# Angular momentum raising lowering operator

## 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.

## The Attempt at a Solution

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

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.

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

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.

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

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:

$[L_+,L_-]=[L_x+iL_y,L_x-iL_y]=i[L_y,L_x]-i[L_x,L_y]=\hbar (L_z+L_z)=2\hbar L_z$