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214 November 7, 2013<br />

Also, in the annihilation of an on-shell quark-antiquark pair, we have seen that<br />

= 0 . (8.38)<br />

Taking all this into account, we see that in the now newly available Feynman<br />

diagram for q¯q → gg :<br />

a<br />

p q<br />

1<br />

1<br />

n j<br />

p<br />

2<br />

b<br />

the gluon propagator effectively reduces to just its g µν term, and the corresponding<br />

expression reads<br />

g µν<br />

M 3 = i¯hgg 3 v(p 1 )γ µ u(p 2 )<br />

(p 1 + p 2 ) 2 Y (q 1, ɛ 1 ; q 2 , ɛ 2 ; −q 1 − q 2 , ν) h jk n(T n ) a b .<br />

(8.39)<br />

with summation over the colour n implied. Note that the lowering of indices<br />

in the h symbol does not have any significance, I do it simply to make the<br />

typography looks nicer. Putting the handlebar on gluon 1 as before, we get<br />

Y (q 1 , q 1 ; q 2 , ɛ 2 , −q 1 − q 2 , ν) = ( ∆(q 2 ) νλ − ∆(q 1 + q 2 ) νλ) ɛ 2λ<br />

so that<br />

q 2<br />

k<br />

= −(p 1 + p 2 ) ν (p 1 + p 2 · ɛ 2 ) + (p 1 + p 2 ) 2 ɛ 2 ν , (8.40)<br />

M 3 ⌋ = i¯hgg 3 v(p 1 )/ɛ 2 u(p 2 ) h jk n(T n ) a b . (8.41)<br />

The total handlebarred amplitude thus becomes<br />

M 1+2+3 ⌋ = i¯hg v(p 1 )/ɛ 2 u(p 2 ) ( g 3 h jk nT n − g[T j , T k ] ) ab<br />

(8.42)<br />

The colour current will therefore be conserved if we choose<br />

and<br />

g 3 = g (8.43)<br />

[T j , T k ] = h jk n T n . (8.44)<br />

Note that since the matrices T are hermitean, the constants h must be purely<br />

imaginary 8 . Moreover, we can compute them, using Eq.(8.3), as<br />

h jkl = 2Tr ( T j T k T l − T l T k T j) . (8.45)<br />

8 It is customary to write [T j , T k ] = i f jk n T n . The f’s are then called the structure<br />

constants, and the set of T matrices are then the generators of the Lie algebra of the group<br />

SU(N). The i is then combined with the overall i of the vertex to give a Feynman rule without<br />

any i. This is of course a matter of taste.

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