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Renormalization of Gauge Theory on the Light Cone World Sheet

Permanent Link: http://ufdc.ufl.edu/UFE0021154/00001

Material Information

Title: Renormalization of Gauge Theory on the Light Cone World Sheet
Physical Description: 1 online resource (106 p.)
Language: english
Creator: Qiu, Jian
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007

Subjects

Subjects / Keywords: ads, bremsstrahlung, light, n, qcd, renormalization, world
Physics -- Dissertations, Academic -- UF
Genre: Physics thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

Notes

Abstract: We calculated the scattering of gluon, scalar and quarks in gauge theory in the light cone gauge. Some computation techniques suited for the light cone gauge are introduced. We observed some inadequacies of the counter terms suggested in our earlier work, and we suggest a new way of xing counter terms using Lorentz invariance as a guide. Gluon scattering with massive matter elds in the loop are presented for completeness. The helicity amplitude method is extensively used in this work and is also modifed to simplify the light cone gauge calculation.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Jian Qiu.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Local: Adviser: Thorn, Charles B.

Record Information

Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0021154:00001

Permanent Link: http://ufdc.ufl.edu/UFE0021154/00001

Material Information

Title: Renormalization of Gauge Theory on the Light Cone World Sheet
Physical Description: 1 online resource (106 p.)
Language: english
Creator: Qiu, Jian
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007

Subjects

Subjects / Keywords: ads, bremsstrahlung, light, n, qcd, renormalization, world
Physics -- Dissertations, Academic -- UF
Genre: Physics thesis, Ph.D.
bibliography   ( marcgt )
theses   ( marcgt )
government publication (state, provincial, terriorial, dependent)   ( marcgt )
born-digital   ( sobekcm )
Electronic Thesis or Dissertation

Notes

Abstract: We calculated the scattering of gluon, scalar and quarks in gauge theory in the light cone gauge. Some computation techniques suited for the light cone gauge are introduced. We observed some inadequacies of the counter terms suggested in our earlier work, and we suggest a new way of xing counter terms using Lorentz invariance as a guide. Gluon scattering with massive matter elds in the loop are presented for completeness. The helicity amplitude method is extensively used in this work and is also modifed to simplify the light cone gauge calculation.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Jian Qiu.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Local: Adviser: Thorn, Charles B.

Record Information

Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0021154:00001


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ce32cad00d1b2f0f0b496ab4be33480133570cfe







RENORMALIZATION OF GAUGE THEORY ON THE LIGHT CONE WORLD SHEET


By

JIAN QIU




















A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA

2007
































S2007 Jian Qiu


































to Lisa Elkenhans









ACKNOWLEDGMENTS

My greatest gratitude toward my advisor Professor Thorn for sparing no effort in

helping me obtain deeper understanding of ]li-,-i, and many a timely encouragement to

forge on.









TABLE OF CONTENTS
page

ACKNOWLEDGMENTS ................................. 4

LIST OF TABLES ....................... ............. 7

LIST OF FIGURES .................................... 8

ABSTRACT ................................. ...... 10

CHAPTER

1 INTRODUCTION ...................... .......... 11

2 COMPUTATION TECHNIQUES IN THE LIGHT CONE ............ 18

2.1 Mini Introduction to Spinor Helicity Amplitude Method .......... 18
2.2 The Light Cone Setup ................... ...... 23
2.3 Brief Description of the Calculational Procedure ....... ..... ... 26

3 BOX REDUCTION ...................... ......... 31

3.1 Box with a Helicity Violating Sub-diagram ......... ....... .. 31
3.2 Box without a Helicity Violating Sub-diagram . . ..... 37

4 MASSLESS AMPLITUDES .................. .......... .. 45

4.1 Two Point Functions .................. ........... .. 45
4.1.1 Gluon Self-Energy .................. ......... .. 45
4.1.2 Fermion and Scalar Self-Energy ............... .. 48
4.2 Three Point Functions .................. .......... .. 48
4.2.1 Gluon Vertex Correction .................. ... .. 49
4.2.2 Fermion Vertex Correction ................ .... .. 50
4.2.3 Scalar Vertex Correction .................. ... .. 51
4.2.4 Four-Point Functions .................. ..... .. 51
4.3 Scattering Amplitudes .................. .......... .. 51
4.3.1 Helicity Violating Amplitudes ............... . .. 52
4.3.2 Helicity Conserving Amplitudes ............... .. 53
4.3.3 Restoring Gauge Covariance ............ .... . .. 56
4.3.4 All 2-2 Processes .................. ......... .. 57

5 BREMSSTRAHLUNG IN LIGHT CONE ............... .. .. .. 61

5.1 A List of Infrared Terms .................. ......... .. 61
5.2 Bremsstrahlung Process .......... . . . .... 63
5.3 Combining with the Infrared Terms from the Virtual Process . ... 74
5.4 The Inclusion of Disconnected Diagrams .................. .. 76









6 GLUON SCATTERING WITH MASSIVE MATTER FIELDS


6.1 Computation Technique


6.2 Self-Energy Diagrams ...........
6.3 Triangle Diagrams .. ..........
6.4 Scattering Amplitudes .. ........
6.5 Photon Photon Scattering ........

7 CONCLUSIONS AND FUTURE WORK ...

7.1 Conclusion ................
7.2 Restoring Gauge Covariance in the Light
7.3 Triangle Anomaly .............
7.4 Two-Loop and n-Point Amplitudes .

APPENDIX

A SPINOR NOTATION IN THE LIGHT CONE

B FEYNMAN RULES .. ...........

REFERENCES ............... ....

BIOGRAPHICAL SKETCH .............


Cone .......


' ' '









LIST OF TABLES
Table page

7-1 List of mismatches in all the amplitudes ................... .. 95

7-2 List of the effect of the old counter terms schemes . . ..... 95

7-3 List of the effect of the new counter terms scheme . . ..... 96









LIST OF FIGURES
Figure page

1-1 Double line notation .. ... ............. ............ 16

1-2 Open string scattering diagram .................. ........ .. 16

1-3 Light-cone parametrization .. ....... . . ... .. 17

2-1 Five-point amplitude .................. .............. .. 28

2-2 Some building blocks that can be computed separately ............. ..29

2-3 Second level building blocks .................. ........ .. .. 29

2-4 Tri-gluon vertex .................. ................. .. 29

2-5 Propagator in the t-channel is contracted ................ ...... 29

2-6 Fermion-gluon vertex ............... ............ .. 29

2-7 Fermion propagator contracted ............... ....... .. 30

2-8 Here we assume only (k2 k4)2 / 0, and k+ > k0 > k+ ............. 30

3-1 Dual momentum assignment with k+ < k+ < k+ < k+ ............. 43

3-2 Model box with a helicity violating subtree ............... .. 43

3-3 Graphical representation of box reduction ............... ... 44

3-4 Model box with alternating helicity ............. .. .... 44

4-1 Two-point function and the dual momentum assignment . ..... 59

4-2 A self-energy diagram embedded in a scattering process . . ..... 60

4-3 Triangle diagram .................. ................ .. 60

4-4 Fermion vertex correction corresponding to Eq.49 . . ..... 60

4-5 Scalar vertex correction corresponding to Eq.4-11 ................ ..60

4-6 Four-fermion scattering corresponding to Eq.4-20 and Eq.4-21 . ... 60

5-1 Dual momentum assignment .................. .......... .. 78

5-2 Two diagrams with an extra 'unseen' gluon ................ 78

5-3 Cancelation of collinear divergence .................. ..... .. 78

5-4 Cancelation of the soft bremsstrahlung radiation against a virtual process . 79










5-5 Phase space integration region of x and y . .......

5-6 Configuration of k, F. and in the CM frame of p, +_ .

5-7 Self-energy bubble on leg 4 . ...............


All non-vanishing Bremsstrahlung processes . .

Disconnected Bremsstrahlung with two extra 'unseen'


gluons .


7-1 Triangle anomaly


Gluon-fermion-fermion 3 point vertex . .

Gluon-scalar-scalar 3 point vertex . .

Two diagrams contribute to the fermion-gluon

Scalar-gluon 4 point vertex . .....

Tri-gluon vertex . ...........

Gluon 4 point vertex . ........

Fermion 4 point vertex . .......

Scalar 4 point vertex . ........

Scalar Fermion 4 point vertex . ...


point


vertex .


B-l

B-2

B-3

B-4

B-5

B-6

B-7

B-8

B-9


. 79

. . 79

. 79

. 79

. . 80

. 92

. 101

. 102

. . 102

. 102

. 102

. 103

. 103

. 103

. 103









Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy

RENORMALIZATION OF GAUGE THEORY ON THE LIGHT CONE WORLD SHEET

By

Jian Qiu

August 2007

Chair: Charles Thorn
M.i,'r: Physics

We calculated the scattering of gluon, scalar and quarks in gauge theory in the light

cone gauge. Some computation techniques suited for the light cone gauge are introduced.

We observed some inadequacies of the counter terms -ii..-I -1. .1 in our earlier work, and

we -ii,.'-. -1 a new way of fixing counter terms using Lorentz invariance as a guide. Gluon

scattering with massive matter fields in the loop are presented for completeness. The

helicity amplitude method is extensively used in this work and is also modified to simplify

the light cone gauge calculation.









CHAPTER 1
INTRODUCTION

Gauge theory is generally thought of as the fundamental theory that governs most of

the important interactions in particle li,--i. The standard model of particle li,--i. is

based on a gauge theory with a gauge group SU(3) 0 SU(2) 0 U(1) minimally coupled

to some matter fields. The electro-weak part of this theory has relatively small ( ili. lii:.-

and the perturbative calculations produced results that agree with the experiments

extremely well. The strong interaction (QCD) has a coupling constant of the size 0.1 at

high momentum (short distance) where the perturbative calculation can still give some

useful predictions. However, the coupling is of order one around 1 GeV and will tend

to infinity at lower moment. The perturbative calculations cease to make sense at this

scale, yet it is at this scale that some interesting phenomena happen. For example, at high

energy QCD can be described by quark and gluon fields, while at low energy the quarks

become confined and the description in terms of pions and baryons is more relevant. At

high energy the quarks enjoy an SU(2)L 0 SU(2)R symmetry, but at low energy the axial

vector part of this symmetry is spontaneously broken down. These phenomena are beyond

the grasp of perturbation theory, and the quantitative results mainly come from lattice

computation. The quest for an analytic solution to the low energy spectrum or a dual

description of QCD that suits the low energy and strong coupling has been the focus of

many l]'r, -i, i- -

Besides its phenomenological importance, gauge theory is also interesting in its

own right because of its close connection with topology. A gauge theory consists of a

curvature field F,, -.1I i-fi-inl-, the Bianchi identity dF = 0. In QED, the entries of F

are simply the electric and magnetic filed E and B, while the Bianchi identity is two of

the four Maxwell equations V B 0 and -9B/0t V x E. In general, the Bianchi

identity says the two form F,,dx dxz is closed, so we cannot simply vary the field strength

to obtain the Euler-Lagrange equation. Instead, we write F as dA for some 1-form A,

known as the connection 1-form, and vary A rather than F. Although not all closed forms









(forms -.,l i-fi, in-; dw = 0) are exact (forms can be written as w = dT), writing F as dA

causes no problem in the perturbative calculations, because we are only interested in

small fluctuations around A = 0. Yet for a non-perturbative computation, a non-trivial

configuration of the A field can give important results such as the anomalous breaking of

the axial U(1) symmetry. In those cases, we have to manually sum over configurations of

different winding numbers. The A field, as its name -ii-.'I -. is a connection in the gauge

bundle, more concretely, the 1-form ata'dx" acting on the tangent vector = O/Ox",

converts it to a vector A"t in the direction of the fibre (elements of the Lie group

translate a field in the 'vertical' direction). With this connection, we can compare two

fields in nearby space-time points using the covariant derivative D, = t A,. The

geometrical nature of the gauge theory is not explored in this dissertation, however, the

problem of gauge fixing is still important for this work.

Putting aside gauge theory for a moment, we also have a possible alternative

description of the strong interaction: string theory. It was initially proposed in order

to model the dynamics of the flux tube which is the explanation of confinement in the

strong interactions. In analogy with the point particle, where the equation of motion is

such that the path traversed by a particle(world line) in space-time is the shortest path,

the bosonic string theory studies a one dimensional extended object and the equation of

motion is such that the two dimensional surface swept by the string (world sheet) is a

minimal surface. An open string has two free ends, they can be given the Chan-Paton

degrees of freedom, which are rather like a label that labels the different states of the end

of the string. These states can be taken to transform in a representation of a Lie group,

say, in the fundamental for one end and anti-fundamental for the other. With this setup,

the open string is like a pion which has presumably a quark and an anti-quark at each

end and connected by the flux tube. In fact, the scattering amplitude of the open strings

reproduces the scattering of gluons in the low energy limit.









The similarity was explored in [1], where 't Hooft looked at the SU(N) gauge theory

as N oo. In this limit, the adjoint representation of SU(N) can be approximated

by N 0 N. The A field transforming in the adjoint representation, now carries two

indices Ak in the large N limit, where i and j transform in the N and N respectively.

Hence a gluon line can be drawn as two lines carrying the i and j index each. With this

representation, a Feynman diagram will look like Fig.1-1. The index corresponds to the

Chan-Paton factors. Any closed index loop such as in the middle of Fig.1-1 will be a trace

Tr6j producing a factor of N. If a Feynman diagram is drawn with crossed lines, we will

lose factors of N, so by keeping only the leading terms in the power of N, we pick out

all the planar diagrams. There is a clear analogy between the planar diagrams and the

open string scattering diagrams Fig.1-2, where the arrows now indicate the Chan-Paton

factors. By summing over all the planar diagrams, we hope to learn something about the

non-perturbative nature of the gauge theory.

't Hooft also pioneered the light-cone parametrization of these planar diagrams in

[1]. The correspondence between the two graphs in Fig.1-3 is that, a rectangle maps to

a propagator, a line (cut) maps to the blank space between two propagators, and the

beginning(end) of a line is the splitting(merging) point of two propagators, or simply

a 3-point vertex. The success of this parametrization is because in the light-cone, a

propagator contains a step function O(x+p+). This factor says if a propagator is to

propagate forward in x+, it has to carry positive p+. This parametrization is also used in

[6, 7, 8, 9].
The correspondence between gauge field theory and string theory was revolutionized

by the AdS/CFT correspondence due to Maldacena [2], a certain superstring theory on an

AdS5 x S5 background is equivalent to the AT = 4 supersymmetric gauge theory. Detailed

proposals of this correspondence were made by Gubser, Klebanov and Polyakov [3] and

by Witten [4]. Their idea is the holographic concept, where the Minkowski space M4 is

the boundary of the AdS space, and the correlators in M4 are computed from the bulk of









AdS. The field theory and its perturbative expansions grasp the weak coupling limit, while

the string theory description naturally grasps the strong coupling limit. For example, in

[5], the author studied the energy of a string hanging between two static sources in the

classical limit (in the context of AdS/CFT duality, classical limit in the string theory

corresponds to the strong coupling limit in the field theory), which shows a force that

obeys inverse square law, in accordance with the fact that N = 4 SYM is a conformal

theory. Despite all the above, a lot of the details of the correspondence still await filling

in.

So the establishment of the detailed correspondence between string theory and gauge

theory becomes urgent. In [6, 7], the authors proposed a local world sheet description

of the (supersymmetric) gauge theory, which maps a Feynman diagram to a world

sheet representation. The vertices in this description become the merging and splitting

of strings, and the summation of Feynman diagrams becomes a path integral on the

(discretized) world sheet. The Feynman rules, or the vertex functions were realized by

inserting a local world sheet operator close to the splitting or merging point. There is no

need for the representation of a four point vertex, because it is automatically generated

when two insertions for three point vertices coincide and produce a contraction term. This

is rather fortuitous, for a four point vertex is very unnatural in a string diagram. The

summation of all the planar diagrams becomes the summation over whether or not there

is a cut in each of the lattice sites. This treatment still has many loose ends to tie up. For

example, the problem of renormalization which requires not only a local representation of

the bare diagrams on the world sheet but also a local representation of the counter terms.

The problem of renormalization for scalar fields on the world sheet was addressed in [8],

but was only partially solved for gauge theory in [9] due to some complications.

It is absolutely necessary therefore, to study the renormalization for gauge fields

on the world sheet, making sure that the world sheet path integral description at

the least will reproduce all the perturbative results before we jump into studying its









non-perturbative side. The checks include, whether all the divergences appearing in the

perturbative expansion can be absorbed into target space local counter terms, and whether

these counter terms can have a local world sheet representation. And most importantly in

the case of (supersymmetric) gauge theory, how the regulators will affect gauge covariance

and supersymmetry.

Since 't Hooft's planar diagram representation was most easily given in the light cone

parametrization, it is natural that we chose to study the gauge theory on the world sheet

also in light cone gauge. In this case, in order to study the gauge invariance property, we

have to deal with some collateral complications due to the light cone gauge choice and also

the issue of infrared divergence which is inherent to a gauge theory. These tasks have been

the main focus of my work. In the two papers [13, 14], we studied the gluon scattering in

the light cone gauge in great detail and found the need for some counter terms that were

quite unexpected. An infrared regulator is also proposed in [14] that is very different from

dimensional regulation yet still respects gauge covariance. In [15], the extension to the

gauge theory coupled to general matter was made, and some results that are peculiar to

M = 4 were observed. Also the cancelation of counter terms between different species was

achieved, whose importance will be explained later.

As an extension, the scattering of quarks and scalars in addition to gluons was

computed, which showed a need for some new counter terms not given in the previous

work. The complete determination of these new counter terms is still a work in progress.

And the usage of Lorentz covariance as a guide to fix counter terms will be initiated in the

future work section.

The organization of the dissertation is as follows, in Chapter 2, the helicity amplitude

method is briefly introduced. This method, when modified to suit the light cone, can

greatly increase the flexibility of light-cone gauge, and offers an alternative way to obtain

light cone Feynman rules. Some computational details are also given in the same chapter.

In Chapter 3, I shall describe the box-reduction technique which occupied a bulk of









the work done in [14], and is essential to our achievement of the cancelation of artificial

divergence and the extraction of infrared divergence. In Chapter 4, I shall simply list the

results, since the details of the computation are complicated and not very illuminating.

Discussion about some known features of supersymmetric gauge theory is given in Chapter

4 too. In Chapter 5, the Bremsstrahlung calculation on the light cone is presented, and

the real processes are combined with virtual processes to obtain an IR finite result.

In Chapter 6, I shall give the scattering of gluon with massive matter in the loop for

completeness, and the photon scattering amplitude is also presented. I shall conclude this

dissertation by pointing out some remaining problems and the outlook for some future

work. In the appendices, I shall spell out the spinor notations used in the paper and all

the Feynman rules obtained by using the method of Chapter 2.












Figure 1-1. Double line notation












Figure 1-2. Open string scattering diagram
































6 7

3 45 3 4

1 2 1 2
Figure 1-3. Light-cone parametrization

Figure 1-3. Light-cone parametrization









CHAPTER 2
COMPUTATION TECHNIQUES IN THE LIGHT CONE

2.1 Mini Introduction to Spinor Helicity Amplitude Method

This method exploits the covering group of SO(3, 1): SL(2, C). Lorentz invariance

will be represented as SL(2, C) invariance. By dotting a momentum into the sigma

matrices, we obtain a 2 by 2 matrix: p, -- p,,a^. If p were light like, we will have

det(p a) = p2 = 0, so the matrix p a can be decomposed into a product of two spinors:



p a""( popo (2-1)


where we have used the same letter for both the momentum and its spinor. p" (p6) is the

left (right) handed spinor that satisfies the massless Dirac equation. The notation using

bracket is commonly used in the literature:



pa p); Pa4 'i/. Pa p6 -' (2-2)

Polarization vectors for the gauge particles are also cleverly chosen to minimize the

computation [10, 21, 22]. The reader can refer to [11] for a review, and to [12] for some

more simplifications and extensions to massive fields.

As an example, a left handed polarization vector e, can be given by (up to a

normalization factor) Ca = rPa Ir /'](i, or equivalently ea" p" a, > p)[rq. The

handedness can be seen by remembering that 7^ or a^ is the Clebsch-Gordon coefficient

for the projection



1- (2 3)
WL /R

and that p) is the spinor of a left handed fermion. We have e rap} = qrl]{p p) 0. This

means that the photon must be left handed so the product of it with a left handed fermion

will not have a right handed component.









