Pictures Paths Particles Processes

November 7, 2013 289
form of Π in the regular case can be written as follows. We have an angle χ
such that p 2 = cosh(χ) 2 and P = sinh(χ), and two vectors k µ and s µ such that
k · k = 1, s · s = −1 and k · s = 0; then p µ = cosh(χ)k µ and q µ = cosh(χ)s µ , and
Π(α, β) = 1 4
(
1 + αβ/kγ 5 /s + α [ cosh(χ)/k + i sinh(χ)γ 5]
+β [ cosh(χ)γ 5 /s − i sinh(χ)/k/s ]) . (10.163)
The two parameters α and β satisfy α, β = ±1, and we have introduced them
here since the set of four elements Π(1, 1), Π(1, −1), Π(−1, 1) and Π(−1, −1)
satisfy Eq.(10.152). The situation can be simplified further by the use of the
equivalence transform based on
Σ = cosh(χ/2) − i sinh(χ/2)γ 5 /k : (10.164)
the equivalent form is then given by the simpler
Π(α, β) = 1 4
(
1 + α/k
) (
1 + βγ 5 /s ) . (10.165)
The only possible way to relate this projection operator to a massive on-shell
Dirac particle of mass m and momentum p µ is to choose k µ = p µ /m, while s µ
then embodies the remaining (spin) degree of freedom. The final result is the
well-known Dirac form
Π(α, β) = 1 ( ) (
m + α/p 1 + βγ 5 /s ) ,
4m
p · p = m 2 , s · s = −1 , p · s = 0 , α, β = ± . (10.166)
Obviously, the sum of any two or three of the above projection operators is also
a resolution to our quest.
10.10.3 Irregular cases
First irregular case
Let us now assume that, in Eq.(10.155), S ≠ 0 and p µ ≠ 0 but p 2 = 0. In that
case q µ must be proportional to p µ , and we write q µ = c p µ . Now the trace
−2Tr (γ µ N) = Sp µ + cɛ ρµαβ p ρ T αβ (10.167)
proves that both T and c must be nonzero. Then, the relation
− ( Sg µκ g νλ + P ɛ µνκλ) Tr (σ µν N) = T κλ ( S 2 + P 2) (10.168)
shows that no solution is possible in this case since T must vanish.