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Living_Dog
Living_Dog is offline
#1
Nov17-10, 09:11 AM
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) [tex]\xi[/tex] does not transform like [tex]\xi^{+}[/tex], but [tex]\[ \left( \begin{array}{c} \xi_1 \\ \xi_2 \end{array} \right)\][/tex] and [tex]\[ \left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)\][/tex] 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: [tex]\xi' \equiv U\xi, \xi^{+}' \equiv \xi^{+}U^{+}, ...[/tex] 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:

[tex]\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[/tex].

  • if [tex]\xi \equiv (\xi_1 \xi_2)[/tex] then [tex]\xi^{+}[/tex] is not what he uses here - he uses that mysterious other spinor that is NOT [tex]\xi^{+}[/tex]!!
  • 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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