• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

193 submissions , 152 unreviewed
4,820 questions , 2,008 unanswered
5,292 answers , 22,482 comments
1,470 users with positive rep
779 active unimported users
More ...

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

+ 5 like - 0 dislike

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
asked Nov 16, 2014 in Theoretical Physics by JakobH (110 points) [ no revision ]
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

1 Answer

+ 4 like - 0 dislike
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 conformal_gk (3,625 points) [ revision history ]
edited Nov 17, 2014 by conformal_gk

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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights