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  (A,B)-Representation of Lorentz Group: Coefficient functions of fields

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I have a question regarding the construction of general causal fields in Weinberg's book on quantum field theory.

In his conventions a field that transforms according to the irreducible (A,B) representation of the Lorentz group is given by (eq.5.7.1) \begin{equation} \psi_{ab}=(2\pi)^{-3/2}\sum_{\sigma}\int d^3p\left[\kappa a(\boldsymbol{p},\sigma)e^{ipx}u_{ab}(\boldsymbol{p},\sigma)+\lambda a^{c\dagger}(\boldsymbol{p},\sigma)e^{-ipx}v_{ab}(\boldsymbol{p},\sigma)\right]\, . \end{equation} Here, $a$ and $a^{\dagger}$ are the usual creation and annihilation operators, $u_{ab}$ and $v_{ab}$ are coefficients carrying an irreducible representation of the Lorentz group, and $\kappa$ and $\lambda$ are coefficients.

The zero-momentum coefficients $u_{ab}(0,\sigma)$ have to fulfill the conditions \begin{equation} \sum_{\bar{\sigma}}u_{\bar{a}\bar{b}}(0,\bar{\sigma})\boldsymbol{J}^{(j)}_{\bar{\sigma}\sigma}=\sum_{ab}\mathcal{J}_{\bar{a}\bar{b},ab}u_{ab}(0,\sigma) \end{equation} \begin{equation} -\sum_{\bar{\sigma}}v_{\bar{a}\bar{b}}(0,\bar{\sigma})\boldsymbol{J}^{(j)*}_{\bar{\sigma}\sigma}=\sum_{ab}\mathcal{J}_{\bar{a}\bar{b},ab}v_{ab}(0,\sigma), \end{equation} where $\boldsymbol{J}^{(j)}_{\bar{\sigma}\sigma}$ are the angular momentum matrices in the$j$- representations of the rotation group, and $\mathcal{J}_{\bar{a}\bar{b},ab}v_{ab}(0,\sigma)$ are the angular momentum matrices in the $(A,B)$ representation of the Lorentz-group.

Weinberg shows that $u_{ab}(0,\sigma)$ is given by \begin{equation} u_{ab}(0,\sigma)=(2m)^{-1/2}C_{AB}(j\sigma;ab)\, , \end{equation} where $C_{AB}(j\sigma;ab)$ is the Clebsch-Gordan coefficient and the normalization was chosen for convenience. However, when I try to calculate the coefficient $u_{ab}$ in the $(1/2,1/2)$ representation and want to relate them to the $u^{\mu}$ obtained when working directly in the vector representation of the Lorentz group I cannot reproduce them. , where \begin{equation} u^{\mu}(0,\sigma=0)=(2m)^{-1/2}\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\qquad u^{\mu}(0,\sigma=1)=-\frac{1}{\sqrt{2}}(2m)^{-1/2}\begin{pmatrix}0\\1\\+i\\0\end{pmatrix} \end{equation} \begin{equation} u^{\mu}(0,\sigma=-1)=\frac{1}{\sqrt{2}}(2m)^{-1/2}\begin{pmatrix}0\\1\\-i\\0\end{pmatrix}\, . \end{equation} What is the procedure of translating from (A,B) to a mixture of Lorentz indices and spinor indices in more general cases, such as in the Rarita-Schwinger field?

This post imported from StackExchange Physics at 2014-08-07 15:37 (UCT), posted by SE-user Lurianus
asked Jul 26, 2014 in Theoretical Physics by Lurianus (15 points) [ no revision ]

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