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.

  • 2
    $\begingroup$ There is clearly no $\delta_{ij}$ term in the result because the commutator is $ij$-antisymmetric while delta is $ij$-symmetric. Please calculate it again and be more careful about the signs and identities. $\endgroup$ Commented May 26, 2013 at 7:46
  • 1
    $\begingroup$ Forget about $\epsilon_{ijk}$. Try to derive $[L_1,L_2]=i\hbar L_3$ etc. $\endgroup$
    – user10001
    Commented May 26, 2013 at 8:30
  • $\begingroup$ @user10001. I prefer to do the more general version. For your way I would then have to make an argument for cyclic permutations, which isn't what I wanted to do. $\endgroup$
    – user27182
    Commented May 26, 2013 at 22:50
  • $\begingroup$ I wanted to comment but i am not able to yet. So @Methusalem when i do not rename the indices I get the result $$ i \hbar \left(\varepsilon_{kij} x_i P_j - x_j P_i\right) + \delta_{ij}\left(x_k P_k - x_m P_m\right) $$ When i request $i \neq j$ for the epsilon tensor the $\delta$ part vanishes and i remain with the $L_k$ what i wanted. What is the intuition for the $\delta$ part except beeing diagonal elements of the whole matrix? $\endgroup$ Commented Dec 30, 2023 at 10:16

2 Answers 2


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.

  • $\begingroup$ This is exactly what I was after, thank you. My problem was that I didn't use the commutator relations, I think I got a 3 i'd monster somewhere too. $\endgroup$
    – user27182
    Commented May 26, 2013 at 22:51
  • $\begingroup$ I tried this out and I think the identity you give is wrong. If we apply [A,BC] = B[A,C] + [A,B]C to [AB,CD], we get the RHS of your first centred equation. $\endgroup$
    – user27182
    Commented May 28, 2013 at 22:23


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.

  • $\begingroup$ how exactly would i argue the cyclic permuation? $\endgroup$
    – user27182
    Commented May 28, 2013 at 22:23
  • $\begingroup$ Argue that (ijk) -> (xyz) and that the same answer holds for (jki) -> (yzx) etc. Changing the indices doesn't change the solution, for any permutation. You could write it out, explaining, or you could actually show the permutations are the same simply. $\endgroup$ Commented May 29, 2013 at 12:43

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