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## Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3)

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.
1. where does he get that (is it magic)??
2. I can choose another spinor and it also transforms "the same way"
3. ...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.)
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