Another way of seeing the same fact is by inspecting whether the field strength F,

constructed out of e~ is anti-selfdual (right handed) or self-dual (left handed).



Fp 1i [Paa [p IPal = K'i tPal'at[P 9 1


The c symbol in the spinor language is given by



PVP(a bb dd [ 4 b b 6d rbd d]
C-aa Cc acc a c c a

so



Sdet gcd"P"F, = i"e""pA, -0 e%,, .I..[, ] iFTM" (2-4)
2

hence *F = iF.

The reference spinor r] can be chosen to our advantage. The most efficient way is to

pick it to be a spinor of one of the moment in the problem. As an illustration, consider

the amplitude M(1, 2, 3, 4, -, -, +, +). Pick



e1 -1)[41; e2 2 2)[4; (3 1)[31; (4 1)[41 (2 5)


They have the property that only the pair C2 and C3 have non-vanishing inner product.

These polarization vectors are not properly normalized yet, their normalization factors will

be put back in at the last step.

Let us look at the t-channel exchange diagram:



S1 6C4( (4 P1) + c(2pi + 4) C4 + (-2P4 -P1) i g )2 (2-6)
S t t t t t t t v i+ P

I did not bother to write down the vertex on the right since the left vertex is already zero.









The four-point contact vertex is in fact zero too. Recall that the four-point vertex

always involves two pairs of polarization vectors dotted into each other, while we only have

one non-zero pair available.

Finally the s-channel exchange diagram:


1 2 2(P1 _) + (21 + ) c1 i + (-2 -6_) 21 C 2 X (
2____+ 2 +_) 1_ (P ++ _

[C3 C4i', p P + (2p4 + I.) 3 + c '( 2( P4) C4


(2 7)


The underlined terms are all zero, and we are left with


4- (C2 3) (P2" 1) (/'. C 4)


i
- [43] (21) [24] 21) [34] (31)
2s


(2 8)


As the last step, we need to put in the normalization factors for the polarization

vectors: -[14][24](31)(41)/4, I get


[34]212)2
st


2( 12)4
(12) (23) (34) (41)


(2-9)


The second presentation is holomorphic in the left handed spinors, which will turn out to

be interesting later on.

As a slightly more non-trivial example, I shall compute the amplitude M(+, +,

Fig.2-1.

I shall define the polarization vectors to be


14](11
[14]

C 3](1
(13)


-14](2
[24]

14](11. 5](11
C4 14 I 5 215
(14) (15)


(2-10)








Here I have used pl and p4 as reference legs. A general observation is that a vertex that
involves both legs that are used as reference vanishes, so diagrams 2,3 and 4 are zero.
A closer look at the products between a pair of polarization vectors shows that all the
diagrams involving four point vertices are zero. We are thus left with only three diagrams
to compute.
I will start with some small building blocks according to Fig.2-2:


A" = [54](41)c4 [45](51) B =- [35](51)c [53](31)c`
S- ,[43](12)(/,_ .) + [43](32)e [32](21)c' DP [42](21)c" [41](12)C"
(2-11)

With these building blocks, we can calculate further larger blocks Fig.2-3:


E" = Ac1, [g"(1 2 -
= 0+ A"[412 + 31) +



F" A Csp [g'P(1+ 2 -

= [ [54](41)[4|2 -

3 [[54](41) [42](21)

C e[54](21)[23](31) -


3)P + gP(2 + 3 4 5)P + gP.(4 + 5
0 = -A,[415](511)



3) + gP"(3 4 5)" + g-"(4 + 5 1

3|1) [45](51)[5|2 31) + A'[3|
2 j

- [45](51)[52](21) A- [3|2|1)
A- [32](21)


G" BC4, [gP((3 5 4) + g+ (4 1 2) + gP(1 + 2 3 5)"]

S0 + [35](51)[514- 211) [53](31)[314- 21) + B[412- 3 51)
2 2 j 2

C[[35](51)[54](41) [53](31) [34](41)( + B[4221)
e[35](41)[42](21) + B'[42](21) (2-12)


-1)"]




2)P]

211)












A,,[11.](5|1) [143](12)(/_- 1.)+ [43](32)c- [32](21)e ]


S[415] (5|1)[43](12) [54](41)[4 2 1)-

-0 [415](511)[32](21)[415](511)[54](21)
2
1 -3
[415] (511) [43] (12) [54] (41) [41 |1)


- [415](5 1)(2 1)2[54]
[]2 2
1 [45]2 (2 1)3 (51) [42] [32]
2


[43] [42] (41)


[


-[45](51)[5 1)


2-3 ]
- [45](51) [5 2 2 1)

43][52](51) + [32][45](51)


C31[54](21)[23](31) A,[32](21) [42](21)e/ [41](12)e/
e2 -3[54][42](21)2[23](31)- 0 + 5 2[45](51)[32][42](21)2 +0
1 1
[34](21)[54][42](21)2[23]31) + [54](21)[45](51) [32] [42] (21)2
2 2
-[54] (21)4[42]2[23]
2


C4l[35](41) [42](21) + B,[42](21) [42](21)c` [41](12)c`


11
0 0 + [42]2(212 IL)5 2[35](51) C3 C2[53](31) -

1[4]221)2 [54]21)[35]5) -[34]{21)[53]{31)

1 [42]{21)4 [35]
2


Assemble them together:


Finally


EICl


(2-13)


G"D,


(2-14)


(2-15)


F"D,










G-D F-D E-C
(21) [12] (35) [53] (21) [12](45) [54] (23) [32] (45) [54]
21)3 [42] [42]2(54) (32) [42] [23] (53) (32)+ [45] [12] (51) (53)
2[12] (53)(54)(32)
2) [] [21][41](51)(13) (216)
= (21)3 [42] (2-16)
2[12] (53)(54)(32)

After putting back the normalization factor:


/25
(-_02,"v (2-17)
[14] [24] (13) (14) (15)

we get


5(218)
2(12)(23)(35)(54) 41)

To go to large Nc, we simply multiply it by (-ig~\ /2)3.
2.2 The Light Cone Setup

The light cone gauge Feynman rules are usually obtained by the lagrangian method,

namely, we first set A_ = 0 and integrate out A-. Then the Feynman rules can be read off
from the lagrangian, which is a function only of A1 and A2.

Here, I combine the covariant vertex functions with the above spinor helicity method

to obtain the Feynman rules in a more flexible way. But this will require us to fix rl

r l [1, 0].
The gluon propagator in light cone gauge is


_(gpv lt+k' g + V
-i(g k k (2-19)


Note the metric is diag(1, -1, -1, -1) throughout the dissertation. The numerator can be

factored into











0 gP+k -.g gV+P k~' g +g+
( e e + k+ kpn v s gn +2b

Where CA,V are light cone gauge polarization vectors given by


1 k' ik2
v v2( +k
Cv -v2p1Ie ;( cA


k ik2 1 k^
kO + k k+2 V
v/-2-1 ^ k q


They satisfy e = -1 and k e 0. These polarization vectors are defined both on-shell

and off-shell. Note from now on, the definition of Ik], Ik), [k| and (k| will be that of the

appendix A, which is slightly different from the conventional ones.

The Feynman rules are obtained by dotting the polarization vectors into the covariant

three or four point vertices.

The tri-gluon vertex Fig.2-4, for example, becomes



V,, -gfa(_ )(-(2,) (-h(3*)[giJ(p +P gP(i + >' gP'j -pi

Setting ei, C2 = A, C3 = V, the above becomes



gfab[ (_ .)+I A .) P) P /'. 1) 2gfab ( + '- 21
++ -2+ K21
Pi P2 P P2
(2-22)

where K" := (pfpj pfp). They are related to spinor products according to



K3 P- Pt [Pi Pj] P Pj+PPJa

The spinor notation here is also different from the conventional one [20]. The reader can

refer to the appendix for an explanation of the spinor notation.


(2 20)


k
, k+
I^iT)


(2 21)








The gluon propagator Eq.2-19 almost factorizes into the product of two polarisation
vectors. While the third term on the rhs of Eq.2-20 will make an extra contribution to
the four point vertex. For example, consider the t-channel diagram Fig.2-5, the first two
terms of Eq.2-20 can be associated with the two tri-gluon vertices, the third term, which
describes the mediation of A-, gives


(PI i + P4 C+1 94 1 (P4
*i C pl +p4) iC2 i 1'4 pt4 )123 p2 (2
S(Pl P4)223P (pl 4+)2 (ptl 2+ + P+4

What's happening here is that the explicit factor of k2 in the third term of Eq.2-20
cancels the propagator, effectively making a four point contact vertex.
The fermion-gluon vertex Fig.2-6 is the usual ig 7' t'. We can set p = A or V by
dotting -e ,, into p.1 Multiplying the spinors to the gamma matrix, we get (assuming
now the fermion is left-handed)


3)


ig(ta)b (- o) [ p 0 4 o ())o6 [C,]


A [
-2g()ba K 2ig K (2 24)
(+ h PPiP [ pA P\
Vq+ +Iq+Ppl plq q'Pl q(2


In Eq.2-24, in order to avoid defining what is /p, I chose to associate p+ instead of

ypj to a vertex. This won't cause any problem, since a fermion line either always
closes, or end up as an external particle (then the phase of the root can be defined
arbitrarily).


1 Calling polarizations by V or A is potentially confusing, especially if you are looking
at the diagram up-side-down. So,sometimes it is clearer to associate A with 'in' and V
with 'out'.









The fermion propagator is given by ipl, /(p2 + ie). We can decompose p 7 according

to




v/2p -P -+ 2p
2[ 21 07] 1 0


pP+
P-^ = p)v^ [p | p2


SpV p- c P+ b ) 2p+
p2
= v g2p+ p]d{p+a b + 2d ] ) (2-25)

Since the fermion propagator almost factorizes too, it is also possible to contract a









3 ( P + P4)2
(p( )+24)




Pi + Pi 4

The scalar Feynman rules have no suspense in them at all, and can be read off

from any field theory book. The Feynman rules that pertain to our calculation will be

summarized in the appendix. The main property of the Feynman rules above is the

absence of p.

2.3 Brief Description of the Calculational Procedure

In the Fig.2-8, the k's and q are the dual moment. They are related to the real

moment coming into the three legs according to pl = kl k2, P2 = k4 k1 and

p k2 k4. The unregulated integrands have a symmetry under ki -- ki + a, which








would ensure that each diagram depends only on the real moment. But here, as in
[13, 14], we use a regulator exp (-26q^qv) that breaks this symmetry. Hence, a regulated
amplitude can depend on the individual dual moment. This seemingly unwieldy regulator
is designed for the world sheet description, but the result doesn't differ too much when a
cut off regulator is used.
The calculation roughly goes as follows,
1. Exponentiate all propagators according to


j- j dT eTP2
P2 + M o
For the schematic diagram Fig.2-8. We have


f d4q i i i
(27)4 (q k22 (q_ -k)2 (q k4)
J d Texp{i W(q- k)2}

2. Integrate out q-, leaving a delta function relating q+ to the Feynman parameters(this
step requires the absence of q- in all Feynman rules).


S/ dq+d2Zq ldTi(Y- Ti+ Tiq+)exp {iyTi( (-_ )2}

3. Integrate q1, q2 using exp [-6q ] as a damping factor.


F 16dq x6( x 1)6( zx k -12 k k 2

where x := Tjl T,.
Here is the rub: we cannot simply integrate over x2,x1 and x4, because the prefactor
of this diagram will have up to second order poles at q+ = k+. In order to show the









cancellation of these poles (gauge artificial divergences), we proceed as follows: first

eliminate one Feynman parameter in favour of q+:


S, x(q+ k) (I x)(q+ k2+)
k+
21 k -kk k2 -k
S^2 4 2
k+ < q+ < k: l : 2 2 : 4 1 xl -- xk + [0, 1]

Now dx2dxldx46(Zi 1)6( xik q+) dq+dxJ. After integrating out x, we are

left with a function of q+ which is defined differently in different regions: k+ < q+ < k+

and k+ < q+ < k+. All these can be visualized very clearly when we represent a Feynman

diagram on the light cone world sheet. The details can be found in [13, 14].

Our observation is that, in each region, all poles cancel.2 Hence we can perform the

final q+ integral and obtain the results.

To summarize the computation procedure, perform the q+ integral last.

5



1 2





Figure 2-1. Five-point amplitude










2 in the case helicity conserving amplitude, all poles cancel up to infrared terms, but
since infrared divergence is always proportional to a tree, they are easy to recognize and
deal with.










5 3

B^\.


3

C P=-v^


Figure 2-2. Some building blocks that can be computed separately


4 5


Figure 2-3. Second level building blocks


p, C, 1,




/I, a, pi b, '_

Figure 2-4. Tri-gluon vertex


o", 4 p, p.


^^7


ft, pi


-, P2


Figure 2-5. Propagator in the t-channel is contracted


ft, a




C, Pi b, p

Figure 2-6. Fermion-gluon vertex


4 5
AP
ft


ft
DP


2 1


4 5



, I Y/--


5 3


















(o, V, P4 p,




b, pi c, p2

Figure 2-7. Fermion propagator contracted


Figure 2-8. Here we assume only (k2 k4)2 / 0, and k0 > k0 > k-









CHAPTER 3
BOX REDUCTION

This chapter contains a lot of technical details, the reader may skip it for first

reading.

Our general procedure for evaluating box diagrams is

1. Evaluate a box diagram that is free of artificial, collinear and infrared divergences

using text book method (combining denominator with Feynman's trick, shift the

momentum, perform momentum and Feynman parameter integrals, etc).

2. Evaluate a box that contains collinear or infrared divergence but is free of artificial

divergence by subtracting those collinear or infrared divergences in the form of

triangle diagrams, and then use method 1.

3. A box that has artificial divergences has to be reduced to triangle diagrams.

In the next section, I will describe how to achieve 2 and 3. The dual moment assignment

given by Fig.3-1 will be used through out all the calculations.

3.1 Box with a Helicity Violating Sub-diagram

The Feynman rules for the gluon, scalar and fermion in the light cone gauge are very

similar. Any three point vertex will have the following form: 7R(k )K;' where R(k+) is

rational function of the + component of external moment and K' = (pi ^ Pi^ )

is introduced in the appendix. This allows us to write the numerator of any box diagram

as



R(kj)KA,' K^' K^' K^' (3-1)
sy kl rn op

We can also label a box by the helicity of the four corners, regardless of the details of the

boxes. For example the following Fig.3-2 will be abbreviated as A A VV box.

R(k+) in general will contain poles of k+. These poles are usually interpreted using

Cauchy principle value or some other prescriptions. We chose to show directly that

these poles are all fake, namely, when we compute a ]li .-i. .1l quantity, these poles will









be cancelled. To show this would entail us to express the amplitude as a holomorphic

functions of k+ and analyze the poles one by one. But the box integrand in general will

have a denominator that is quadratic in the Feynman parameters and k+. Integrating it

by force will introduce logarithm and di-logarithms with various arguments. Although we

know of certain relations between (combinations of) di-logarithms and double logarithms,

the application of these relations is hard to automate.

Fortunately, for the case at hand, all box diagrams can be reduced to triangle-like

integrands, whose manipulations are considerably easier. Here I describe the box reduction

technique in detail.

The lower half of the box Fig.3-2 is the 't channel' of (+ + +-) scattering process,

which is zero. So we can replace this 't channel' with minus the 's channel', which will

result in a triangle diagram. But since the amplitude is only truly zero on-shell, while the

two internal lines are certainly off-shell, I will get some more terms that are proportional

to the virtue .il vr of the two internal lines (1j' and p2). Schematically, I have Fig.3-3:

More concretely, Fig.3-3 is written as


1
+(P1.)2 16 52
1 2 2
(- K A K66 K KA +2A KA K______
S(p + _)2 21 65 ( + 1'.)2(p )2 52 21 ( 1')2(p 2 21 16
(3-2)


Clearly, in this case, the box is reduced into three triangle-like diagrams.

Here I list its contribution: First, define some functions that occur ubiquitously:











(kI q+)(-k +2 q+)s6
(k kf)2
(k+ -q+)(q+-k k)t6ec
(-+ k)2
(kj q+)(-kj +q+)s6
(k+ k+) (k k+)
(q+ kf)(k+ q+)sbe
(kA kA ) (kA kA )


(-kf +q+)( k+ q+)t

(k_ k )(-k + k +)


e7
k < q+ < k

, k+ < q+ < k+
e7
k < q+ < k+

k ^+ < q+ < k+




k +; < q+ < k+


(- kf + q)2t6kk
(k+ k+)(k+ k+) 1 k 4
4 1 4 31


k+ < q+ < k+
2 1


(q+ kA)(-k+ + q )s6e
(k+ kf) (k kf) 1 4
(-kf + q )2t6C7
2+ k+)e + < q+ < k+
(kA kA)(kA k' 2 3


k+ < q+ < kF


Next, I list the results for the two model boxes V A AV and A V VA.

* VAAV:


i
83Tr [tatbtctd] X
87 2


(q1 k) (q4 k2) K
+ 1/4 (k kf)s 431 log(H) +

(q+ k+)(q+ +))
1/4 (k- kfs K log(H,)


(q+ k)2 (k+ k+)(kf kf)
1/q 0kj + k log(H,)
k~ k2~~


Ht:


Hi:


(3-3)


(k+ q+)(q+ k2)sSe
(k+ k)k -kf) '


H, .


(k+ q+)(q+ k+)t6e-
(_-k+ +k2)(_ -+ + k)'













(q+ k+)(q+ k)2+)
+ 1/4 (+- K k+)s 43K241 log(H)
(/ 2 kf)S
(k- kf +(k 1 + +
S1/8 3 )(+ k ) (-k+ k+ k k++ k + + 2q k-
k1k- g
(ks k )(k+2 k ) 3 ) 3
+ k+k k+ + q+2k 2q+k+ q+ + + q+2 +) log(nH)


4 2
)2(+kf _
+)2

2k+-


(q+ k )(q+ k2+) (q+
+ 1/4 (k+ k+)s K43K2 log(Ht) 1/8-

(q+ k+)(q+ k+) (q+
- 1/4 4 K43K log(Hi) + 1/8
2q 4
(q+ kf+)(q+ k2+)
- 1/4 (+ -+) K43K log(H,)
2(k < k)s



k+ < q+ < k+


+2 +
q )kl



kq2( k? k,)
- 2 3 log(H i)


k4 (k3
4k+"


k+) log(H,)


(q+ k f+)(q+ ) A
+ 1/4 ( -k)s K4K log(H)
2 443- 21


(q+ kf)2(k +
1/8


k log(Hi)


(q+ k+) (q+ k+)
1/4 -(k4/-+)s K43-21 log(HI)
(12 kf)s

Now assume that its coefficient is C+ A/(q+ k +)2- 1/( + k+)1, what to do next is

to combine this with the coefficients of the log H terms, then extract the polynomial part

through partial fraction and perform the integrals to get its contribution.


k k+ k + k
1/36 (k+ k )(k+ kf+)(k+ k+)(k1 k +)+ 1/8. 3 K
S
(k12 Ik+2 k k++ k+k+ k+k+ k+k+ +2k+k+)CK l
1/24 K43K21


-k/8o( k+ + +2A
+ 1/8 log(s) 4 2 K43K21
8


1/24 log(s) (k k43K)


(3-4)


kf+) (k
k+


+ 1/24log(t). (kf kf+)(k+ kf+)(k+ k+)(k+ k+)C
-k- + k + 2A (k+ k +)2c
- 1/8 log(t) 4 2 K4^ K2 + 1/24 log(t) 4 K4^ K2








If its coefficient were C + A/(q+ k)2 /(q+ kf1, then the result can be obtained
from the above by replacing k+ +- -k+, k+ (k+) +- -k1 (-k+).

