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I have been working throught Weinberg´s book The quantum theory of fields volume I. In page 250 he derives an equation (5.9.22) for the transformation of the polarization vector for a standard vector k: \begin{equation} D\left(W\left(\theta,\alpha,\beta\right)\right)_{\,\nu}^{\mu}\epsilon^{\nu}\left(\mathbf{k},\sigma\right)=S\left(\alpha,\beta\right)_{\,\gamma}^{\mu}R\left(\theta\right)_{\,\nu}^{\gamma}\epsilon^{\nu}\left(\mathbf{k},\sigma\right)=e^{i\sigma\theta}\left[e^{\mu}\left(\mathbf{k},\sigma\right)+\frac{\left(\alpha+i\sigma\beta\right)}{\sqrt{2}\left|\mathbf{k}\right|}k^{\mu}\right] \end{equation} then in page 251, we says that the transformation for arbitrary momentum p is \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)+p^{\mu}\Omega_{\sigma}\left(\mathbf{p},\Lambda\right)=e^{i\sigma\theta}e^{\mu}\left(\mathbf{p}_{\Lambda},\sigma\right) \end{equation} I was trying to obtain this result but I get an extra $\Lambda$ in the second term of the left hand side. The procedure is the following. In this representation we have \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)=\Lambda_{\,\nu}^{\mu} \end{equation} applying this to the polarization vector with momentum p \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=\Lambda_{\,\nu}^{\mu}\epsilon^{\nu}\left(\mathbf{p},\sigma\right) \end{equation} now I use the standard boost \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=\Lambda_{\,\alpha}^{\mu}L\left(\mathbf{p}\right)_{\,\nu}^{\beta}\epsilon^{\nu}\left(\mathbf{k},\sigma\right) \end{equation} and I insert a idendity \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=L\left(\Lambda\mathbf{p}\right)_{\,\gamma}^{\mu}\left(L^{-1}\left(\Lambda\mathbf{p}\right)_{\,\epsilon}^{\gamma}\Lambda_{\,\alpha}^{\epsilon}L\left(\mathbf{p}\right)_{\,\nu}^{\beta}\right)\epsilon^{\nu}\left(\mathbf{k},\sigma\right) \end{equation} so we get the element of the little group \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=L\left(\Lambda\mathbf{p}\right)_{\,\nu}^{\mu}\left(W\left(\Lambda,\mathbf{p}\right)_{\,\alpha}^{\nu}\right)\epsilon^{\alpha}\left(\mathbf{k},\sigma\right) \end{equation}

\begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=L\left(\Lambda\mathbf{p}\right)_{\,\nu}^{\mu}D_{\,\delta}^{\nu}\left(W\left(\alpha,\beta,\theta\right)\right)\epsilon^{\delta}\left(\mathbf{k},\sigma\right) \end{equation}

at this point I use the equation (5.9.22) of Weinberg´s boook \begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=e^{i\sigma\theta}L\left(\Lambda\mathbf{p}\right)_{\,\nu}^{\mu}\left[e^{\nu}\left(\mathbf{k},\sigma\right)+\frac{\left(\alpha+i\sigma\beta\right)}{\sqrt{2}\left|\mathbf{k}\right|}k^{\nu}\right] \end{equation}

\begin{equation} D_{\,\nu}^{\mu}\left(\Lambda\right)\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=e^{i\sigma\theta}\left[e^{\mu}\left(\mathbf{p}_{\Lambda},\sigma\right)+\frac{\left(\alpha+i\sigma\beta\right)}{\sqrt{2}\left|\mathbf{k}\right|}\left(\Lambda p\right)^{\mu}\right] \end{equation} \begin{equation} \Lambda_{\,\nu}^{\mu}\epsilon^{\nu}\left(\mathbf{p},\sigma\right)=e^{i\sigma\theta}\left[e^{\mu}\left(\mathbf{p}_{\Lambda},\sigma\right)+\frac{\left(\alpha+i\sigma\beta\right)}{\sqrt{2}\left|\mathbf{k}\right|}\left(\Lambda p\right)^{\mu}\right] \end{equation} as you can see there is an extra $\Lambda$ in the term proportional to p. ¿What am I doing wrong?

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