# Dirac spinors under Parity transformation or what do the Weyl spinors in a Dirac spinor really stand for?

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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

$$\chi_L \rightarrow \chi_L' = {\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L$$

$$\chi_L^P \rightarrow (\chi^P_L)' = ({\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L)^P = {\mathrm{e }}^{ - \frac{\vec{\theta}}{2} \vec{\sigma}} \chi_L^P,$$ 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
Why do you expect the parity transformation on the Dirac spinor to be given by the parity transformation on the Weyl spinors?

This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user ACuriousMind
Because a Dirac spinor consists of two Weyl spinors. Analogously, we derive how a Dirac spinor transforms under, for example Lorentz boosts: $\Psi \rightarrow \Psi'= \begin{pmatrix} {\mathrm{e }}^{ \frac{\vec{\theta}}{2} \vec{\sigma}} &0 \\ 0& {\mathrm{e }}^{ \frac{-\vec{\theta}}{2} \vec{\sigma}} \end{pmatrix} \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}$ from the transformation behaviour of the Weyl spinors.

This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user JakobH
But is "parity" a Lorentz transformation? The statement "consists of Weyl spinor" is, more formally, the statement that the Weyl spinors constitute subrepresentations of $\mathrm{Spin}(1,3)$ of the space of Dirac spinors (for $m \neq 0$). As parity is not in that group, there is no reason to believe that parity on the Dirac spinor is the direct sum of parity on the Weyl spinors.

This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user ACuriousMind
In my understanding parity is a Lorentz transformation. The Lorentz group consists of four components that are connected by parity and time inversion: $O(1,3) = \{ SO(1,3)^{\uparrow}, \Lambda_P SO(1,3)^{\uparrow} , \Lambda_T SO(1,3)^{\uparrow} , \Lambda_P \Lambda_T SO(1,3)^{\uparrow} \}$. I'm pretty sure there must be some sort of connection, obviously not the direct sum, between the parity transformations of Weyl and Dirac spinors (that can used to motivate the parity transformation of a Dirac spinor, apart from looking at the Dirac equation) and I'm trying to understand this a little better.

This post imported from StackExchange Physics at 2014-11-17 09:07 (UTC), posted by SE-user JakobH

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The proper special orthogonal group SO(1,3) (proper orthochronus Lorentz group) does not include parity. Parity is a discrete symmetry. The generators of the SO(1,3) follow a Lie algebra. Furthermore, note that Parity acts as $$P\psi(x)P^{-1} = \gamma^0 \psi(t,-\vec{x}), \,\,\,\,\,\,\, P\bar{\psi}(x)P^{-1} = \bar{\psi}(t,-\vec{x})\gamma^0,$$ where $\psi$ is a Dirac spinor. Now you ask what happens for a Weyl spinor. Well, write down the Dirac spinor in terms of two Weyl spinors and write down the $\gamma^0$ matrix in terms of the Pauli matrices. The behavior you are asking about is usually visualized in terms of helicity and not chirality. The important thing you miss in your post is that it is not apparent the sign change in the spacial part of the 4-vector. Parity is just a change in the sign of $\vec{x}$, $$P: \vec{x} \mapsto -\vec{x}.$$ Now let me also add the definition of Charge Conjugation which is no other than $$C \psi(x) C^{-1}=C\bar{\psi}^{t}(x), \,\,\,\,\,\,\, C\bar{\psi}(x) C^{-1} = -\psi^t(x)C^{\dagger},$$ where we have $C=i\gamma^2\gamma^0$ and it is really easy to check that $C^{-1}=C^{\dagger}$, $C^t=-C$ and $C\gamma^{\mu}C=(-\gamma^{\mu})^t$. Thus we see that charge conjugation is just producing transformed gamma matrices. It is really easy to see what happens to the Weyl spinors now just by singling out the transformation of the upper (or lower) parts. As for the direct-productness I ve seen seen somewhere above, I think it is true only for helicity. A simpler way to think about it is to consider what parity does to the rotations and to boosts. It is easy to see that for axial vectors like angular momentum parity does nothing (and from this we understand that parity commutes with rotations). On the other hand parity acts as above to the polar vectors. I have repeated stuff you have already written or can find easily in the web. Let us know if something is not clear and we (I) can try again.
answered Nov 17, 2014 by (3,625 points)
edited Nov 17, 2014

I think maybe you didn't quite understand OP's question. OP is not asking how Weyl spinors, as components of a Dirac spinor, transforms under parity. He's thinking like this: let's forget all about Dirac spinor, just consider, say, a left handed Weyl spinor, and think how it transforms, and he reaches the conclusion that it shall transform into a right handed one; same shall be true to a right handed Weyl spinor. Then he put the two together into a Dirac spinor and discovered it does not obey the conventional parity transformation rule described in textbooks. This is all faulty from the first line of his thinking, stand-alone Weyl spinors simply do not furnish a representation of Lorentz transformation that includes parity, all the rest of his logic is not legitimate mathematics since you are not allowed to talk about how a stand-alone Weyl spinor transforms under parity . I think OP fell victim to "blur thinking", and user ACuriousMind nailed it in the comments.

Well, I think that my answer can help him to find out his mistakes. Maybe I should supplement it with the fact that you can write any Weyl spinor as a projection of a Dirac spinor. Then all should be clear.

"This is all faulty from the first line of his thinking, stand-alone Weyl spinors simply do not furnish a representation of Lorentz transformation that includes parity, all the rest of his logic is not legitimate mathematics since you are not allowed to talk about how a stand-alone Weyl spinor transforms under parity . I think OP fell victim to "blur thinking", and user ACuriousMind nailed it in the comments."

I'm reading through this thread, and trying to understand what you mean by saying you're "not allowed" to ask how a Weyl spinor transforms under parity.  Ordinarily in mathematics, if you're not allowed to do something, it's either because it's not well-defined or because you introduce some axiom that disallows it from the set of valid operations.

My thought process:  every 2-component spinor has a uniquely defined direction in Euclidean space, which depends on its direction but not its phase (eg, if $\chi$ points in the $\vec{z}$ direction, then so do all spinors of the form $e^{i\theta}\chi$.  The parity operation on the corresponding vector in Euclidean space we know is defined as $\vec{x} \rightarrow -\vec{x}$.  That uniquely determines the parity operation on a Weyl spinor up to an arbitrary phase.

That seems well-defined enough for me, we can just choose the phase to be 1 for simplicity.  So if it's not that it's not well-defined, there must be an axiom somewhere that says you "just can't do it"... what is it?

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