematical formulation without reaching a definitive conclusion.

rwooduk
Messages
757
Reaction score
59
when calculating the momentum expectation value the term i(h-bar)d/dx goes inbetween the complex PSI and the 'normal' PSI, so do you differentiate the normal PSI and then multiply by the complex PSI? or do you differentiate the product of the two PSI's i.e. the modulus of PSI?

thanks for any direction on this.
 
Physics news on Phys.org
rwooduk said:
when calculating the momentum expectation value the term i(h-bar)d/dx goes inbetween the complex PSI and the 'normal' PSI, so do you differentiate the normal PSI and then multiply by the complex PSI?

Yes. The expectation value of ##\hat A## is ##\int \Psi^* \hat A \Psi \, dx## where any differential operators in ##\hat A## works to the right.
 
  • Like
Likes   Reactions: 1 person
Another hint:
[tex]\hat{p}=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x}.[/tex]
Note the sign!
 
  • Like
Likes   Reactions: 1 person
Brilliant! Many Thanks CC!

vanhees71 said:
Another hint:
[tex]\hat{p}=-\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} x}.[/tex]
Note the sign!

hm, not sure what you're getting at, the negative sign just comes to the front of the integral, I am probably missing something. but thanks also!
 
I just wanted to make sure that you get that sign right, because you got it wrong in your original posting!
 

Similar threads

Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K