22.2 Explicit representation of the Dirac algebra
From the definition ofγ^5 , we have, tr(γ^5 ) =−itr(γ^0 γ^1 γ^2 γ^3 ). By using the cyclic property of the
trace, this quantity equals−itr(γ^1 γ^2 γ^3 γ^0 ). On the other hand, by anti-commutingγ^0 through the
matricesγ^1 γ^2 γ^3 , it also equals +itr(γ^1 γ^2 γ^3 γ^0 ), and thus must vanish,
tr(
γ^5)
= 0 (22.13)Since we also have (γ^5 )^2 =I, it follows thatγ^5 must have equal numbers of eigenvalues +1 and
−1. Thus, the representation space must have even dimension 2n, forninteger. In a basis where
γ^5 is diagonal, we thus have
γ^5 =(
In 0
0 −In)
(22.14)The matrixγ^0 anti-commutes withγ^5 , and squares to−I 2 n; it may be chosen as follows,
γ^0 =(
0 In
−In 0)
(22.15)Finally, the combinationsγ^0 γifori= 1, 2 ,3 commute withγ^5 , and must thus be block-diagonal in
the basis whereγ^5 is diagonal,
γ^0 γi=(
ai+ 0
0 ai−)
(22.16)From the Clifford algebra relations, we also have
{
γ^0 γi,γ^0 γj
}
= 2δijI 2 n (22.17)so that
{
ai±,aj±
}
= 2δijIn (22.18)But this last relation is the defining equation for the Pauli matrices, of which the lowest-dimensional
representation hasn= 2, andai±=ε±σi, with (ε±)^2 = 1. We may chooseε+= 1 without loss
of generality. The valueε− = +1 then makesai+ =ai−and results in [γ^0 ,γi] = 0, which is
unacceptable. Thus, we are left with the unique solutionε−=−1. Theγμ-matrices now take the
following explicit form in this basis,
γ^0 =(
0 I 2
−I 2 0)
γi=(
0 σi
σi 0)
(22.19)We check that
γ^5 =−iγ^0 γ^1 γ^2 γ^3 =−i(
σ^1 σ^2 σ^30
0 −σ^1 σ^2 σ^3)
=(
I 2 0
0 −I 2)
(22.20)