Rotational generators in an arbitrary direction

Question

In the Ernest Abers Quantum mechanic's book (p.107-p.108) he proved:

$D(\vec{J}\cdot\hat{n})=\sum_{i}n_iD(J_i)=\hat{n}\cdot D(\vec{J})$ by using:

$$ \hat{R}=e^{-i\theta\vec{J}\cdot\hat{n}}=\exp{(-i\theta\ n_x J_x)}\exp{(-i\theta\ n_y J_y)}\exp{(-i\theta n_z J_z)}. $$

$\vec{J}=(J_x,J_y,J_z)$ means the generators of rotation matrices along x, y, z axes. And in p.12 he wrote: $$[J_i,J_j]=i\sum_{k}\epsilon_{ijk}J_k$$

So it means three components of $\vec{J}$ are not commutative. Why can he split $\hat{R}$ as a product of three exponent term directly?

Note: $\hat{R}$ is a rotation operator with axis $\hat{n}$. The representation of $\hat{R}$ is denoted as $D(\hat{R}(\hat{n},\theta))$.

$$ D(\hat{R})=e^{-i\theta D\big(\vec{J}\cdot \hat{n}\big)} $$

Answer 1

You dramatically misunderstood and misquoted Ernie's formula (4.17)!

The whole point of his formula is the term you skipped, $𝑂(𝜃^2)$ on the right-hand side, $$e^{-i\theta\vec{J}\cdot\hat{n}}=\exp{(-i\theta\ n_x J_x)}\exp{(-i\theta\ n_y J_y)}\exp{(-i\theta n_z J_z)} +𝑂(𝜃^2), $$ which is to say his rotations only compose like this for infinitesimally small rotations, a point he stresses in (4.28)! This is the essence of Lie groups, namely that the non commutativity first emerges at $𝑂(𝜃^2)$.

You may see this by two sequential applications of the CBH composition of exponentials of noncommuting operators, (4.25), namely $$ \exp{(-i\theta\ n_x J_x)}\exp{(-i\theta\ n_y J_y)}\exp{(-i\theta n_z J_z)} \\ = \exp \bigl( -i\theta ~\hat n\cdot \vec J -i (\theta^2 /2)(n_xn_y J_z- n_x n_zJ_y + n_y n_z J_x )+ O(\theta^3)\bigr) \\ = \exp(-i\theta ~\hat n\cdot \vec J ) + O(\theta^2). $$ cf. (4.29).

As far as non commutativity goes, he threw out the baby with the bathwater, but he is illustrating mappings of sums of matrices, not their noncommutative multiplication...