quantum mechanics - 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.

Answer

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.