Orbital angular momentum and kinetic energy

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
bon
Messages
547
Reaction score
0

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!
 

Attachments

  • question..jpg
    question..jpg
    37.9 KB · Views: 497
Physics news on Phys.org
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!