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In 3+1 Minkowski spacetime, we can use the fact that the Lie algebra of the Lorentz group decomposes (I am omitting some details here to keep it short) into $su(2) \oplus su(2)$ and then use the fact that spinors are the fundamental representations of $su(2)$ to start building representations of the Lorentz group$^\dagger$ of the form $(j_+, j_-)$ where $j_\pm$ label representations of $su(2)$.

My question is: since there are systems that include spinors and live in 2+1 or 1+1 dimensions being studied at the moment, how do spinors come into play? How do we get them as representations of the (universal cover of the) Lorentz group in 2+1 and 1+1 dimensions? Can their Lie algebras be decomposed in similar ways or maybe there is a different definition of spinors in these dimensions (i.e. not as the fundamental representation of $su(2)$)?

$^\dagger$or rather its universal cover since we're interested in projective reps of the Lorentz group

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There is a systematic way to define spinor representations in any number of dimensions. The key idea is the following. The SO$(d-1,1)$ algebra can be written

$$i[ \Sigma^{\mu\nu}, \Sigma^{\sigma\rho}]=\eta^{\nu\sigma} \Sigma^{\mu\rho} + \eta^{\mu\rho}\Sigma^{\nu\sigma} - \eta^{\nu\rho} \Sigma^{\mu\sigma} -\eta^{\mu\sigma} \Sigma^{\nu\rho}$$ where $\eta^{\mu\nu}=$diag(-1,+1,...,+1) and $\Sigma^{\mu\nu}$ is an antisymmetric matrix which packages the rotation and boost generators $J_i,K_i$. For instance in $d=3+1$ we have $$\Sigma^{\mu\nu}=\begin{pmatrix}0 &K_1 & K_2 & K_3\\-K_1 & 0 & J_3 & -J_2\\-K_2 & -J_3 & 0 & J_1 \\-K_3 & J_2 & -J_1 & 0\end{pmatrix}. $$

Moreover, whenever we have $d$ Dirac matrices $\Gamma^{\mu}$ satisfying

$$\{ \Gamma^\mu,\Gamma^\nu\} = 2\eta^{\mu\nu}$$

we automatically have a representation of SO($d-1,1$) using

$$\Sigma^{\mu\nu} = -\frac{i}{4} [\Gamma^\mu,\Gamma^\nu]$$ as one can check. So all we need are suitable $\Gamma$s and we are done; the spinors can be identified by building raising/ lowering operators out of the $\Gamma$s as usual. In $d=1+1$ we can use

$$\Gamma^0 = \begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}\quad\Gamma^1 = \begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$$ and in dimension 2+1 we can tack on $\Gamma^2 = \Gamma^0\Gamma^1$. There's a systematic way to get higher dimensions, too, and to find out which representations are irreducible. (Note the number of components of the spinor goes like $2^{d/2}$ for $d$ even or $2^{(d-1)/2}$ for $d$ odd).

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  • $\begingroup$ Great! Just a question: this seems pretty straightforward: introduce gamma matrices such that their commutator satisfies the appropriate Lie algebra and you're basically done; so why do textbooks on QFT usually get into the trouble of decomposing the Lorentz algebra into $su(2)\oplus su(2)$ in order to explain spin states? Maybe because in this way we can clearly see the motivation behind labelling reprenestations of the Lorentz group by the notation $(j_+ , j_-)$? Through what was shown in the answer, the labelling $(j_+ , j_-)$ due to the decomposition $su(2)\oplus su(2)$ is not apparent. $\endgroup$ Nov 14, 2020 at 18:11
  • $\begingroup$ Not sure, but if I were to speculate, I'd guess it's a pedagogical tool. Students learn QFT after learning the quantum mechanics of SU(2), so it's hard to pass this up. Such a splitting can't work for $d=6,7,10,11$, for instance, where the Lorentz group has odd dimension. $\endgroup$
    – Dwagg
    Nov 14, 2020 at 19:22

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