In Appendix 2B of Weinberg's QFT Vol I, he provided a proof for a theorem that the phases of the operators $U(T)$ for finite symmetry transformations $T$ may be chosen so that these operators form a representation of the symmetry group, rather than a projective representation.

Once he constructed a non-projective representation $U[\theta]$, he proceeds to prove any projective representation $\tilde U[\theta]$ of the same group with the same representation generators $t_a$ can only differ from $U[\theta]$ by a phase $$\tilde{U}[\theta]=e^{i \alpha(\theta)} U[\theta]$$ such that the phase $\phi$ in $$\tilde{U}\left[\theta^{\prime}\right] \tilde{U}[\theta]=e^{i \phi\left(\theta^{\prime}, \theta\right)} \tilde{U}\left[f\left(\theta^{\prime}, \theta\right)\right]$$ can be removed by a change of phase for $\tilde U[\theta]$

To prove this, he first considered the operator $$U[\theta]^{-1} U\left[\theta^{\prime}\right]^{-1} \tilde{U}\left[\theta^{\prime}\right] \tilde{U}[\theta]=U\left[f\left(\theta^{\prime}, \theta\right)\right]^{-1} \tilde{U}\left[f\left(\theta^{\prime}, \theta\right)\right] e^{i \phi\left(\theta^{\prime}, \theta\right)}\tag{1}$$ and he asserted that because $U[\theta]$ and $\tilde{U}[\theta]$ have the same generators, the derivative of the left-hand side with respect to $\theta^{\prime a}$ vanishes at $\theta^{\prime}=0$, and so $$0=\frac{\partial}{\partial \theta^b}\left\{U[\theta]^{-1} \tilde{U}[\theta]\right\}+i \phi_b(\theta) U[\theta]^{-1} \tilde{U}[\theta]\tag{2}$$ where $$\phi_b(\theta) \equiv h_b^a(\theta)\left[\frac{\partial}{\partial \theta^{\prime b}} \phi\left(\theta^{\prime}, \theta\right)\right]_{\theta^{\prime}=0}$$ and $h^a_b$ is defined implicitly as $$\left[h^{-1}\right]_b^a(\theta) \equiv\left[\frac{\partial f^a(\bar{\theta}, \theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0} \tag{2.B.4}$$ Finally, by differentiating this result with respect to $\theta^c$ and antisymmetrizing in $b$ and $c$ we have $$0=\frac{\partial \phi_b(\theta)}{\partial \theta^c}-\frac{\partial \phi_c(\theta)}{\partial \theta^b}\tag{3}$$

Notice in $(1)$, the partial derivative is with respect to $\theta$, rather than the described $\theta'$, could this be a misprint? Also the the introduction of $h^a_b$ in the definition of $\phi_b(\theta)$ feels unmotivated, the indices in the definition of $\phi_b$ does not even match the summation convention. I wonder how one should go from $(1)$ to $(2)$ then to $(3)$, any help is appreciated.


1 Answer 1


OK, it seems I managed to work out the missing steps myself. First, bearing in mind the definition for $h^{-1}$ and by Taylor's theorem we have $$U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]\approx U[\theta^a+(h^{-1})^a_b\theta'^b]^{-1}\tilde U[\theta^a+(h^{-1})^a_b\theta'^b]$$ Let $\delta \theta^a=(h^{-1})^a_b\theta'^b$, and using again the Taylor's theorem we have $$U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]\approx U[\theta^a]^{-1}\tilde U[\theta^a]+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]\tilde U[\theta^a])$$ Thus, $$\begin{align} \frac{\partial}{\partial \theta'^b}\left(U[f(\theta',\theta)]^{-1}\tilde U[f(\theta',\theta)]e^{i\phi(\theta',\theta)}\right)=U[\theta^a]^{-1}\tilde U[\theta^a]\frac{\partial}{\partial \theta'^b}e^{i\phi(\theta',\theta)}+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]^{-1}\tilde U[\theta^a)e^{i\phi(\theta',\theta)}+(h^{-1})^a_b\theta'^b\frac{\partial}{\partial \theta^a}(U[\theta^a]^{-1}\tilde U[\theta^a])\frac{\partial}{\partial \theta'^b}e^{i\phi(\theta',\theta)} \end{align}$$ Now, take the limit as $\theta'\rightarrow0$, so the third term vanishes, simplify and cancelling the phases, then multiply through by $h^a_b$ yields the desired result. Notice there is a typo in Weinberg's book, in the definition of $\phi_b$, $\partial/\partial\theta'^b$ should have been $\partial/\partial\theta'^a$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.