Physics Asked by TheQuantumMan on March 1, 2021
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
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
$$i[ Sigma^{munu}, Sigma^{sigmarho}]=eta^{nusigma} Sigma^{murho} + eta^{murho}Sigma^{nusigma} - eta^{nurho} Sigma^{musigma} -eta^{musigma} Sigma^{nurho}$$ where $eta^{munu}=$diag(-1,+1,...,+1) and $Sigma^{munu}$ is an antisymmetric matrix which packages the rotation and boost generators $J_i,K_i$. For instance in $d=3+1$ we have $$Sigma^{munu}=begin{pmatrix}0 &K_1 & K_2 & K_3-K_1 & 0 & J_3 & -J_2-K_2 & -J_3 & 0 & J_1 -K_3 & J_2 & -J_1 & 0end{pmatrix}. $$
Moreover, whenever we have $d$ Dirac matrices $Gamma^{mu}$ satisfying
$${ Gamma^mu,Gamma^nu} = 2eta^{munu}$$
we automatically have a representation of SO($d-1,1$) using
$$Sigma^{munu} = -frac{i}{4} [Gamma^mu,Gamma^nu]$$ 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
$$Gamma^0 = begin{pmatrix} 0 & 1 -1 & 0end{pmatrix}quadGamma^1 = begin{pmatrix} 0 & 1 1 & 0end{pmatrix}$$ and in dimension 2+1 we can tack on $Gamma^2 = Gamma^0Gamma^1$. 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).
Answered by Dwagg on March 1, 2021
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