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

conservation that

h13i[32] = h1|3̂|2] = h1 ��1̂� 2̂� 4̂�2] = �h14i[42]. (5.18)

Then,

M

1

(q1L , g2R , g3L , q̄4L) =

2h13i3h43ih12ih23ih34ih41i . (5.19)

Next, we will compute the second color-ordered amplitude M(q1L , g2L , g3R , q̄4L). We will

again go through the same exercise of trying to force as many terms as possible to vanish.

We will do it slightly di↵erently this time. The amplitude reads

M = �"h1|✏̂

2L

(1̂ + 2̂)✏̂3R

|4]s

12

� h1�µ4]s

14

(✏2L

· ✏3R

(p2

� p

3

)µ

+✏

3Rµ

(2p3

+ p

2

) · ✏2L

+ ✏

2Lµ

(�2p2

� p

3

)✏3R

)

#.

(5.20)

Lets focus on the “non-abelian” contribution to this amplitude. There are three terms that

involve

p

3

· ✏2L

, p

2

· ✏3R

, ✏

2L

· ✏3R

. (5.21)

– 19 –

Since

✏

2L

· ✏3R

⇠ [r2

3]hr3

2i, (5.22)

we can ensure that all terms in Eq.(5.21) vanish if r2

⇠ p

3

and r

3

⇠ p

2

. With these choices

of reference vectors, we find

h1|✏̂2L

= �p2h12i[3|[32]

,

✏̂

3R

|4] =p2|2i[34]h23i .

(5.23)

Therefore, we find

M(q1L , g2L , g3R , q̄4L) =

2h12i[34][31]h12is

12

[32]h23i =2[34][31]h12i[21][32]h23i . (5.24)

For further simplifications, multiply both numerator and denominator with h12ih42i. Then

M(q1L , g2L , g3R , q̄4L) =

2[34][31]h12i2h42i[21][32]h23ih12ih42i . (5.25)

Now, in the denominator use

[32]h42i = �[32]h24i = �[3|2̂|4i = [3|1̂|4i = [31]h14i, (5.26)

so that

M(q1L , g2L , g3R , q̄4L) =

2[34]h12i2h42i[21]h14ih23ih12i . (5.27)

To simplify it further, note that since s

12

= s

34

, we have

[34]h43i = [21]h21i $ [34]

[21]=

h12ih34i . (5.28)

M(q1L , g2L , g3R , q̄4L) = � 2h12i3h42i

h12ih23ih34ih41i . (5.29)

Amplitudes for other helicity configurations can be obtained from the computed ones

using complex conjugation.

– 20 –

6 Lecture 5: gluon scattering amplitudes and the idea of color ordering

As the next example, we consider gluon scattering 0 ! g

p1 + g

p2 + g

p3 + g

p4 . The color labels

will be chosen as a1

, .., a

4

. There are four diagrams, – three with three-gluon vertices and one

with the four-gluon vertex, see Fig. 1. Feynman rules for three-gluon vertices have already

been shown; the Feynman rules for four-gluon vertex are (clockwise (µ, a), (⌫, b)(�, c), (�, d) ):

� ig

2

hf

abe

f

cde

⇣(gµ�g⌫� � g

µ�

g

⌫�

⌘+ f

ace

f

bde

⇣g

µ⌫

g

�� � g

µ�

g

µ�

⌘

+f

ade

f

bce

⇣g

µ⌫

g

�� � g

µ�

g

⌫�

⌘i.

(6.1)

Each of the three Feynman diagrams with three-gluon vertices and each term in Eq.(6.1)

contains a color structure that reads fabe

f

cde, where a, b, c, d are choosen from a set {a1

, a

2

, a

3

, a

4

}and e is a dummy index. We will try to simplify the color structure.

First, lets write f

abe

f

cde is a canonical form. To do so, we will combine two equations

for generators of SU(3) algebra

[T a

, T

b] = i

p2fabe

T

e

, TrhT

a

T

b

i= �

ab

. (6.2)

Then,

�2fabe

f

cde = Trh[T a

, T

b][T c

, T

d]i= Tr

hT

a

T

b

T

c

T

d

i� Tr

hT

b

T

a

T

c

T

d

i

� TrhT

a

T

b

T

d

T

c

i+Tr

hT

b

T

a

T

d

T

c

i.

(6.3)

Since we can do the same in all diagrams that contribute to four-gluon scattering ampliutde,

we conclude that the full amplitude can be represented as

M = M1

Tr [T a1T

a2T

a3T

a4 ] +M2

Tr [T a1T

a2T

a4T

a3 ] +M3

Tr [T a1T

a3T

a2T

a4 ]

+M4

Tr [T a1T

a3T

a4T

a2 ] +M5

Tr [T a1T

a4T

a2T

a3 ] +M6

Tr [T a1T

a4T

a3T

a2 ] ,(6.4)

whereM1..6

are functions of momenta and polarization vectors. We will now try to understand

which Feynman diagrams contribute to those functions. Consider first Feynman diagram in

Fig.1. The diagram has two triple-gluon vertices. The color factor is fa1ea4f

a2a3e and this is

4

1 2

34

1 2

3

1 3

4 2

1 2

34

Figure 1. Four-gluons scattering diagrams

– 21 –

expressed through traces of SU(3) generators, as

f

a1ea4f

a2a3e ⇠ Tr [T a1T

a4T

a2T

a3 ]� Tr [T a1T

a2T

a3T

a4 ]

� Tr [T a1T

a4T

a3T

a2 ] + Tr [T a1T

a3T

a2T

a4 ] .(6.5)

We conclude that this diagram does not contribute to two color-structures, Tr [T a1T

a3T

a4T

a2 ]

and Tr [T a1T

a2T

a4T

a3 ], i.e. where gluons 1 and 4 are not adjacent. Also, signs of various

terms in Eq.(6.5) are di↵erent. To understand the meaning of this, consider a diagram with

two three-gluon vertices where gluons appear clockwise as ((4, 1), (2, 3)). This diagram is

proportional to

V

3g

(1,�14, 4)V3g

(2, 3� 23)fa1ea4f

a2,a3,e

⇠ V

3g

(1,�14, 4)V3g

(2, 3,�23)hTr (T a1

T

a4T

a2T

a3)� Tr (T a1T

a2T

a3T

a4)

� Tr (T a1T

a4T

a3T

a2) + Tr (T a1T

a3T

a2T

a4)i.

(6.6)

Now, we can use the fact that the color-stripped three-gluon vertex functions V

3g

(i, j, k) is

anti-symmetric w.r.t. permutations of any pair of gluons, e.g. V

3g

(i, j, k) = �V

3g

(j, i, k).

Hence, if we synchronize the order of arguments in functions V3g

with the order of arguments

in color traces, the signs of all terms in Eq.(6.6) are the same. Indeed,

V

3g

(1,�14, 4)V3g

(2, 3,�23)Tr (T a1T

a4T

a2T

a3)

= (�1)V3g

(1, 4,�14)V3g

(2, 3,�23)Tr (T a1T

a4T

a2T

a3) ,

V

3g

(1,�14, 4)V3g

(2, 3,�23)Tr (T a1T

a2T

a3T

a4)

= V

3g

(4, 1,�14)V3g

(2, 3,�23)Tr (T a4T

a1T

a2T

a3) ,

(6.7)

etc. Hence, we conclude that all terms in Eq.(6.5) will enter with the sign minus provided

that the color-stripped diagram is drawn with the appropriate order of gluons in each of the

three cases; namely, the order should be identical to the order of generators in the color trace.

One can perform a similar analysis for other diagrams that contribute to luon scattering

and come to the conclusion that color-ordered amplitudes are, in fact, one and the same

function that di↵er only by permutations of its arguments

M = M(1, 2, 3, 4)Tr (T a1T

a2T

a3T

a4) +M(1, 3, 2, 4)Tr (T a1T

a2T

a4T

a3) + · · · (6.8)

These color-ordered ( or “color-stripped”, or “partial”) amplitudes are obtained from “color-

stripped” Feynman rules shown in Fig. 2.

We note that all graphs that are properly ordered and only such graphs should be included

when color-ordered amplitude for a particular process is computed. As we will see on a simple

example of four-gluon scattering, this implies that diagrams for which external gluons are

improperly ordered can be dropped.

– 22 –

Figure 2. Color stripped Feynman rules.

We now turn to the computation of the four-gluon color-ordered amplitude M(1, 2, 3, 4).

Three diagrams, shown in Fig. 4 contribute. The result reads

M(g1

, g

2

, g

3

, g

4

) =

✓igp2

◆2

(�i

s

14

[✏1

· ✏4

(4� 1)�

+ ✏

1�

(1 + 14) · ✏4

+ ✏

4�

(�14� 4) · ✏1

]⇥

[✏2

✏

3

(2� 3)�

+ ✏

3�

(3 + 23) · ✏2

+ ✏

2�

(�23� 2) · ✏3

]

+�i

s

12

[✏1

· ✏2

(1� 2)�

+ ✏

2�

(2 + 12) · ✏1

+ ✏

1�

(�12� 1) · ✏2

]⇥[✏3

· ✏4

(3� 4)�

+ ✏

4�

(4 + 43) · ✏3

+ ✏

3�

(�3� 34) · ✏2

]

� i [2✏1

· ✏3

✏

2

· ✏4

� ✏

1

· ✏2

✏

3

· ✏4

� ✏

1

· ✏4

✏

2

· ✏3

]

).

(6.9)

This expression looks rather complex and we will now discuss how to simplify its computation

dramatically. We will start with a very special case, taking helicities of all colliding gluons

to be the same. It follows from Eq.(6.9) that every term in the amplitude contains a scalar

product of polarization vectors. From Lecture 2 we know that if two polarization vectors have

the same helicity and the same reference vector, their scalar product vanishes since

✏

L

(pi

, r

i

) · ✏L

(pj

, r

j

) ⇠ [ri

r

j

], ✏

R

(pi

, r

i

) · ✏R

(pj

, r

j

) ⇠ hri

r

j

i. (6.10)

Hence, to compute M(g1L

, g

2L

, g

3L

, g

4L

) or M(g1R

, g

2R

, g

3R

, g

4R

), we just need to choose

identical reference vectors for all gluons and observe that both amplitudes vanish

M(g1L

, g

2L

, g

3L

, g

4L

) = M(g1R

, g

2R

, g

3R

, g

4R

) = 0. (6.11)

– 23 –

4

1 2

34

1 2

3

1 2

3 4

Figure 3. Four-gluons scattering diagrams

Note that the above argument generalizes to the case of n-gluon scattering in a straightforward

way.

As the next step, consider the amplitude where three helicities are the same and one is

di↵erent. For definitiness, we take helicities of g1

, g

2

, g

3

to the R ( “plus”) and the helicity

of g4

to be L ( “minus”). If we choose equal reference vectors for g

1,..,3

, all scalar products

between their polarization vectors vanish. The scalar product of a right-handed polarization

vector and left-handed polarization vector reads

✏

R

(pi

, r) · ✏L

(pj

, s) =hr�µi]p2hrii

(�1)[s�µ

jip2[sj]

= �hrji[si]hrii[sj] . (6.12)

In our case p

j

= p

4

and so choose the reference vector for all the left-handed polarization

vectors to be p

4

, forces all the scalar product that involve ✏

4L

to vanish. Hence, we conclude

that

M(g1R

, g

2R

, g

3R

, g

4L

) = 0. (6.13)

Clearly, by a similar argument, any other amplitude for three equal helicities and one di↵erent

helicity vanishes. Also, similar to all-equal helicity case, the argument generalizes to the case

of n-gluon scattering in a straightforward way.

Next, we consider the case when there are two gluons with equal helicities and two gluons

with di↵erent helicities. We need to consider two cases, that we take to beM(g1R

, g

2R

, g

3L

, g

4L

)

and M(g1R

, g

2L

, g

3R

, g

4L

).

Let us begin by considering M(g1R

, g

2R

, g

3L

, g

4L

). From our previous studies, we know

that scalar products of same-helicity polarization vectors vanish if reference vectors are the

same and scalar product of di↵erent helciity vectors vanish if a reference vector for one

gluon is the momentum of the other. Therefore, if we choose identical reference vectors

for g

1

and g

2

and identical reference vectors for g

3

and g

4

, we will have ✏

1R

· ✏2R

= 0 and

✏

3L

· ✏4L

= 0. Next, by choosing the reference vector for left-handed polarizations to be p

µ

2

,

we obtain ✏

2R

· ✏3L

= ✏

2R

· ✏4L

= 0. Finally, by choosing the reference vector for right-handed

polarizations to be pµ3

, we have ✏1R

·✏3L

= 0. Therefore, the only non-vanishing scalar product

is ✏1R

· ✏4L

.

– 24 –

Also, thanks to our choices of reference vectors, we have

p

µ

2

✏

3Lµ

= p

µ

2

✏

4Lµ

= 0, p

µ

3

✏

1Rµ

= p

µ

3

✏

2Rµ

= 0. (6.14)

It follows from Eq.(6.9) that only the second term in the sum contributes and the result reads

M(g1R

, g

2R

, g

3L

, g

4L

) =�2g2

s

12

(✏1R

· ✏4L

) (p1

· ✏2R

) (p4

· ✏3L

) . (6.15)

It is straighforward to compute the scalar products in Eq.(6.15). Using Eq.(6.12), we

find

✏

1R

(p1

, p

3

) · ✏4L

(p4

, p

2

) = �h34i[21]h31i[24] , (6.16)

and

p

1µ

✏

µ

2R

=h31̂2]p2h32i , p

4µ

✏

µ

3L

= � [24̂3ip2[23]

. (6.17)

Now, putting everything together, we obtain

M(g1R

, g

2R

, g

3L

, g

4L

) =g

2

s

12

h34i[12]2h32i

h43i[23]

= �g

2

[12]2h34i2s

12

s

23

= �g

2

[12]2h34i2s

12

s

23

� g

2

[12]2h34i2s

34

s

14

= �g

2

[12]2h34i2h34i[43]h14i[41] = �g

2

[12]3h34i[12][43]h14i[41]

(6.18)

We would like to get rid of “wrong” brackets in this expression h34i/h14i. We can further

simplify the last expression if we use momentum conservation

[23̂4i = �[21̂4i. (6.19)

This implies [23]h34i = �[21]h14i, so that h34i/h14i = [12]/[23]. Using this in Eq.(6.18), we

obtain

M(g1R

, g

2R

, g

3L

, g

4L

) = g

2

[12]4

[12][23][34][41]. (6.20)

The second helicity amplitude that we need to consider is M(g1R

, g

2L

, g

3R

, g

4L

). The

calculation is similar to what we already did: taking reference vectors for right-handed polar-

ization vectors to be p2µ

and for left-handed polarizations to be p3µ

implies that the only scalar

product of polarization vectors that survives is ✏

1R

· ✏4L

. The expression for the amplitude

reads

M(g1R

, g

2L

, g

3R

, g

4L

) =�2g2

s

12

(✏1R

· ✏4L

) (p1

· ✏2L

) (p4

· ✏3R

) . (6.21)

To compute the scalar products, we use

✏

1R

(p1

, p

2

) · ✏4L

(p4

, p

3

) = �h24i[31]h21i[34] , (6.22)

and

p

1µ

✏

µ

2L

= � [31̂2ip2[32]

, p

4µ

✏

µ

3R

= � h24̂3ip2h23i . (6.23)

– 25 –

We find

M(g1R

, g

2L

, g

3R

, g

4L

) =g

2[13]2h24i2s

12

[32]h23i =g

2[13]2h24i2s

12

s

23

=g

2[13]2h24i2s

34

s

14

. (6.24)

To simplify, we multiply this expression with [13]2, use

h24i2[13]2h34ih41i =

h24i2h41̂3]h34i[13] = � h24i2

h42̂3]h34i[13]=

h24i[23]h34i[13] =

h24i[23][13̂4i = � h24i

[23][12]h24i =�1

[23][12]

(6.25)

to find

M(g1R

, g

2L

, g

3R

, g

4L

) =g

2[13]4

[12][23][34][41]. (6.26)

As we see, use of spinor-helciity methods allows us to find very compact expressions

for scattering amplitudes for four-gluon scattering. This completes the calculation of spinor-

helicity ampltudes for gluon scattering. Every amplitude that we have not computed explicitly

can be obtained from the complex conjugation.

– 26 –

7 Lecture 6: gluon scattering cross-sections

In this Lecture we will discuss how to use the color-ordered helicity amplitudes calculated

in the prevoius lecture to compute scattering cross-sections. A di�cult part here is the sum

over colors. Recall that the scattering amplitude is written as

M(ga11

, g

a22

, g

a33

, g

a44

) =X

�2P (2,3,4)

M(g1

, g

�2 , g�3 , g�4)⇥ Tr [T a1T

a�2T

a�3T

a�4 ] , (7.1)

where �

i

is an element of the permutation set of three numbers a

2

, a

3

, a

4

. To compute the

cross-section, we need to square the amplitude and sum over colors and helicities. The

helicity sums are easy but the sum over colors seems complicated. We will discuss how it can

be performed.

To sum over colors, we need to deal with products of traces, summed over color indices

Tr [T a1T

a�1T

a�2T

a�3 ]⇥ Tr [T a1T

a⇠2T

a⇠3T

a⇠4 ]⇧�a�ia⇠j

. (7.2)

In general, products of traces can be computed with the help of the following identity

X

ij;km

=N

2�1X

a=1

(T a)ij

(T a)km

= �

im

�

kj

� 1

N

�

ij

�

km

. (7.3)

To prove Eq.(7.3), we use transformation properties of the right-hand side under SU(N), the

fact that T

a’s are traceless and that Tr⇥T

a

T

b

⇤= �

ab

. We can use these identities to, e.g.,

transform the product of traces into trace of products and simple terms. Indeed, consider

Tr [A1

T

a

A

2

] Tr [B1

T

a

B

2

] = A

1,i1i2A2,j2,i1

B

1,k1k2B2,m2k1

�

i2m2�k2j2 �1

N

�

i2j2�k2m2

�

= Tr [A1

B

2

B

1

A

2

]� 1

N

Tr [A1

A

2

] Tr [B1

B

2

] .

(7.4)

The result will be a complicated collection of traces to compute. It can be done but it is not

easy. However, it turns out that there is a simpler way to do it and we will describe it now.

We have so far considered the group SU(N) where S tells us that group elements should

have determinant 1. This implies that SU(N) generators are traceless; indeed, an element of

SU(N) is

g ⇡ e

iT

a✓

a ⇡ 1 + iT

a

✓

a ) 1 = det(g) ⇡ 1 + i✓

aTr[T a] ) Tr[T a] = 0. (7.5)

Let us imagine now that we extend the SU(N) group to U(N). This amounts to the intro-

duction of a “phase” generator TN

2 which commutes with all generators of SU(N) and that

is normalized as Tr[TN

2TN

2 ] = 1. Hence, we take T

N

2 to be a diagonal matrix with elements

1/pN . With this extension, equation for X

ij;km

simplifies

X

U(N)

ij,km

=N

2X

a=1

T

a

ij

T

a

km

=N

2�1X

a=1

T

a

ij

T

a

km

+ T

N

2

ij

T

N

2

km

= �

im

�

kj

. (7.6)

– 27 –

We would like to use this result to compute color factors to describe scattering of gluons,

e↵ectively changing SU(N) ! U(N). Are we allowed to do that? The answer is yes, because

U(1) gluons do not couple to other, SU(N) ones, sine f

abc = 0 if a, b or c is N2. Therefore,

we can use a simple U(N) formula to sum over colors.