What is not included in Eq.3-4 is of the form:



(q k)2 q -- log H (3-5)

where ... only depends on the external moment. The second order poles will have to
cancel eventually, the first order pole always combines to become proportional to a tree
amplitude with a universal structure.
AVVA
k+ < q+ < k+:


(q+ kf+)2(-k + k+)(k+f k +)
1/8 k klog(H,)
Y O----------- ,+ ,;+------t g s
k2 k4
(q+ k )2(k k+)(-kf + k +)
+ 1/8- 3 1+ 1 + 2 log(Hd)
S-k) lo(
(-k+ + q)(q+ k)
1/4 ( KK2 log(Hd)
S(k+ ( k+)(-t) 14 32
(-k+ + q+)2(k^ + kf)(kf k)o(
+~ ~~~~~~~; 1------ p--p ----1 -i+{u
1/8 klog(H)

S(-k+ + q)(q k+)
1/4 K A4K2 log(H,)
(k3+ (-1+)(-t) *414o32

k+ < q+ < k+:











,-1- 1/8 k k3 (-k+k + 2q+k+ k+k + k+k+ q2)k



1/8 k- 1 ) )
(k+ k+)(kl+ k I) \( 3 2 2 2 3 1
+ kk + q +2 -_ 2kgq+k+ + g+2k q k+ klog(H,)

(-k+ q+)(q k3+ ) (- + ( ko )
1/4 -) KK log(H,) 1/4 K14 log(Ht)
(k+ k )(-t) 14 32 :(-t
(q+ kf+)2(k+ kf(+)( k+ k) (-k + q+)(q+ kf3+)






1/8/++ )1 1g+ -/ +K) T + -k4 1(
1/ lo g( H)
k k- K
/4 (-k + q)(( k )
1/44( K+ K log(H,)
((k k))((-t) 14 1332


k+ < q+ < kf+:


( a t )2 (k + k + )( k-) )
S1w/8od be log(H)
k2 k4
(-kf + _q +)2(k+ k+)(k )
1/8 log(Hd)
k -k
+ 1/4I 3+ A + A WV log(Hd)
(k+ k()(-t) 14-32
(-k+ + q+)2(kf+ k+)(k+ k2)
1/- log(H,)
3 1
(-k+ + q+)(-k + q)
1/4-( KA4 K log(H,)
(k+ k)(-t) 14 32


Now assume that its coefficient is C + A/(q+ k +)2 I/(q+ k+)', its contribution
would be










-k +k+ +k k+
1/36 (kf k+)(k+ k+)(k k)( k+)C + 1/8 -4 3 1 K K

(k+k+ 2k9k++ k+k+ + kfk +k+k+ k+2 k2
1 /24. -- K14K32

+ 1/241og(s) (k. kff+)(kk k)( kf )(k kf C
-k+ 2A ( + (k+ k+)2C
+ 1/8 log(s) KKA^ K 1/24 log(s) 3 1 K K
(_0 14 32 (_0 14 32
-k + 2A + k+ (k+ k )2c
1/8log(t) -t 14K2 + 1/241og(t) (t3 K14K32 (3 6)

Just as in Eq.3-4, the pole terms are dropped here.
3.2 Box without a Helicity Violating Sub-diagram
But of course, not all boxes are going to have a helicity violating subtree, I now turn
to a second, much more complicated case Fig.3-4.
Assuming this box diagram can be written as


A B A V
(q+ k )2 (+ ) K16K 2K3,-5K-6,4 (3 7)

Here I have used the notation of dual moment to make the formalism more symmetric
looking. Write










A B
(q+ kf+)2 (q+ K- K) K K52
A B
/A V KA A V A V fV A B A A V 4 A AV
(q+ ]1+)2(/162 16 / 52 /W16/) @ (_+ 71+) (/ 16/ 52 1 5 16161652
A B
2 (x16K + C) (q+ ) (16^ .)
A B
( ) -K^ K
((q+ k1+)2 (q+ k- )K)16 52
A 1
( (k+ q)-(-k + q)s

A 1
+ (q+7-k)2 [(k 1 (])(2+ q ) + (k? k2)(q+ k})] (q k1)2

+ (q )2 2(- + k)(-l+ + q)(q k4)2
A 1
+ A k2(-+ + kl+)(k+ q+)(q k2)2
(q+ -_ k +)22 ) 1
+ (- ) (K 15 c.c) (q+ +- )
(q+ K5) (k4 k+)
A B
( )K2^_ K (38)
(q(+ k)2 (q+ kk) 16 52

Notice that the first four terms in Eq.3-8 contain covariant products of moment: s,

(q k1)2, (q k4)2 (q k2 )2. The coefficient of s is simply -sA/2 (the annoying

1/(q+ k+)2 has disappeared, the fifth term also has this property). Terms like (q ki)2
will simply cancel one of the four propagators, effectively making a triangle-like terms.

Now Let us look at the sixth term of Eq.3-8. Putting back the factor of K3,_sK 6,4

gives [A/(q k+)2 B/+ k)^ 52K K-5-6,4. Now that two A's (or V) are next
to each other, this term is in fact nothing but the previous model diagram.

In summary, the evaluation of the second, fifth and sixth terms of Eq.3-8 goes

through without further twist. However, the first(hence forth called half scalar box), third

and fourth term have collinear divergence individually (on top of the infrared divergence),

some more manipulations are needed.









Subtraction of Collinear Divergence: A collinear divergence in a virtual process

happens when the moment flowing through two internal propagators connected to a

common massless external leg become parallel. For a box diagram, there can be a collinear

divergence at each massless corner. When only transverse components propagates in the

loop, the vertex function will vanish due to helicity conservation. So, a bona fide box in

light cone gauge will have no collinear problem. However, in the manipulations of Section

3.2, we have tampered with the vertex structures of the lower sub-tree, so there will be

collinear divergences in some of the resulting pieces. But they will cancel when all the

pieces are collected.

The collinear divergence in a box diagram can be located into two triangle-like

diagrams. This process is rather like doing a partial fraction. Consider


i i i i
KV KA (3-9)
3,-5 -6,4 (q ki)2 (q k4)2 (q k3)2 (q k2)2

Take limit q k4 (so the right propagator is soft). Eq.3-9 becomes


i i i i
KV K^ (3-10)
3,-5 4,3 (q k1)2 (q k44)2 (q k3)2 S

Take limit q k1 (so the bottom propagator is soft). Eq.3-9 becomes


i i i i
K- K^ (3-11)
3,-5 1,4 (q k1)2 (q k4)2 t (q k2 )2

Naturally, if these two 'poles' are subtracted from Eq.3-9, we should have a term free of

collinear divergence, which will be demonstrated next.

According to Chapter 2, the Schwinger representation of Eq.3-9 is (with H given by

x1x3t + x2x4s + /2, where p is a small mass used here to regulate temporarily the possible

collinear divergences)










ST3dT((x, 1)d4xi
2T(2-)3 exp (-iTq, + iTH)

(k kf)q + X2K43 + xK2] (k k+)qA + 4K43 1^4]

f T3dT(zx, )d4xi 2
S 2T(27)3 exp (-iTq + iTH)

(kf k3+)(k+ k~2)q2 X24K43 K4 i43 x2K43 x1K43K4 + xK32K14
12 4 3 3 2 1 14 1 14],


ST3dT6(x, 1)d4X exp
162T expTH

S1(k k)(k2 k+) 4
2 (iT)2 iT 43 43


X1X4 KK K3K
SV X V K
+ i 32 43 + i 43 14 + T 3214


Integrate over T:


S 6(zx 1)dxi
j 87T


1-
4H[-


The first two terms won't have collinear divergence, the p can be set to zero, but the rest
will have to be evaluated with a small p to regulate the collinear divergences.

There are basically two types of Feynman parameter integrals in this case. The others
are either finite or could be obtained from these two.


S(x 3t + 24S 2)2
1xi

1
Slog2
2st


S 28
2t(s t)2 g
2t(s + t)2 t


i (xx3t 124S 2 )2
Jxi,<1 (XXl3t + X2X4S + [2)2


1 2
l2og2
2(s + t)2 t


2 lo 2
st s


2 2
2 log +


St S
Slg( )
t(s+ t) t


log( )
S t) t


4
st
St


2 2S
2t(s t)2


7t2
2(s t)2


(3-12)


Collecting all the divergent terms from Eq.3-9:


2
X2X4KA XX4 Kv X2Xl
k++)(k+ -k-+) + 22K43K 3 K 4314 2 H 32 1 K K
3 3 23243 2H2 43 14 2H2 32 14











iA 2 1 12 2 1 i t2
s 1 log2 /2 + 10g T 2
8 8(-t) 2(-t)


4(-t) lgs g 2 4


iA 1
i I log f2 (K4 KA + K43 K2)


Eq.3-10 and Eq.3-11 simply give


i i i i
Kv K^ A
3-5 4,3 (q k1)2 (q k4)2 (q k3s)2
ST2dTdzld3dx46a(x 1)
2T(27r)3


- k+ k+
S(-t) q
1 2
2 [log 2 + 2
87 2


x tK
(-0) 32


0 V1 KA
(-t) 43 43


1
log (-t) K4^ K32
2s(-t)


Vi i i i
K3v K)
1,4 (q ki)2 (q k4 )2 t (q )2
[ T2dTdxidx2dx4(Xi 1)


[ k+ k+
1
(-0t)
2 (log S -
87 2


2T(27r)3
V_ K 2 2
v V
(-t 32 (-t)

2) log /2 1g2 2
2


K V A
K3] K^ exp(
1
- ( log2 s 4
2


iq 2 + iH)
1
2 log s) K3 K2
2s(-t) 14


12
+ [log [2 + 2
87 2


1
log ] 2s(-t)43 14


This, multiplied by -sA/2, is the same as the half scalar box contribution Eq.3-13.

The complete contribution of Eq.3-9 with Eq.3-10 and 3-11 subtracted is (pole terms

omitted)


(3-9)di


(3-13)


exp (-iTq 2 + iTH)










1 s I s(-k+ + kf)(-k + kf+)
A log !
872 [- s (-) -

4 ((-t) )(-t) 2 4 ((--t) )(-t) 4( ) -
1 log2 1 S(-C + +kf)(- k + f3+)(-t)
+ -Aog + 2 3
8Al2 t 16 ((-t) s)2
1 s2 1 s 1 s ]1
8((-t) )2(_t)K^4 32 + 8 (( 8)2143 ((t)- 8)2 K4 K32 (314)

So, when evaluating Eq.3-9, which is collinearly divergent by itself, I shall subtract
Eq.3-10 and 3-11 from it, to make 3-9 finite.
Next I will show that Eq.3-10 and Eq.3-11 also cancel the collinear divergence in the
third and fourth term of Eq.3-8. Putting the triangle-like terms in Eq.3-8 together with
Eq.3-10, 3-11:


Suppose q


[(-k + k+)(- k+ + q+)(q k4 )2 + (-kf+ k+ )(k+ q+)(q k2")2]
1 i i i i
Kv KA
(q+ k)2 (q k)2 (q k2)2 (q_ 2 3)2 (q k4 )2 3,-5 -6,4
i i i i
+ K K ^ -s
3,-5 4,3 (q k1)2 (q k4)2 (q )2
i i ii
+ Kv^ (3-15)
3,-5 1,4 (q ki)2 (q k4)2 t (q k2 2)2

- ki = A(p1 1-_), q k2 (1 (+ A)(Pl 1._). The above expression becomes


1 1
Kv
(q k1)2 (q k2)2 3,-5
[ (1(1 t A)K14 0 0 + -K 4 0
SupposeA ( ), 4 ). The above expression becomes
Suppose q 1 =- (P4 Pl), q k4 (XA- 1)(P4 pl). The above expression becomes










1 1
KV
(q k1)2 (q k4)2 3,-5
[ (-k+ lt) 1 k) 1 1 11]KA 0
0 + A(-+ )( 1) (K14 + AK24+ 4,3 (- + ^'4 -S 0
-A(k k) (A )t ( A)t s t As

Thus all the collinear divergences cancel.

The complete contribution of 3-10 and 3-11 and third fourth term of Eq.3-8 are quite

complicated. I observed that the rational part of their contribution is zero, so if there is a

way to get the logarithmic parts through some other methods (such as unitarity), we can

be spared all these ordeals.

To summarize, I have subtracted 3-10 and 3-11 from Eq.3-9 and then added them

back to the triangle-like terms in Eq.3-8 to make both parties collinearly finite. I want to

add that the introduction of ft will not mess up gauge covariance as people would normally

think.

4\ 3


1 2



Figure 3-1. Dual momentum assignment with k+ < k+ < k+ < k+







iA A



Figure 3-2. Model box with a helicity violating subtree



























Figure 3-3. Graphical representation of box reduction


\p4 i, /
A V

1 I" II

V A

-Pi P i


Figure 3-4. Model box with alternating helicity









CHAPTER 4
MASSLESS AMPLITUDES

I will be following [21] and decompose an n-particle amplitude into



MI,= > Tr(talta2...ta, )M(pl,i; p2,(c2;...;pn, n) (4 1)
perm'

where perm' is over non cyclic permutations for complex representations, non cyclic

and non reflexive permutations for real representations. In the following results, the

representation is assumed to be the adjoint representation.

4.1 Two Point Functions

A factor of -ig2/(16T2)fcadfdbc ig2/(16r2)Tr [tatb] will be omitted in the following

list.

4.1.1 Gluon Self-Energy

Refer to Fig.4-1



II(g+, g+;s) [k2 + k k + k2]
3 1 1 3 3

II(g+, g+; q) -= [k2 + k 3 + k2]

I(g+, g+;g) -= [k 2 + kkk + k2] (4-2)

where the first two arguments in II tell the species of particle and the helicity, and

the third denotes the particle in the loop. For the above assignment of helicity, the

contribution should be zero due to Lorentz covariance. They are only nonzero because the

regulator used here doesn't respect Lorentz covariance. This is purely an artifact, and it

has to be cancelled by a counter term. Also, it can be observed that II(g+, g+, s) x N, +

I(g+, g+; q)F x Nf + II(g+, g+; g) x N, = 0 in the A = 4 SYM case.










1 + A
6
-p p
3
7
-p p + IR terms
3


(4-3)


The notation g+ simply means that I am using rl](rl or g+ as polarization vector. The
first two terms should also be zero due to Lorentz covariance, so they have to be set to
zero with counter terms.
Here are the helicity conserving 2 point functions:


H(g+, g-;s)



I(g+, g-; q)



I(g+, g-, g)


/1
Sdxx(1 x)p2 log x(1l x)p2 6

p2 Ilog p26
18 6
2 dx [x2 +(1 ) 2 log x(1 )p26l-

p2 [2-4logp26e

2 dx + -- + x(1 x)p2logx(1 x)p26e
o [ -x xl
67 11 2 2
[p2 7+ log p266Cy dx f&[ g 1 2)26]o l-4)
9 3 J 1-x x I


Here x is in fact (q+ kt)/l(k kl).
Several comments are in order:
In the last line of Eq.4-4, the x integral is certainly divergent, this is due to the
artificial divergence. This divergent integral is what we call the infrared term (with x
interpreted as (q+ k+)/(k+ k+). Temporarily forgetting about these infrared terms, it
is easily observed that the sum of these diagrams is zero with the A = 4 field content. I
shall comment on this later.


n(g+, g+; s)

n(g+, g+; q)

I(g+, g+; g)









In the above list, all quadratic divergence 1/6 are omitted. Due to the way the light
cone world sheet computation is setup, all the tadpoles are being dropped. These tadpoles
will only contribute to the quadratic 1/6 divergence. As a result, the 1/6 terms can
actually have non-trivial p+ dependence. When the p+ dependence is of the type 1/p+, it
can be interpreted as a world sheet cosmological constant as in [13]. While in other cases,
the p+ dependence could be of the form (1/p+) logp+, whose interpretation has not been
fully understood yet.
There is one small point about numerics that I want to point out here. The 2 point
function naturally comes with a color structure of fcadfd -Tr [t at], when we try to fit
the 2 point functions into the big picture as in Fig.4-2, the trace factor becomes (assuming
that ta is in a real representation):


fabeiTr[tetf ffcd T Tr[(- i)[t, tb](-i) [, td]]
-Tr[tatbtctd] + Tr[tatbtdtc] + Tr[tbtatctd] Tr[tbtatdtc]
-2Tr[tatbftd] (45)

Also, there is always another factor of i2 coming from the two propagators connecting to
the 2 point function.
As a comparison, I shall give the two point function that describes the 'propagation'
of A-. The argument g+ means that I am taking the polarization vector to be g"+.



n(g+ ,g+; s) = p+2 [_ log p2

n(g+,,g+;q) = p+2[ 9 4 logp2JC]

II(g+,g+,g) = +2 j 1 log p26C

We observe that the logarithmic piece that will give a cut matches between II(g+, g-)

and II(g+, g+) when the loop particle is fermion or scalar. This is a hint that we shall use
counter terms to enforce the total agreement between II(g+, g-) and II(g+, g+), due to the








consideration of Lorentz covariance. This observation shall be a guide to fix counter terms
later on.
Clearly, there is a mismatch when the gluon is in the loop. This mismatch is due to
the presence of infrared divergences in I(g+, g-, g), while II(g+, g+, g) is free of infrared
divergence.
4.1.2 Fermion and Scalar Self-Energy



n(q+, q+) j- dx2 3x 3 22log (1 x)p266
J (1 x)x
p- 2 -6 + 3 log p2 + d --- 2 log x(1 )p26~e

The divergent x integral is likewise interpreted as infrared term.
I shall also give the result when I use rq}) or I| ] as spinors instead of Ip) or I p].


II(q+, q+) = p [2- logp26]

where p flows with the direction of the fermion line. We again observe a mismatch of
log p2 term, due to the infrared divergence.
The scalar self-energy diagram is given by


-2 4x 4x2
II(s, s) = dx 4p2 log x(l x)p26e
o ( x)x
p- 2 8 + 4 logp2 + / dx 2+ ] log x(1- x)p26e


4.2 Three Point Functions
The general structure of 3 point function Fig.4-3 is


(const + logpo) tree + a term + const(k21 + k-11 + k41)









The a term will arise if the off-shell leg has unlike helicity.
A factor of g3/872fdaefebfffed i g3/ 2Tr[tatbtc] is omitted.
4.2.1 Gluon Vertex Correction


r(g+ ,g, -;s) 2P 1 [ log_ 1 ap+p+
P(g+,g+,g -;q) gK2 1 j +g + a-- t- a +2
S 6 9 6P3 P

F(g+, g+, g-; q) = K2A 4 lo-g p 32 +a2p
-2p, 11 70 1 p2
/(g++, g -; g) = pK2 o1 gp6 l ea p+ +a IR terms (4-6)

where a 1= if leg 3 is off shell and 0 otherwise, and po is the momentum of the only
off-shell leg. The infrared terms present in the gluon triangle diagram will eventually be
combined with infrared terms from other diagrams to become proportional to a tree. They
are given in [14] and are not repeated here.
All of the above also contain an anomalous term:


11 4
P(g+, g-, -; s),,o = ;(an ^ + k^ + /^)
4
F(g+,g+,g-; q)0 -2 (k^ + k^ + ^A)
2 1


here the k's are dual moment, they arise because 6 is an exponential damping factor
of the transverse dual moment. Had we used a cut off regulator ((qi kii)2 < A for
example) instead of 6, the anomalous terms would change according to


(k2L + i21 + /74) (A721 kiL) + (441L 1I)11

These polynomial terms must be canceled by counter terms.
The MHV triangles (with all three legs having the same helicity) give











F(g+, g+, g

F(g+, g+, g

F(g+, g+, g


Here, we again observe that with Af


(K21 2
+; + + + 2
PlP2 P3+ 3po
AK )3 -8
+; g) (K21+
(^ ) 43
Pl P2 P3 3po

4 SYM field content, the gluon vertex correction


vanishes.

As a comparison, I shall give the result when the off shell leg takes on + component.


-(p p) I log p2C^I-
6 18



(p p+) -log P ^ + + IR terms
(P 2 3 0


F(g+, g-, g+; s)

F(g+, g-, g+; q)

F(g+, g-,.g+; g)

r(g+,.g+, g+)


7)


(4-8)


Although F(g+, g-, g+) are not directly used in the computation of the scattering

amplitudes, they are a guide to fix the counter terms

4.2.2 Fermion Vertex Correction

The fermion vertex correction Fig.4-4 is given by


F(q+, q+,g+)


K qK [3 logp+e + 6] IR terms


The notation here is that: q means an incoming fermion line while q means an outgoing

fermion line; q+ means that the fermion is right handed; g+ means that the gluon

is right handed which corresponds to A. The reader can refer to Fig.4-4. The factor

ig3/(872)Tr[tatbt,] is omitted as usual. The other combination is


(4-9)











(q+,q +,g-) 2p K [-3 logpje + 6] + IR terms (4 10)
q+ pi

While r(q-, q-, g+), r(q+, q+, g-); and r(q-, q-, g-), r(q+, q+, g+) are related by

charge conjugation' These results up to the IR terms are the same with any one of the

legs being off shell.
We can also have the case when a gluon is off shell but taking on + component.



r(q+, q+, g+) 2p+ [- log p2e7 + 2] + IR terms

4.2.3 Scalar Vertex Correction

Scalar vertex Fig.4-5 is given by



F(s, s, g+) = K2 -4 log p2e7 + 1] + (k^ + k^ + k) + IR terms (4-11)
q(s, s, g2)2 2 1 4


with F(s, s, g-) equal to F(s, s, g+) with the obvious change of A to V. This result is also

the same up to IR terms with any one of the legs being off-shell.
4.2.4 Four-Point Functions

The box diagrams are too cumbersome to present here, but I do want to point out

that, up to the infrared terms, the total contribution of the box diagrams in a specific
amplitude is surprisingly simple. I shall list the rational part of the contribution of some

boxes in conjunction with the scattering processes in which they occur.

