Some questions about Dirac spinor transformation law

Question

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

Comments

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

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@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

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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

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@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

Answer 1

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

Comments

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

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@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

Answer 2

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

Comments

$b_{\mu}\sigma^{\mu}_{a \dot {b}}$ 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