1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutation of Angular and Linear Momentum

  1. Oct 17, 2011 #1
    If I have a relation such as [tex][L_{j} , \vec{p}^2]=0[/tex] where j=x,y,z.

    Can I re-write it as [tex][L_{j}, \vec{p} \vec{p}]=0[/tex] and then evaluate it as though it were an identity? e.g. [tex][A,BC]=[A,B]C+[B,A]C=...[/tex]
  2. jcsd
  3. Oct 17, 2011 #2
    If unsure, you are better off explicitly writing out the dot product:
    [itex][ L_j, \vec{p}^2] = [L_j, \sum_{k} p_k^2] = \sum_{k} [L_j, p_k^2][/itex]

    Now you can apply the ABC rule. The one you quote is wrong, by the way. It should be:
    [itex][A, BC] = B[A,C] + [A,B]C[/itex]
  4. Oct 18, 2011 #3
    Thanks for the help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Commutation of Angular and Linear Momentum