Deriving the Angular Momentum Commutator Relations by using $\epsilon_{ijk}$ Identities I've been trying to derive the relation
$$[\hat  L_i,\hat  L_j]  = i\hbar\epsilon_{ijk} \hat  L_k $$
without doing each permutation of ${x,y,z}$ individually, but I'm not really getting anywhere. Can someone help me out please?
I've tried expanding the $\hat L_i = \epsilon_{nmi} \hat x_n \hat p_m$ and using some identities for the $\epsilon_{ijk} \epsilon_{nmi}$ which gives me the LHS as something like $-\hbar^2\delta_{ij}$ but I've got no further than this.
 A: User27182,
I'll answer the question user10001 posited; leaving out the Levi-Civita symbol, do the calculation for just $x$ and $y$. 
$$[\hat{L}_x , \hat{L}_y]=[\hat{p}_z \hat{y} - \hat{p}_y \hat{z}, \hat{p}_x \hat{z} - \hat{p}_z \hat{x}]$$
$$=[\hat{p}_z \hat{y},\hat{p}_x \hat{z} - \hat{p}_z \hat{x}]-[\hat{p}_y \hat{z},\hat{p}_x \hat{z} - \hat{p}_z \hat{x}]$$
$$=[\hat{p}_z \hat{y},\hat{p}_x \hat{z}]-[\hat{p}_z \hat{y},\hat{p}_z \hat{x}]-[\hat{p}_y \hat{z},\hat{p}_x \hat{z}]+[\hat{p}_y \hat{z},\hat{p}_z \hat{x}]$$ Must be careful of the signs in the above step! Next I use the rule $[\hat{A}\hat{B},\hat{C}]=\hat{B}[\hat{A},\hat{C}]+[\hat{A},\hat{B}]\hat{C}$ repeatedly. Before I can conclude, a quick reminder what everything is. So we know $\hat{p}_i=-i\hbar \frac{\partial}{\partial q_{a}}$, $[\hat{q}_i,\hat{q}_j]=0, [\hat{p}_i,\hat{p}_j]=0], [\hat{q}_i,\hat{p}_j]=i \hbar \delta_{ij}$
We can now easily see that $$[\hat{L}_x,\hat{L}_y]=\hat{p_{x}}[\hat{p}_z,\hat{z}]\hat{y}-0-0+\hat{x}[\hat{p}_z,\hat{z}]\hat{p}_x$$ Note that $\hat{x}$ and $\hat{p}_y$ commute
$$=-i\hbar\hat{y}\hat{p}_x+i\hbar\hat{x}\hat{p}_y$$
$$=i\hbar L_{z}$$
The other commutators need not be calculated; they are inferred by cyclic permutation! This is where the Levi symbol comes in to say that. Note, I may have mixed up the order of things, but I think its still right. Someone correct me if I have a mistake! Hope this helps. 
A: Since $L_i = \epsilon_{ijk} x_jp_k$ (operators) one has
$$ [L_i,L_j] = \epsilon_{iab}\epsilon_{jcd}[x_ap_b,x_cp_d] = \epsilon_{iab}\epsilon_{jcd}(x_a[p_b,x_c]p_d + x_c[x_a,p_d]p_b) $$
This first step relies on the following property of the commutator: [AB,CD] = A[B,CD] + [A,CD]B + C[AB,D] + [AB,C]D, and then performing the expansion again. The only terms that 'survive' are those involving the canonical conjugate variables. terms like $[x_a,x_b] =0$. So,
$$ [L_i,L_j] = \epsilon_{iab}\epsilon_{jcd}(x_a\underbrace{[p_b,x_c]}_{-i\hbar \delta_{b,c}}p_d + x_c\underbrace{[x_a,p_d]}_{i\hbar \delta_{ad}}p_b) = i\hbar \epsilon_{iab}\epsilon_{jcd}(-x_ap_d \delta_{bc} + x_cp_b\delta_{ad})  $$
Because of the definition of levi-civita tensor, you can absorb a minus sign by just permuting any two neighboring indices. Furthermore, after carying out the deltas, I like to rename $x_cp_b$ to $x_ap_d$ in the second term. This leads to
$$ [L_i,L_j] =i\hbar(\epsilon_{iab}\epsilon_{bjd} + \epsilon_{dib}\epsilon_{bja})x_ap_d$$
Keep in mind that any index apart from i and j are summed over
$$[L_i,L_j] = i\hbar(\delta_{ij}\delta_{ad} - \delta_{id}\delta_{aj} + \delta_{dj}\delta_{ia} - \delta_{da}\delta_{ij})x_ap_d = i\hbar(x_ip_j - x_jp_i) = i\hbar \epsilon_{ijk}L_k $$
I suggest you work out the missing parts to understand how this levi-civita business works.
