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?


2 Answers 2

  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).

  • $\begingroup$ But... this trivial Gamma rep will never connect the L and R components, no? $\endgroup$ Jun 30, 2017 at 16:15
  • $\begingroup$ I was more so asking if there existed a $(1,0) \oplus (0,1)$ representation of the gamma matrices. This post seems to say that the irreducible complex representations look like V. $\endgroup$
    – Joe
    Jul 1, 2017 at 19:16
  • 1
    $\begingroup$ @Joe I see... as long as you appreciate this rep, e.g., the 0 component in the mass term, never mixes the upper 4 with the lower four components ... what I understood to be the defining feature of L and R.... Look at $\Gamma_5$; its chiral projectors eliminate two doublets in tandem. $\endgroup$ Jul 1, 2017 at 19:41
  • $\begingroup$ Hi @Qmechanic, can you refer to some book getting to some depth about this rep? $\endgroup$
    – schris38
    Oct 23, 2023 at 6:51

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

  • $\begingroup$ I understand this part. In the Weyl representation of $\gamma^\mu$ that you gave, the generated $S^{\mu\nu}$ are block diagonal, and each block corresponds to the $SL(2,C)$ Lie Algebra. My question was asking whether we can pick the gamma matrices to be 8x8 real matrices so that $S^{\mu\nu}$ are block diagonal and correspond to the $SO(3,1)$ Lie Algebra. If this were true, we could treat each component, $\psi_L, \psi_R$ as transforming like a 4-vector, by the expression for $S[\Lambda]$ given above. $\endgroup$
    – Joe
    Jun 27, 2017 at 19:53
  • $\begingroup$ ?? What's wrong with this representation? Of course the algebra of the 6 Ss is the algebra of SO(3,1); how could it fail that? It is a bi-spinor rep, acting on 4D: $\psi_L\oplus \psi_R$ rep. You want to add the two spin 1/2s of each block into a spin 1 of each? $\endgroup$ Jun 27, 2017 at 21:11
  • $\begingroup$ There's nothing wrong with the representation per-say. I meant that I wanted the 6 Ss as the same matrices as the Lorentz Algebra as regularly viewed in 4d space, or as a direct sum of them. I want to know if we can express the spinor as two spin 1's instead of two spin 1/2's by some other choice of gamma matrices that aren't in $C^{4x4}$, but in $R^{8x8}$. I haven't seen anywhere that it can't be done explicitly, but I haven't seen its impossibility stated, and I'm having trouble constructing them myself, which is why I'm asking on stackexchnge $\endgroup$
    – Joe
    Jun 27, 2017 at 21:28
  • $\begingroup$ I see. Your want the spinor map extended to the Lorentz group. It maps 2x2 complex matrices to real vectors. It is the basis of twistor theory, but a very different question. $\endgroup$ Jun 27, 2017 at 21:56
  • $\begingroup$ You seem to be trying to reinvent APS, a sorry tale.... $\endgroup$ Jun 27, 2017 at 22:08

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