Rotation operator as discussed in S. Weinberg Following the discussion on rotation operators in Sakurai made clear sense, however, due to coursework, I need to also understand the discussion provided in Quantum Mechanics by S. Weinberg. 
In Chapter 4, page 100, he writes the infinitesimal rotation matrix as
$$\tag{1} \mathbb{R}=\mathbb{1}+\mathbb{W}$$
where $\mathbb{W}$ is a matrix. The orthogonality condition readily gives $\mathbb{W}^T=-\mathbb{W}$. So far so good. 
Then, Weinberg goes on to write
$$\tag{2} U(1+\mathbb{W})\rightarrow 1+\frac{i}{2\hbar}\sum_{ij}\mathbb{W}_{ij}\mathbb{J}_{ij}+\mathcal{O}(\mathbb{W}^2)$$
where $\mathbb{J}$ is a matrix of operators. He doesn't really explain much how we can write this down.


*

*We can write an infinitesimal unitary operator as $U(\epsilon)=1+i\epsilon\  \hat{T}$ for some Hermitian operator $\hat{T}$. Applying this to Eq (2), why don't we write $$U(1+\mathbb{W})\rightarrow 1+\frac{i}{2\hbar}\sum_{ij}(\delta_{ij}+\mathbb{W}_{ij})\mathbb{J}_{ij}+\mathcal{O}(\mathbb{W}^2)\ ?$$

*I have seen vector of operators, but I have never encountered a matrix of operators until now. What allows us to guess that we actually need to introduce a matrix of operators in Eq (2)?

*On page 101, he says that in three-dimensions, we can define a vector of operators $\mathbf{J}$ with components 
$$J_1=\mathbb{J}_{23} \quad J_2=\mathbb{J}_{31} \quad J_3=\mathbb{J}_{12}.$$
How? Why these entries of the matrix of operators $\mathbb{J}$?


Understanding rotations in QM made a lot of sense when reading Sakurai or other references! Weinberg's treatment of the problem seems too convoluted and there's a great lack of contextual justification in steps taken. Any help in understanding the concept behind a matrix of operators that shows up in Eq (2) and picking the $J_x,J_y,J_z$ components would be appreciated. 
 A: *

*What you mean by $U(\epsilon)$ Weinberg here seems to write as $U(1+\epsilon)$ - it's a difference in notation, but you both mean the same thing.

*Weinberg is writing $\mathbb{W} = \sum_{i,j} \mathbb{W}_{ij}\mathbb{J}_{ij}$, where the $\mathbb{J}_{ij}$ are the basic antisymmetric matrices with a $1$ at the $ij$-th position, a $-1$ at the $ji$-th position and zeroes elsewhere. Since $\mathbb{W}$ is an antisymmetric matrix and these form a basis for the vector space of antisymmetric matrices, $\mathbb{W}$ can be expressed as this linear combination of them.

*Due to $\mathbb{W}^T = -\mathbb{W}$, these are the independent non-vanishing components of $\mathbb{W}$ in three dimensions.
A: I've had similar doubts with that chapter, and I think that I've been able to solve them, so here are my takeaways:
$U(1+\omega)$ is the unitary operator associated to the infinitesimal rotation $R=1+\omega$, and that does not mean that $1+\omega$ itself is the parameter of the unitary operator. Actually, the parameters are the elements of the $\omega$ matrix, the generator of rotations in $\mathbb{R}^n$, $\omega_{ij}$. According to this, we could write $U(1+\omega)$ without loss of generality as
\begin{equation}\tag{1'}
U(1+\omega)=1+\frac{i}{2\hbar}\sum_{ij} \omega_{ij} J_{ij} + O(\omega^2),
\end{equation}
where each $J_{ij}$ is an hermitian operator (so as for $U$ to be unitary). The subscripts of $J_{ij}$ are just one way of labeling each of them, and they indicate which parameter they are related to. For example, the $J_{23}$ operator is the generator of a transformation made by an infinitesimal $\omega_{23}$.
Now, since $\omega_{ij}=-\omega_{ji}$, we have
$$\sum_{ij}\omega_{ij}J_{ij}=\sum_{ij}(-\omega_{ji})J_{ij}=\sum_{ij}(-\omega_{ij})J_{ji}=\sum_{ij}\omega_{ij}(-J_{ji})\Longrightarrow J_{ij}=-J_{ji},\tag{2'}$$
where I have performed $i\leftrightarrow j$ in the second step. Note that this does not say anything about tensors yet, it only says that the $J_{ji}$ operator is equal to the negative of the $J_{ij}$ operator.
Following what Weinberg says in the book, it can be proved that the $J_{ij}$ operators are the components of a tensor, which turns out to be antisymmetric due to $(2')$.
To sum up, $J_{ij}$ are hermitian operators, and they are the components of an antisymmetric tensor $\mathbb{J}$. The dimension of $\mathbb{J}$ will be the same as the space where we perform the rotation, so if it is in $\mathbb{R}^3$, then
$$\mathbb{J}=\begin{pmatrix}0&J_{12}&J_{13}\\-J_{12}&0&J_{23}\\-J_{13}&-J_{23}&0\end{pmatrix},\tag{3'}$$
where each $J_{ij}$ is an hermitian operator. Now, in $\mathbb{R}^3$, since there are only $3$ independent $J_{ij}$, we can define a three-component vector $\mathbf{J}=(J_1,J_2,J_3),$ where
$$J_k=\frac{1}{2}\sum_{ij}\epsilon_{ijk}J_{ij}\tag{4'}.$$
I know that I'm a bit late in the discussion, but I hope this could help someone in the future.
