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 P: 102 I have been struggling with this for a long time. I gave up to review GR and came back to Ryder. I started in Chapter 2 and the material was easier and more intuitive on this second pass. But the same topic, "SU(2) and the rotation group" has trapped me yet again. I am bogged down specifically on pages 34 - 36. So here are my questions: 1) From pg. 35: Under SU(2) $$\xi$$ does not transform like $$\xi^{+}$$, but $$$\left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)$$$ and $$$\left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)$$$ do.where does he get that (is it magic)?? I can choose another spinor and it also transforms "the same way" ...what does he mean by "the same way"??? I tried all four possibilities: $$\xi' \equiv U\xi, \xi^{+}' \equiv \xi^{+}U^{+}, ...$$ and the other two with the second spinor. None of them look the same, and they all transform "the same way"! 2) From pg. 36: $$\xi \xi^{+} \equiv $\left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)$ $\left( \begin{array}{cc} -\xi_2 & \xi_1 \end{array} \right)$ \equiv -H$$. if $$\xi \equiv (\xi_1 \xi_2)$$ then $$\xi^{+}$$ is not what he uses here - he uses that mysterious other spinor that is NOT $$\xi^{+}$$!! what's the point of calling this thing 'H'?? why does he construct 'h' - as if from nowhere then he says 'h' is 'H'! since 'h' is a 2x2 matrix, how can it act on r?? If one thing is clear from my post it is that I have no clue what Ryder is talking about on these two short pages. Thanks in advance to anyone for anything anywhere at any time!!! (PS: I can't get that column vector times a row vector to equal the 'H' matrix to display properly. Sorry.)