4.3 Scattering Amplitudes

I shall start with gluon scattering, with a factor of ig4/(8 72)Tr[tatbtctd omitted.



1 up to the external line factors which are defined ..i- mmetrically, see the explanation
below Eq.2-24









4.3.1 Helicity Violating Amplitudes

The tree level amplitude for four gluons with the same helicity is zero. At one loop,

the amplitude is


The tree level amplitude

amplitude is


4 K43 K K^21 K^4
A(g+, g+, g+, g+; s) = 3 14
3 p 2 P3 P4 st
16 K4A3K2K21K 1A
A(g+, g+, g+, g+; q) = +- 21 14
3 p2 P3 P4 pst
8 K4 K^2 K2 K^4
A(g+, g+, g+, g+; g) = 3 2 P 14 (4-12)
3 t s t t t

with only one unlike helicity is zero too. At one loop level, the


I K' I
A(g+, g+, g+, g-; s) (s+t) '4 l A
0 -v4A3 K3Vv2 tv K

A(g+, g+, g+, g-; q)= ( + )
3 IK43 K32 K21 K14
1 K, -' +
A(g+, g+, g+, g- g) -(s+ t) (4-13)
3 K43 Kv K32 21K14

It is easily observable that the helicity violating amplitude is zero if there is any amount of

supersymmetry [17]. For example:


16 1 8 2 1
./-=1 : --x-+-xl 0 --x-
3 2 3 3 2
16 8 4
AC 2: --xl+-xl+-x2 0
3 3 3
16 8 4
NA 3(4): x2+-x1+-x6
3 3 3


1


--xl+-x+-x2 0
3 3 6
2 1 1
--x2+ xl+ x6 0(4-14)
3 3 6


Since for gluon scattering up to one loop, adding supersymmetry is simply taking into

account the multiplicity of each species, hence A = 3 is the same as A = 4.

In the list of amplitudes, the fermions are assumed to be dirac, this explains the

strange looking 1/2 multiplicity for the fermions in the first line of Eq.4-14.









4.3.2 Helicity Conserving Amplitudes

The helicity conserving amplitude is non-zero at tree level. They are given by [20]



A(g+, g+, g-, g-) ig2fae fd -2K j P+p
4 43K ,32K21K14 p

A(g+, g-, g+, g-) ig2fabe fd -2K (4-15)
K 43 K^32 K21 14 pi p

The factor fabe fec can be converted to -1/C(G)Tr [[t, tb][t, td]] -2/C(G)Tr [tatbttd].

Here I have chosen ta to be the structure constants, while most literature picks ta to be in
the fundamental representation. So, instead of the one loop amplitude gaining a factor of
NA, here the tree amplitude is down by a factor of N,.

At one loop level, the amputated Green's function is (with external leg corrections
omitted)











A(g+,g+, g-,g-; s)


A(g+, g+, g-, g-; q)


A(g+, g+, g-, g-; g)


-2K^ j~Jp
K43K32K21 14p1p


1 1 1 1
[ + log 6 -
18 6 6 3


2 -2K^ 1 .p+ 19
l43 ^3221 K14pi1 9


-2K^p4 P+ +
K43 K32 K21 K14p p
1 2
3 3
3


2
- log Se6t
3 j


log2 S t 2)


1X+2
3 3


11 o 73
log t -
3 9


(4-16)


-2K"^ J + S2 2t
K43 A 4p 2(s + t4 log2 + 2)
K4A K3A K2, KI 4 2 P3
s(2t2 5st s2) S 1 ts 1
+ 6 -8-+-t)3 log- + -log 6e s + t +-
6(s +t)3 t 6 2(s t)2 18
1 1
X +
6 3
- 2 lo 1g+S W2+ -2+ 2 )
S UI + [st(t2 s 2) (log2 S
K- KK2pA +2(s t)4 +t7
K 43 K32K21A p 2 14 1 43
s(5t2+ t + 22) s 2 ts 19
+ log- log Ge s + +t
3(s + tt 3 ( + t)2 9
1x12}
3 3


AK,^+4 + + 2
-2__A_ _i_ [ (s2 P
K43K32K21K 14p (t
s (14t2 + 19st +112) S
3 (s + t)3 t
1 2
-3-
3 3


st + t2)2 (l2 S 2)
- 11 log2 + 2
11 ts 73
logy es+ -+
3 (s + t)2 9


(4-17)


The symbol X above is the relevant four point vertex: -2(pFp3 + p+p+)/[(p+ +
p+)(p + p+)] or 2(pp +p+p+p)[(p +p)(p + p)] + 2(p +p )/[(pt + )h(p +p)].
Here I also give the rational contribution of the corresponding box diagrams:
Here I also give the rational contribution of the corresponding box diagrams:


A(g+, g-,g+, g-;s)






A(g+, g-, g+, g-; q)






A(g+, g-, g+, g-; g)











4
B(g+, g+, g-,g-; s) = all a terms
9
16
B(g+, g+, g-, g-; q) all a terms
9
8
B(g+, g+, g-, g-; g) = all a terms
9
1 st -2KI;o 12 P4+
B(g+, g-, g+, g-; s) + A A KKp++
9 2(s +2 43 K 32K 21K141 p3
4 2st -2K4+ +
B(g+, g-, g+, g-; q) -13 24
9 (s )243 32K21 14p 1p3
2 st -2K1^ IJp
B(g+, g-, g+, -; g) 2 + 2 (4 18)
S9 (s t)2 K43^323A K21K~14p p

So, the rational part of a box diagram vanishes in N = 4 SYM together with the

earlier observation that two and three point functions vanish in N = 4 SYM. This agrees

with the 'no-triangle' assertion and some other technical observations that are commonly

used ii. .-.l.1.rvs to simplify the computation of A = 4 SYM amplitudes. Using a box

reduction procedure [18], an integrand with d powers of moment in the numerator and n

propagators can be reduced to tensor box integrals of degree up to d + 4 n. For gauge

interactions, all three point vertices have one power of moment, so there is n powers

of moment in the numerator of an n-gon diagram. Then the result of reduction is a

combination of degree four box integrals, and some of the degree four box integrals can be

further reduced to triangle and bubble integrals. In N = 4 SYM, due to the ultraviolet

cancelation between different species, the degree of an n-gon integrand will be n-4, so

the result of reduction is thus scalar box integrals. This essentially is the 'no triangle'

assertion, which we did observe in Section 4.2 of vertex corrections. While it can also be

shown that the scalar box integrals will not in any way produce rational terms, this I have

explicitly shown in the above list.

The gluon scattering in N = 4 SYM is very simple (up to infrared terms):











2K ,^ + lpg2 s 2)
A(g+, g+, g-, g-; SYM) I -log2 2

A(g+, g-, g+, g- SYM) 2K1iJjp log2 + 2) (4 19)
'K43 32 K21 14'l p3

without any need for counter terms.

4.3.3 Restoring Gauge Covariance

So far, we have only studied the pure gluon scattering, and we have encountered some

non-gauge-covariance (non-Lorentz-Covariance) such as the hanging four point vertex.

We have to complete everything to a tree in order to maintain Lorentz covariance. At

four point level, the spinor structure of the leading order amplitude is the unique one that

agrees with all the helicity assignment. So for the result to be Lorentz covariant, it has to

be proportional to the leading order. This remains true to all orders if supersymmetry is

present according to [17], which says MHV amplitudes are proportional to tree amplitudes.

We cannot just simply use a counter-term to cancel the hanging four point vertex,

because they are not polT ii. iiials in external moment. However, we can adjust the

relative strength of the exchange diagram to the contact diagram by adding a term

proportional to p2 to the self-energy term Eq.4-4. This modification only changes the

field strength renormalization by a constant, hence is perfectly allowed. With this term,

the coefficient of s and t channel exchange diagram is shifted. So if we pick the numerical

factor in front of p2 to be -1/6, -1/3, -1/3 for scalar, fermion and gluon respectively,

then they will match the coefficient of the lone four point vertex, completing it to a full

tree. This brings about a change in the numerical factor: 1/18 -- -5/18, 19/9 -- 13/9,

73/9 67/9 [14].

I have also done the computation with dimension regulation. The procedure was

to use dimension regulation to regulate the transverse momentum integral, and as

soon as this is done, e will be set to zero. Hence the infrared regulator is still k+. The

computation shows that the hanging four point vertex and the pure number will vanish,








this is expected since I have used a gauge invariant regulator. The numerical factor in this
scheme will come out in agreement with [23], who used dimension regulation through and
through, (not that the numerical factor is any thing important, as it can be altered by a
redefinition of coupling constant).
This type of counter-term will be put to a more severe test later on when we consider
the scattering of not just gluons but quarks and scalars. These counter terms had better
be universal in the sense that they are only the property of the self-energy bubble, and
should not depend upon what process it is embedded into.
4.3.4 All 2-2 Processes


8K^22 K 2(s+ 2t) 2s
A(s, s, g, g-) 8K13 14 9 -) log 6ecs log aect
spfptp. 2 (s t) (s t)
log!+2) 2-P+ + P3
( + t)2 (p2 p p
t(p2 (P i '


2(lA2V 2 /+
A(g+, s, g-, s)= -821 K 9 2 og es 2og e6t log2 2)
stp+p3+P2 9 log s 2ogtat


592 +77st 59t2 3(2s 3t) lg
A(s, s, s, s) = log e s
[ 3st s
3(2t + 3s) log at 2(s2 t2) log2 S + 2)
t st t
l ) \(P -P4 log P3
+_+P+ 2 _+ + P+2
2 (I p )+ (Y + )2 3

while A(s, s, s, s)o = 2(s2 + t + t2)/(st).The above are the processes only involving
bosons, the scheme for picking counter term described in Section 4.3.3 remains valid,
namely, we can find a universal set of numerical factors that will complete all the
above amplitudes into trees. This set can be chosen as -1/3 for II(g+, g-; g), -1/3 for
I(g+, g-; q), -1/6 for I(g+, g-; s) and -1/2 for f(s, s).









However, when fermions are involved, I failed to find a universal set of counter terms
that will fix the problem. I shall list the result of the computation first.


-4K21K 67 11 l t 2 S 1 -4p+ +p
A(q-, q-, q-, q-) 21 43 log et log2 + 72 -
S9 3 t 3 (pt + )2
(4-20)

A(q-, q-, q-, q-) 13 24\
stP3 p
67 1 1 11 s 2s2 st 2t2 lo S 2)1
log Ge's y log Get -log
9 3 s+t 3 s+t 2(s+ t)2 t og
1 4p2 P 4p +(4
3 [p(+ pf)2 (p++)2] (

The notation here is that a 'q' represents an incoming fermion line, a 'q' represents
an outgoing one, '+' corresponds to right handedness while '-' corresponds to left
handedness. So the process corresponding to Eq.4-20 and Eq.4-21 is Fig.4-6:


-4K^ K'2 67 11 s 1 -4ptpt
A(q-, q-, q+, q+) 4K 67 11 log log2 72)] x 2 P4
sp+p+ 9 3 t 3 (p +p)2
I (P+2)2

In fact, by charge conjugation, A(q-, q-, q + q+) can be related to A(q-, q-, q-, q-),
which then can be obtained from rotating A(q-, q-, q-, q-) clockwise by one notch (up

to the external line factors). As was explained in Section 2.2, instead of associating
fp to each fermion line, I only associate a factor of p+ to an outgoing line, but
nothing to an incoming line. This is simply because a square root always causes
troubles in automated computation. For example, -4K4K32/(sptpt) should really
be -4K^4 K32/ (s /pJ + p pf ), which is quite clear as the latter is Lorentz covariant (can

be written as spinor products), but the former is not. The phase of the square root here
can be given arbitrarily.











A(s, s, q-, q-)
4K^134 [92 29 2t + S 2 S 2 2 2p- (p+ -p )
log es o + 22
+sp p 9 63 2(s P t) t 3 (p+ pf)2

So far, the effect of using a non-covariant regulator is the mismatch between the

exchange vertex and 4 point vertex in the final result. However, in the case of gluon

fermion scattering, the mismatch is much worse.


8K^2KV KV
A(g+,g-, q-, q-) 13I+2 2 6 3 log es log2 _- +2
stp P2 P3 t
1
x s channel tree
3


A V\ V 2 A V2\
A(g+, g-, q+, q+) 8K14K1332 [6 3log s (log2 +72)
stptpjp3 L
1tYl p2 p3 p4
1
x s channel tree
3

Here we see that the mismatch is between s and t channel. By s-channel tree, I mean the

s-channel exchange diagram and also its descendent four point vertex.


-8K212K43K32 log2 2)
A(g+, q-, g-, q-) tp+p+2 6 3log t log + 72
stpipj p L 2 t / j

Here, it is mere coincidence that everything matches (the planar condition eliminates the

u-channel tree).

The problem of fixing counter terms will be revisited in the future work discussion.

P k3

a k b

Figure 4-1. Two-point function and the dual momentum assignment









a d


b c

Figure 4-2. A self-energy diagram embedded in a scattering process

V, c, p,


k2 k4
q


A, a, pi A, b, _

Figure 4-3. Triangle diagram

A, a



k2 k4


r, b, p kl r c,

Figure 4-4. Fermion vertex correction corresponding to Eq.4-9

A, a q


k2 k4



b, pi c, p2

Figure 4-5. Scalar vertex correction corresponding to Eq.4-11

pp44 l ], ,, l





pi, 1 i_, _,

Figure 4-6. Four-fermion scattering corresponding to Eq.4-20 and Eq.4-21









CHAPTER 5
BREMSSTRAHLUNG IN LIGHT CONE

In this chapter, I shall deal with the infrared terms.

5.1 A List of Infrared Terms

All the amplitudes given in Section 4.3.4 are amputated Green's function, the

external leg corrections are not included. As we can see that the self-energy diagram

always contains a term log 6ep2x(1 x), which gives a multi-particle branch cut on

the positive real axis, stopping us from doing wave function renormalization. This can

be cured by summing over collinear emissions or absorptions from the external legs.

The analysis of [14] showed that doing so is equivalent to replacing log e7p2x(1 x)

with log JSeA2x(1 x), A being the jet resolution. For H(g, g, g, g; s) or H(g, g, g, g, q),

this substitution alone is enough to regulate the infrared divergence (the triangle or

box diagrams involving fermions or scalars are devoid of further infrared divergences).

While all the other processes (whenever there is a gluon propagator juxtaposed between

two massless external legs), the computation is fraught with IR terms. But these IR

terms when combined finally, are universal, which shows that they are ]li,,-i, .,1 infrared

divergences, instead of light cone gauge artifices. Here I list all of them first.

All moment appearing below are dual moment, with the real moment given by

pi ~ k k+, p k. k- p k+ k pf k+ k+ + and kl+ < k+ < k+ < k, see

Fig.5-1.









.4
g4 Tr[tatbtctd] x
87r2
k-+ < q+ < k
g (kf q)(-k+ +i)se




log (k7+ kfl(k2 kfl L q
log ( 2^ )2 2 1
(k+ k)+2 q+ kf+

(k k+)(k+ k+) q+ k+

(k+ q+)(-k+ _q+)s~e[ 2
log- 3 + +
(k k kf)(q2- kq te + k
g(-k + q+)26 I -2
log 2
l (k kq)(q k) q+ k

kl+ < q+ < k+
og(k q+)(-k+ q+)se [
log [2]
(k ) +)2 q+ I + k2
(k+ q+)(q+ k+ )t6eY
+ log -+(- [ + J
So (+- + k+)2 -+ k4 _


(k+ q+)(+ kq+)s~e[ 2


Slog ( 2 qq







(k-k kfl)(k-kf) Lq+-k
(q+-kf)(kf-q) se? 2 1
log (+ k+)(k- kf) [q+- kf
(q+- tq+)(-+ q+))Ge 2




+ log (- 2 [ 2 ~ 1
(k4 kf)(2 k ) [q+ kf
(-k + q) (k+ q+)t6+e 2
log 4 3
(k+ -k) +) -k++ k+) q+ k+


(kf q+)(q(- k +)sb7 2
(kf k+) 2+ k+ q+ -









The above is always multiplied by the corresponding tree amplitude (by a tree, I mean

whatever I wrote in front of the square bracket in the list of amplitudes). As long as the

process has infrared divergence, the infrared divergence will be of the above form.

The infrared terms above will be combined with soft Bremsstrahlung and collinear

emission (absorptions) along with the self-energy insertions on the external legs to give a

finite result. The basic idea is the same as the standard treatment of infrared divergence.

That is, we insist on measuring jets (within a certain resolution A) rather than gluons or

quarks.

5.2 Bremsstrahlung Process

I will be focusing on final state bremsstrahlung radiation (region between leg 3 and 4)

Fig.5-2, and study the four gluon core scattering process as an example.

pi, i = 1... 4 and k are assumed to be incoming, and k will be called the extra one

for now. The moment of the 2-2 core process will be denoted as ji, i = 1... 4 (j for jet).

Possible sources of divergence associated with region 34 are

1. k is collinear with 1'. or p4

2. k is soft.

We can look at roughly how each type of divergences get canceled. From Fig.5-3, it is

quite natural to guess that the collinear divergence arising from k being parallel with p3 is

going to be cancelled by the corresponding self-energy bubble on leg 3. The cancellation

will work as long as k is not too soft. We can make sure of this by setting k+ away from

zero as a cut off. What happens when k -> 0 is that we lose coherence. More concretely,

when k is not too soft, we only need to worry about the case when it is attached to pS.

When k ~ 0, it can be attached to either p3 or p4. So we will be considering the following

cancelation of Fig.5-4.

There is no natural boundary as to when k is soft enough. In fact, we can first impose

an artificial boundary A, such that when k+ > A we use the scheme in Fig.5-3, when

k+ < A we use the scheme in Fig.5-4. When the dust settles the A dependence disappears.









The method described above was used in [14], it involves a careful analyzing of phase

space to avoid double counting. However the beautiful Parke-Taylor expression for helicity

amplitudes greatly simplifies the computation: the first diagram in Fig.5-2 is given by (in

the large N, limit)



(g N/3)P 3 ,P\P2) 2>
(pi p2)4 2


2 (p ip p2 /'.)( /. | k)(1|* )(p4 Pi)

It's quite clear from Eq.5-1 that when k is collinear with p3,4 or soft, there will be

divergences.

As stated before, we have to set k+ away from zero in order to cut off infrared

divergence, but this does not regulate collinear divergence, so a temporary cut-off C will be

used tentatively to cut off collinear divergence.

In order to parameterize the phase space, define (refer to Fig.5-5)


2Q (-1,.) 2Q (-p4)
x 2 ,2 Q = P1 + P2 (5-2)
Q2 2

The integration region of x and y is depicted in Fig.5-5, we will only be integrating x and

y in the 'L' shaped region. R is related to detector resolution according to R := A2/s.

The li, -i .I1 meaning of these cuts is quite clear: when x ~ 1, k p4 = 1/2(k + p4)2

1/2(Q + .)2 1/2Q2(1 x) = 0; when x, y ~ 1, k Q ~ 0 (since Q is time-like, k ~ 0).
Here and after, s and t will denote the Mandelstam invariants for the 2-2 core process.

A trick used in [24] turned out useful. Write


1 1 1
(' (- )(~ p4) ('. ( )(k. (P. +p) (p4 ( )(k- (. +P4))
2 2 2 2
( y)s(2- -y)+ s( x)s(2 (5 3)
s(1 y) s(2 x y) s(1 ) s(2 y)









The first term will only diverge when 1., and k are collinear and(or) k is soft. The

second term will only diverge when p4 and k are collinear and(or) k is soft. For the second

term, the integration of x and y in area A: 1 R < x < 1 e; 1 x < y < 1 is divergent,

but finite in the area B : 1 R < y < 1; 1 y < x < 1 R. These two areas will be

treated slightly differently.

