- #1
mr_sparxx
- 29
- 4
Homework Statement
Prove that
## [L_a,L_b L_b] =0 ##
using Einstein summation convention.
Homework Equations
[/B]
## (1) [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##
## (2) \epsilon_{abc} \epsilon_{auv} = \delta_{bu} \delta_{cv}- \delta_{bv} \delta_{cu}##
## (3) \epsilon_{abc} = \epsilon_{bca} = \epsilon_{cab} ##
## (4) L_a = \epsilon_{abc} x_b p_c ##
## (5) [AB,C] = A [B,C] + [A,C] B ##
## (6) [x_a,p_b] = i \hbar \delta_{ab}##
The Attempt at a Solution
Well, using (1) and (5):
##[L_a,L_b L_b] = [L_a,L_b ] L_b+L_b [L_a,L_b] = i \hbar \epsilon_{abc} L_c L_b + L_b i \hbar \epsilon_{abd} L_d = \\
= i \hbar \epsilon_{abc} (L_b L_c +[L_c,L_b ] ) + i \hbar \epsilon_{abd} L_b L_d = i \hbar (\epsilon_{abc} L_b L_c + \epsilon_{abc} i \hbar \epsilon_{cbe} L_e + \epsilon_{abd} L_b L_d ) ##
Then, using (2) and (3), and putting the two equivalent terms together, I get:
##i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar \epsilon_{bca} \epsilon_{bec} L_e ) =
i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar (\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) =2 i \hbar \epsilon_{abc} L_b L_c ##
This does not look well to me... Anyway, I have continued by using (4):
## 2 i \hbar \epsilon_{abc} \epsilon_{buv} x_u p_v \epsilon_{ckj} x_k p_j ##
Then (2) and (6):
## 2 i \hbar (\epsilon_{ckj} x_c p_a x_k p_j - \epsilon_{ckj} x_a p_c x_k p_j) = 2 i \hbar \epsilon_{ckj} [x_c, p_a] x_k p_j = 2 i \hbar \epsilon_{ckj} i \hbar \delta{ca} x_k p_j = \\ = -2 \hbar^2 \epsilon_{akj} x_k p_j = -2 i \hbar L_a ##
I cannot find my mistake(s)... and my result is absurd.