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I learned recently that if you have the Dirac spinor represented in the Weyl (chiral) basis $\Psi = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$, then given a Lorentz Transformation $\Lambda = exp[\frac{1}{2}\Omega_{\rho \sigma}M^{\rho\sigma}]$, the corresponding transformation $S[\Lambda]$ for $\Psi \rightarrow S[\Lambda]\Psi$ looks like $S[\Lambda] = exp[\frac{1}{2}\Omega_{\rho \sigma}S^{\rho\sigma}]$. In the chiral basis, this looks like each of the $\psi_L$ and $\psi_R$ transforming as the different spin-1/2 representations of $so(3,1)$.

Here, the $so(3,1)$ Lie algebra is represented by the standard Lorentz matrices $M^{\rho\sigma}$ for the $x^\mu$ transformation, and the other matrices $S^{\rho\sigma}$ are the algebra generated by the gamma matrices $S^{\rho\sigma} = \frac{1}{4}[\gamma^\rho,\gamma^\sigma]$.

My question is (in probably imprecise phrasing), is it possible to choose $\gamma^\mu$ to be 8x8 matrices in such a way that $S^{\rho\sigma} = \frac{1}{4}[\gamma^\rho,\gamma^\sigma] = M^{\rho\sigma} \oplus M^{\rho\sigma} $? i.e. can we pick $\gamma^\mu$ so that the chiral components $\psi_L, \psi_R$ each transform like 4-vectors would?

Joe
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2 Answers2

4

  1. OP apparently wants to discuss reducible representations of the Clifford algebra $Cl(1,3;\mathbb{R})$.

  2. Concretely, it seems that OP is asking about an 8-dimensional direct sum representation $$W~:=~V\oplus V$$ of 2 copies of the 4-dimensional Dirac spinor representation $$V~:=~(\frac{1}{2},0) \oplus (0,\frac{1}{2}).$$ See also this Phys.SE post.

  3. The $8\times 8$ gamma matrices $$\Gamma^{\mu} ~=~\begin{pmatrix} \gamma^{\mu} & 0 \cr 0 & \gamma^{\mu} \end{pmatrix}$$ in the $W$-representation are block matrices with 2 copies of $4\times 4$ gamma matrices in the $V$-representation.

  4. The $W$-representation of the Lorentz generators take a similar block diagonal form, cf. OP's last question (v1).

Qmechanic
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Without full appreciation of the gist of your question (8×8?), let me just review the chiral basis expressions for $\gamma_5$ and the Lorentz generators $S^{\mu\nu}$, which are both block diagonal with respect to the chiral projections, so they do not mix $\psi_L$ with $\psi_R$, unlike the γs:

$$\gamma^0 = \begin{pmatrix} 0 & I_2 \\ I_2 & 0 \end{pmatrix},\quad \gamma^k = \begin{pmatrix} 0 & \sigma^k \\ -\sigma^k & 0 \end{pmatrix},\quad \gamma^5 = \begin{pmatrix} -I_2 & 0 \\ 0 & I_2 \end{pmatrix},$$

$$ S^{0j} = \tfrac{1}{2}\begin{pmatrix} -\sigma^j & 0 \\ 0 & \sigma^j \end{pmatrix}, \qquad S^{jk}=\frac{i\epsilon^{kjm}}{2} \begin{pmatrix} \sigma^m & 0 \\ 0 & \sigma^m \end{pmatrix} = i \epsilon^{kjm} \gamma_5 S^{0m} . $$

So, generators (in the algebra), manifestly constitute a reduced representation of non-interacting 2×2 blocks.

More formally, the reduced 2 ⊕ 2 rep is $$ 2S^{0j}=(-\sigma^j)\oplus \sigma^j, \qquad 2S^{jk}=i\epsilon^{kjm} (\sigma^m\oplus \sigma^m). $$

Cosmas Zachos
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