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.


1 Answer 1


The tricky part is the interpretation of
$$ U\left(R_{\theta}, 0\right)=\exp (i \mathbf{J} \cdot \theta) $$

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.

  • $\begingroup$ 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. $\endgroup$
    – Jiahao Fan
    Sep 5, 2020 at 1:12

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