Region A:

Since x ~ 1, p4 and k are either collinear or k is soft. All x at innocuous places can be

set to be 1. Further define L to be a reference (light-like, Lo > 0) 4-vector, such that

L p = p+. Now the 3 body phase space integral is given by


sx
6 dQ3dxdQ4k
64(27)5

dQ3 is the angular integral of I'.. dQ4k includes the angular integral of p4 and k, dx

represents the integral of the norm of 1.. Note that if the phase space integral is isotropic,

it can be written as the standard form sdx dy/1287r3 for a 3-body final state. We can

factorize the phase space integral into [1/(8(2r)2)d 3] [s2/(8(2 r)3', ,1., ]. Here r. is

picked out as 'the special one', since in region A, P3 is always hard, and is almost equal to

j3 of the 2-2 core process. I will integrate out [sx/(8(27r)3)dQ4k], while the [1/(8(27)2)dQ3]

part is what should be compared to the 2-2 process.

The second term of Eq.5-3 is parameterized as


g6 N (p )4
c (lP2 (5-4)
2 (Pl P2)(P2 '.)(P4 k) (k (P '))(p4 p4 P)
6N3s4 S
4s(/._ + 1..)2(4 + p)2 (P4 k)(k (p4
g6N3s4 1
s(p2 + .)2(p4 + p)2 s(1 x)s(2 x y)








As x ~ 1, (p2 + .)2 t, (pi p 4)2 ~ yt. The first approximation is always good in
region A, the second is good as long as p4 is not too soft. Plug in the phase space integral,
thus arriving at


f sx g6Ns2 S dQ4k
64r3 st(,,) (1 x)(2- x y)-

S6 cA2 N-A + d1,"1, (5-5)
64 (1 x)(2 x) y 2 x y

where Acore g 4N2(s/t)2, let us drop the factor g2A' 2or hereafter.

dQ4k can be expressed as


2dudv
/(ul u)-(u uo)
S: 2(1- y)

21|jk 2|1k 1
v:= (1k+l < Alj)

: 1 2+(l (1- x) 2b(1 x) Ijb
1 + 1 ijI (56)
Q++pt |x x J11

We can first pick a particular value of p3, then go to the CM frame of p, + 1_ + p, to
evaluate the invariant expression 6(k + p4 +1 PI 2 + 1.)6(k2)(p2)d41k d4p4. Notice in this
frame, u is in fact cos(/~.k), v is cos(Lk) and w is cos .,L. Then after some exercises in
Euclidean geometry in Fig.5-6, we can obtain the expression for d(4k.

o, := vw (1 2)( 2) are the lower/upper limit of u (y) integral. The limit
placed upon k+ is for breaking the Bose symmetry between p4 and k and to make sure
that the approximation (pi + p4)2 ~ yt works (the A dependence will drop out).
Performing the u integral using










1 (a > b > 0)
Sa r bx x2 a2 b- a> b2
a (-a>b>0)


/d1 l l- Sr 1 2

-2A -R 64 core 1- x)(2 x) x
2x 2x
( ) (5-7)
S1 U)( -1 no) ( -1 Ui)( -1 -o)

The first square root can be obtained from the second one by replacing v by -v. So I will

only compute the second square-root. The following indefinite integral will perhaps be

useful:


1 1 1 -b + ax + /(+b(2 +)
dx log
J -x + 1 a + bx va + b2 a + bx

The second term of Eq.5-7 becomes


I1 dv i- v


-[(1 v) + (1 + 2b)R] + ) + (1+ 2b)R]2 + 8bR2(1 + V)
+ log 5-8)
4bR

here we have treated the two cut-off's R and c differently: c is set to zero with fixed v

(fixed k+) while R is taken to be small but non-zero. This treatment agrees with the

strategy given in the paragraph above Eq.5-1. Notice the second term in the square

bracket is finite (when k+ is kept away from zero), while the first will be combined with

self mass contributions to cancel the c dependence.

We can make a change of variable (the y below is not the previous y):











[(1 v) + (1 + 2b)R] + [(1 v) + (1+ v 2b)R]2 + 8bR2(1+ v) 4bRy
1 7- 1 y(l b) dv b( R)y2 + 2bRy + bR (9)
1 v = 2R = -2R (5-9)
y(l R) + R dy [R + y(1 R)]2

then the integral becomes


1 [l b(1 R)y2 +2bRy + bR l f1 dv 1 v
162 [[+ y(l -R)][1- ,,- y(1 -b)] 16log 1 2A (1 v)
1 /1[" 1--R 1 1 ] 1 11 dv 1o -v
YRII log y + 2 d log I
162 0 [R +y(1 )] 1 y y 1/b lo 16 1-A lg C(lt )
(5-10)

where yo R(1 A)/A; yi = 1. The above integral becomes


1 [ 1- R 1 72 1 dl' d og -V
1 [f -R logy -dilog(l+ b6) + 1 log
162 0 [R + y( R)] 6 167 1- V (lo + v)

r2 log2 R dilog + log R log A dilog(l + b)
1672 2 A A 3

+ -- log (-- v) (5-11)
1-2Al- V I

We can do one more thing to make it more symmetric: as b always comes in pairs with
1/b, we may replace dilog(l+ Ij+/j'+) with -1/4 log2 Ij+/j+ -_ 2/12 by using the formula
dilog[l/(l x)] + dilog[1/x] = -72/6 1/2 log2 (I/ 1).
The first term of Eq.5-7 is obtained from the above by substituting v to -v and the
limit of y integral becomes y E [0, RA/(l A)]. We get










1- dilog + log R log + log 1
I IA I I 1 + v o v
162 dilog logR log -A) A log2 + log2 (1- A)
162 A 6 2 2
1 X2
+ [ -2dilogA log2 (1-)+ + 2 loglog 1 ( A)
2 3
log log (1 A) (5-12)
R

In fact, by noticing that du//(u u) (u uo) behaves like (7r times) 6 function as x ~ 1,
I can set y = 1 x(1 u)/2 to be 1 x(1 v)/2 ~ (1 + v)/2 in the first term of Eq.5-5,
which will lead to the same result in a much quicker way.
Region B
Now p4 is the hard momentum, (p4 + p)2 is equal to t, while (. + P2)2 is approximated as
xt. The three body phase space integral should likewise be factored as dQ4dydQ3k. Since
there is no need here to hold k+ fixed, and the restriction Ik+l < A'IQ + p4+ can be lifted,
sy/(8(27)3)dydQ3k is simply -,1,'1/i/(1672). Here I have used a different partition: A',
because the partition associated to leg 3 need not be the same as leg 4.


Sg6N, 3 11 3k 4 1
S (8(27)3) s(._ + '..)2(p4 p1)2 s(1 x)s(2 x y)
f g6 N l,1, s4 1 1 g2
-- 62NcA2 (5-13)
1672 s(xt)t s(l x)s(2 x y) 167212 core

I have dealt with the first term of Eq.5-3, the second is similar:


1 1 1 -' A 2
S- log2 R dilog + log R I log A' dilog(l + j/j )
16[2 2A' A' 3
X 'lJIl dk+ k+ r2
+ k+log + -
ko l+ k+) 12
1 A' 1 /J3o dk+ 1+ k14)
+ dilog A'+ log R log +A' / log
A- I A' I1- A' o '1 V+ ck+









So far, I have covered the first diagram of Fig.5-2 when k does not dominate over

neither p, nor p4 (k- < A lj l, k-" < A'|jfl ). To complete the region k+ > AI\j we need to

do a computation in region 14.


g6N3S4 1
4s(p2 1' .)2(p4 +1)2 (P4 k)(k (p4 '))
g6Ns3 31 1 1
I +(5-15)
8(_ +j .)2(k (p4 ,'.)) [ p4 kp4. (k p) p4. p44-p(k ()

with (k + p4)2 < Rs and k+1 > Al +I. One point that I should have emphasized earlier is

that in the presence of 1/(p4 k), we can only make approximations controlled by O(p4 k),

and similarly in the presence of 1/(p4 pi), we can only make approximations controlled by

O(p4 -pi), otherwise there will be errors of the type R log e. This rule makes the evaluation
of the first term much easier: we can set k to be parallel to p4 in all the irrelevant terms.


g6N. S3 1 1 g6Ns3 1 1
8(p2 '..)2(k (p4 .)) p4 kPp4 (k p) 8t(k (p4 +.)) 4 k p4 (k pi)

Now set up a parametrization for region 14:


2pi (pi p4 k) -2k (pi p4 k)
(pi p4 k)2 (pi p4 k)2

Abbreviate (pi p4 k) as T. Then k p4 can be worked out as 1/2(1 x)T2. With this

parametrization:


g6NCs3 1 1 96N3s3 1 1
8t(k (p4 ))p4 k p4 (k pi) tys (1 x)T22 (- x- )T2

In the limit x= 1, T2 = It. In fact, the above expression can be calculated without effort,

simply by identifying A as 1 A and b as pl/|p4 (pi and p4 here are the core values), we

can borrow the result from Eq.5-11.










g6N,3T2d xd4k 3 1 1
I 647 3 tys (1 )T2 (2 )T2
22 cA2 r,1 I A2 *+
4 J 4
core log2 R dilog l R log(1 A) dlog(1 + )
167 kl | l \ lg j+*
+ log L- ]

1 [ 1 -1 A 1 dk+ V
+ dilog- + l og + log k) (5-16)
1672 A A A A|(+I k+)

Comparing this to Eq.5-11, 5-12, we see that the A dependence cancels.
Now we look at the second term of Eq.5-15, it in fact can be interpreted as a false
jet, since it will make a contribution when P4 is parallel with pi, thus the final state will
be a three well separated jets unless p4 is soft. Our task is to compute this term in the
region 2k p4 < Rs with I t+ > Ik+I > Alj 4+. This region can be dissembled into a part

{1 x < R} n{y < 1- R}, which produces a 1/(167r2)r2/12 (the restriction on k+ makes
no difference).
There is the other part: {1- x < R} n{1 y < R} n{Ik+ > Al '+l}. Here I only lay
out the strategy if one were to compute it honestly. First compute in region {1 y < R}
with no restriction on k+, then subtract the entire region of {x < 1 R} t{1 y < R}
again with no restriction on k+. The validity of this strategy lies in the fact that there
exists an upper limit of order 0(R) for |p| beyond which there is no intersection with
the region {1 K < R} n{1 y < R}, while the limit Ik+ > Aj+4 I translated to

p4+ < (1 A)ll is well beyond the said upper limit.
Yet in practice, the condition that we can only make approximation of order p4 pi = 0
makes the evaluation (and interpretation) quite hard. So we make a compromise, and
allow for errors of R log e, trusting that it will go away with a complete calculation. I shall
set not only p4 p =P 0 but also p4 k = 0. Then, this term will cancel against some
disconnected diagrams which will be explained later.
So the completed result for Eq.5-4 is










1 [ 1 72 jl+ ifJ dk~+ |j+ +1
log2 R dilog(1 + j+l + log 4 -
167 2 4 J+ 0 1j V Uk+
1 1 2 3 4 i, dk+ kV
_____ L dilog(1 +^ _lo ______- 1
+ 16 log2 R dilog( + l + log
42J4 log 1(|j| k+)j
(5-17)

It makes a similar contribution associated to leg 3, with some suitable substitutions.
Let us take a look at the self-energy bubble on leg 4 Fig.5-7, calculated with a cut off

K4,kK,k/pk4+ > p to regulate the on shell divergence. e should be in this case identified
with pt2/s and E with pt2/ltl.
The wave function renormalization of these diagrams are (with the 1/2 from '/Z' and
another 2 since it enters as a cross term):



left: 16 + log x(1 )p2
o1672 Ia I -
1 1 ( 3
middle: j 2 -1 ) logx(l x)p2
Jo 1672 x
right: l log x(1- x)t2 (5-18)
1 1672 1 X7

where x is k1" / j+ Not all of Eq.5-18 will contribute to region 34. For now, I only take
the first line. Combine Eq.5-18 to the expression above:


1 1 1 _2 3o
I log2 R- log2 R- dilog(l+ )-dilog(l + )
167 2 [2 2 2 J4 J4
i dk+ k+2s /4lj Idk (j\ k+l)2t]
+ + log + log + 2
40 I4J 4+- '4 J4

Thus, the effect of a self-mass insertion is to replace c by 1/sx(1 x) or 1/|t|x(l x).
Fig.5-8 represents all the diagrams that contribute to the Bremsstrahlung process.
The second diagram of Fig.5-2 or Fig.5-8, when combined with the second line of Eq.5-18,
will contribute to leg 4:










1 1 T2 i3
S log2 R dilog(1 )
162 2 4 1 +
/"3 r]]' + +2o ]I
+ /li d+ log k-+ + (-3 3x- 2) log x(1 x)RS (5-19)
o k J4 I0
Note its contribution here should be attributed to region 34. This diagram won't
contribute to region 14, because it remains finite when p4 0.
The third is similar to the second:


1- log2 dilog(1+ )
1672 2 4 J4+
/4~l k*+ k+2 14
+ k+ log --+2+ (-3+3x -x2)logx(1 -x)R|t| (5-20)
O V J4 0
Note that it is credited to region 14.
In summary, the contribution from Bremsstrahlung plus self-mass insertion to region
34 is



16A 2 2 2 +2 J1
2 log2 2 log2 i3
167 2 S 6 2 |J'4
/?lj+l4 d+ k+2s +l d+ k+2 67 11
+2 logf +2 logg k- + 9 3 log A2
0 V j4 V k2 9 3
And a symmetric contribution to region 14:


1 [ 2 log2 2 2 1lo2 2P
-2 log 1 2 log i lo


4o 2k+ j 2 k+ :+2 9

As a summary, although the kinematics in 14 and 34 region are very different, the results
are almost symmetric up to the false jets that I haven't included. Next, we study how to
combine these results to the virtual process, and defer the discussion of the false jets.








5.3 Combining with the Infrared Terms from the Virtual Process
The list of infrared terms in Section 5.1 can be rewritten in a more symmetric form
that is independent of the relative size of the dual moment:
Region 34:


dk+ k+2s
log
k+ |pf|l'i |


0I
o ,


dk+ k+2s
+ log
k+ 'I. ||pf|


dk+ k+2s
+log +11I
k' |p1 ||pf|


dk+ k+2s
Slog+ 1 I
k+' |p1 ||p


I' dk+ (Ip+l k+)lp +
+ i k+ (Plo k+) I j
fp+1 (1~ + 2+)/


Region 41:


-1[ I dk+ k+2(-t)
8[J k log lIm

+ 1 dk og (p k)lp4
I k+ (k+ |+ -pM


-1


dk+ k+2(-t)
+log l ++
k+' |p1 ||p|


Region 23:


_8-1 I dk+ k+2(-t) P dkl k+2(-t)
sl I_ log + Jo +g logI
8 0 k+ P2 1P3 k+ p2 p
+ lo dk (1pI- k+)'J I
p k+l (k+ )p
og P j P


8 2 0o


Region 12:


I'I" dk+ log(Il' k+)lp4 1
,i k (|p | + k+)j' |


(5-21)


8"2 O









The first two terms of Eq.5-21 cancels the divergence of Eq.5-19, while the last term
integrates to be -1/(167r2) [-2 log2 (pf/p)]. Finally, all the non-covariance pieces

cancel and everything falls together nicely:


1 A2 72 67 11
[- log2- + logA 2] (522)
872 S 3 18 6

is the total contribution of virtual and Bremsstrahlung processes to region 34. While for

region 14 we have similarly:


1 log2 A2 2 r2 67 11
[- log2 + log A2] (5-23)
872 Itl 3 18 6

Thus, we can write down the total scattering probability:



P(g+, g+, g-, g-; g) =
gs2
P(g+, +,g -, g-; g)o 1 I+

[ a 2 A2 2 2 67 11 4e 6A4 2
s Itl 3 9 3 Itl It



P(g+, g-, 9+, g-; g)i

P(g+, g-, g+,g-g)o 1 + N x

S2 log2 2 og2-- + log A
s |t| 3 9 3 s
(s2 +t +t2)2 2 s S (14t2 +19t + 11S2) s ts]
log" log
(t + s)4 gtl 3 (t )3 lotl (t + s)2

And the probability with i = 4 SYM is











P(g+, g+, g-, g-; SYM)I
g2 e 2 2 2 2 67 5 26
P(g+, g+, g-, g-)o 1+ 82 21g 2 1g + [ Ns -Nf]
8 Itl 3 9 18 9
+ log [- + -_N + -Nf] + log2
|t| 3 6 3 tj



P(g+, g-, g+, g-; SYM)i
g2Ne 2 2 2 67 5 26
P(g+, g+, g-, g-)o 1+ 2g lo2 + [- -N -Nfl
87 Itl 3 9 18 9
+ o A4 11 1 4
+log [- + N + -Nf]
3 6 3
1 log2 S t2N + 2(2 + t2)Nf + 2(82 + St + t2)2]
2(s + t)4 Itl
+ log [-s(2t2 5st s2)N +4s(5t2 + st + 22)Nf 2s(14t2 + 19st + 11S2)
6(s +t)3 |t

+ t [-N + 4Nf 2] (5-24)
2(s + t)2

5.4 The Inclusion of Disconnected Diagrams

In this section I will discuss certain disconnected diagrams Fig.5-9, whose importance

was explained in [16].

Only the cross term between the first and others in Fig.5-9 has the correct power in

coupling constant, and the factor 2Ek(27r)33(k k') = (27)434(k k') corresponding to

the forward particle line will force the two extra legs to have the same momentum.

The fourth and fifth diagram of Fig.5-9 look rather like self-energy diagrams, and

indeed they will receive the same factor of 1/2 discount just as self-energy diagrams on the

external legs do. The fourth (resp. fifth) diagram will have to be evaluated first with p4

(resp. pl) off shell. When the dust settles, any term that do not contain p2 or p2 will be
dropped just the same as we are wont to drop the tadpole diagrams.

We have insisted on having two partons scattering into two jets, so the two extra

gluons have to be soft. The Feynman rules give (with the approximation k ~ 0)











2 8pl 4 6 2
2 core ( p)2( )2 t2 2(k pi)(k p4)

to be evaluated in the region (k p)2 < A2 and (k p4)2 < A2.

Next I show that it will cancel the false jet terms. Setting p4 Pi = 0 and p4 k

the second term of Eq.5-15:


g6N s3 1 1 g6N3S3 1 1
8(p2 + ..)2(k (p 4 '.))4 pi P4 (k pi) 4ts p4 Pi P4 (k pi)

The third diagram of Fig.5-8 will double the above. To make comparison with the

disconnected diagrams, we need to identify k with p4 and p4 with -k:


g6N3 3 6N s3 1
2x C _-
4ts p4 -pi P4 (k pi) 2ts (-k pi)(-k (p4 P))

The fifth and seventh diagram will give a similar contribution:


g6N s3 1
2x -
8(p2 ,.)2(k (Pl + p2)) 1 -P4 P1 (k p4)

Now identify k with pi and pi with k:


f6Ne3s3 1 1
2ts pi p4 pi1 (k P4)


g6Ns3 I 1 6N3 1
6 (5-27)
2ts pi p4 p1 (k p4) 2ts (k p4)((k (pi p4))

The sum of Eq.5-26 and Eq.5-27 cancels Eq.5-25. The cancelation above is ad hoc

and certainly not the prettiest. For one thing, we have made approximations of order

O(R log e), for another, the interpretation of the disconnected diagrams as false jets is not

intuitive. However we can see the similarity between Fig.5-9 and the false jet terms: 5-26

(resp. 5-27) would be a lot more natural had p4(resp. pi) be an incoming(resp. outgoing)
particle. While if we could change the sign of ko in Fig.5-9, they would all become true


(5-25)


= 0 in


(5-26)









loop diagrams. So both the false jets and Fig.5-9 have the 'incoming-outgoing-reversed'

problem. I believe a better way of dealing with them is possible.

In [14], the false jets come more naturally. They come from the diagrams with leg

4 absorbing a gluon collinear with itself, or leg 1 emitting a gluon collinear with itself.

These processes suffer collinear divergences, and they are cured by the fourth and fifth

diagram in Fig.5-9, while the cross term between them is canceled by the second and third

in Fig.5-9.

4\ 3

1+




Figure 5-1. Dual momentum assignment


p4 k '. p4 k p.