Hence, we need to compute products of traces summed over colors. First, we calculate

(T q

T

a)ij

= N�

ij

. (7.7)

Then

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

b

T

c

T

d

i+

= TrhT

a

T

b

T

c

T

d

iTr

hT

d

T

c

T

b

T

a

i

= T

a

ik

⇣T

b

T

c

T

d

⌘

ki

⇣T

d

T

c

T

b

⌘

jm

T

a

mj

= �

ij

�

mk

⇣T

b

T

c

T

d

⌘

ki

⇣T

d

T

c

T

b

⌘

jm

= TrhT

b

T

c

T

d

T

d

T

c

T

b

i= NTr

hT

b

T

c

T

c

T

b

i= N

3Tr [1] = N

4

.

(7.8)

The second product of traces that we need to compute is more complicated

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

b

T

d

T

c

i+

= TrhT

a

T

b

T

c

T

d

iTr

hT

c

T

d

T

b

T

a

i

= TrhT

b

T

c

T

d

T

c

T

d

T

b

i= NTr

hT

c

T

d

T

c

T

d

i.

(7.9)

To compute the last trace note that, for any matrix A, we have

(T c

AT

c)ij

= T

c

ik

T

c

mj

A

km

= �

ij

�

km

A

km

= Tr [A] �ij

. (7.10)

This implies that

T

c

T

d

T

c|d=N

2 =pN 1̂, and T

c

T

d

T

c|d2N2�1

= 0. (7.11)

Therefore, the last term in Eq.(7.9) becomes

NTrhT

c

T

d

T

c

T

d

i= N

3/2TrhT

N

2i= N

2

. (7.12)

Hence,

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

b

T

d

T

c

i+

= N

2

. (7.13)

The remaining contirbutions are

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

c

T

b

T

d

i+

= N

2

,

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

c

T

d

T

b

i+

= N

2

,

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

d

T

c

T

b

i+

= N

2

,

TrhT

a

T

b

T

c

T

d

iTr

hT

a

T

d

T

b

T

c

i+

= N

2

.

(7.14)

– 28 –

Suppose now we square Eq.(7.1) and sum it over gluon color indices. Then, in the result-

ing sum over colors, there will be terms proportional to squared of color-ordered amplitudes

and terms that are interferences. According to our calculation above, color-factors that mul-

tiply squares of amplitudes are N

4, while color factors that multiply all interference terms

are N

2. Hence, we find

X

colors

|M|2 = N

4

6X

I=1

|MI

|2 +N

2

X

I 6=J

MI

M⇤J

. (7.15)

We will simplify this formula using the following identity for color-stripped amplitudes

6X

I=1

MI

= 0. (7.16)

This equation is not obvious and we will explain shortly why it is valid. But let us see first

how it helps. We find X

I 6=J

MI

M⇤J

= �X

I

|MI

|2, (7.17)

so that Eq.(7.15) becomes

X

colors

|M|2 = N

2(N2 � 1)6X

I=1

|MI

|2. (7.18)

We conclude that for four-gluon scattering, the full amplitude squared is given by the sum of

squares of color-ordered amplitudes.

Let me now explain why the sum of color-ordered amplitudes vanishes. For our purposes,

it can be viewed as a consequence of the fact that the U(1) gluon can not interact with SU(N)

gluons. This statement is obvious as long as the color-information is kept. However, once we

use color-ordered amplitudes, the color information disappears and the “non-interaction” of

certain types of gluons with the rest manifests itself in a complex way. This is the meaning

of Eq.(7.16). To see how it works in detail, consider a four-gluon scattering amplitude and

take one of those gluons to be the U(1) gluon and the other three SU(N) gluons. Then

0 = M⇣g

N

2

1

g

a2g

a3g

a4

⌘=

X

�2PM(g

1

, g

�2 , g�3 , g�4) TrhT

a

2NT

a�2T

a�3T

a�4

i. (7.19)

There are six di↵erent traces on the right hand side of the above equation, but they can

grouped into two groups of equal traces since T

N

2generator is proportional to an identity

matrix. Thefore

TrhT

a

2NT

a2T

a3T

a4

i= Tr

hT

a

2NT

a4T

a2T

a3

i= Tr

hT

a

2NT

a3T

a4T

a2

i,

TrhT

a

2NT

a2T

a4T

a3

i= Tr

hT

a

2NT

a3T

a2T

a4

i= Tr

hT

a

2NT

a4T

a3T

a2

i.

