5 Lecture 4: quark-gluon scattering and color ordering Consider the amplitude for quark-gluon scattering process 0 ! q(p 1 )+¯ q(p 4 )+ g(p 2 )+ g(p 3 ). There are three diagrams that contribute; two “abelian” and one “non-abelian”, that involves triple gluon couplings. We will take the left-handed spinor h1| for the outgoing quark with momentum p 1 and the left-handed spinor |4] for the outgoing (right-handed) anti-quark with momentum p 4 . The expression for the matrix element is iM = -ig 2 h1| ⇢ ˆ ✏ 2 (ˆ p 1 +ˆ p 2 )ˆ ✏ 3 s 12 ⇣ t a t b ⌘ ij + ˆ ✏ 3 (ˆ p 1 +ˆ p 3 )ˆ ✏ 2 s 13 ⇣ t b t a ⌘ ij |4] - g 2 f abc t c ij h1γ λ 4] s 14 (✏ 2 · ✏ 3 (p 2 - p 3 ) λ + ✏ 3λ (2p 3 + p 2 ) · ✏ 2 + ✏ 2λ (-2p 2 - p 3 )✏ 3 ) . (5.1) Here, t a,b are the SU (3) Lie algebra generators in the fundamental representation and i, j, a, b refer to color indices of quarks and gluons. The SU (3) generators are normalized Tr ⇥ t a t b ⇤ = δ ab /2 and, as generators of a Lie algebra, they satisfy the commutation relation t a t b - t b t a = if abc t c . (5.2) We can use this relation to remove the SU (3) structure constants from the expression for the amplitude. Also, we rescale t a = T a / p 2, to have Tr ⇥ T a T b ⇤ = δ ab . As the result of this, the amplitude is written as the sum of two terms M = ✓ g p 2 ◆ 2 ⇣ M 1 (T q T b ) ij + M 2 (T b T a ) ij ⌘ , (5.3) where M 1 = - " h1|✏ 2 ( ˆ 1+ ˆ 2)✏ 3 |4] s 12 - h1γ μ 4] s 14 (✏ 2 · ✏ 3 (p 2 - p 3 ) μ +✏ 3μ (2p 3 + p 2 ) · ✏ 2 + ✏ 2μ (-2p 2 - p 3 )✏ 3 ) # . M 2 = - " h1|✏ 3 ( ˆ 1+ ˆ 3)✏ 2 |4] s 13 - h1γ μ 4] s 14 (✏ 2 · ✏ 3 (p 3 - p 2 ) μ +✏ 3μ (-2p 3 - p 2 ) · ✏ 2 + ✏ 2μ (2p 2 + p 3 )✏ 3 ) # . (5.4) If we write M 1 = M (1, 2, 3, 4), then M 2 = M (1, 3, 2, 4), so it is sufficient to compute one function of external momenta to get the full result. We note that out of three diagrams that contribute to the amplitude M only two contribute to the function M 1 . The diagram that does not contribute has its external particles arranged in such a way that they can not be ordered (clockwise) as p 1 ,p 2 ,p 3 ,p 4 . The amplitude M (1, 2, 3, 4) is called “color-ordered”. It is transversal ( gauge-invariant) and independent of color indices of colliding particles. We now calculate the color-ordered – 17 –
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5 Lecture 4: quark-gluon scattering and color ordering
Consider the amplitude for quark-gluon scattering process 0 ! q(p1
) + q̄(p4
) + g(p2
) + g(p3
).
There are three diagrams that contribute; two “abelian” and one “non-abelian”, that involves
triple gluon couplings. We will take the left-handed spinor h1| for the outgoing quark with
momentum p
1
and the left-handed spinor |4] for the outgoing (right-handed) anti-quark with
momentum p
4
. The expression for the matrix element is
iM = �ig
2h1|⇢✏̂
2
(p̂1
+ p̂
2
)✏̂3
s
12
⇣t
a
t
b
⌘
ij
+✏̂
3
(p̂1
+ p̂
3
)✏̂2
s
13
⇣t
b
t
a
⌘
ij
�|4]
� g
2
f
abc
t
c
ij
h1��4]s
14
(✏2
· ✏3
(p2
� p
3
)�
+ ✏
3�
(2p3
+ p
2
) · ✏2
+ ✏
2�
(�2p2
� p
3
)✏3
) .
(5.1)
Here, ta,b are the SU(3) Lie algebra generators in the fundamental representation and i, j, a, b
refer to color indices of quarks and gluons. The SU(3) generators are normalized Tr⇥t
a
t
b
⇤=
�
ab
/2 and, as generators of a Lie algebra, they satisfy the commutation relation
t
a
t
b � t
b
t
a = if
abc
t
c
. (5.2)
We can use this relation to remove the SU(3) structure constants from the expression for
the amplitude. Also, we rescale t
a = T
a
/
p2, to have Tr
⇥T
a
T
b
⇤= �
ab. As the result of this,
the amplitude is written as the sum of two terms
M =
✓gp2
◆2 ⇣
M
1
(T q
T
b)ij
+M
2
(T b
T
a)ij
⌘, (5.3)
where
M
1
= �"h1|✏
2
(1̂ + 2̂)✏3
|4]s
12
� h1�µ4]s
14
(✏2
· ✏3
(p2
� p
3
)µ
+✏
3µ
(2p3
+ p
2
) · ✏2
+ ✏
2µ
(�2p2
� p
3
)✏3
)
#.