Pi Pi
Figure 5-2. Two diagrams with an extra 'unseen' gluon the arrows indicate helicities,all
moment are incoming









Figure 5-3. Cancelation of collinear divergence













Figure 5-4. Cancelation of the soft bremsstrahlung radiation against a virtual process


y= 1


x 1


Figure 5-5. Phase space integration region of x and y




F --
Figure 5-6. Configuration of



Figure 5-6. Configuration of k, j*, and L in the CM frame of pi +I-_ + '.


Figure 5-7. Self-energy bubble
Bremsstrahlung





34,14,23





12, 14, 23


on leg 4, they will cancel the c dependence in the


34





12


14





14


23





23


Figure 5-8. All non-vanishing Bremsstrahlung
are the regions they contribute to


processes, the numbers underneath them


U





























P4 '. P 4 '. p4 '.





Pi P2 k pi P2 Pi P2
k/ P4 '. P4 1',





k k
P1 P2 P91 P2

Figure 5-9. Disconnected Bremsstrahlung with two extra 'unseen' gluons









CHAPTER 6
GLUON SCATTERING WITH MASSIVE MATTER FIELDS

I only computed gluon scattering amplitude with massive matter in the loop, for these

are the only processes without any internal gluon propagators. If it were not so, we have

to devise a way to regulate the infrared divergences. Dimension regulation certainly works,

and the entire problem is reduced to calculating the 'box coefficients'. But this is simply

bringing coals to Newcastle, as there are people who have already developed the technique

and can do it a hundred times faster than me. The new thing about our calculation is

the pl, l,-i, .,1 infrared cut off that does have a meaning, yet I haven't worked out how to

incorporate this cut off into massive amplitudes.

6.1 Computation Technique

The main difficulty for the massive field calculation is the Feynman parameter

integrals. Indeed, I can only reduce all Feynman parameter integrals into a set of three

definitive integrals. They are



(s) : dx sinh-
Ssx( x) + M Vs2 + 4Ms 4M
1 1 SX1(1 X) + M 2
x M
J(s) X jdxilogSX x)-M 2yslnh 4M)

K )1 st (s(1 x) + M)(tx(1 ) + M)
Sstx( x) + M(s +t) /-

Where M is in fact -m2 + ie. As M 0:



I(s) 2 log
1 s
J() log2
2 M
s t
K(s, t) ~ -2 2 + 2og -log
M M
K(s, t) 2J(s) 2J(t) -2 log2 (6-2)


For example, a simple integral:











Six2(l 2)
S6(xi +X2 +3 4 1)dxidx2dx3dx4
o ( arir3t + r2r4S + M)2

can be reduced to



M J(s)- J(t) K(s, t) +
s2(8 t) tQ(8 t) 2st(S t) 2st

The definition of Eq.6-1 is in accord with [27, 31]. The last integral of Eq.6-1 can

be expressed as dilogarithm and logarithms. In [30], the author gave two equivalent

expressions of the box integrals, one in terms of dilogarithms the other in terms of

hypergeometric functions. But for the purpose of this paper, I find it most convenient to
use Eq.6-1 to spell out the results.

The Feynman rules for scalar fields remain good, but the decomposition Eq.2-25

does not, since it introduces 1/q+ factors into the Feynman rules. This will complicate

the already complicated Feynman parameter integral. Also q+ and q- will be treated

the same in contrast to the massless case, where q- is integrated out and q+ is given by


In order to organize the gamma matrix algebra in the fermion part of the calculation,

we make use of the factorizability of gluon polarisation vectors Eq.2-21 to reduce products
of gamma matrices to products of K = pp pfp", which had been proved to be quite
Pi Pi j- Pj Pi
handy. For example, if we are to calculate the diagram Fig.4-1. We would write down:



n(g+, g-, q) =(ig)2Ttb C ( 2'. 2 (6-3)
TI(g+, g-, q) (ig)2Tr(tatb) Tr ([(q kl) + m]'yj[(q k3) + r ) (6-()

(-0) [(q k)2 m2] (-0) [(q- ks )2 m2]

The numerator can be written as











Tr m (q ki) a 0
[ (q- k i).a 2i k3](ri
S (q k3) -a 0 2
(q- k3) m 2qr]](k3 kll

S2['|7q k I)[k3 kIq k3kj1 k3) + 2(k1 k3jq
+(ki k3\mjrq) [k3 ki|m|rq] + [r\mnk3S ki]({r\mki


v l1) [k ki| x
k- k I

0
-kl k3)[O ]q-
-k1k3-1 ki('7q-k3171


k3)


For the notation of spinor, see the appendix. The standard procedure of momentum
integration tell us to shift q -+ q + xk + (1 Z)k3, and to replace qP'q by q2g //4, etc:


[r1q kil )[k3
[( x)(k3 -

(1- x) v2(k3-


- klq k3ski k3)
2
ki) lk)[k3 kllx(kik k3) Iki k3) + 7 [l| a7)[i k- kj1 b ki k3) q 2
bb4
kl)+ -x(ki k) [l 3 k] (ki k3 q2
2(ki k3)+ 2


x(1- x)(ki k3)2 q
2


here we have used an identity:


a V 2 ;ab bb p ab ab
adaaUgPv dabcab;, p, gPV


For a more complicated example, a string of spinor products becomes (with q shifted
to q + x2k2 + x1kI + X4kA4 + X3k3)


[Piq k2 p4)[r1q k31'
S[pi|q x+ x(ki k2) + x4(k4 k2) x3(k3 k2) 4)
[r]\q + z2(k2 k3) + xi(ki k3) + x4(k44 k3)'j.)

S[Pllrl]{ 1)4}- +[p1 X4p2lp4)[rjlZ2p4 xIp1/'.
1 q2 2X4 2X2 2x1
S(-1) K4^ K2K4/2( K^ K^) (6-7)
1) 43 + K4 12 ( 34 + 32) (
3 P4 2 p P2 P4 P3 P3


(6-4)


(6-5)


(6-6)









Basically, all of the spinor products can be reduced to one of the K-y's, and the reader

can find in the appendix some practical details as to how to organize the products of Kij's.

After the momentum integral is done, we can perform the Feynman parameter integrals

using Eq.6-1.

However, there is one more complication due to the 6 regulator: the integration over

q^(V) is different from q(-). Since the momentum integration is no longer homogeneous,

the replacement of qP"q by q2g/"/4 is problematic. But fortunately, when doing the

contraction of q"q"V g"q2/4, qA(V) and q+(-) will never coexist. For example, in the first

line of Eq.6-5, only q+ will appear in the first bracket while only qA, qV, q- appears in the

second bracket. So, when we make the replacement q"q" g""q2/4, we should remember

that q2 actually means q2. Another example, if we have a spinor product such as in the

first line of Eq.6-7, qA, q q- will appear in both brackets, then q2 means q_ this time.

6.2 Self-Energy Diagrams

A factor of -ig2/(1672)fcadfdbc g2/(167r2)Tr [tatb] will be omitted.

For Fig.4-1, the results are



H(g+, g-; s)m [2 log 6m2 + 2 (2) _- +2 2

4 2 (p2 -4m2(p2 2m2) 26 16
H(g+, g-; q)m = 2 log 6 2 m2 2 () p2) ) p2 + 2
3 3 p 9 3
(6-8)

The gluon mass counter term in the expressions above has been removed already.

6.3 Triangle Diagrams

A factor of g3/(872)fdaefebfffcd 2g3/(8 72)Tr[tatbt ] is omitted.

For Fig.4-3:












F(g+, g+, g-; s)m




F(g+, g+, g-; q),.


-2pKA 1 -
+ K21 1log m26e (p
PI P2 [6


(p2 4 2)
4 2)( 12p4


+ i2
a +2 4)I(p)
P3 pO


3a J(pO)
p2 p0


-2p+
P2 P2


[ 2
log rn26C
1-3


+(p2 4r2)( (P 2+ ) 22
3p

Sp+ m2 16 8m2
-2a P+ J (po) ++
P3 PO 9 3Po


p-I p+ M2
+2 4 0
P3 PO
1pjIp( 24M2]
+2(1 2
p3 PO


where a 1= if leg 3 is off shell and 0 otherwise, and po is the off shell momentum. Their

anomalous terms are identical with the massless result, which agrees with the fact that

anomalous terms are UV effects.

For an MHV triangle:


F(g+, g+, g+; s)m

F(g+, g+, g+; q),m


(K21)3

2
2 (!2, )3


4m2(o2 4m2) 8m2 2

Po PO
8n2PO POn


2( 2421

4(po 24+M2)
3p4
(6-10)


6.4 Scattering Amplitudes

A factor of ig4/(8 2)Tr[tatbIttd] is omitted.


A(g+, g+, g+, g+; s)m

A(g+, g+, g+, g+; q)m.


p.p pfp st 6
16 43 K32 K21 14
p p2 p3 P4 st


4
- K(s, t)
st
2M4 K(s, t)
st


(6-11)


1 4m2
9 3p-


a +2(1
6 p32


24m2)
+ 2
POn


(6-9)











A(g+, g+, g+, g-; s)m
K13 ^24 (2t S)tm2 2_
K43K32K21K4 2(s t)s
(2s + t)t2m2 J (2t s)s22 (
+ J(s)+ J(t)+
(s + t)28 (S t)2t
2(2S2 t + 2t2) 2 ( S+ t)
st 6


A(g+, g+, g+, g-; q),
K- i-Jp [ (2t+ )tm2 42
-2 (1
K43K32K2V1K (S +t)s S
2(2s + t)t2m2 2(2t + s)s2m2
+ 2J(s) J(t) +
(s + t)2 (S +t)2t
4(22 St + 2t2 ) M2 ( + t)
st 3


I(s)


stm2
2(s+ t)2


)I(s) -

stm2
(s t)2


(2s + t)sm2
2((1
2(s t)t


4m2
)I(t)
t


2 2( t)K(s,t)
st


(2s+ t)sm2 4m2
(1- )I(
(s +t)t t

(1 2 2 )K(s, t)
st


(6-12)


In contrast to the massless case, the external leg factors are included in the results

below for the helicity conserving amplitudes, since we can perform wave function

renormalization now.


A(g+, g+, g-, g-; s)
-2K^ 1p [i mn2
KA KA KA KA ++ 2s
K43~32K21 14PI p 2s
2m2 4m2 1 1
--+ +- -lo
s 3t 18 6


2m2
3t

g 6e'm2


2m4
st
at


4m4 1
-2+ )I(t) +
3t2 1


1 X
6 3


A(g+, g+, g-, g-; q),
-2K^ 1 + m2
-2 -1
-2. KA K A K A K A ,,+ (
K43 32K21 14p p1 2
4m2 8m2 19 2
--+ ++- log
S 3t 9 3


2m2
3t

7ym2 -_


4m4 8m4
-^+
st 3t2
2 4
+ -X
3 3


)I() (-
3


m2 2m4
---+ )K(s, t)
s s2


(6-13)


t)


M74
-K(s, t)
S











A(g+, g-, g+, g-; s)m = ^2KIo 1P +
'K43 K32K( 21'p 14p 13
m t2(17t + s2 + 2) 2tm4(2t + 5s) t(5st 2s2 + t2
6(s + t)3 3(s +t)282 12(s +t)3
sm2(17st+ t2 482) 2sm4(5t + 2s) s(5st 2t2 + s2)
6(s+ t)3t 3(s+ t)2t2 12(s t)3
2stm2 s2t2 2stm2 m4 s2t2
+(t )(J(s) + J(t)) + (- + t+ )K(s, t)
(s + t)3 (s + t)4 (s + t)3 (s + t)2 2(s + t)4
2m2(2t2 2s2 + 3st) s2 +t2 1+ 1 1 2 1 1
log em2 X -
3(s + t)st 18(s +t)2 6 3 6 3

A(g+, g-, g+, g-; q), = -2 -^2K^i P4
K43K32 K 21 Ki4Pp 3
tm2(5s2 2t 5st) 4tm4(2t + 5s) t(5s2 2t2 + st)
3(s + t)3s 3(s + t)2s2 6(s +t)3 )(
(tm2(5t2 2S2 5st) 4sm4(5t + 2s) s(5t2 + 22 +st)
3(s +t)3t 3(s +t)2t2 6(s+ t) )
24stm2 stS2_ 2-)2 24 St( +t 2))K 2 ,t
+( 4t + T (J(s) + J(t)) + ( ( t) + TM + t)K t)
(s t+ )3 (s + t)4 (s t)3 (s t)2 2( t t)
4m2(2t2 82 + 3st) + 1982 + 19t 47t log e + (614)
3(s t)st 9(s + t)2 3 3 3 3

The last numerical factors -1/3 and -2/3 inside each square bracket is due to the
amputation of external legs: [limp20o In(p2)/p2]1/2.
6.5 Photon Photon Scattering
The amplitude of photon-photon scattering can be obtained from the above results
fairly easily, all we need to do is to replace g4Tr[tatbtctd] with e4 and sum all the crossings.
The reader might think that we should also remove the triangle diagrams from the
amplitudes since these diagrams involve tri-gluon vertices which are absent in an abelian
gauge theory. But, these diagrams will automatically cancel each other when we sum over
all crossings. This cancellation goes by the name of U(1) decoupling. The amplitudes are
listed in the appendix.
When we sum over all the crossings, the counter term that is proportional to the
four point vertex will vanish, while the pure number 2/3 will become 2, which has to









be subtracted to restore gauge covariance. We obtain the following photon scattering

amplitude:


A(+, +,+,


2m2
u


ie4 4K4^ K K^K^ KA 4
4 43 K32 21 14 2 4
S472 p P2 P3 Pt S[
ie4 stKi .l p +
472 2u K43 IK32V K2` K4 +
4m4- ( t 2m2 4 m4
SK(s, t) + -- --
st I I t su


ie4 -t ^ p+
A(+, -) +
2 SK43 K32K21K14P1 P

s t s Is u
+8m2 42 4st 2s2 (t) + ,J()r
S S 2 [ ) ] [


4m4 2t2 + 2st +
tu+ s2


ie4 -stKI^J p
it 4 -- ___ b 'i o 1'4 P4+
472 u2K4KA KKA A +
t 43 V32 ) 21 [14n 1 43
2t + [8m2 4m2
1 I(t + +
u a s


A(+,-,+,-)


S8m2
u +


4m2
t


a4K
K(s, t)
t


4 4
- K(s, u)
su


4M4 K(t, u)]
tu I


4m2 4m2 4m2] ( J(t)
+ +- [J() + J( + J)]
S t u I


K(s, u) +


2m2
s


---1 I(u)


2t 4st I ,

K(t, u) 2


+ -[ t2 4ut 2U2[J J(s) +] K(u, t) +
u a2 t ut
[2(4t2+4ut + 2)m2 4m4 2t2 2ut+ ,2] 1
+ stu ts + 2 K(t, s) -2
stu is at


2m2r 4m4] K(S U)


(6-15)


The spinor structure above has been set up to be uni-modular and invariant under

crossings of two legs. The results here agree with [27] and [31]. Note that in the first

term of Eq.(127.18) of [31], the authors seemed to have left out terms of -4/s + 2/t

and -4/s + 2/u in the coefficient of B(t) and B(u) (their B function is effectively our

I function) respectively, as Eq.(127.18) will not lead to Eq.(127.20) without those two

terms.


2(4t2 4st + 2)m2
stu


44 ]K(t, u)
tu I


2m2 4m4 ]K(
u su


1]I(s)









CHAPTER 7
CONCLUSIONS AND FUTURE WORK

7.1 Conclusion

To conclude, I have studied the renormalization of gauge theory on the light cone

world sheet. In order to do so, I computed all the four point amplitudes in gauge theory.

The box reduction technique was developed to extract the artificial divergences and IR

divergences. The artificial divergences are the rational functions containing 1/q+ poles,

they come from the light cone gauge propagators and were shown to cancel in a gauge

invariant quantity. The IR divergences, regulated by setting q+ away from zero, were

combined with the Bremsstrahlung contributions to give a Lorentz covariant scattering

cross section. The calculation of Bremsstrahlung contributions was also done in the light

cone fashion to facilitate the comparison with the virtual processes. A mismatch in the

rational parts of the amplitudes were prevalent, the restoration of gauge covariance was

only partially addressed. Finally, the scattering of gluon by gluon with general massive

matter was computed for completeness, and the light by light scattering amplitudes were

obtained along the way.

Next I give some discussion on some unresolved issues and an outlook for the future

work.

7.2 Restoring Gauge Covariance in the Light Cone

In [13, 14, 15], we insisted on only allowing counters terms that are polynomials in

the target space. Thus, when we saw that there was a hanging four-point vertex in an

amplitude, we could not put in a four point vertex as a counter term, but instead we

modified the self-energy by a term constxp2 to adjust the strength of the exchange diagrams.

But this scheme does not restore the gauge covariance for all the itimplitilde-,.

Here I -1.._ -I a new system of putting in counter terms. Diagrams such as II(g+, g) enter

the amplitudes as 'double quartic' graphs, similarly F(g+, g-,g+) enters as quarticc s\-, i.]-li'

diagrams, both of which are treated as 1PIR graphs in the canonical light cone fltrmili.itm.









Treating them as 1PIR graphs forbids us to adjust their strength, for they usually contribute a
multiple of four point vertices, which is not a polynomial in the moment.
In Chapter 4, I compared I(g+, g-; s) with 1(g+, g+; s) and I(g+, g-; q) with n(g+, g+; q).
Here I give the result again:



I(g+, g-; s) p2 1--- logp2ej


\9 9 1-
n(g+, g+; s) = +2 0 -- log 2


1(g, g-; q) = 2 r 0log p2e
Ir(gog;eq) = p22

The non-rational parts of the self-energy contributions are the same regardless of the index on
HIP". We have also reason to .. ii '. .- that, when there is no infrared divergence, the rational part
should also match due to Lorentz covariance. More -1"' ii. illy, vacuum polarization should be
of the form JP" = (gPp2 p/p,) H(p2). This tells us that


I(g+,g-) = evp2n2p2) -p2np2)
H(g+,g+) 0
1(9g+, g+) = _p+2 (p2)

Hence, apart from the factor of p2 and p+2, 1(g+, g-) should be equal to HI(g+, g+). Therefore,
we have to invoke counter terms to force this equality. I chose to associate to each I(g+, g+; s)
a term -1/6, and to n(g+, g+; q) a term 2/3, I cannot quite find what is the correct value for
I,(g+,g+;g) because of the infrared divergence. So I simply defined it to be -1/3, chosen such
that these counter terms vanish with A = 4 SYM field content. These counter terms are going
to affect the four point vertex that is derived from the exchange diagrams.
The gluon vertex correction diagram has a similar problem:












F(g, g-, g; s) = K2 1 10-ogp 7-



2+ + 4 32
F(g+, g-, g+;) K2Pa 0log po 6e -
P 13 9
F(g+, g-, g+;q) -3 K2,1 logp PO


F(g+,g-,g+;q) = -(p p2) log P e6 9

The logarithmic piece matches, so I will associate to F(g, g, g; s) a term -1/6, to F(g, g, g; q) a

term 2/3 and to F(g, g, g; g) a term -1/3 to enforce the total agreement between F(g+, g-, g)

and F(g+, g-, g+). Again, the number for F(g, g, g; g) is hand picked so that there is no need

for counter terms in = 4 SYM. These counter terms will affect the strength of the exchange

vertex.

There are also mismatches between F(s,s, g+) and F(s, s, g ), which we cannot easily

determine due to the IR divergence, so we throw in two arbitrary numbers and adjust these

two numbers to make all the amplitudes work. The numbers are determined to be -1/2 for

F(s, s, g+) and 0 for F(s, s, g+) and F(q, q, g).

Note that all of the modifications above can be achieved by ]" l1, 1,m i ,ls in the external

moment1 I shall report how this counter term system is working out. First, I list all am-

plitudes in Table.7-1, note that by s-pt I mean the 4 point vertex that is derived from the

s-channel exchange diagram.

The effect of the old regularization scheme is listed in Table.7-2.

We see by comparing 7-1 and 7-2 that all the bosonic amplitudes are fixed as long as there

is only one species of scalar, while fermionic amplitudes generally have problems.

The new scheme gives Table.7-3.

By comparing 7-1 and 7-3, all the mismatches are fixed. But of course, this scheme answers

as many questions as it raises. The loose threads include how to determine the counter term



1 the structure (-2p, /p p) K2 doesn't look like a polynomial, but actually its p+
dependence comes from the polarization vectors









for those diagrams that have IR divergences, and whether they vanish in A = 4 SYM and
how to realize them on the light cone world sheet. Finally, I want to point out an interesting
observation that whenever a genuine four point vertex (meaning not derived from an exchange
diagram) exists, the amplitude will have a constant mismatch.

