Angular momentum raising lowering operator

[v_pino]

Homework Statement

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

Homework Equations

L_{\pm}=L_x \pm iL_y

The Attempt at a Solution

[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])

Is this right so far? If so, how do I proceed from this? The books and websites I'll read always just state [L_\pm , L^2]=0 instead of proving it.

 

[CanIExplore]

L^{2}=L^{2}_{x}+L^{2}_{y}+L^{2}_{z}

Then

[L_{±},L^{2}]=[L_{x}±iL_{y},L^{2}_{x}+L^{2}_{y}]

which is going to give you some cross terms. Using the fact that [L_{i},L_{j}]=ih\epsilon_{ijk}L_{k}, 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.

 

[v_pino]

I'm having some trouble expanding the commutators due to the +/- sign. Is it still [A,B]=AB-BA?

 

[CanIExplore]

Is it still [A,B]=AB-BA?

This is always true.

After writing the commutation relation, and expanding, the ± just becomes \mp when multiplied by -. If you have two like terms, one of which has the sign ± and the other of which has the sign \mp, they will cancel.

 

[v_pino]

Is the following ok?

[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)

[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

= 0

 

[v_pino]

I tried to do the same as what I did above to prove [L_+,L_-]=2\hbar L_z but I got zero instead:

[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

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

[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