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!

Homework Help: Angular momentum raising lowering operator

  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Derive [itex] [L_\pm , L^2]=0 [/itex]

    2. Relevant equations

    [itex] L_{\pm}=L_x \pm iL_y [/itex]

    3. The attempt at a solution

    [itex] [L_\pm , L^2]=[L_x,L_x^2] \pm i[L_y,L_y^2]=[L_x,L_x]L_x + L_x[L_x,L_x] \pm i([L_y,L_y]L_y+L_y[L_y,L_y]) [/itex]

    Is this right so far? If so, how do I proceed from this? The books and websites I'll read always just state [itex] [L_\pm , L^2]=0 [/itex] instead of proving it.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 18, 2011 #2



    which is going to give you some cross terms. Using the fact that [itex][L_{i},L_{j}]=ih\epsilon_{ijk}L_{k}[/itex], you can move some stuff around and cancel terms to arrive at the final answer. I'm not sure how you got to where you are.
  4. Nov 18, 2011 #3
    I'm having some trouble expanding the commutators due to the +/- sign. Is it still [A,B]=AB-BA?
  5. Nov 18, 2011 #4
    This is always true.

    After writing the commutation relation, and expanding, the ± just becomes [itex]\mp[/itex] when multiplied by -. If you have two like terms, one of which has the sign ± and the other of which has the sign [itex]\mp[/itex], they will cancel.
  6. Nov 19, 2011 #5
    Is the following ok?

    [itex] [L_\pm,L^2]=(L_x \pm iL_y)(L_x^2+L_y^2)-(L_x^2+L_y^2)(L_x \pm iL_y) [/itex]

    [itex] [L_\pm,L^2]=L_x^3+L_xL_y^2\pm iL_x^2L_y\pm iL_y^3 -L_x^3 \mp i L_x^2 L_y -L_x L_y^2 \mp iLy^3 [/itex]

    = 0
  7. Nov 19, 2011 #6
    I tried to do the same as what I did above to prove [itex] [L_+,L_-]=2\hbar L_z [/itex] but I got zero instead:

    [itex] [L_+,L_-]=(L_x +iL_y)(L_x-iL_y)-(L_x-iL_y)(L_x+iL_y)=(L_x^2+L_y^2)-(L_x^2+L_y^2)=0 [/itex]

    I found the following derivation but I don't understand how the last three equalities were arrived at:

    [itex] [L_+,L_-]=[L_x+iL_y,L_x-iL_y]=i[L_y,L_x]-i[L_x,L_y]=\hbar (L_z+L_z)=2\hbar L_z [/itex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook