Does the Angular Momentum Commutator [L_a, L_b L_b] Equal Zero?

[mr_sparxx]

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.

 

[TSny]

$[L_a,L_b L_b] = [L_a,L_b ] L_b+L_b [L_a,L_b] = i \hbar \epsilon_{abc} L_c L_b + L_b i \hbar \epsilon_{abd} L_d$

In the first term on the far right you have two summation indices: b and c. In the second term you have summation indices b and d.

Summation indices are "dummy" indices. You can rename them if you wish. What happens if in the second term you re-label the b index as c and re-label the d index as b?

 

[TSny]

$i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar (\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) =2 i \hbar \epsilon_{abc} L_b L_c$

Note that $(\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) \neq 0$.

What is value of $\delta_{cc}$? Remember, repeated subscripts are summed.

 

[mr_sparxx]

So $i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{abd} L_b L_d = i\hbar \epsilon_{abc} L_c L_b+i \hbar \epsilon_{acb} L_c L_b = 0$

So simple now... b is a dummy index! It is still difficult for me to realize what is "fixed" and what is a dummy index without writing down the summation symbols... Now I see that $L_b L_b$ stands for the square of the modulus in classic mechanics: I lost this somewhere in my work.

Note that $(\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) \neq 0$.

What is value of $\delta_{cc}$? Remember, repeated subscripts are summed.

I think it is 3.

Ok so the first term is actually, $\delta_{ae}L_e = L_a$ and the second term is $3 \delta_{ae}L_e = 3 L_a$
so:

$i \hbar (2 \epsilon_{abc} L_b L_c + i \hbar (\delta_{ce}\delta_{ac}- \delta_{cc}\delta_{ea}) L_e ) =2 i \hbar \epsilon_{abc} L_b L_c + 2 \hbar^2 L _ a$

And according to my previous attemp (in which I corrected the last step):
$2 i \hbar \epsilon_{abc} L_b L_c + 2 \hbar^2 L _ a = -2 \hbar^2 L_a + 2 \hbar^2 L _ a = 0$

I think I got it... Thank you once again!