(7.20)

– 29 –

Using these equalities and the fact that T

N

2is proportional to the identity matrix, we can

re-write the right-hand side of Eq.(7.19) as

0 = Tr [T a2T

a3T

a4 ]�M(g

1

, g

2

, g

3

, g

4

) +M(g1

, g

4

, g

2

, g

3

) +M(g1

, g

3

, g

4

, g

2

)�

+Tr [T a2T

a4T

a3 ]�M(g

1

, g

2

, g

4

, g

3

) +M(g1

, g

3

, g

2

, g

4

) +M(g1

, g

4

, g

3

, g

2

)�.

(7.21)

Now, we can choose any values for remaining color indices. For example, take a

2

= N

2.

Then, since Tr [T a

2

T

a

4

] = Tr [T a

4

T

a

2

], we obtain

0 = M(g1

, g

2

, g

3

, g

4

) +M(g1

, g

4

, g

2

, g

3

) +M(g1

, g

3

, g

4

, g

2

)

+M(g1

, g

2

, g

4

, g

3

) +M(g1

, g

3

, g

2

, g

4

) +M(g1

, g

4

, g

3

, g

2

),(7.22)

which is Eq.(7.16).

One the other hand, Eq.(7.21) contains more information than what we have in Eq.(7.16).

This is because, for a general group SU(N), Tr [T a2T

a4T

a3 ] and Tr [T a2T

a3T

a4 ] are linear-

independent. Therefore, the right hand side in Eq.(7.21) can vanish if an only if the coe�cients

of two color traces there vanish independently of each other

0 = M(g1

, g

2

, g

3

, g

4

) +M(g1

, g

4

, g

2

, g

3

) +M(g1

, g

3

, g

4

, g

2

),

0 = M(g1

, g

2

, g

4

, g

3

) +M(g1

, g

3

, g

2

, g

4

) +M(g1

, g

4

, g

3

, g

2

),(7.23)

These two equations can be understood as a consequence of a simple identity. We can rewrite

the first equation in (7.23) ( using cyclic symmetry of the color-stripped amplitudes) as

0 = M(g1

, g

2

, g

3

, g

4

) +M(g2

, g

1

, g

3

, g

4

) +M(g2

, g

3

, g

1

, g

4

), (7.24)

which shows that the sum of all color-ordered amplitudes where the position of one gluon is

changed and the position of all other gluons are kept fixed vanishes. This is an example of a

more general set of “abelian” color identities that reduce the number of independent color-

ordered amplitudes. To give you an idea about the reduction in the number of independent

amplitudes, let me note that “naive” estimate of the number of independent color-ordered

amplitudes for n-gluon scattering is obviously (n�1)!. However, the abelian identities and the

so-called Bern-Johannson-Carrasco identities, bring the number of independent amplitudes

can be brought down to (n� 3)!.

For now, we will complete the calculation of the ampltiude squared for gluon scattering,

focusing now on the sum over gluon heliciites. We have computed two helicity ampltiudes in

the previous lecture; they are given in Eq.(6.20) and Eq.(6.26). We will use s = s

12

= s

34

,

t = s

13

= s

24

and u = s

23

= s

14

to denote kinematic invariants. We have

|M(g1R

, g

2R

, g

3L

, g

4L

)|2 =����

g

2[12]4

[12][23][34][41]

����2

=g

4

s

4

12

s

12

s

23

s

34

s

41

=g

4

s

2

u

2

,

|M(g1R

, g

2L

, g

3R

, g

4L

)|2 =����

g

2[13]4

[12][23][34][41]

����2

=g

4

t

4

s

2

u

2

.

(7.25)

– 30 –