# Quantum Mechanics - Angular Momentum part 2

• Feb 23rd 2011, 01:13 PM
bugatti79
Quantum Mechanics - Angular Momentum part 2
Consider

$
[\hat L^2, \hat L_x]=[\hat L_x^2+\hat L_y^2+\hat L_z^2, \hat L_x]=[\hat L_x^2,\hat L_x]+[\hat L_y^2,\hat L_x]+[\hat L_z^2,\hat L_x]
$

$
=\hat L_x [\hat L_x,\hat L_x]+ [\hat L_x,\hat L_x] \hat L_x +\hat L_y [\hat L_y,\hat L_x]+ [\hat L_y,\hat L_x] \hat L_y+\hat L_z [\hat L_z,\hat L_x]+ [\hat L_z,\hat L_x] \hat L_z
$

$
= i \hbar (- \hat L_y \hat L_z -\hat L_z \hat L_y +\hat L_z \hat L_y +\hat L_y \hat L_z)=0
$

On the second line I am wondering how the first two terms go to 0 and how the third term

$
\hat L_y [\hat L_y,\hat L_x]=- i \hbar \hat L_z
$
etc?
• Feb 23rd 2011, 01:27 PM
Ackbeet
Any operator commutes with itself, which is another way of saying its commutator is zero. As for your second question, note that

$\left[\hat{L}_{x},\hat{L}_{y}\right]=i\hbar\hat{L}_{z},$

$\left[\hat{L}_{y},\hat{L}_{z}\right]=i\hbar\hat{L}_{x},$

$\left[\hat{L}_{z},\hat{L}_{x}\right]=i\hbar\hat{L}_{y}.$

Also, note that, in general, $\left[\hat{A},\hat{B}\right]=-\left[\hat{B},\hat{A}\right].$

Hence, you have

$\hat{L}_{y}\left[\hat{L}_{y},\hat{L}_{x}\right]=\hat{L}_{y}\left(-i\hbar\hat{L}_{z}\right)=-i\hbar\hat{L}_{y}\hat{L}_{z}.$

Does that make sense?