Orbital angular momentum and kinetic energy

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Homework Help Overview

The discussion revolves around the concepts of orbital angular momentum and kinetic energy in quantum mechanics, specifically focusing on the commutation relations involving position and momentum operators.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the derivation of the expression for the momentum operator in spherical coordinates and questions the origin of certain terms. They also inquire about the commutation relation between position and momentum operators, expressing confusion over the calculations involved.

Discussion Status

Some participants have provided insights into the calculations, particularly regarding the use of the product rule in differentiation. However, the original poster is still seeking clarification on earlier parts of their question, indicating that the discussion is ongoing and further exploration is needed.

Contextual Notes

The original poster has referenced specific equations and terms from quantum mechanics, suggesting a focus on the mathematical foundations of the topic. There may be assumptions about prior knowledge of quantum mechanics that are not explicitly stated.

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Homework Statement



Please see attached.

I have two questions:

how does pr get to be -ihbar (D/Dr + 1/r) ? where does the extra factor of D/Dr come from? one comes from r.del, but what about the other? surely div (r/r) = 2/r?

Also, why does [r,pr]=-ih[r,D/Dr] = ih?

surely -ih[r,D/Dr] = -ih(rD/Dr - 1)?

am i missing something?

Homework Equations





The Attempt at a Solution



See my attempt above. thanks!
 

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To calculate [r, d/dr], you need to use the product rule in calculating the second term:

[tex]\left[r,\frac{\partial}{\partial r}\right]\psi = r\left(\frac{\partial}{\partial r}\psi\right) - \frac{\partial}{\partial r}(r\psi) = r\left(\frac{\partial}{\partial r}\psi\right) - \left[\left(\frac{\partial}{\partial r}r\right)\psi + r\left(\frac{\partial}{\partial r}\psi\right)\right] = -\psi[/tex]
 
Great thanks!

any ideas on the previous bit of my question?

Thanks!
 
Probably a similar reason. Try working it out again.
 

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