samalkhaiat
Science Advisor
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vanhees71 said:Indeed, in the standard treatment, you have
$$S^{-1}(\Lambda)\gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu},$$
where ##S(\Lambda)## is the bi-spinor representation of the Lorentz group and ##\Lambda## a Lorentz-transformation matrix. Written the 6 parameters of the Lorentz group in the usual way as ##\omega_{\mu \nu}=-\omega_{\nu \mu}## it turns out that
$$S(\Lambda)=\exp \left (-\frac{1}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right) \quad \text{with} \quad$$
$$\sigma^{\mu \nu}=\frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
All this should be consistent with what samalkhaiat wroten in #41, maybe in a somewhat different convention, but I haven't checked this carefully.
Yes, I used 2-spinor notation. The group-theoretic treatment is generally clearer in the 2-spinor notation. This is because \chi_{A} \in (1/2 , 0) and (\varphi_{A})^{*} \equiv \bar{\varphi}_{\dot{A}} \in (0 , 1/2) are the irreducible representations of the Lorentz group, while Dirac bispinors are reducible \Psi = \chi_{A} \oplus \ \epsilon^{\dot{A}\dot{B}}\bar{\varphi}_{\dot{B}} . The transformation laws for the fundamental spinor and it’s conjugate are given by \chi_{A} \to S_{A}{}^{B} \ \chi_{B} = \left( e^{- \frac{i}{2}\omega_{\mu\nu}s^{\mu\nu}} \right)_{A}{}^{B} \ \chi_{B} ,
\bar{\varphi}^{\dot{A}} \to \bar{S}^{\dot{A}}{}_{\dot{B}} \ \bar{\varphi}^{\dot{B}} = \left( e^{- \frac{i}{2}\omega_{\mu\nu}\bar{s}^{\mu\nu}} \right)^{\dot{A}}{}_{\dot{B}}\ \bar{\varphi}^{\dot{B}} , where (s^{\mu\nu})_{A}{}^{B} = \frac{i}{4} ( \sigma^{\mu}\bar{\sigma}^{\nu} - \sigma^{\nu} \bar{\sigma}^{\mu})_{A}{}^{B} , and (\bar{s}^{\mu\nu})^{\dot{A}}{}_{\dot{B}} = \frac{i}{4}(\bar{\sigma}^{\mu}\sigma^{\nu} - \bar{\sigma}^{\nu}\sigma^{\mu})^{\dot{A}}{}_{\dot{B}} , are the corresponding Lorentz generators. From these transformations, and with no reference to Weyl or Dirac equations, we obtain the transformation law for the Dirac bispinor \Psi \to ( S \otimes \bar{S} ) \Psi = (e^{-\frac{i}{2}\omega_{\mu\nu}(s^{\mu\nu} \oplus \bar{s}^{\mu\nu})}) \Psi , with the generators in block diagonal form
\sigma^{\mu\nu} = \begin{pmatrix} s^{\mu\nu} & 0 \\ 0 & \bar{s}^{\mu\nu} \end{pmatrix} = \frac{i}{4}[\gamma^{\mu},\gamma^{\nu}] , where \gamma^{\mu} \equiv \begin{pmatrix} 0 & \sigma^{\mu} \\ \bar{\sigma}^{\mu} & 0 \end{pmatrix} .
Sam
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