Pictures Paths Particles Processes

290 November 7, 2013
Second irregular case
Let us now assume S ≠ 0 and p µ = 0. From
2Tr ( γ 5 γ µ N ) = S q µ (10.169)
we find that also q must vanish. Eq.(10.168) then says that the tensorial term
must also be absent, upon which
2iTr ( γ 5 N ) = SP (10.170)
proves that also P = 0. The only possibilities left are the trivial ones Π = 1
and Π = 0.
10.10.4 The second regular case
We have now examined all consequences of the assumption S ≠ 0. The remaining
case S = 0 gives a projection operator that can be written as
Π = 1 2
(
1 + /p + γ 5 /q + iP γ 5 + T αβ σ αβ) . (10.171)
The relation
− 1 8 ɛµνκλ Tr (σ µν N) = P T κλ − 1 2
(
q κ p λ − p κ q λ) (10.172)
allows us to distinguish two cases, P = 0 and P ≠ 0.
The case P ≠ 0
In this case the vectors p and q are not necessarily related to one another. The
projection operator reads
Π = 1 (
1 + /p + γ 5 /q − i
)
2
2P (/p/q − /q/p) + iP γ5 , (10.173)
under the single condition (from Tr (N)) that
1
P 2 (
p 2 q 2 − (p · q) 2) + p 2 − q 2 − P 2 = 1 . (10.174)
Now, we can always find a vector r µ with p · r = q · r = 0 and r 2 = −1. The
equivalence transform
Σ = 1 √
2
(
1 − iγ 5 /r ) (10.175)
will then eliminate both the axial-vector and the pseudoscalar term, so that we
actually arrive at a special case of the situation for P = 0.