- #1

- 29

- 4

## Homework Statement

Prove that

## [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##

using Einstein summation convention.

I think I have achieved the solution but I am not sure of my last steps, since this is one of my first excersises using this convention.

## Homework Equations

[/B]

## (1) [L_a,p_b] = i \hbar \epsilon_{abc} p_c ##

## (2) [L_a,x_b] = i \hbar \epsilon_{abc} x_c ##

## (3) L_a = \epsilon_{abc} x_b p_c ##

## (4) [AB,C] = A [B,C] + [A,C] B ##

## (5) \epsilon_{abc} = \epsilon_{bca} = \epsilon_{cab} ##

## (6) \epsilon_{abc} \epsilon_{auv} = \delta_{bu} \delta_{cv}- \delta_{bv} \delta_{cu}##

3. The Attempt at a Solution

3. The Attempt at a Solution

Well, using (3) and (4)

## [L_a,L_b] = [\epsilon_{auv} x_u p_v, L_b] = \epsilon_{auv} x_u [p_v, L_b] + \epsilon_{auv} [x_u , L_b] p_v ##

using (1) and (2), I get:

## \epsilon_{auv} x_u (- i \hbar \epsilon_{bvw}p_w)+\epsilon_{auv} x_u ((- i \hbar \epsilon_{but} x_t) p_v = - i \hbar (\epsilon_{auv} \epsilon_{bvw} x_u p_w + \epsilon_{auv} \epsilon_{but} x_t p_v )##

And using (5) and (6) :

## - i \hbar (\epsilon_{vau} \epsilon_{vwb} x_u p_w + \epsilon_{uva} \epsilon_{utb} x_t p_v ) = - i \hbar (x_b p_a - x_u x_u + x_t x_t - x_a p_b) ##

This expression reduces to

## i \hbar (x_a p_b - x_b p_a) ##

I see this is the expression for the "c" component of the angular momentum but I would like to get the final expression using Einstein's notation and the Levi-Civita, so I guess this could be written as

## i \hbar \epsilon_{cab} x_a p_b ##

Using (3) I get :

## i \hbar L_c ##

Which is different to what I want to prove.

## [L_a,L_b] = i \hbar \epsilon_{abc} L_c ##

Then my assumption is that the Levi Civita symbol in the original equation makes explicit that c is different from a and b and that

## [L_b,L_a] = - [L_a,L_b] ##

So would it be correct to do the following in order to arrive to the solution?

## i \hbar \epsilon_{cab} x_a p_b = i \hbar \epsilon_{abc} \epsilon_{cab} x_a p_b = i \hbar \epsilon_{abc} L_c##