# Commutator between Angular Momentum J and Cross Product

I'm trying to show that a cross product $$C$$ of two vectorial operators (let's say $$A$$ and $$B$$) it's a vector by it's own, which means, I want to show $$\left [J_i,C_j \right ] = i\hbar \epsilon_{ijk}C_k.$$

Attempt of solution:

$$A$$ and $$B$$ are vectorial operators, so $$\left [J_i,A_j \right ] = i\hbar \epsilon_{ijk}A_k,$$ $$\left [J_i,B_j \right ] = i\hbar \epsilon_{ijk}B_k.$$

$$C$$ is the cross product of $$A$$ and $$B$$, so $$C_i = \epsilon_{ijk}A_jB_k$$ $$C_j = \epsilon_{jik}A_iB_k$$ $$C_k = \epsilon_{kij}A_iB_j.$$

Building the commutator of $$J_i$$ and $$C_j$$, I found this: $$\left [ J_i,C_j \right ] = \left [ J_i,\epsilon_{ijk}A_iB_k \right ] = \epsilon_{ijk}(\left [ J_i,A_i\right ]B_k + A_i\left [ J_i,B_k\right ]).$$

The term with $$\left [ J_i,A_i\right ]$$ is zero, so we have $$\left [ J_i,C_j \right ] = \epsilon_{ijk}(A_i\left [ J_i,B_k\right ]) = \epsilon_{ijk} A_i (\epsilon_{ikj}B_j) = i\hbar\epsilon_{ikj}\epsilon_{ijk} A_iB_j = i\hbar \epsilon_{ikj}C_k = -i\hbar \epsilon_{ijk}C_k.$$

I was not supposed to find this minus signal, but I cannot find the first mistake in my calculations.

• You're not writing out the correct expressions for $C_j$ and $C_k$, where are you getting those from? Mar 21, 2020 at 18:56
• Theres is a general statement that for any vector operator commutator will be $i\epsilon_{ijk}C_k$. You should just contract one index, $\epsilon_{ijk}\epsilon_{krs}$ and obtain the answer Mar 21, 2020 at 19:06
• $C_j=\epsilon_{ijk}A_iB_k$ can be rewritten by changing alphabetical labels $C_i=\epsilon_{jik}A_jB_k = -\epsilon_{ijk}A_jB_k$ which contradicts your expression for $C_i=+\epsilon_{ijk}A_jB_k$.
– JEB
Mar 21, 2020 at 19:22
• What follows "I found this:" is worse than wrong: you are abusing i as both a free index and a saturated dummy index! Use another letter. Mar 21, 2020 at 21:55
• Jacob, Artem and Cosmas, thank very much for your comments, I found out what was wrong. I posted the solution below Mar 21, 2020 at 22:56

Reading the comments of jacob1729, Artem Alexandrov and Cosmas Zachos, I found out the so many mistakes of my calculations. Here comes the complete solution in case anyone needs this proof in the future:

$$C_j$$ can be writen as $$C_j = \epsilon_{jqr}A_qB_r$$, so we have

$$[J_i,C_j] = [J_i,\epsilon_{jqr}A_qB_r] = \epsilon_{jqr} [J_i,A_qB_r] = \epsilon_{jqr} ([J_i,A_q,]B_r + A_q[J_i,B_r])$$

Using the comutation relations for A and B $$\left [J_i,A_q \right ] = i\hbar \epsilon_{iqp}A_p$$ $$\left [J_i,B_r \right ] = i\hbar \epsilon_{irs}B_s$$

we can found

$$[J_i,C_j] = \epsilon_{jqr} (i\hbar \epsilon_{iqp}A_pB_r + A_qi\hbar \epsilon_{irs}B_s) = i\hbar\epsilon_{qjr}\epsilon_{qip}A_pB_r - i\hbar\epsilon_{rjq}\epsilon_{ris}A_qB_s$$

Remembering the levi-civita "multiplication" identity, we obtain

$$[J_i,C_j] = i\hbar(\delta_{ji}\delta_{rp}-\delta_{jp}\delta_{ri})A_pB_r - i\hbar(\delta_{ji}\delta_{qs}-\delta_{js}\delta_{qi})A_qB_s = i\hbar(A_rB_r - A_jB_i-A_sB_s+A_iB_j)$$

Using the Einstein Sum Notation we can see the equality between $$A_rB_r$$ and $$A_sB_s$$, and because of this, these terms annihilate each other.

In the end of the day, we have this:

$$[J_i,C_j] = i\hbar(A_iB_j- A_jB_i) = i\hbar\epsilon_{ijk}\epsilon_{klm}A_lB_m = i\hbar\epsilon_{ijk}C_k$$

There is the general statement about commutator between momenta operators, $$J_i$$ with scalar, vector and tensor operators. In your case the answer is $$[J_i,C_j]\sim\epsilon_{ijk}C_k,$$ where you substitute $$C_k=\epsilon_{kab}A_aB_b$$, so $$\epsilon_{ijk}\epsilon_{kab}A_aB_b=\epsilon_{ijk}\epsilon_{abk}A_aB_b=(\delta_{ia}\delta_{jb}-\delta_{ib}\delta_{ja})A_aB_b=A_iB_j-A_jB_i.$$

I am not sure what do you want to show, but let me scetch the derivation of general fact, $$[J_i,C_j]=\epsilon_{ijk}C_K$$, where $$C_j$$ is an arbitrary vector operator. Consider coordinate system $$r'$$ and coordinate system $$r$$, which differs in rotation, $$R_{\phi}$$. Two coordinate systems relates to each other by the relation $$r'=R_{\phi}r,$$ where $$R_{\phi}$$ is rotation matrix (and it will be written explicitly below). Then consider an arbitrary vector operator $$\hat{C}$$. Its components should transform by the similar relation, $$\hat{C'}=e^{-i\phi(n\cdot J)}\hat{C}e^{+i\phi(n\cdot J)},$$ which comes from the fact that operator $$J$$ describes rotations. For small angle $$\delta\phi$$, we can write down $$\hat{C'}\approx (1-i\delta\phi(J\cdot n))\hat{C}(1+i\delta\phi(J\cdot n))\approx\hat{C}-i\delta \phi [(n\cdot J),\hat{C}]\approx\hat{C}-\delta\phi[n\times\hat{C}],$$ where the last term comes from the Euler relation for rotation by small angle and this term can be rewritten in tensor denotations as $$n_i[J_i,\hat{C}_j]=-ie_{kil}n_i\hat{C}_l=i\epsilon_{ikl}n_i\hat{С}_l,$$ where you should understand unit vector $$n_i$$ as a direction of rotation. The last expression gives exactly the desired general statement for an arbitrary vector operator. Hope that this will help.

• I want to show exactly the first statement. I tried to write $C_j = \epsilon_{jab}A_aB_b$ and plug this in the statament Mar 21, 2020 at 19:19
• As I understand, you want to show that for any vector operator $C_j$ commutator will be $[J_i,C_j]=\epsilon_{ijk}C_k$? Mar 21, 2020 at 21:17
• Exactly! I want to show that $C_j$ obeys this statement Mar 21, 2020 at 22:25
• @CauêEvangelista I have updated the answer and hope that it will help. Mar 21, 2020 at 23:13