Ryder: QFT: 1985: pp. 3436: 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 [itex]\xi[/itex] and [itex]\xi^\dagger[/itex] 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)
[tex]\xi^\prime = U\xi[/tex]
[tex](\xi^\dagger)^\prime = \xi^\dagger U^\dagger[/tex]
These are the two different ways. For them to transform the same way would be
[tex]\xi^\prime = U\xi[/tex]
[tex](\xi^\dagger)^\prime = U\xi^\dagger[/tex]
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:
[tex]\xi^\prime = U\xi[/tex]
and the second one says
[tex]\chi^\prime = U\chi[/tex]
where
[tex]\chi = \[ \left( \begin{array}{c} \xi^*_2 \\ \xi^*_1 \end{array} \right)\][/tex]
This is what the author means by transform the same way. I will address H and h in my next post.
