Parity transformations and Dirac Spinor

Question

I'm reading "No-Nonsense quantum field theory" and I have some doubts about the transformation law for Dirac Spinors as explained by the author. In the book the left chiral spinors $\chi$ and right chiral spinors $\xi$ are introduced as objects that have two components and behave under rotations $R$ around $x$-axis and boosts along $z$-axis $B$ as follows:

$$\chi_a \rightarrow R_{ab}^{(\chi x)} \chi_b \\ \chi_a \rightarrow B_{ab}^{(\chi z)} \chi_b $$

where $$R_{ab}^{\chi z} = \begin{pmatrix} \cos(\theta/2) & i\sin(\theta/2)\\\ i\sin(\theta/2) & \cos(\theta/2)\end{pmatrix} \\ \\ B_{ab}^{(\chi z)} = \begin{pmatrix} e^{\phi/2} & 0\\\ 0 & e^{-\phi/2}\end{pmatrix}$$

and

$$\xi_a \rightarrow R_{ab}^{(\xi x)} \xi_b \\ \xi_a \rightarrow B_{ab}^{(\xi z)} \xi_b $$

where $$R_{ab}^{\xi z} = \begin{pmatrix} \cos(\theta/2) & i\sin(\theta/2)\\\ i\sin(\theta/2) & \cos(\theta/2)\end{pmatrix} \\ \\ B_{ab}^{(\xi z)} = \begin{pmatrix} e^{-\phi/2} & 0\\\ 0 & e^{\phi/2}\end{pmatrix}$$

Then the author introduces the Dirac spinor: $$\Psi = (\chi, \xi)^T$$ which tranforms under boosts as

$$(\chi, \xi)^T \rightarrow \begin{pmatrix} B^{(\chi z)} (\phi) & 0\\\ 0 & B^{(\xi z)} (\phi)\end{pmatrix} (\chi, \xi)^T$$. So far I'm following the argument, but then the author claims the equation just above becomes:

$$(\chi, \xi)^T \rightarrow \begin{pmatrix} B^{(\xi z)} (\phi) & 0\\\ 0 & B^{(\chi z)} (\phi)\end{pmatrix} (\xi, \chi)^T$$ because under parity transformation we have $B^{(\xi z)} (\phi) \rightarrow B^{(\xi z)} (-\phi) = B^{(\chi z)} (\phi)$ and $B^{(\chi z)} (\phi) \rightarrow B^{(\chi z)} (-\phi) = B^{(\xi z)} (\phi)$. And then asserts that this implies that the Dirac Spinor $\Psi$ transforms under parity transformations as $$ \Psi = (\chi, \xi)^T \rightarrow (\xi, \chi)^T$$ I'm confused about why the last statement follows from the discussion above. I've also attached a picture of the section of the book where I got this from:

Answer 1

Under parity in spherical coordinate we have,

$$\textbf{P} \theta = \pi - \theta \\ \textbf{P} \phi = \pi + \phi $$

This explains why ,

$$ B^{(\xi z)} (\phi) \rightarrow B^{(\chi z)} (\phi)$$

Now we need to know what do you mean by a left-chiral spinor. Left-chiral spinor is an object which transforms like this boost,

$$ \chi' \rightarrow B^{(\chi z)} \chi \;\;\;\;\;\; \;\;\;\;\;\; eq.1$$

similar thing is true for the right-handed spinor.

Let's start with ,

$$ \xi' \rightarrow B^{(\xi z)} \xi \;\;\;\;\;\; \;\;\;\;\;\;$$

Now we apply parity to both sides.

$$ \textbf{P} \xi' \rightarrow \textbf{P} ( B^{(\xi z)} \xi ) \;\;\;\;\;\; \;\;\;\;\;\; \\ \textbf{P} \xi' \rightarrow \textbf{P} B^{(\xi z)} \textbf{P} \xi \;\;\;\;\;\; \;\;\;\;\;\; \\ \textbf{P} \xi' \rightarrow B^{(\chi z)} \textbf{P} \xi \;\;\;\;\;\; \;\;\;\;\;\; eq. 2$$

Now we need to ask what is $\textbf{P} \xi$. In order to answer this question, you should compare eq.1 with eq. 2. $\textbf{P} \xi$ is an object which transform like a left-handed spinor so it must be a let-handed spinor.