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 P: 2,179 Ryder: QFT: 1985: pp. 34-36: SU(2) and O(3) I have the second edition in front of me. The text you refer to is on page 33 of this edition. The text below eqn (2.40) says: We we see from (2.39) that $\xi$ and $\xi^\dagger$ transform in different ways. Here is eqn (2.39). Note that the arrow means 'transforms to'. I prefer the following notation which the author also uses in eqns (2.41) and (2.42) $$\xi^\prime = U\xi$$ $$(\xi^\dagger)^\prime = \xi^\dagger U^\dagger$$ These are the two different ways. For them to transform the same way would be $$\xi^\prime = U\xi$$ $$(\xi^\dagger)^\prime = U\xi^\dagger$$ but this is not how they transform. However, if you look carefully at eqns (2.41) and (2.42), you will see that the first one says: $$\xi^\prime = U\xi$$ and the second one says $$\chi^\prime = U\chi$$ where $$\chi = $\left( \begin{array}{c} -\xi^*_2 \\ \xi^*_1 \end{array} \right)$$$ This is what the author means by transform the same way. I will address H and h in my next post.