7.3 Triangle Anomaly

Although I have stuck to the adjoint representation, in which the cubic invariant dabc is
zero, it is still necessary to look at how anomaly calculation turn out in the light cone. The first

b,p2^ k b, p2
k3 I k3 I
k2 ck k2 ck
1 1
a, p, A a, pl,A
Figure 7-1. Triangle anomaly


diagram of Fig.7-1 is given by


Tr [bitaic] Tr [62 o(q k 2) ( 2 )(q ki) a(k3 ki) (q k3) ]
(-i)(q k22(i)(q )2(i)(q k3)2
fTr [tbtac 2(rql(q k2)lki k2](r(q ki)(k3 ki)(q k3)lk2 k3]
iTr ] (-i)(q k22(-i)(q k)2(-i)(q k)2
-2Tr bac (71(q k2)|ki k2]
[tbtat,] K~(V k2) k1 k2]
S (rq(q ks)k2 k3] (rq(q k1)k2 k] (71)
(-i)(q k2)2(-i)(q k3)2 (-i)(q k2)2(-i)(q k1)2

After the usual steps of momentum integrals:



S- { (162 X3(k3 k2)lk k2](riX2k2 + 3k3 1Lk2 k3]

+(l|2(k2 kl)lk2 k3](7?lxiki 22k2 |k2 k3]
i 2K1
2p (p+(k2 + 2ks3)V p+(ki + 2k2)v) (7 2)
The second diagram of Fig.7-1 is similar:
The second diagram of Fig.7-1 is similar:











iTir btac] ,Tr [62 7(k3 q) (k3 1k) (ki q) l a(k2 q) ]
S(-i)(q k)2(-i)(q k)2(-i)(q k3)2
SiTr [ l 2(q|(k3 q)(k3 ki)(ki q)lkl k2](lq(k2 q)lk2 k3]
S(-i)(q k2)2(-i)(q k)2(-i)(q k3)2
2 Tr [t' j (l(k2 q)k2 -k3]
(q l|(ki q)lki k2] (lq(k3 q)k k (73)
(-i)(q k2)2(-i)(q ki)2 (-i)(q k2)2(-i)(q k3)2 (73)

After the momentum integral:


-idx:1-
1672{ (1Xl(ki k2)k2 k3](likil X2k2 |kl k



i _2K
+(Kx2(k2 k3)lkl k2](rqJx2k2 + X3k3 _Ik2 k3]

S2K 2(p (k2 + 2kl) + p(k3 + 2k2)) (7-4)
167r2 3p+P 1

So the sum of these two diagrams gives


i r { 97 2K 2
T- r {LK 12 (p (k2 + ki) p + p(k3 + k2)V) (7-5)
8- L i P1 p2

The right handed fermion gives the same contribution as above (the Feynman diagram

itself gives a negative sign relative to the left handed contribution, while the trace factor gives
a second negative sign). But if we put t' to be 1 and study the axial U(1) current (commonly
known as the abelian anomaly), the divergence of this current will be twice the above result.
Or we can keep the theory chiral, and look at the divergence of the chiral current (know as the
non-abelian anomaly) only. Note that the difference between these two cases is of order A3,

which will not show up here.
*F A F is of the form K12K12/(p1 p ) and 0 if the two gluons have different helicity. So

some subtractions have to be made for Eq.7-5 to be of the correct form (again, due to the
regulator we used, Eq.7-5 depends on each dual momenta. I haven't shown the calculation
for the case when the two gluons having different 1, 11. iv, the result is non-zero, so it has to be
subtracted by counter terms too. Here, we see that the counter terms associated to three-point










function are populating fast, there may or may not be an economical choice of counter terms)

that will take care of all these problems.

7.4 Two-Loop and n-Point Amplitudes

The commonly adopted method for calculating one loop n-point amplitude is to use

(generalized) unitarity [25, 26] to calculate box-coefficients, and to use recursion relations to

recycle old results. And the results can be checked partially by looking at its collinear limit soft

limit and multi-particle factorization properties.

We have basically calculated the four particle amplitude in the light cone gauge by

brute force, should there be a need to obtain amplitudes with more legs or even higher loop

Lnmplitlude.,, to way to proceed is certainly not brute force.

Let us focus on the A = 4 SYM theory as a first step. We know that in this theory,

all integrands can be reduced to scalar boxes. This is very useful if we are using dimension

regulation since the IR (or collinear) divergences are regulated. But we'd like to stick to our IR

regulator, so scalar boxes have to go through some more subtractions to become infrared safe. I

don't know yet how is this going to tell on the procedure of computing box coefficients.

The IR divergence is local in the sense that it is present whenever there is a four-point

MHV subtree in the loop diagram, so the one loop IR structure we found in Chapter 5 will

persist into a higher point amplitude. This feature perhaps can help us to define an IR safe part

in an amplitude and 'bootstrap' it to a larger amplitudes.











Table 7-1. List of mismatches in all the amplitudes


Amplitude


A(g+,g+, g-, g-;s)
A(g+, g+, g-,g-; q)
A(g+,g+, g-, g-; g)
A(g+, g-,g+, g-; s)
A(g+, g-,g+,g-; q)
A(g+, g-, g+, g-;g)
A(s, s, g+, g-)
A(g+, s, g-, s)
A(s, s, s, s)
A(q-,q-,q-,q-)
A(q-,q-,q-,q-)
A(s, s, q-, 4-)
A(g+,g-,q-,q-)
A(g+, g-, q+, q+)
A(g+, q-,g-, q-)


s-exch
0
0
0
0
0
0
0
0
0

0
0
-1/3
-1/3
0


t-exch
0
0
0
0
0
0
0
0
0
0
0

0
0
0


s-pt t-pt
-1/6
2/3
-1/3
-1/6 -1/6
2/3 2/3
-1/3 -1/3
-1/2

-1/2 -1/2
-1/3
-1/3 -1/3
-2/3 -
-1/3 0
-1/3 0
0


Table 7-2. List of the effect of the old counter terms schemes


Amplitude


A(g+,g+, g-, g-;s)
A(g+, g+, g-,g-; q)
A(g+,g+, g-, g-; g)
A(g+, g-,g+, g-; s)
A(g+, g-,g+,g-; q)
A(g+, g-,g+, g-;g)
A(s, s, g+, g-)
A(g+, s, g-,s)
A(s, s, s, s)
A(q-,q-,q-,q-)
A(q-,q-,q-,q-)
A(s, s, q-, 4-)
A(g+,g-,q-,q-)
A(g+, g-,q+,q+)
A(g+,q-,g-,q-)


s-exch
-1/6
2/3
-1/3
-1/6
2/3
-1/3
-1/2
-1/2
-1/2

1/3
1/6
1/3
1/3
0


t-exch
-1/6
2/3
-1/3
-1/6
2/3
-1/3
-1/2
-1/2
-1/2
1/3
1/3

0
0
0


s-pt t-pt
- 0
- 0
- 0
0 0
0 0
0 0
0 -

0 0
- 0
0 0
0 -
0 0
0 0
- 0
0


const
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0


const
1/2
-2
1
0
0
0
1/2
2
0
0
0
0
0
0
0

























Table 7-3. List of the effect of the new counter terms scheme


Amplitude


A(g+,g+, g-, g-;
A(g+, g+,g-, g-;,
A(g+,g+,g-,g-;:
A(g+, g-,g+,g-;
A(g+, g-, g+,g-;
A(g+, g-, g+, g-;
A(s, s, g+, g-)
A(g+, s, g-,s)
A(s, s, s, s)
A(q-,q-,q-,q-)
A(q-,q-,q-,q-)
A(s, s, q-, 4-)
A(g+, g-, q-,q-)
A(g+,g-,q+,q+)
A(g+,q-,g-,q-)


s-exch
-1/3
4/3
-2/3
-1/3
4/3
-2/3
-1
-1
-1

0
-1/2
1/3
1/3
0


t-exch
-1/3
4/3
-2/3
-1/3
4/3
-2/3
-1


0
-1
-1
0

0
0
0
0


s-pt t-pt
-1/6
2/3
-1/3
-1/6 -1/6
2/3 2/3
-1/3 -1/3
-1/9


-i

-1/2

1/3
1/6
1/3
1/3


-1/2
1/3
1/3

0
0
0


const
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0









APPENDIX A
SPINOR NOTATION IN THE LIGHT CONE


0 (TP
a 0
ab ab
Cabpb pa
:p- \P}:


a := (I, a) a := (I,-a')


ba
pa r/' :-P (/ =Pa


(A-l)


So far the spinor notations are common to all, and in the last line of Eq.A-1, I have

conformed to the 'hep-ph' notation: Ip] is assigned a lower index while Ip) an upper

index.

In light cone, unlike what we had in Section 2.1, the reference spinor is fixed to be


1 0
0 -(A2)

We can define the light cone version spinor as


p^ 1
p= P P = p (A-3)
1 p

The spinors satisfy the Dirac equation if p is light like. Note that they don't have the

correct normalization, namely p a"" 7 pc'p, but they have the merit that p = (-p)".

The polarization vectors of gluon can be written as


pA p\P (j Pij Pi

[ l I i I pj> j I p i


-V 2\P ]{I( \, C vd]-- -I --v / ]
Sp[pj' [p Ipj
v /2j p Ap[p IPiv

rl] -+ Kji, rl\PiP I J [PJ Pi I } + --Kp i
Pu Pu


(A-4)



(A-5)


The Kij's satisfy










o
i+P 1 0
pfK +p K p K =
KlKk + KlkK + K Kk = 0
K A "V 2
z KUjk + +Y (
pE pp (A-6)

The third line of Eq.A-6 is called the Schouten identity: [ij][kl] + [jk][il] + [ki][jl] = 0.
In the current case we are dealing with, i, j run from 1 to 4, but only two of the six
Kiy's are independent, say K43 and K32. And any product of Kyi's with total helicity 4
can be reduced to either (K43)4 or (K4A3)K32. Product of helicity 2 can be reduced to

(K43)2 and K43K32. Product of helicity 0 can be reduced to 1 and KA2K43. The reduction
is in general a formidable task for human, but quite a piece of cake for computers, as all
our calculations are done with computers.









APPENDIX B
FEYNMAN RULES

We remind the reader that g + (-) corresponds to a A (V) on the gluon line, q(q)

corresponds to an incoming (outgoing) fermion line, q + (-) corresponds to right(left)

handedness, s corresponds to a scalar.

The fermion gluon vertex Fig.B-1 is given by


-2g(ta)bc P K p ,q V(q+, q+, g+)

-2ig(ta)bc-Kp, V(q+, q+, g-)
q+ 2,q


1
S-2ig(t)bc--!K

-2ig(ta)bc~ Kq(B -1)
q~ p1~


The gluon scalar vertex Fig.B-2 is given by


1
V (g + (-))= 2ig(ta)b K

Here we have used real scalar fields and hence it transforms in a real representation.

The gluon fermion 4 point vertex Fig.B-3 is given by


V(q-, q-,g+, g-)

V(q+,q+, g+, g-)

V(q-, q-,g-, g+)

V(q+,q+, g-, g+)


-2 2(td) c(ta) eb 22f dae(e)cb P4 )P2
p+ + 2g f (t )cb (p +-)2
S+( + ++
2fdaete (P3 ~ P4 )P2
2g fdae (e)b (+ +)2




2(td) cea eb 22fdae P4 )P2 (B-3)
-21g+pt + + 2 (te)cb (+ + P ()2
Pi 4 W + P4p~)


The gluon scalar four point vertex Fig.B-4 is given by



V(g+, g-, s, s) V(g-, g, s, s)
ig2 [(ta)ce(tb)ed (tb)ce(ta)ed] 2fabe (t)cd +)( +

V(g+,g+, s, s) = V(g-, g-, s, s = 0


V(q-,q-,g+)

V(q-, q-,g-)


(B-2)


(B-4)









For tri-gluon vertex Fig.4-3



V(g, g9 ) -gfabc[. (P1


Setting p, v = A, p = V, the above becomes


V(g+, g+, g-)


gfabc[(- -p )+
+ Pi iK


The gluon four point vertex receives contribution from two sources: the left diagram

in Fig.B-6 is simply the covariant four point vertex:


Vi = ig2 {abe fecd [gpqp


*q Pgv] fdaefebc[JVpy(P


qpgV] + fcaefebd[ypVaPJ


(B-6)


The second is obtained by shrinking a propagator.


V2 = 2 fdae ebc[ 4
V2 9 1e C4(P4


p ](p+ I+ p+)2(p 2 2 p, (
1 4 +pmP4p^' 1'


I' ,)1]


. 2fdaefebc (1 4*)(* )(P4 Pl)(P2 '
(pt + t+)2


There are two cases in which the gluon four point vertex is nonzero:


V(g+, g+, g-, g-)

V(g+, g-, g+, g-)


2ig2

2ig2 [


+ +
f dae ebc P + PiP
(p + +P) (P + pt)
fabe fecd 2 + P+ P4
( 2 Pa + )3++ 4)


Sface febd P3P P 1 1Pi
+ + ++) (2 4
fdaefebc P + P P3 P4
(P +P4)(P2 +P3 7)
(B-7)


When using these vertices, we need to watch the indices of structure constants closely: not

all terms are going to make contributions to Tr[tatbtctd].


pi) C]


Pi)
P2


('. 1)A]


(B-5)


'_-) +3 2 3(/'-


*. ) 1 + C ',.


(/.-_ )A + (.









The fermion four point vertex comes from contracting a pair of three point vertices

connected by a gluon propagator. It is given by (for configuration of Fig.B-7)


V(q, q, q, q)


++
P3 P4
-4i2 (t)da(t),b (- p+ +)2


(B 8)


with the obvious restriction that fermion line 1 and 4 having the same handedness while 2

and 3 having the same handedness.

The scalar four point vertex Fig.B-8 comes from contracting a pair of three point

vertices connected by a gluon propagator. It is given by


V(S, s, S, S)


(p+ + p +)
- (t)ah ) cd ---- ( ++ + -
(pTi 2)


The scalar fermion four point vertex Fig.B-9 also comes from contracting a gluon

propagator:


(B-10)


2pp p+ )
V(s, s, q, q) = ig2(te)be )cd 2 p+


with the restriction that the fermion lines having the same handedness.

A, a V a
q



c, Pi b, P2 c, pi b, P2

Figure B-1. Gluon-fermion-fermion 3 point vertex


(B-9)









p a, q



b, j ^/ c, Pc



Figure B-2. Gluon-scalar-scalar 3 point vertex

a, p4
a, p4
a, 4 1' C


b, pi


C, /*'


b, Pi C, '-


Figure B-3.


Two diagrams contribute to the fermion-gluon 4 point vertex. Note these
are not exchange diagrams but rather four point vertices gotten through
cancelling a 1/p2 factor in a fermion propagator or a gluon propagator, see
Eq.2-25


A, b, p4 V, a, r,


c, pl


A, b, p4 V, a, ',



\JI


d, p2


Figure B-4. Scalar-gluon 4 point vertex. Again, the second diagram is not an exchange
diagram but rather a four point vertex obtained through shrinking a gluon
propagator

p, C, 1,





i-, a, pi l, l '_

Figure B-5. Tri-gluon vertex











a, d, P4 p, c, P. a, d, p4 p, c, ',





/I, a, pl v, b, p2/ a, Pi V, b, 1'


Figure B-6. Gluon 4 point vertex


a, pi


b, '.


Figure B-7. Fermion 4 point vertex


P4 C, ,
"\ /,"


a, pi


b, I.


Figure B-8. Scalar 4 point vertex


C, ,


a, pi


b, i/


Figure B-9. Scalar Fermion 4 point vertex








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[1] G. 't Hooft, Nucl. Phys. B72 (1974) 461. J. M.

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G. Chalmers and W. Siegel Phys.Rev.D59:045013,1999. [arXive:hep-ph/9801220]
G. Chalmers and W. Siegel Phys.Rev.D59:045012,1999 [arXive:hep-ph/9708251]
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C.F. Berger, Z. Bern, L. Dixon, D. Forde, D.A. Kosower, Phys Rev D74, 036009
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[27] R. Karplus and M. Neuman Phys.Rev. 83, 776 (1951)[arXive:hep-ph/0604195]

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[31] V.B.Berestetskii, E.M.Lifshitz and L.P.Pitaevskii Quantum Electrodynamics Landau
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BIOGRAPHICAL SKETCH

Jian Qiu was born in 11.2.1979 in Shanghai, China. He was raised up in a small

town Le-Ping in the Jiang-Xi province, where he finished the first two years of elementary

school. He moved back to Shanghai in 1988 and went to elementary school, middle

school and high school there. From the sixth to the twelfth grade, he also took part in

competitions in remote controlled race car.

After graduating from Shi-Er high school in 1998, he was admitted to Fudan

University to study 11li, -i. He was first interested in Nuclear 1.li, -i. -. then slowly

shifted his interest to theoretical p lli-i -, and mathematics. He then graduated from Fudan

University in 7.2002 with the bachelor's degree.

In the same month, he left China and came to the University of Florida to pursue his

graduate study in pil -i. And he has been in the Physics Department ever since.





PAGE 1

RENORMALIZA TION OF GA UGE THEOR Y ON THE LIGHT CONE W ORLD SHEET By JIAN QIU A DISSER T A TION PRESENTED TO THE GRADUA TE SCHOOL OF THE UNIVERSITY OF FLORID A IN P AR TIAL FULFILLMENT OF THE REQUIREMENTS F OR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORID A 2007 1

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c r 2007 Jian Qiu 2

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to Lisa Elk enhans 3

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A CKNO WLEDGMENTS My greatest gratitude to w ard m y advisor Professor Thorn for sparing no eort in helping me obtain deep er understanding of ph ysics and man y a timely encouragemen t to forge on. 4

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T ABLE OF CONTENTS page A CKNO WLEDGMENTS . . . . . . . . 4 LIST OF T ABLES . . . . . . . . . 7 LIST OF FIGURES . . . . . . . . . 8 ABSTRA CT . . . . . . . . . . 10 CHAPTER 1 INTR ODUCTION . . . . . . . . 11 2 COMPUT A TION TECHNIQUES IN THE LIGHT CONE . . . 18 2.1 Mini In tro duction to Spinor Helicit y Amplitude Metho d . . 18 2.2 The Ligh t Cone Setup . . . . . . . 23 2.3 Brief Description of the Calculational Pro cedure . . . 26 3 BO X REDUCTION . . . . . . . . 31 3.1 Bo x with a Helicit y Violating Sub-diagram . . . . 31 3.2 Bo x without a Helicit y Violating Sub-diagram . . . . 37 4 MASSLESS AMPLITUDES . . . . . . . 45 4.1 Tw o P oin t F unctions . . . . . . . 45 4.1.1 Gluon Self-Energy . . . . . . . 45 4.1.2 F ermion and Scalar Self-Energy . . . . . 48 4.2 Three P oin t F unctions . . . . . . . 48 4.2.1 Gluon V ertex Correction . . . . . . 49 4.2.2 F ermion V ertex Correction . . . . . 50 4.2.3 Scalar V ertex Correction . . . . . . 51 4.2.4 F our-P oin t F unctions . . . . . . 51 4.3 Scattering Amplitudes . . . . . . . 51 4.3.1 Helicit y Violating Amplitudes . . . . . 52 4.3.2 Helicit y Conserving Amplitudes . . . . . 53 4.3.3 Restoring Gauge Co v ariance . . . . . 56 4.3.4 All 2-2 Pro cesses . . . . . . . 57 5 BREMSSTRAHLUNG IN LIGHT CONE . . . . . 61 5.1 A List of Infrared T erms . . . . . . . 61 5.2 Bremsstrahlung Pro cess . . . . . . . 63 5.3 Com bining with the Infrared T erms from the Virtual Pro cess . . 74 5.4 The Inclusion of Disconnected Diagrams . . . . 76 5

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6 GLUON SCA TTERING WITH MASSIVE MA TTER FIELDS . . 81 6.1 Computation T ec hnique . . . . . . . 81 6.2 Self-Energy Diagrams . . . . . . . 84 6.3 T riangle Diagrams . . . . . . . 84 6.4 Scattering Amplitudes . . . . . . . 85 6.5 Photon Photon Scattering . . . . . . 87 7 CONCLUSIONS AND FUTURE W ORK . . . . . 89 7.1 Conclusion . . . . . . . . . 89 7.2 Restoring Gauge Co v ariance in the Ligh t Cone . . . . 89 7.3 T riangle Anomaly . . . . . . . . 92 7.4 Tw o-Lo op and n-P oin t Amplitudes . . . . . 94 APPENDIX A SPINOR NOT A TION IN THE LIGHT CONE . . . . 97 B FEYNMAN R ULES . . . . . . . . 99 REFERENCES . . . . . . . . . 104 BIOGRAPHICAL SKETCH . . . . . . . . 106 6