M
2
= �"h1|✏
3
(1̂ + 3̂)✏2
|4]s
13
� h1�µ4]s
14
(✏2
· ✏3
(p3
� p
2
)µ
+✏
3µ
(�2p3
� p
2
) · ✏2
+ ✏
2µ
(2p2
+ p
3
)✏3
)
#.
(5.4)
If we write M
1
= M(1, 2, 3, 4), then M
2
= M(1, 3, 2, 4), so it is su�cient to compute one
function of external momenta to get the full result. We note that out of three diagrams that
contribute to the amplitude M only two contribute to the function M
1
. The diagram that
does not contribute has its external particles arranged in such a way that they can not be
ordered (clockwise) as p1
, p
2
, p
3
, p
4
.
The amplitude M(1, 2, 3, 4) is called “color-ordered”. It is transversal ( gauge-invariant)
and independent of color indices of colliding particles. We now calculate the color-ordered
– 17 –
amplitude M(1L
, 2, 3, 4L
). As the first step, we consider equal photon helicities, starting from
right-handed photons. The relevant formula reads
�
µ
✏
µ
R
=
p2
hrki (|k]hr|+ |ri[k|) , (5.5)
so that
h1|✏̂3R
=
p2
hr3
3ih1r3i[3|
h1|✏̂2R
=
p2
hr2
2ih1r2i[2|.(5.6)
Also, scalar products of polarization vectors with same helicities vanishes if the two vectors
have identical reference momenta
✏
3R
· ✏2R
⇠ hr2
r
3
i. (5.7)
It is then easy to see that if we choose r
2
= r
3
= p
1
, the amplitude vanishes
M
1
(q1L , g2R , g3R , q̄4L) = 0. (5.8)
Similar argument can be used to prove that amplitude M
1
(q1L , g2L , g3L , q̄4L) vanishes as
well. Indeed,
✏̂
L
= �p2
[rk](|r]hk|+ |ki[r|) , (5.9)
so that
✏̂
3L
|4] = �p2
[r3
3]|3i[r
3
4],
✏̂
2L
|4] = �p2
[r2
2]|2i[r
2
4].
(5.10)
So, we choose r
2
= r
3
= p
4
and find M
1
(q1L , g2L , g3L , q̄4L) = 0.
Next, we will study the color-ordered amplitude where the two photon polarizations
are di↵erent. Specifically, we consider M(q1L , g2R , g3L , q̄4L). The explicit expression for the
amplitude reads
M = �"h1|✏
2R
(1̂ + 2̂)✏3L
|4]s
12
� h1�µ4]s
14
(✏2R
· ✏3L
(p2
� p
3
)µ
+✏
3Lµ
(2p3
+ p
2
) · ✏2R
+ ✏
2Rµ
(�2p2
� p
3
)✏3L
)
#.
(5.11)
To understand how to simplify computations, we will study contributing terms in Eq.(5.11)
separately. The first term reads
h1|✏̂2R
(1̂ + 2̂)✏̂3L
|4] = �2h1r2
i[2|(1̂ + 2̂)|3i[r3
4]
hr2
2i[r3
3]= �2h1r
2
i[21]h13i[r3
4]
hr2
2i[r3
3].
(5.12)
– 18 –
The third and the fourth terms in Eq.(5.11) contain the following spinor products
h1�µ4]✏3Lµ
= h1|✏̂3L
4] = �p2h13i[r
3
4]
[r3
3],
h1�µ4]✏2Rµ
= h1|✏̂2R
4] =
p2h1r
2
i[24]hr
2
2i ,
(5.13)
Hence, we conclude that if we choose r
3
= p
4
and r
2
= p
1
all contributions in Eq.(5.12) and
Eq.(5.13) vanish; therfore, only the second term in Eq.(5.11) contributes. We find
M
1
(q1L , g2L , g3L , q̄4L) =
h1|(2̂� 3̂)|4]s
14
✏
2R
· ✏3L
. (5.14)
To simplify it further, we use momentum conseration
h1|(2̂� 3̂)|4] = �2h13i[34], (5.15)
and compute the product of two polarization vectors
✏
2R
· ✏3L
=hr
2
�
µ2]p2hr
2
2i(�1)[r
3
�
µ
3ip2[r
3
3]= �h1�µ2][4�
µ
3i2h12i[43] = �h13i[42]
h12i[43] . (5.16)
We therefore find ( use s
14
= s
23
= h23i[23] )
M(q1L , g2R , g3L , q̄4L) =
2h13i[34]h23i[23]
h13i[42]h12i[43] = � 2h13i2[42]
h12ih23i[23] . (5.17)
We can simplify this expression by multiplying it by 1 = h13i/h13i. It follows from momentum