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284 November 7, 2013
10.9 The fundamental theorem for Dirac matrices
10.9.1 Proof of the fundamental theorem
In this appendix we prove the following statement : if we have two sets of four
matrices, γ µ and ˆγ µ (µ = 0, 1, 2, 3) , satisfying Dirac’s anticommutation relation
γ µ γ ν + γ ν γ µ = 2 g µν , ˆγ µ ˆγ ν + ˆγ ν ˆγ µ = 2 g µν , (10.138)
then there is a matrix S such that
ˆγ µ = S γ µ S −1 . (10.139)
To this end, we first set up a basis of the Clifford algebra as follows :
Γ 0 = 1 , Γ 1 = γ 0 , Γ 2 = iγ 1 , Γ 3 = iγ 2 , Γ 4 = iγ 3 ,
Γ 5 = γ 0 γ 1 , Γ 6 = γ 0 γ 2 , Γ 7 = γ 0 γ 3 , Γ 8 = iγ 1 γ 2 ,
Γ 9 = iγ 1 γ 3 , Γ 10 = iγ 2 γ 3 , Γ 11 = iγ 0 γ 1 γ 2 , Γ 12 = iγ 0 γ 1 γ 3 ,
Γ 13 = iγ 0 γ 2 γ 3 , Γ 14 = γ 1 γ 2 γ 3 , Γ 15 = iγ 0 γ 1 γ 2 γ 3 , (10.140)
which we denote by Γ k , k = 0, 1, 2, . . . , 15 ; and using the ˆγ µ we construct an
analogous set ˆΓ k in the same way. These have a few interesting properties. In
the first place, Γ k 2 = 1 for all k. Secondly, for every pair j and k there is a
number c n such that
Γ j Γ k = c n Γ n , c n = 1, −1, i or − i. (10.141)
From these properties it follows that simultaneously
Γ k Γ j = 1
c n
Γ n (10.142)
We can thus construct the multiplication table given below 22 , where the possible
values of j define the rows, and those for k the columns: the corresponding entry
is then the value of n. For instance,
(in this case c 13 happens to be 1).
Γ 6 Γ 4 = Γ 13
22 Kids! Don’t do this at home, since constructing this multiplication table is extremely
tedious. The numbers c n are not given: they are anyhow only defined up to a sign, since we
can always replace Γ j by −Γ j (using γ 2 γ 0 instead of γ 0 γ 2 , say) without changing the Dirac
anticommutation relation.