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In PCT, spin and statistics, and all that book, the following example is given: Let $S(A)$ be a representation of $SL(2,C)$ given as :

$$S(A)=\frac{1}{2}\left(a^{0} \mathbf{1}+\mathbf{a} \cdot \mathbf{\sigma}\right)\left(1+i \gamma^{5}\right)+\frac{1}{2}\left(\bar{a}^{0} \mathbf{1}-\overline{\mathbf{a}} \cdot \mathbf{\sigma}\right)\left(1-i \gamma^{5}\right) $$ where

$\gamma^{5}=\gamma^{0} \gamma^{1} \gamma^{2} \gamma^{3}$

$\boldsymbol{\sigma}=-i \gamma^{5} \gamma^{0} \boldsymbol{\gamma}$

$A=\sum_{\mu=0}^{3} a^{\mu} \tau^{u}, \quad \operatorname{det} A=a^{2}=1$, with $\tau^{0}$ the identity, and $\tau^{i}$ are the Pauli matrices.

then this representation satisfies:

$${S}(A)^{-1} \gamma^{\mu} S(A)=\Lambda_{v}^{\mu}(A) \gamma^{v}$$

And the book states that the proof requires a certain amount of $\gamma$-gymnastics, but I can't see a way to do this. any help or resources would be appreciated.

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  • $\begingroup$ $S(A) = \exp(-\frac{i}{2} \omega_{\mu \nu} S^{\mu \nu})$ and $\Lambda = \exp(-\frac{i}{2} \omega_{\mu \nu} J^{\mu \nu})$ where $\omega_{\mu \nu}\ll 1$, $S^{\mu \nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu]$ using the Weyl representation of $\gamma^\mu$ this can be easily calculated. $J^{\mu \nu}$ is the representation of the generator of the Lie-algebra of the Lorentz group. The idea is to look at $\Psi(x) \mapsto S(A) \Psi(\Lambda^{-1}x)$ where $\Psi$ is a Dirac spinor field. Looking at how $\bar{\Psi} \gamma^\mu \Psi$ transforms under this at first order in $\omega_{\mu \nu} $ should give this. $\endgroup$ – Mathphys meister May 25 at 17:09
  • $\begingroup$ For a full reference of this, see e.g. David Tong's lecture notes on quantum field theory, chapter 4 (in particular 4.2) $\endgroup$ – Mathphys meister May 25 at 17:11

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