# Homework Help: Commutator Relation [L_i,L_j]

1. Mar 21, 2014

### unscientific

1. The problem statement, all variables and given/known data

Find $[L_i,L_j]$.

2. Relevant equations

$$[x_i,p_j] = \delta_{ij}i\hbar$$

3. The attempt at a solution

$$[L_i,L_j] = \epsilon_{ijk}\epsilon_{jlm} [x_jp_k,x_lp_m]$$
$$= \left( \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\right)[x_jp_k,x_lp_m]$$
$$= [x_jp_k,x_jp_k] - [x_jp_k,p_kx_j]$$
$$= p_kx_jx_jp_k - x_jp_kp_kx_j$$

Since $x_jp_k = p_kx_j$ for j≠k, we swap the two in the second term:

$$= p_kx_jx_jp_k - x_jp_kx_jp_k$$
$$= [p_k,x_j]x_jp_k$$
$$= -i\hbar \delta_{jk}x_jp_k$$

I think the answer is something like $i\hbar L_k$.

2. Mar 21, 2014

### TSny

You've used j as a dummy summation index on the right, which leads to trouble because on the left j is a fixed index. You'll need to use a dummy index other than j when you express $L_i$.

3. Mar 21, 2014

### unscientific

$$[L_i, L_l] = \epsilon_{ijk}\epsilon_{lmn}[x_jp_k,x_mp_n]$$
$$= (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km})[x_jp_k, x_mp_n]$$
$$= -[x_jp_k, x_kp_j]$$
$$= -\left( [x_jp_k,x_k]p_j + x_k[x_jp_k,p_j]\right)$$
$$= -\left( x_j[p_kx_k]p_j + x_k[x_j,p_j]p_k\right)$$
$$= i\hbar \left( x_jp_j - x_kp_k\right)$$
$$= 0$$

Last edited: Mar 21, 2014
4. Mar 21, 2014

### TSny

$$\epsilon_{ijk}\epsilon_{lmn} \neq (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km})$$
(Note that indices $i$ and $l$ appear on the left but not on the right.)

5. Mar 21, 2014

### unscientific

Ah that's right, for that identity to work they must have the same leftmost letter.

6. Mar 21, 2014

### TSny

Yes. Then you would be summing over the leftmost index of each Levi-Civita symbol. But, in your case there is no summation (yet).

7. Mar 21, 2014

### unscientific

Ok I've given it another go!

$$[x_jp_k, x_mp_n] = [x_jp_k, x_m]p_n + x_m[x_jp_k, p_n]$$
$$= \left( x_j[p_k, x_m]p_n + x_m[x_j, p_n]p_k\right)$$
$$= -i\hbar \left( \delta_{km}x_jp_n - \delta_{jn}x_mp_k\right]$$

Now applying $\epsilon_{ijk}\epsilon_{lmn}$:

$$= -i\hbar \left( \epsilon_{ijk}\epsilon_{lkn}x_jp_k - \epsilon_{ijk}\epsilon_{lmj}x_mp_k\right)$$

$$= -i\hbar \left( -\epsilon_{kij}\epsilon_{kln}x_jp_n + \epsilon_{jik}\epsilon_{jlm}x_mp_k\right)$$

$$= i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right]$$

Now, $i=m=n$ and $j=k=l$:

$$= i\hbar \left( x_jp_j - x_jp_i -x_jp_j+x_ip_j\right)$$

$$= i\hbar (x_ip_j - x_jp_i)$$

$$[L_i,L_j] = i\hbar \epsilon_{kij}x_ip_j$$

$$[L_i,L_j] = i\hbar L_k$$

8. Mar 21, 2014

### TSny

OK. Looks good.

From your post #3 you are evaluating $[L_i, L_l]$. So the indices $i$ and $l$ are fixed.
Have another go at simplifying $$i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right]$$

9. Mar 22, 2014

### unscientific

$$i\hbar \left[ (\delta_{il}\delta_{jn} - \delta_{in}\delta_{jl})x_jp_n - (\delta_{il}\delta_{km} - \delta_{im}\delta_{kl})x_mp_k\right]$$

$$= i\hbar [\delta_{im}\delta_{kl}x_mp_k - \delta_{in}\delta_{jl}x_jp_n]$$

$$= i\hbar [(\delta_{im}\delta_{kl} - \delta_{ik}\delta_{ml})x_mp_k]$$

10. Mar 22, 2014

### TSny

OK. You can keep going and simplify further. Note that you are summing over m, k, [STRIKE]j[/STRIKE], and [STRIKE]n[/STRIKE].

[EDIT: Sorry. I see you've reduced it to just summing over m and k]

Last edited: Mar 22, 2014
11. Mar 22, 2014

### unscientific

I have simplified it till it's a sum over m and k only. But that doesn't reduce to 2 epsilons

12. Mar 22, 2014

### TSny

OK. What is your final expression after summing over m and k?

Last edited: Mar 22, 2014
13. Mar 22, 2014

### unscientific

the deltas don't seem to combine to give an epsilon

14. Mar 22, 2014

### TSny

We'll see how to get an epsilon factor later. For now, what expression do you get after summing over m and k?

