Pictures Paths Particles Processes

288 November 7, 2013
where S, p µ , q µ and P are real, and T µν is real and antisymmetric. The requirement
is now that N ≡ Π 2 − Π vanish, and so its trace with any Clifford
element must also vanish. We can immediately find
2Tr ( γ 5 /pN ) = (p · q)S , 2Tr ( (γ 5 /q − /p)N ) = (p 2 + q 2 )S . (10.155)
There are now several possibilities, the first of which is the regular case : it is
the case where S ≠ 0 and p 2 ≠ 0. We see that it implies that q 2 = −p 2 and
p · q = 0, so that p and q are linearly independent and one of them must be
timelike. In that case we may form a Vierbein by finding two additional vectors
e 1,2 µ with
p · e 1,2 = q · e 1,2 = e 1 · e 2 = 0 , e 1,2 2 = −1 , (10.156)
so that the tensor T can be decomposed 24 as follows:
T αβ = c pq p [α q β] + c 12 e [α 1 e β] 2 + ∑ (
) c pj p [α e β] j + c qj q [α β]
e j , (10.157)
j=1,2
where the coefficients are all real and the square brackets indicate antisymmetrization
over the indices. We can now find two more conditions:
1
p 2 S Tr ((c p1p µ e ν 1 − c q2 q µ e ν 2 )σ µν N) = c 2 p1 + c 2 q2 ,
1
p 2 S Tr ((c p2p µ e ν 2 − c q1 q µ e ν 1 )σ µν N) = c 2 p2 + c 2 q1 , (10.158)
which tells us that c p1 = c p2 = c q1 = c q2 = 0. The tensorial part can therefore
only consist of /p/q and /pγ 5 /q, and we may write
Π = 1 4
(
(2 − S) + /p + γ 5 /q + iP γ 5 + ia/p/q + b/pγ 5 /q ) (10.159)
with a and b real. Then, the results
− 2 p 2 Tr (/pN) = S − p2 b , 2iTr ( γ 5 N ) = SP + p 4 ab (10.160)
fix the values of a = −P/p 2 and b = S/p 2 . Continuing, we evaluate
− 1 p 2 ɛ αβµνp α q β Tr (σ µν N) = S 2 + P 2 − p 2 , (10.161)
which proves that p µ must actually be the timelike vector, and fixes |P |. Using
all the relations obtained, we finally have
Tr (N) = S 2 − 1 , (10.162)
which tells us that if S ≠ 0 we can take S = 1 (without loss of generality since
both Π and 1 − Π satisfy Eq.(10.151)), and we must have p 2 ≥ 1. The generic
24 No matter that the vectors e 1,2 are not unambiguous : the point is that a decompisition
is possible.