My question follows chapter 10 of Tung's Group Theory book, in particular definitions 10.11 and 10.12. Let $\mathcal{R}[\Lambda]$ be an $n\times n$ matrix representation of $L_+^{\uparrow}$ and a wave-function $\{\Psi^a\}$, which transforms (actively) under Lorentz as

$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\equiv\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x).\tag{1}$$

I think I understand this: under active Lorentz transformation $x\mapsto\Lambda^{-1}x:$, $\mathcal{R}[\Lambda]\Psi(x)\overset{\text{passive}}{\equiv}{\Psi}'(\Lambda x)$ becomes $\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x)\equiv{\Psi}'(x)$, where the equivalence is due to passive transformations involving changes of frames, as you can see from the image below


On the other hand, when we consider operator-fields Tung writes

$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\overset{\text{op. tr.}}{:=}U[\Lambda]\Psi(x)U[\Lambda^{-1}]\equiv\mathcal{R}[\Lambda^{-1}]\Psi(\Lambda x).\label{eq2}\tag{2}$$

In think the last expression should be

$$\Psi\xrightarrow{\Lambda}{\Psi}'\quad {\Psi}'(x)\overset{\text{op. tr.}}{:=}U[\Lambda]\Psi(x)U[\Lambda^{-1}]\equiv\mathcal{R}[\Lambda]\Psi(\Lambda^{-1}x).\tag{3}$$

I know I am missing something, because \eqref{eq2} is similar to 5.1.6-7 in Weinberg's book QFT vol. 1, but involving also translations.


1 Answer 1


I don't know if it helps, but if $\varphi(\boldsymbol{a},\mathbb{L})$ is an unitary realization (carrier space is infinite because acts on operators) of the Poincaré group (external semi-direct product of translations and Lorentz), then field operators transform by definition like \begin{gather*} \tilde{\phi}^\alpha(\tilde{\boldsymbol{x}}) = \varphi^\dagger(\boldsymbol{a},\mathbb{L}) \phi^\alpha(\tilde{\boldsymbol{x}}) \varphi(\boldsymbol{a},\mathbb{L}) = {{\mathbb{S}(\boldsymbol{\vartheta})}^\alpha}_\beta \phi^\beta\left(\mathbb{L}^{-1}(\tilde{\boldsymbol{x}}-\boldsymbol{a})\right) \end{gather*} where $\mathbb{S}$ is a matrix belonging to the irreducible unitary representation (dimension finite: the same number of the field components) of the Poincaré group on the components of the field. Hope this helps even if I have the suspect it doesn't.


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