My problem is understanding the transformation behaviour of a Dirac spinor (in the Weyl basis) under parity transformations. The standard textbook answer is
$$\Psi^P = \gamma_0 \Psi = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix}, $$ which I'm trying to understand using the transformation behaviour of the Weyl spinors $\chi_L $ and $\xi_R$. I would understand the above transformation operator if for some reason $\chi \rightarrow \xi$ under parity transformations, but I don't know if and how this can be justified. Is there any interpretation of $\chi $ and $\xi$ that justifies such a behaviour?
Some background:
A Dirac spinor in the Weyl basis is commonly defined as
$$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}, $$ where the indices $L$ and $R$ indicate that the two Weyl spinors $\chi_L $ and $\xi_R$, transform according to the $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$ representation of the Lorentz group respectively. A spinor of the form
$$ \Psi = \begin{pmatrix} \chi_L \\ \chi_R \end{pmatrix}, $$ is a special case, called Majorana spinor (which describes particles that are their own anti-particles), but in general $\chi \neq \xi$.
We can easily derive how Weyl spinors behave under Parity transformations. If we act with a parity transformation on a left handed spinor $\chi_L$: $$ \chi_L \rightarrow \chi_L^P$$ we can derive that $\chi_L^P$ transforms under boosts like a right-handed spinor
\begin{equation} \chi_L \rightarrow \chi_L' = {\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L \end{equation}
\begin{equation} \chi_L^P \rightarrow (\chi^P_L)' = ({\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L)^P = {\mathrm{e }}^{ - \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L^P, \end{equation} because we must have under parity transformation $\vec \sigma \rightarrow - \vec \sigma$. We can conclude $ \chi_L^P = \chi_R$ Therefore, a Dirac spinor behaves under parity transformations $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^P= \begin{pmatrix} \chi_R \\ \xi_L \end{pmatrix} , $$ which is wrong. In the textbooks the parity transformation of a Dirac spinor is given by
$$\Psi^P = \gamma_0 \Psi = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} = \begin{pmatrix} \xi_R \\ \chi_L \end{pmatrix}. $$
This is only equivalent to the transformation described above of $\chi = \xi$, which in my understand is only true for Majorana spinors, or if for some reason under parity transformations $\chi \rightarrow \xi$. I think the latter is true, but I don't know why this should be the case. Maybe this can be understood as soon as one has an interpretation for those two spinors $\chi$ and $ \xi$...
Update: A similar problem appears for charge conjugation: Considering Weyl spinors, one can easily show that $ i \sigma_2 \chi_L^\star$ transforms like a right-handed spinor, i.e. $i \sigma_2 \chi_L^\star = \chi_R $. Again, this can't be fully correct because this would mean that a Dirac spinor transforms under charge conjugation as
$$ \Psi= \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^c = \begin{pmatrix} \chi_R \\ \xi_L \end{pmatrix},$$ which is wrong (and would mean that a parity transformation is the same as charge conjugation). Nevertheless, we could argue, that in order to get the same kind of object, i.e. again a Dirac spinor, we must have
$$ \Psi= \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \rightarrow \Psi^c = \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix},$$
because only then $\Psi^c$ transforms like $\Psi$. In other words: We write the right-handed component always below the left-handed component, because only then the spinor transforms like the Dirac spinor we started with.
This is in fact, the standard textbook charge conjugation, which can be written as
$$ \Psi^c = i \gamma_2 \Psi^\star= i \begin{pmatrix} 0 & \sigma_2 \\ -\sigma_2 & 0 \end{pmatrix} \Psi^\star = i \begin{pmatrix} 0 & \sigma_2 \\ -\sigma_2 & 0 \end{pmatrix} \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}^\star= \begin{pmatrix} -i\sigma_2 \xi_R^\star \\ i\sigma_2 \chi_L \end{pmatrix}= \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} .$$ In the last line I used that, $i \sigma_2 \chi_L^\star$ transforms like a right-handed spinor, i.e. $i \sigma_2 \chi_L^\star = \chi_R $. The textbook charge conjugation possible hints us towards an interpretation, like $\chi$ and $\xi$ have opposite charge (as written for example here), because this transformation is basically given by $\chi \rightarrow \xi$.
Answer
You are looking for a unitary representation of partity on spinors. That it should be unitary can be seen from the fact, that partity commutes with the Hamiltonian. Compare this to time-reversal and charge conjugation, which anticommute with $P^0$ and hence need be antiunitary and antilinear. They involve complex conjugation.
As demonstrated parity transforms a $(\frac{1}{2},0)$ into a $(0,\frac{1}{2})$ representation. Hence it cannot act on any such representation alone in a meaningful way. The Dirac-spinors in the Weyl-Basis on the other hand contain a left- and right-handed component $$ \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$
As a linear operator on those spinors - a matrix in a chosen basis - it mixes up the spinor components. After what has been said before, left- and right-handed components should transform into each other. The only matrix one can write down that does this is $\gamma^0$. There could in principle be a phase factor. In a theory with global $U(1)$-symmetry this may be set to one however.
Edit: Statements like $\chi_L \rightarrow P\chi_L=\chi_R $ for a Weyl-Spinor $\chi_L$ are not sensible. The Weyl-Spinors are reps. of $\mathrm{Spin(1,3)}$, whereas $P\in \mathrm{Pin(1,3)}$. One cannot expect that some representation is also a representation of a larger group. Dirac-Spinors on the other hand are precisely irreps. of $\mathrm{Spin(1,3)}$ including parity, which cannot act in any other sensible way than by exchanging the chiral components.
Think of what representation means. It's a homomorphism from a group to the invertible linear maps on a vectorspace. $$ \rho: G \rightarrow GL(V)$$ Particulary, for any $g\in G$ and $v\in V$, $\rho(g)v\in V$. Now set $V$ to be the space of say left-handed Weyl-Spinors and $g=P\in\mathrm{Pin(1,3)}$ the parity operation. As you have shown above, the image of a potential $\rho(P)$ is not a left-handed Weyl-Spinor, hence is not represented.
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