I'm supposed to show that $\left[\mathbf A,\mathbf B\right]=0$ (for two vector operators $\mathbf A$ and $\mathbf B$) if and only if all components of $\mathbf A$ commute with all components of $\mathbf B$. What I have so far is this: \begin{align*} \mathbf A \otimes\mathbf B &= \mathbf A\mathbf B^T \Rightarrow \left[\mathbf A,\mathbf B\right] = \mathbf A\otimes \mathbf B - \left[\mathbf B \otimes \mathbf A\right]^T=\mathbf A\mathbf B^T - \left[\mathbf B \mathbf A^T\right]^T = \mathbf A\mathbf B^T - \mathbf B^T\! \mathbf A = 0. \end{align*} So much for the basic algebra... what am I supposed to do now? I can expand into components, but the next question "give an expression for $\left[\mathbf A,\mathbf B\right]_{ij}$ in Einstein notation" kind of suggests I should avoid saying $\mathbf A = a^i_j$ and go from there. I tried doing so anyway, with the obvious result of: \begin{align*} \mathbf A\mathbf B^T - \mathbf B^T\! \mathbf A = a^i_j\left(b^i_j\right)^T - \left(b^i_j\right)^T a^i_j = a^i_jb^j_i - b^j_i a^i_j = 0. \end{align*} Should I be interpreting the result as "One of these vectors will produce a matrix, the other will produce a scalar (an inner product), which means that this expression can only be zero when the matrix equals zero" (?). I'm at a loss as to how to approach this..

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    $\begingroup$ A and B are vector operators so, ${\bf A} \equiv a_i$ and ${\bf B} = b_i$. Then you result reads $[a_i , b_j ] = 0$ implying that each component of A must commute with every component of B. $\endgroup$
    – Prahar
    Nov 7, 2015 at 17:06
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    $\begingroup$ @Prahar How did you go from $\left[\mathbf A,\mathbf B\right]=0$ to $\left[a_i,b_j\right]=0$? Since $\mathbf B \equiv b_i \neq b_j$? $\endgroup$
    – user55789
    Nov 7, 2015 at 17:09
  • $\begingroup$ Why do your vectors have two indices? And where did you get that definition for $\otimes$ from? That looks very odd.. I guess you could write $[\vec A_i,\vec B_j]=(\vec A\otimes\vec B)_{ij}-(\vec B\otimes\vec A)_{ij}=[a_i,a_j]$ $\endgroup$
    – Daniel
    Nov 7, 2015 at 17:12
  • $\begingroup$ @Daniel I'm quite certain that this post screams "CONFUSION!!!" and I quite honestly can't debate it. The more in-depth your (or anyone's) explanation on the matter, the better. :-) $\endgroup$
    – user55789
    Nov 7, 2015 at 17:15
  • $\begingroup$ First of all, "direct product" $\neq$ "algebraic product" (as in algebra of operators), so $A\otimes B \neq AB$ and $A\otimes B \neq AB^T$. Then ${\bf A} = {A_i}_i$, meaning vector operators are labeled by one index each, not two. Third, see Prahar's first comment above. Don't complicate things just because you think this is a complicated thing. $\endgroup$
    – udrv
    Nov 7, 2015 at 17:46

1 Answer 1


I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate.

${\bf A}$ is a vector of operators. For example $$ {\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3} $$ We can denote this collectively as $A_i$. Now, note that each of these $A_i$'s are themselves operators. In other words, they are matrices $(A_i)_{ab}$. Thus, each element of $A$ has three indices. One index is the vector index and the other two are the matrix operator indices.

Finally, the very definition of $[{\bf A} , {\bf B}]$ is the following $$ [ {\bf A} , {\bf B} ]_{ij} = [ A_i , B_j ] $$ The latter can further be written in matrix notation as $$ [ A_i , B_j ]_{ab} = (A_i)_{ac} (B_j)_{cb} - (B_j)_{ac} (A_i)_{cb} $$ Thus, $[{\bf A} , {\bf B} ] = 0$ is precisely the statement that all components of ${\bf A}$ commute with those of ${\bf B}$.

Let me finally point out that this question is really poorly worded. I would say that $[ {\bf A} , {\bf B} ] = 0$ is by definition the statement that the components commute. More clearly, (if it is not already) I mean that $[ {\bf A} , {\bf B} ] = 0$ already means that the components commute. There is nothing to derive here.

  • $\begingroup$ This was very illumating! $\endgroup$
    – user55789
    Nov 7, 2015 at 18:23
  • $\begingroup$ @Prahar "Thus, each element of A has three indices..." although quite clear from the context, notice that indices of a vector operator are not the same as vector indices, as there is no basis they expand on. "Vector" operator is an unfortunate terminology given to a collection $A_i$ of single operators that happen to all together transform like a tensor under the action of the angular momentum operator. $\endgroup$
    – gented
    Nov 7, 2015 at 21:22
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    $\begingroup$ Minor remark: although $[\boldsymbol A,\boldsymbol B]$ is a rank-2 tensor, in three dimensions it is sometimes (implicitly) mapped into a vector by Hodge duality. This justifies formula $[\boldsymbol L,\boldsymbol L]=i\boldsymbol L$ sometimes find in the literature. $\endgroup$ Jul 9, 2018 at 21:38

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