# Commutation relations of angular momentum operators

On page 54 of Weinberg's QFT I, he says that an element $$T(\theta)$$ of a connected Lie group can be represented by a unitary operator $$U(T(\theta))$$ acting on the physical Hilbert space. Near the identity, he says that $$U(T(\theta)) = 1 + i\theta^a t_a + \frac{1}{2}\theta^a\theta^bt_{ab} + \ldots. \tag{2.2.17}$$ the group multiplication law says that $$T(\bar{\theta}) T(\theta)=T(f(\bar{\theta}, \theta))$$ and we then have $$U(T(\bar{\theta})) U(T(\theta))=U(T(f(\bar{\theta}, \theta)))$$ we could expand $$f^a(\bar{\theta},\theta)$$ to second order $$f^{a}(\bar{\theta}, \theta)=\theta^{a}+\bar{\theta}^{a}+f_{b c}^{a} \bar{\theta}^{b} \theta^{c}+\cdots$$ ... After calculation we have the commutation relations of $$\{t_a\}$$ $$\left[t_{b}, t_{c}\right]=i C_{b c}^{a} t_{a}$$ where $$C_{b c}^{a} \equiv-f_{b c}^{a}+f_{c b}^{a}$$ so we know that if $$f$$ of a symmetry group has the form $$f^{a}(\theta, \bar{\theta})=\theta^{a}+\bar{\theta}^{a}$$ so that $$f_{c b}^{a} = 0$$, then the generators have the commutation relations $$\left[t_{b}, t_{c}\right]=0$$

Now consider the rotation $$R_{\theta}$$ by an angle $$|\theta|$$ around the direction of $$\theta$$, it is clear that the rotation is additive, so that $$f_{c b}^{a} = 0$$, so it seems that the generators of the rotation group commute with each other. And furthermore the rotation can be written from the generators $$U\left(R_{\theta}, 0\right)=\exp (i \mathbf{J} \cdot \theta)$$ But we know that the angular momentum generators have non-trivial commutation relations $$\left[J_{i}, J_{j}\right]=i \epsilon_{i j k} J_{k}$$ I want to know at what point do this line of argument go wrong to have a different conclusion of the commutation relations of angular momentum generators.

$$U\left(R_{\theta}, 0\right)=\exp (i \mathbf{J} \cdot \theta)$$
If by $$\mathbf{J}$$ you mean a single operator $$n_xJ_x+n_yJ_y+n_zJ_z$$ then your operation generates a one-parameter Abelian subgroup, and you're not gonna get any commutation out of that other than $$\mathbf{J}$$ as the operator $$n_xJ_x+n_yJ_y+n_zJ_z$$ commutes with itself.
If by $$\mathbf{J}$$ you mean a triple $$(J_x,J_y,J_z)$$, then your $$\theta$$ should really be a vector $$(\theta_x,\theta_y,\theta_z)$$. In this case you'd have something like $$\mathbf{J} \cdot \boldsymbol{\theta}=\theta^aJ_a$$ and you can use the the general direction suggested by Weinberg: the expansion will produce cross terms in $$J^a J^b$$ and computing $$\bar T(\boldsymbol{\theta}) T(\boldsymbol{\theta})=\mathbb{I}$$ will lead to commutation relations.
Note that $$e^{i \mathbf{J} \cdot \boldsymbol{\theta}}$$ is a rotation by a single axis but the net rotation angle is not so trivial to work out in terms of the "components" $$(\theta_x,\theta_y,\theta_z)$$. Indeed one usually writes \begin{align} e^{-i\omega \hat{\mathbf{n}}\cdot \mathbf{J}} \end{align} for a rotation by an angle $$\omega$$ about an axis specified by the unit vector $$\hat{\mathbf{n}}$$, itself specified by two polar angles $$\Theta,\Phi$$ and that \begin{align} e^{-i\omega \hat{\mathbf{n}}\cdot \mathbf{J}}\mathbf{r}:= \mathbf{r}' =\mathbf{r}\cos\omega+\hat{\mathbf{n}} (\hat{\mathbf{n}}\cdot \mathbf{r})(1-\cos\omega)+(\hat{\mathbf{n}}\times \mathbf{r})\sin\omega\, , \end{align} with $$n_x=\sin\Theta\cos\Phi$$ etc.
• Thanks! @ZeroTheHero My confusion: If we assume $e^{J \cdot \theta}$ = $e^{\theta_x J_x}$, then combining the $\{\theta_x\}$ with $\{\theta_{x’}\}$ rotation symmetry operations should produce a rotation with parameters $\{ \theta_y\}$ such that each component of $\{ \theta_y\}$ is the sum of $\{\theta_x\}$ and $\{\theta_{x’}\}$, namely $f^a(\theta, \bar{\theta}) = \theta^a + \bar{ \theta}^a$. (Since rotation around a fixed axis is just component adding) Then according to the discussion of Weinberg, the generators themselves should commute with each other. Sep 5, 2020 at 1:12