Deriving commutator for angular momentum components

[mr_sparxx]

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
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?

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

 

[TSny]

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

... I guess this could be written as

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

These two expressions are not equivalent. The first expression makes no reference to index c, whereas the second expression does explicitly reference c. Also, the first expression contains fixed values of a and b. But the second expression is a summation over all values of a and b.

I will let you think about this a bit more.

 

[mr_sparxx]

I knew I was guessing...
Thanks for pointing that out!

Now I am very uncomfortable with the previous steps (and with this notation in general, for now at least).:)

I was trying to avoid summation symbols but I think I have to. When I wrote

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

This expression reduces to

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

Am I really doing
$- i \hbar ( \sum_u \sum_v \sum_w \epsilon_{vau} \epsilon_{vwb} x_u p_w + \sum_u \sum_v \sum_t \epsilon_{uva} \epsilon_{utb} x_t p_v ) = \\ =- i \hbar ( \sum_u \sum_v \sum_w (\delta_{aw}\delta_{ub}-\delta_{ab}\delta_{uw})x_u p_w + \sum_u \sum_v \sum_t (\delta_{vt}\delta_{ab}-\delta_{vb}\delta_{at}) x_t p_v ) = \\ = - i \hbar ( \sum_v x_b p_a - \sum_v \sum_u x_u p_u + \sum_u \sum_v x_v p_v - \sum_u x_a p_b) = - i \hbar ( \sum_v x_b p_a- \sum_u x_a p_b)= - i \hbar (3 x_b p_a- 3 x_a p_b)$

?

 

[TSny]

Am I really doing
$- i \hbar ( \sum_u \sum_v \sum_w \epsilon_{vau} \epsilon_{vwb} x_u p_w + \sum_u \sum_v \sum_t \epsilon_{uva} \epsilon_{utb} x_t p_v ) = \\ =- i \hbar ( \sum_u \sum_v \sum_w (\delta_{aw}\delta_{ub}-\delta_{ab}\delta_{uw})x_u p_w + \sum_u \sum_v \sum_t (\delta_{vt}\delta_{ab}-\delta_{vb}\delta_{at}) x_t p_v )$

?

In going from $\sum_u \sum_v \sum_w \epsilon_{vau} \epsilon_{vwb} x_u p_w$ to $\sum_u \sum_v \sum_w (\delta_{aw}\delta_{ub}-\delta_{ab}\delta_{uw})x_u p_w$, should there still be a sum over $v$?

Note, your original work looked good to me all the way to the result of $i \hbar (x_a p_b- x_b p_a)$ (except for a couple of typos where you had $x_u x_u$ instead of $x_u p_u$ and $x_t x_t$ instead of $x_t p_t$). The problem was when you said that $i \hbar (x_a p_b- x_b p_a) = i\hbar \epsilon_{cab} x_a p_b$.

 

[mr_sparxx]

True. I was a bit in panic. Those 3's made no sense at all. I'll think a bit more about it.

 

[mr_sparxx]

This expression reduces to

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

Well I think I've got it by exploring all possible values:

$(x_a p_b - x_b p_a) = \left\{\begin{array}{cc}0,&\mbox{ if } a=b \\L_3,&\mbox{ if } (a,b) =(1,2) \\ -L_3,&\mbox{ if } (a,b) =(2,1)\\ L_2,&\mbox{ if } (a,b) =(1,3)\\ -L_2,&\mbox{ if } (a,b) =(3,1) \\ L_1,&\mbox{ if } (a,b) =(2,3) \\ -L_1,&\mbox{ if } (a,b) =(3,2) \end{array}\right . = \epsilon_{abc} L_c$

However, I still wonder if I have done it correctly, and if I have, if there is a more elegant algebraic solution for this (with no numbers).

Moreover, as I think about it I am not sure if I even understand the original identity:

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

Is here a summation over all possible values of c but none over a and b because they are fixed on the left hand side?

My apologies but as I said before this is the first time I use Einstein's convention.

 

[TSny]

However, I still wonder if I have done it correctly, and if I have, if there is a more elegant algebraic solution for this (with no numbers).

I think it is correct. You just have to realize that $x_a p_b - x_b p_b$ is equivalent to $\epsilon_{abc} L_c$.

Moreover, as I think about it I am not sure if I even understand the original identity:

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

Is here a summation over all possible values of c but none over a and b because they are fixed on the left hand side?

Yes, that's right. For example $[L_1,L_2] = i \hbar \epsilon_{12c} L_c$ where c is summed over the values 1, 2, 3. So, the right hand side reduces to$i \hbar L_3$.

 

[mr_sparxx]

I think I understand Einstein's convention a little more. I will keep working on this topic: $[L_i, L^2] = 0$ next. I hope I manage to prove this alone.

Thanks for your help!

 

[mr_sparxx]

I will keep working on this topic: $[L_i, L^2] = 0$ next. I hope I manage to prove this alone.

Thanks for your help!

I can't get this one correct either... I'll post a new thread anyway. Thanks a lot!