Demonstration [L_i,x_j]= ε_ijk x_k

1. Jun 27, 2012

ebol

Hi!
I have to show that
$[L_i,x_j]= i \hbar \varepsilon_{ijk} x_k$

but my result is different, I'm definitely making a mistake
ok I wrote
$L_i = \varepsilon_{ijk} x_j p_k$
then
$[L_i,x_l]= \varepsilon_{ijk} ( [x_j p_k , x_l] ) = \varepsilon_{ijk} ( {x_j [p_k , x_l] + [x_j , x_l] p_k } ) = \varepsilon_{ijk} ( {x_j [p_k , x_l] } ) =$

$= \varepsilon_{ijk} ( {x_j \frac{\hbar}{i} δ_{kl} } ) = \frac{\hbar}{i} \varepsilon_{ijk} {x_j }$

can anyone tell me where I'm wrong?
thanks anyway!

2. Jun 27, 2012

tiny-tim

welcome to pf!

hi ebol! welcome to pf!
but $\frac{1}{i} \varepsilon_{ijk} {x_j } = i \varepsilon_{ijk} x_k$

3. Jun 27, 2012

ebol

ah!
and why? :)
Because the indices $j$ and $k$ commute and changes the sign?

4. Jun 27, 2012

tiny-tim

yup!

that's what ε does!​

5. Jul 3, 2012

ebol

thank you very much!
I arrived at the solution but I did not know :D

6. Jul 3, 2012

he he