Dirac spinors under Parity transformation or what do the Weyl spinors in a Dirac spinor really stand for? 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 $\chi \rightarrow \xi$.
 A: You are looking for a unitary representation of parity on spinors. That it should be unitary can be seen from the fact, that parity 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 vector space. $$ \rho: G \rightarrow GL(V)$$
Particularly, 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.
A: I find things are clearer using the dotted and undotted spinor notation. The L-spinors $\chi_{L}$ are dotted vectors $\chi^{\dot{A}}$ and the R-spinors $\xi_{R}$ are undotted vectors $\xi^{A}$ with index $A=1,2$. The parity operator has to be a tensor $P^{\dot{A}}_{B}$ and another tensor $P^{A}_{\dot{B}}$ in order to change the way each type of spinor transforms. The action of parity on  $\chi^{\dot{A}}$ is to make $P^{A}_{\dot{B}}\chi^{\dot{B}}$ which transforms as an undotted spinor. Similarly, the action of parity on $\xi^{A}$ is to make $P^{\dot{A}}_{B}\xi^{B}$ which transforms as a dotted spinor. It turns out that (presumably in the rest frame of the particles) the parity tensors are $P^{\dot{A}}_{B}=i\delta^{\dot{A}}_{B}$ and $P^{A}_{\dot{B}}=i\delta^{A}_{\dot{B}}$. The action of parity is then,
$$
\chi^{\dot{A}}\rightarrow P^{A}_{\dot{B}}\chi^{\dot{B}}=i\delta^{A}_{\dot{B}}\chi^{\dot{B}}=i\chi^{A}
$$
$$
\xi^{A}\rightarrow P^{\dot{A}}_{B}\xi^{B}=i\delta^{\dot{A}}_{B}\xi^{B}=i\xi^{\dot{A}}
$$
and this means that the components of the spinors get a phase and the way the components transform is changed. The dots are a reminder of how each component transforms. The action of parity on a Dirac spinor is obtained  from the above transformations by stacking the Weyl spinors.
\begin{equation}
  \left[
    \begin{array}{c}
      \chi^{\dot{1}}\\
      \chi^{\dot{2}}\\
      \xi^{1}\\
      \xi^{2}
    \end{array}
  \right]
     \rightarrow
 i\left[
    \begin{array}{c}
         \xi^{\dot{1}}\\
         \xi^{\dot{2}}\\
         \chi^{1}\\
         \chi^{2}
    \end{array}
  \right] 
\end{equation}
Edit : Clarification.
The Dirac spinor has four components. Components one and two transform as the two components of a  dotted Weyl spinor  and components three and four transform as the components of an undotted Weyl spinor. If we remember that is how the Weyl spinors stack into a Dirac spinor then we can remove the dots and the L and R labels and then the last equation on the action of parity on a Dirac spinor is,
\begin{equation}
  \left[
    \begin{array}{c}
      \chi^{1}\\
      \chi^{2}\\
      \xi^{1}\\
      \xi^{2}
    \end{array}
  \right]
     \rightarrow
 i\left[
    \begin{array}{c}
         \xi^{1}\\
         \xi^{2}\\
         \chi^{1}\\
         \chi^{2}
    \end{array}
  \right] 
\end{equation}
In matrix notation this is,
\begin{equation}
\left[
  \begin{array}{cc}
    i & 0\\
    0 & i
  \end{array}
\right]
\left[
  \begin{array}{c}
    \chi\\
    \xi
  \end{array}
\right]=i\left[
  \begin{array}{c}
    \xi\\
    \chi
  \end{array}
\right]
\end{equation}
Modulo the phase  factor $i$ (because I think that the parity operator on spinors has to be PP=-1), this agrees with the action of parity given by the gamma matrix $\gamma_{0}$ as in the standard texts such as equation (1.4.42) on page 19 of Ramond's "Field Theory", Second Edition.
