# 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.

• 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. Commented May 26, 2013 at 7:46
• Forget about $\epsilon_{ijk}$. Try to derive $[L_1,L_2]=i\hbar L_3$ etc. Commented May 26, 2013 at 8:30
• @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. Commented May 26, 2013 at 22:50
• 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? Commented Dec 30, 2023 at 10:16

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.

• 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. Commented May 26, 2013 at 22:51
• 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. Commented May 28, 2013 at 22:23

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.

• how exactly would i argue the cyclic permuation? Commented May 28, 2013 at 22:23
• 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. Commented May 29, 2013 at 12:43