Quantcast
  • 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.

News

PO is now at the Physics Department of Bielefeld University!

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

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Some questions about Dirac spinor transformation law

+ 3 like - 0 dislike
1598 views

I have perhaps meaningless question about Dirac spinors, but I'm at a loss.

The transformation laws for for left-handed and right-handed 2-spinors are $$ \tag 1 \psi_{a} \to \psi_{a}' = N_{a}^{\quad b} \psi_{b} = \left(e^{\frac{1}{2}\omega^{\mu \nu}\sigma_{\mu \nu}}\right)_{a}^{\quad b}\psi_{b}, \quad \psi^{b}{'} = \psi^{a}(N^{-1})_{a}^{\quad b}, $$ $$ \tag 2 \psi_{\dot {a}} \to \psi_{\dot {a}}' = (N^{*})_{\dot {a}}^{\quad \dot {b}} \psi_{\dot {b}} = \left(e^{\frac{1}{2}\omega^{\mu \nu}\tilde {\sigma}_{\mu \nu}}\right)_{\dot {a}}^{\quad \dot {b}}\psi_{\dot {b}}, \quad \psi^{\dot {b}}{'} = \psi^{\dot {a}}(N^{*^{-1}})_{\dot {a}}^{\quad \dot {b}}, $$ where $$ (\sigma_{\mu \nu})_{a}^{\quad b} = -\frac{1}{4}\left(\sigma_{\mu}\tilde {\sigma}_{\nu}-\sigma_{\nu}\tilde {\sigma}_{\mu}\right), \quad (\tilde {\sigma}_{\mu \nu})_{\quad \dot {a}}^{\dot {b}} = -\frac{1}{4}\left(\tilde {\sigma}_{\mu} \sigma_{\nu}- \tilde {\sigma}_{\nu}\sigma_{\mu}\right), $$ $$ (\sigma_{\mu})_{b\dot {b}} = (\hat {E}, \sigma_{i}), \quad (\tilde {\sigma}_{\nu})^{\dot {a} a} = -\varepsilon^{\dot {a}\dot {b}}\varepsilon^{b a} \sigma_{\dot {b} b} = (\hat {E}, -\sigma_{i}). $$ Why do we always take the Dirac spinor as $$ \Psi = \begin{pmatrix} \varphi_{a} \\ \kappa^{\dot {b}} \end{pmatrix}, $$ not as $$ \Psi = \begin{pmatrix} \varphi_{a} \\ \kappa_{\dot {b}} \end{pmatrix}? $$ According to $(1), (2)$ first one transforms as $$ \delta \Psi ' = \frac{1}{2}\omega^{\mu \nu}\begin{pmatrix}\sigma_{\mu \nu} & 0 \\ 0 & -\tilde {\sigma}_{\mu \nu} \end{pmatrix}\Psi , $$ while the second one - as $$ \delta \Psi ' = \frac{1}{2}\omega^{\mu \nu}\begin{pmatrix}\sigma_{\mu \nu} & 0 \\ 0 & \tilde {\sigma}_{\mu \nu} \end{pmatrix}\Psi , $$ so it is more natural than first, because the first one has both covariant and contravariant components, while the second has only covariant (contravariant components).

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams
asked Mar 8, 2014 in Theoretical Physics by Andrew McAddams (340 points) [ no revision ]
Are the indices a b correct in 1 and 2 ?

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Love Learning
@LoveLearning : did you ask about the horisontal position of the indices? If yes, I think so.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams
Maybe I'm too tired but you use b as a summation and as an index etc.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Love Learning
@LoveLearning : yes, thank you. I fixed it.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams

2 Answers

+ 1 like - 0 dislike

I think it is convention to write the conjugate Weyl fermion in, \begin{equation} \left( \begin{array}{c} \phi _\alpha \\ \bar{\kappa} ^{\dot{\beta }} \end{array} \right) \end{equation} (it is common to put a bar over the conjugate representation), with a raised index in order to comply with the ${} _{ \dot{\alpha} } ^{ \,\, \dot{\alpha} } $ contraction of spinor indicies. Recall that we write, \begin{equation} \phi \chi \equiv \phi ^\alpha \chi _\alpha , \quad \psi \bar{\chi} \equiv \phi _{\dot{\alpha}} \bar{\chi} ^{\dot{\alpha}} \end{equation} Thus having the particular index structure for the Dirac spinor gives, \begin{align} \bar{ \Psi } \gamma ^\mu \Psi & = \left( \begin{array}{cc} \kappa ^{\beta } &\bar{ \phi} _{\dot{\alpha}}\end{array} \right) \left( \begin{array}{cc} 0 & ( \sigma ^\mu ) _{ \beta \dot{\beta} } \\ ( \bar{\sigma} ^\mu ) ^{ \dot{\alpha } \alpha } & 0 \end{array} \right) \left( \begin{array}{c} \phi _\alpha \\ \bar{ \kappa} ^{\dot{\beta}} \end{array} \right) \\ & = \kappa \sigma ^\mu \bar{\kappa} + \bar{\phi} \bar{\sigma} ^\mu \phi \end{align} where all the dotted indices contract with an "upwards staircase", ${}_{ \dot{\alpha} } ^{ \,\, \dot{\alpha} } $, and undotted with a "downwards staircase", $ {} ^\alpha _{ \,\, \alpha } $.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user JeffDror
answered Mar 9, 2014 by JeffDror (650 points) [ no revision ]
Thank you! Your answer satisfied me fully.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams
@AndrewMcAddams: No problem, sorry I didn't get a chance to read your previous concern.

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user JeffDror
+ 0 like - 0 dislike

I suspect the origin of this might have to do with the bi-spinor notation. Given a four-vector $b_\mu$, one defines the corresponding bi-spinor, $b\!\!/_{\alpha\dot{\beta}}=b_\mu (\sigma^\mu)_{\alpha\dot{\beta}}$. In this convention, bi-spinors have both lower indices (or upper indices if one uses $(\bar{\sigma}^\mu)^{\beta\dot{\alpha}})$. Once such a choice is made, the index structure of $4\times 4$ gamma matrices is fixed leading to what seems a strange choice for the index structure for a Dirac spinor. In order to avoid such details, I usually use a single meta index $A=(\alpha,\dot{\alpha})$ (capital letters) to denote the combination leaving the finer detail only when I need to work explicitly with gamma matrices. I recommend appendix A of the article by M. Sohnius titled "Introducing Supersymmetry" (Physics Reports 128 (1985) 39-204).

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user suresh
answered Mar 8, 2014 by suresh (1,545 points) [ no revision ]
$b_{\mu}\sigma^{\mu}_{a \dot {b}}$ refers to the $\left( \frac{1}{2}, \frac{1}{2} \right)$ representation, have diffenent transformation law and another equation comparing to bispinor rep. Maybe it is impossible to move from the direct sum of $\left(\frac{1}{2}, 0 \right) + \left( 0, \frac{1}{2}\right)$ rep to the 4-vector one (however, it would be possible if rep is $(1, 0)$ or $(0, 1)$). Can you comment it?

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams

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:
p$\hbar$ys$\varnothing$csOverflow
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).
Please complete the anti-spam verification




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

Your rights
...