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 $\chi \rightarrow \xi$.
This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user JakobH
Q&A (4901)