15. Mar 22, 2014

### unscientific

Actually, check out post #7. I solved it already by expanding. I thought you meant that there was a way of making use of the identity $(\epsilon_{1}\epsilon_{2} = \delta\delta - \delta\delta)$ to get the epsilon more explicitly..

In post #7, the epsilon was drawn out by observation of a cross product $x_ip_j - p_ix_j$

16. Mar 22, 2014

### TSny

Yes, in post 7 you are getting the right type of result. It's confusing, because you started out post 7 as a continuation of post 3 where you were calculating $[L_i, L_l]$, but somehow ended up with an expression for $[L_i, L_j]$.

But if you now feel that you understand all the steps to get from $[L_i, L_j]$ to $i \hbar (x_i p_j - p_i x_j )$, you are essentially done.

If you let $k$ denote the direction perpendicular to the $i$-$j$ plane,then you have shown that $[L_i, L_j] = i \hbar (x_i p_j - p_i x_j) = i \hbar L_k$

Can you show that this is equivalent to $[L_i, L_j] = i \hbar \epsilon_{ijm} L_m$ with summation over $m$?

17. Mar 22, 2014

### unscientific

It's $[L_i, L_l] = i\hbar (x_ip_l - p_ix_l) = i\hbar L_m$ I'm still missing an $\epsilon$ though..

Last edited: Mar 22, 2014
18. Mar 22, 2014

### TSny

OK, that's correct with the understanding that $m$ refers to the index that defines the direction perpendicular to the $i$-$l$ plane and that the indices $i, l$ and $m$ are in cyclic order.

The nice thing is that the one expression $$[L_i, L_j] = i \hbar \sum_{k=1}^{3} \epsilon_{ijk}L_k = i \hbar \epsilon_{ijk}L_k$$ automatically takes care of everything. In the last expression the summation over $k$ is assumed (using the Einstein summation convention for repeated indices).

19. Mar 22, 2014

### unscientific

I seem to be missing an $\epsilon$ in my final answer..

20. Mar 22, 2014

### TSny

We have $[L_i, L_j] = i \hbar (x_ip_j - x_jp_i)$.

The right hand side contains $(x_ip_j - x_jp_i)$, which is the component of angular momentum that is perpendicular to the $i$-$j$ plane. A convenient way to express this component of angular momentum is $\epsilon_{ijk}L_k$ (where summation over the index $k$ is assumed). So the epsilon is "put in by hand" as a way of expressing the angular momentum component that is perpendicular to the $i$-$j$ plane.

Thus, we can write

$[L_i, L_j] = i \hbar \epsilon_{ijk}L_k$

As an example, if $i = 1$ and $j = 2$, then $[L_1, L_2] = i \hbar \epsilon_{12k}L_k = i \hbar L_3$.

21. Mar 23, 2014

### unscientific

Yeah, but matematically $(x_ip_j - x_jp_i)$ simply means $\vec{x}$x$\vec{p}$. So by putting an epsilon sign, it becomes $\epsilon_{ijk}x_jp_k$ which is simply $L_k$. By putting another epsilon, won't you get an extra factor of epsilon?

22. Mar 23, 2014

### TSny

$(x_ip_j - x_jp_i) = (\vec{x} \times \vec{p})_k$ where $k$ refers to the component of the cross product that is perpendicular to the i-j plane.

I'm not following. $\epsilon_{ijk}x_jp_k$ represents the $i^{th}$ component of the cross product $\vec{x} \times \vec{p}$. (Note the repeated indices j and k in your expression, which means you are summing over j and k.)

I don't see how you would get two epsilon factors. For fixed i and j, $(x_ip_j - x_jp_i) = \epsilon_{ijk}L_k$ with just one epsilon factor (and summation over the index k).

In summary, you have for fixed i and j, $$[L_i, L_j] = i \hbar (x_ip_j - x_jp_i) = i\hbar \epsilon_{ijk}L_k$$

23. Mar 23, 2014

### unscientific

Because $L_k$ is defined as $\epsilon_{kij}x_ip_j$

Last edited: Mar 23, 2014
24. Mar 25, 2014

### unscientific

any input?

25. Mar 25, 2014

### Fredrik

Staff Emeritus

You want to find an equality that holds for all $i,l\in\{1,2,3\}$ such that $i\neq l$. If m=1, then $[L_i,L_l]=i\hbar L_m$ can't possibly hold for all $i,l\in\{1,2,3\}$ such that $i\neq l$. Same thing if m=2, or if m=3.

The right-hand side (and every intermediate step) must depend on i and j, and nothing else.

Yes, we have $L_k=\epsilon_{kij} x_i p_j$. Now what does that imply about $\epsilon_{ijk}L_k$?

Depending on the values of i and j, $x_ip_j - x_jp_i$ is either equal to 0, or to one of the components of $\vec x\times\vec p$. Which component that is depends on the values of i and j. If i≠j, then it's the kth component, where k is the unique element of {1,2,3} such that $\epsilon_{ijk}\neq 0$.

Last edited: Mar 25, 2014