How is the commutator of two vectors of operators in Quantum Mechanics defined?

Question

I'm in a quantum mechanics class, and I have some questions about how the commutator in some cases; what I know is that: if $\hat{A}$ and $\hat{B}$ are operators, their commutator is defined as $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$. (A hat: $\hat{}$, is used to denote an operator, and $\vec{\hat{}}$ to represent a vector of operators).

1. Given the generalized angular momentum $\vec{\hat{J}}=\hat{J_1} e_1 + \hat{J_2} e_2 + \hat{J_3 }e_3$, prove that $[\hat{J_i},\hat{J}^2]=\hat{0}$, where $\hat{J}^2 = \hat{J_1}^2 + \hat{J_2}^2 +\hat{J_3}^2$. They use the fact that $[\hat{J_i},\hat{J_j}] = i\hbar\epsilon_{ijk}\hat{J_k}$. There is no problem with that, but how do I get $[\hat{J_i},\vec{\hat{J}}]$? . What I think is $$ [\hat{J_i},\vec{\hat{J}}] =[\hat{J_i},\hat{J_1} e_1 + \hat{J_2} e_2 + \hat{J_3 }e_3] = [\hat{J_i},\hat{J_1}]e_1 + [\hat{J_i},\hat{J_2}]e_2 + [\hat{J_i},\hat{J_3}]e_3 . $$ In the matrix representation, this looks like $$ \begin{bmatrix} & \\ \hat{J_i}, & \begin{pmatrix} \hat{J_1} \\ \hat{J_2}\\ \hat{J_3} \end{pmatrix} \\ & \end{bmatrix} = \begin{pmatrix} [\hat{J_i},\hat{J_1}]\\ [\hat{J_i},\hat{J_2}]\\ [\hat{J_i},\hat{J_3}] \end{pmatrix}. $$ Is this correct?

2. Given the vectors $$ A = \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix} a_i \in \mathbb{C} , \vec{\hat{P}}= \begin{pmatrix} \hat{P_1}\\ \hat{P_2}\\ \hat{P_3} \end{pmatrix}, \vec{\hat{X}}= \begin{pmatrix} \hat{X_1}\\ \hat{X_2}\\ \hat{X_3} \end{pmatrix}. $$ If I want to compute $[\hat{X_i}, \vec{A}\circ\vec{\hat{P}}]$, where $\circ$ is the usual dot product, what I would do is $$ [\hat{X_i}, \vec{A}\circ\vec{\hat{P}}] = [\hat{X_i}, a_1\hat{P_1}+a_2\hat{P_2}+a_3\hat{P_3}] = a_1[\hat{X_i},\hat{P_1}]+a_2[\hat{X_i},\hat{P_2}] + a_3[\hat{X_i},\hat{P_3}]. $$

But the same result is found if I do $$ [\hat{X_i}, \vec{A}\circ\vec{\hat{P}}] = \vec{A}\circ[\hat{X_i}, \vec{\hat{P}}], $$ Since, if (1) is true, then $$ \vec{A}\circ[\hat{X_i}, \vec{\hat{P}}] = \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix}\circ \begin{pmatrix} [\hat{X_i},\hat{P_1}]\\ [\hat{X_i},\hat{P_2}]\\ [\hat{X_i},\hat{P_3}] \end{pmatrix}. $$ So, my conclusion is that a vector of complex numbers, when dotted on a vector of operators acts as a scalar in the sense that it can be taken out of the bracket. The last makes me ask what would be $[\vec{A}\hat{X_i},\vec{\hat{P}}]$? Since $\hat{X_i}$ acts as a scalar, concerning $\vec{A}$, that would be $$ [\vec{A}\hat{X_i}, \vec{\hat{P}}] = \begin{bmatrix} & \\ \begin{pmatrix} a_1\hat{X_i}\\ a_2\hat{X_i}\\ a_3\hat{X_i} \end{pmatrix}, & \begin{pmatrix} \hat{P_1}\\ \hat{P_2}\\ \hat{P_3} \end{pmatrix} \\ & \end{bmatrix}. $$ Since it must be the same as the previous result, this means that $$ \begin{bmatrix} & \\ \begin{pmatrix} a_1\hat{X_i}\\ a_2\hat{X_i}\\ a_3\hat{X_i} \end{pmatrix}, & \begin{pmatrix} \hat{P_1}\\ \hat{P_2}\\ \hat{P_3} \end{pmatrix} \\ & \end{bmatrix} = a_1[\hat{X_i},\hat{P_1}]+a_2[\hat{X_i},\hat{P_2}] + a_3[\hat{X_i},\hat{P_3}] . $$ But I'm not even sure of what the actual definition of $[\vec{\hat{X}},\vec{\hat{P}}]$ is. Your help is appreciated, as any reference where I can find the definitions for general cases.

Answer 1

1. Yes.
2. Use the Einstein convention of summing over repeated indices. Define $\vec A \cdot \vec {\hat X}=a_i \hat X_i$, $\vec B \cdot \vec {\hat P}=b_j \hat P_j$, so that, mindful that the commutator is a bilinear function of its two arguments,
   $$ [\vec A \cdot \vec {\hat X}, \vec B \cdot \vec {\hat P}]=a_i b_j ~~[\hat X_i, \hat P_j], $$ the trace of a dyadic with a matrix of operators (commutators), $$ \begin{pmatrix} [X_1,P_1] & [X_1,P_2] & ...\\ [X_2,P_1] & [X_2,P_2] &...\\ ...&...&...\end{pmatrix}. $$