Orbital angular momentum and kinetic energy

[bon]

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!

 

[vela]

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

$$\left[r,\frac{\partial}{\partial r}\right]\psi = r\left(\frac{\partial}{\partial r}\psi\right) - \frac{\partial}{\partial r}(r\psi) = r\left(\frac{\partial}{\partial r}\psi\right) - \left[\left(\frac{\partial}{\partial r}r\right)\psi + r\left(\frac{\partial}{\partial r}\psi\right)\right] = -\psi$$

 

[bon]

Great thanks!

any ideas on the previous bit of my question?

Thanks!

 

[vela]

Probably a similar reason. Try working it out again.

 

[paranormal]

Hello, thank you for your questions.

To answer your first question, the extra factor of D/Dr comes from the fact that the momentum operator is acting on the radial coordinate, which is represented by r. This is why the derivative operator, D/Dr, appears in the expression for the orbital angular momentum, pr.

As for your second question, the commutation relation [r,pr]=-ih[r,D/Dr] comes from the commutation relation between the position and momentum operators, [x,p]=ih. By substituting r for x, and pr for p, we get [r,pr]=-ih[r,D/Dr]. And since [r,D/Dr]=1, the final result is -ih.

I hope this helps clarify things for you. Let me know if you have any further questions.