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LIST OF T ABLES T able page 7-1 List of mismatc hes in all the amplitudes . . . . . 95 7-2 List of the eect of the old coun ter terms sc hemes . . . . 95 7-3 List of the eect of the new coun ter terms sc heme . . . . 96 7

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LIST OF FIGURES Figure page 1-1 Double line notation . . . . . . . . 16 1-2 Op en string scattering diagram . . . . . . 16 1-3 Ligh t-cone parametrization . . . . . . . 17 2-1 Fiv e-p oin t amplitude . . . . . . . . 28 2-2 Some building blo c ks that can b e computed separately . . . 29 2-3 Second lev el building blo c ks . . . . . . . 29 2-4 T ri-gluon v ertex . . . . . . . . . 29 2-5 Propagator in the t-c hannel is con tracted . . . . . 29 2-6 F ermion-gluon v ertex . . . . . . . . 29 2-7 F ermion propagator con tracted . . . . . . 30 2-8 Here w e assume only ( k 2 k 4 ) 2 6 = 0, and k + 4 > k + 1 > k + 2 . . . 30 3-1 Dual momen tum assignmen t with k + 2 < k + 3 < k + 1 < k + 4 . . . 43 3-2 Mo del b o x with a helicit y violating subtree . . . . . 43 3-3 Graphical represen tation of b o x reduction . . . . . 44 3-4 Mo del b o x with alternating helicit y . . . . . . 44 4-1 Tw o-p oin t function and the dual momen tum assignmen t . . . 59 4-2 A self-energy diagram em b edded in a scattering pro cess . . . 60 4-3 T riangle diagram . . . . . . . . 60 4-4 F ermion v ertex correction corresp onding to Eq. 4{9 . . . . 60 4-5 Scalar v ertex correction corresp onding to Eq. 4{11 . . . . 60 4-6 F our-fermion scattering corresp onding to Eq. 4{20 and Eq. 4{21 . . 60 5-1 Dual momen tum assignmen t . . . . . . . 78 5-2 Tw o diagrams with an extra 'unseen' gluon . . . . . 78 5-3 Cancelation of collinear div ergence . . . . . . 78 5-4 Cancelation of the soft bremsstrahlung radiation against a virtual pro cess . 79 8

PAGE 9

5-5 Phase space in tegration region of x and y . . . . . 79 5-6 Conguration of ~ k ~ p 3 and ~ L in the CM frame of p 1 + p 2 + p 3 . . 79 5-7 Self-energy bubble on leg 4 . . . . . . . 79 5-8 All non-v anishing Bremsstrahlung pro cesses . . . . . 79 5-9 Disconnected Bremsstrahlung with t w o extra 'unseen' gluons . . 80 7-1 T riangle anomaly . . . . . . . . 92 B-1 Gluon-fermion-fermion 3 p oin t v ertex . . . . . . 101 B-2 Gluon-scalar-scalar 3 p oin t v ertex . . . . . . 102 B-3 Tw o diagrams con tribute to the fermion-gluon 4 p oin t v ertex . . 102 B-4 Scalar-gluon 4 p oin t v ertex . . . . . . . 102 B-5 T ri-gluon v ertex . . . . . . . . . 102 B-6 Gluon 4 p oin t v ertex . . . . . . . . 103 B-7 F ermion 4 p oin t v ertex . . . . . . . . 103 B-8 Scalar 4 p oin t v ertex . . . . . . . . 103 B-9 Scalar F ermion 4 p oin t v ertex . . . . . . . 103 9

PAGE 10

Abstract of Dissertation Presen ted to the Graduate Sc ho ol of the Univ ersit y of Florida in P artial F ulllmen t of the Requiremen ts for the Degree of Do ctor of Philosoph y RENORMALIZA TION OF GA UGE THEOR Y ON THE LIGHT CONE W ORLD SHEET By Jian Qiu August 2007 Chair: Charles Thorn Ma jor: Ph ysics W e calculated the scattering of gluon, scalar and quarks in gauge theory in the ligh t cone gauge. Some computation tec hniques suited for the ligh t cone gauge are in tro duced. W e observ ed some inadequacies of the coun ter terms suggested in our earlier w ork, and w e suggest a new w a y of xing coun ter terms using Loren tz in v ariance as a guide. Gluon scattering with massiv e matter elds in the lo op are presen ted for completeness. The helicit y amplitude metho d is extensiv ely used in this w ork and is also mo died to simplify the ligh t cone gauge calculation. 10

PAGE 11

CHAPTER 1 INTR ODUCTION Gauge theory is generally though t of as the fundamen tal theory that go v erns most of the imp ortan t in teractions in particle ph ysics. The standard mo del of particle ph ysics is based on a gauge theory with a gauge group S U (3) n S U (2) n U (1) minimally coupled to some matter elds. The electro-w eak part of this theory has relativ ely small coupling, and the p erturbativ e calculations pro duced results that agree with the exp erimen ts extremely w ell. The strong in teraction (QCD) has a coupling constan t of the size 0.1 at high momen tum (short distance) where the p erturbativ e calculation can still giv e some useful predictions. Ho w ev er, the coupling is of order one around 1 GeV and will tend to innit y at lo w er momen ta. The p erturbativ e calculations cease to mak e sense at this scale, y et it is at this scale that some in teresting phenomena happ en. F or example, at high energy QCD can b e describ ed b y quark and gluon elds, while at lo w energy the quarks b ecome conned and the description in terms of pions and bary ons is more relev an t. A t high energy the quarks enjo y an S U (2) L n S U (2) R symmetry but at lo w energy the axial v ector part of this symmetry is sp on taneously brok en do wn. These phenomena are b ey ond the grasp of p erturbation theory and the quan titativ e results mainly come from lattice computation. The quest for an analytic solution to the lo w energy sp ectrum or a dual description of QCD that suits the lo w energy and strong coupling has b een the fo cus of man y ph ysicists. Besides its phenomenological imp ortance, gauge theory is also in teresting in its o wn righ t b ecause of its close connection with top ology A gauge theory consists of a curv ature eld F satisfying the Bianc hi iden tit y dF = 0. In QED, the en tries of F are simply the electric and magnetic led E and B while the Bianc hi iden tit y is t w o of the four Maxw ell equations r ~ B = 0 and @ ~ B =@ t = r ~ E In general, the Bianc hi iden tit y sa ys the t w o form F dx dx is closed, so w e cannot simply v ary the eld strength to obtain the Euler-Lagrange equation. Instead, w e write F as dA for some 1-form A kno wn as the connection 1-form, and v ary A rather than F Although not all closed forms 11

PAGE 12

(forms satisfying d! = 0) are exact (forms can b e written as = d ), writing F as dA causes no problem in the p erturbativ e calculations, b ecause w e are only in terested in small ructuations around A = 0. Y et for a non-p erturbativ e computation, a non-trivial conguration of the A eld can giv e imp ortan t results suc h as the anomalous breaking of the axial U (1) symmetry In those cases, w e ha v e to man ually sum o v er congurations of dieren t winding n um b ers. The A eld, as its name suggests, is a connection in the gauge bundle, more concretely the 1-form A a t a dx acting on the tangen t v ector @ = @ =@ x con v erts it to a v ector A a t a in the direction of the bre (elemen ts of the Lie group translate a eld in the 'v ertical' direction). With this connection, w e can compare t w o elds in nearb y space-time p oin ts using the co v arian t deriv ativ e D = @ A The geometrical nature of the gauge theory is not explored in this dissertation, ho w ev er, the problem of gauge xing is still imp ortan t for this w ork. Putting aside gauge theory for a momen t, w e also ha v e a p ossible alternativ e description of the strong in teraction: string theory It w as initially prop osed in order to mo del the dynamics of the rux tub e whic h is the explanation of connemen t in the strong in teractions. In analogy with the p oin t particle, where the equation of motion is suc h that the path tra v ersed b y a particle(w orld line) in space-time is the shortest path, the b osonic string theory studies a one dimensional extended ob ject and the equation of motion is suc h that the t w o dimensional surface sw ept b y the string (w orld sheet) is a minimal surface. An op en string has t w o free ends, they can b e giv en the Chan-P aton degrees of freedom, whic h are rather lik e a lab el that lab els the dieren t states of the end of the string. These states can b e tak en to transform in a represen tation of a Lie group, sa y in the fundamen tal for one end and an ti-fundamen tal for the other. With this setup, the op en string is lik e a pion whic h has presumably a quark and an an ti-quark at eac h end and connected b y the rux tub e. In fact, the scattering amplitude of the op en strings repro duces the scattering of gluons in the lo w energy limit. 12

PAGE 13

The similarit y w as explored in [ 1 ], where 't Ho oft lo ok ed at the S U ( N ) gauge theory as N 1 In this limit, the adjoin t represen tation of S U ( N ) can b e appro ximated b y N n N The A eld transforming in the adjoin t represen tation, no w carries t w o indices A ij in the large N limit, where i and j transform in the N and N resp ectiv ely Hence a gluon line can b e dra wn as t w o lines carrying the i and j index eac h. With this represen tation, a F eynman diagram will lo ok lik e Fig. 1-1 The ij index corresp onds to the Chan-P aton factors. An y closed index lo op suc h as in the middle of Fig. 1-1 will b e a trace T r i j pro ducing a factor of N If a F eynman diagram is dra wn with crossed lines, w e will lose factors of N so b y k eeping only the leading terms in the p o w er of N w e pic k out all the planar diagrams. There is a clear analogy b et w een the planar diagrams and the op en string scattering diagrams Fig. 1-2 where the arro ws no w indicate the Chan-P aton factors. By summing o v er all the planar diagrams, w e hop e to learn something ab out the non-p erturbativ e nature of the gauge theory 't Ho oft also pioneered the ligh t-cone parametrization of these planar diagrams in [ 1 ]. The corresp ondence b et w een the t w o graphs in Fig. 1-3 is that, a rectangle maps to a propagator, a line (cut) maps to the blank space b et w een t w o propagators, and the b eginning(end) of a line is the splitting(merging) p oin t of t w o propagators, or simply a 3-p oin t v ertex. The success of this parametrization is b ecause in the ligh t-cone, a propagator con tains a step function ( x + p + ). This factor sa ys if a propagator is to propagate forw ard in x + it has to carry p ositiv e p + This parametrization is also used in [ 6 7 8 9 ]. The corresp ondence b et w een gauge eld theory and string theory w as rev olutionized b y the AdS/CFT corresp ondence due to Maldacena [ 2 ], a certain sup erstring theory on an AdS 5 S 5 bac kground is equiv alen t to the N = 4 sup ersymmetric gauge theory Detailed prop osals of this corresp ondence w ere made b y Gubser, Klebano v and P oly ak o v [ 3 ] and b y Witten [ 4 ]. Their idea is the holographic concept, where the Mink o wski space M 4 is the b oundary of the AdS space, and the correlators in M 4 are computed from the bulk of 13

PAGE 14

AdS. The eld theory and its p erturbativ e expansions grasp the w eak coupling limit, while the string theory description naturally grasps the strong coupling limit. F or example, in [ 5 ], the author studied the energy of a string hanging b et w een t w o static sources in the classical limit (in the con text of AdS/CFT dualit y classical limit in the string theory corresp onds to the strong coupling limit in the eld theory), whic h sho ws a force that ob eys in v erse square la w, in accordance with the fact that N = 4 SYM is a conformal theory Despite all the ab o v e, a lot of the details of the corresp ondence still a w ait lling in. So the establishmen t of the detailed corresp ondence b et w een string theory and gauge theory b ecomes urgen t. In [ 6 7 ], the authors prop osed a lo cal w orld sheet description of the (sup ersymmetric) gauge theory whic h maps a F eynman diagram to a w orld sheet represen tation. The v ertices in this description b ecome the merging and splitting of strings, and the summation of F eynman diagrams b ecomes a path in tegral on the (discretized) w orld sheet. The F eynman rules, or the v ertex functions w ere realized b y inserting a lo cal w orld sheet op erator close to the splitting or merging p oin t. There is no need for the represen tation of a four p oin t v ertex, b ecause it is automatically generated when t w o insertions for three p oin t v ertices coincide and pro duce a con traction term. This is rather fortuitous, for a four p oin t v ertex is v ery unnatural in a string diagram. The summation of all the planar diagrams b ecomes the summation o v er whether or not there is a cut in eac h of the lattice sites. This treatmen t still has man y lo ose ends to tie up. F or example, the problem of renormalization whic h requires not only a lo cal represen tation of the bare diagrams on the w orld sheet but also a lo cal represen tation of the coun ter terms. The problem of renormalization for scalar elds on the w orld sheet w as addressed in [ 8 ], but w as only partially solv ed for gauge theory in [ 9 ] due to some complications. It is absolutely necessary therefore, to study the renormalization for gauge elds on the w orld sheet, making sure that the w orld sheet path in tegral description at the least will repro duce all the p erturbativ e results b efore w e jump in to studying its 14

PAGE 15

non-p erturbativ e side. The c hec ks include, whether all the div ergences app earing in the p erturbativ e expansion can b e absorb ed in to target space lo cal coun ter terms, and whether these coun ter terms can ha v e a lo c al w orld sheet represen tation. And most imp ortan tly in the case of (sup ersymmetric) gauge theory ho w the regulators will aect gauge co v ariance and sup ersymmetry Since 't Ho oft's planar diagram represen tation w as most easily giv en in the ligh t cone parametrization, it is natural that w e c hose to study the gauge theory on the w orld sheet also in ligh t cone gauge. In this case, in order to study the gauge in v ariance prop ert y w e ha v e to deal with some collateral complications due to the ligh t cone gauge c hoice and also the issue of infrared div ergence whic h is inheren t to a gauge theory These tasks ha v e b een the main fo cus of m y w ork. In the t w o pap ers [ 13 14 ], w e studied the gluon scattering in the ligh t cone gauge in great detail and found the need for some coun ter terms that w ere quite unexp ected. An infrared regulator is also prop osed in [ 14 ] that is v ery dieren t from dimensional regulation y et still resp ects gauge co v ariance. In [ 15 ], the extension to the gauge theory coupled to general matter w as made, and some results that are p eculiar to N = 4 w ere observ ed. Also the cancelation of coun ter terms b et w een dieren t sp ecies w as ac hiev ed, whose imp ortance will b e explained later. As an extension, the scattering of quarks and scalars in addition to gluons w as computed, whic h sho w ed a need for some new coun ter terms not giv en in the previous w ork. The complete determination of these new coun ter terms is still a w ork in progress. And the usage of Loren tz co v ariance as a guide to x coun ter terms will b e initiated in the future w ork section. The organization of the dissertation is as follo ws, in Chapter 2 the helicit y amplitude metho d is briery in tro duced. This metho d, when mo died to suit the ligh t cone, can greatly increase the rexibilit y of ligh t-cone gauge, and oers an alternativ e w a y to obtain ligh t cone F eynman rules. Some computational details are also giv en in the same c hapter. In Chapter 3 I shall describ e the b o x-reduction tec hnique whic h o ccupied a bulk of 15

PAGE 16

the w ork done in [ 14 ], and is essen tial to our ac hiev emen t of the cancelation of articial div ergence and the extraction of infrared div ergence. In Chapter 4 I shall simply list the results, since the details of the computation are complicated and not v ery illuminating. Discussion ab out some kno wn features of sup ersymmetric gauge theory is giv en in Chapter 4 to o. In Chapter 5 the Bremsstrahlung calculation on the ligh t cone is presen ted, and the real pro cesses are com bined with virtual pro cesses to obtain an IR nite result. In Chapter 6 I shall giv e the scattering of gluon with massiv e matter in the lo op for completeness, and the photon scattering amplitude is also presen ted. I shall conclude this dissertation b y p oin ting out some remaining problems and the outlo ok for some future w ork. In the app endices, I shall sp ell out the spinor notations used in the pap er and all the F eynman rules obtained b y using the metho d of Chapter 2 Figure 1-1. Double line notation Figure 1-2. Op en string scattering diagram 16

PAGE 17

PSfrag replacemen ts 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 x+p+Figure 1-3. Ligh t-cone parametrization 17

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CHAPTER 2 COMPUT A TION TECHNIQUES IN THE LIGHT CONE 2.1 Mini In tro duction to Spinor Helicit y Amplitude Metho d This metho d exploits the co v ering group of S O (3 ; 1): S L (2 ; C ). Loren tz in v ariance will b e represen ted as S L (2 ; C ) in v ariance. By dotting a momen tum in to the sigma matrices, w e obtain a 2 b y 2 matrix: p p If p w ere ligh t lik e, w e will ha v e det( p ) = p 2 = 0, so the matrix p can b e decomp osed in to a pro duct of t w o spinors: p = p p (2{1) where w e ha v e used the same letter for b oth the momen tum and its spinor. p ( p ) is the left (righ t) handed spinor that satises the massless Dirac equation. The notation using brac k et is commonly used in the literature: p a j p i ; p a j p ]; p a h p j ; p a [ p j (2{2) P olarization v ectors for the gauge particles are also clev erly c hosen to minimize the computation [ 10 21 22 ]. The reader can refer to [ 11 ] for a review, and to [ 12 ] for some more simplications and extensions to massiv e elds. As an example, a left handed p olarization v ector can b e giv en b y (up to a normalization factor) a a = a p a j ] h p j or equiv alen tly a a = p a a j p i [ j The handedness can b e seen b y remem b ering that r or is the Clebsc h-Gordon co ecien t for the pro jection 1 n 1 2 L 1 2 R (2{3) and that j p i is the spinor of a left handed fermion. W e ha v e j p i = j ] h p j p i = 0. This means that the photon m ust b e left handed so the pro duct of it with a left handed fermion will not ha v e a righ t handed comp onen t. 18

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Another w a y of seeing the same fact is b y insp ecting whether the eld strength F constructed out of is an ti-selfdual (righ t handed) or self-dual (left handed). F = i [ p a a b p b p bb a p a ] = i h p a p b p [ a b ] i = ip a p b a b [ p j ] The sym b ol in the spinor language is giv en b y = a a b b cc d d = 4 i d a b a b c d c b a b c d c d a so 1 2 det g F = i p A = 0 a b p b p a [ p j ] = iF (2{4) hence F = iF The reference spinor can b e c hosen to our adv an tage. The most ecien t w a y is to pic k it to b e a spinor of one of the momen ta in the problem. As an illustration, consider the amplitude M (1 ; 2 ; 3 ; 4 ; ; ; + ; +). Pic k 1 j 1 i [4 j ; 2 j 2 i [4 j ; 3 j 1 i [3 j ; 4 j 1 i [4 j (2{5) They ha v e the prop ert y that only the pair 2 and 3 ha v e non-v anishing inner pro duct. These p olarization v ectors are not prop erly normalized y et, their normalization factors will b e put bac k in at the last step. Let us lo ok at the t-c hannel exc hange diagram: 1 4 ( p 4 p 1 ) + 1 (2 p 1 + p 4 ) 4 + 4 ( 2 p 4 p 1 ) 1 ig ( p 1 + p 4 ) 2 ::: (2{6) I did not b other to write do wn the v ertex on the righ t since the left v ertex is already zero. 19

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The four-p oin t con tact v ertex is in fact zero to o. Recall that the four-p oin t v ertex alw a ys in v olv es t w o pairs of p olarization v ectors dotted in to eac h other, while w e only ha v e one non-zero pair a v ailable. Finally the s-c hannel exc hange diagram: 1 2 ( p 1 p 2 ) + 2 (2 p 2 + p 1 ) 1 + 1 ( 2 p 1 p 2 ) 2 ig ( p 1 + p 2 ) 2 3 4 ( p 3 p 4 ) + 4 (2 p 4 + p 3 ) 3 + 3 ( 2 p 3 p 4 ) 4 (2{7) The underlined terms are all zero, and w e are left with 4 i s ( 2 3 ) ( p 2 1 ) ( p 3 4 ) = i 2 s [43] h 21 i [24] h 21 i [34] h 31 i (2{8) As the last step, w e need to put in the normalization factors for the p olarization v ectors: [14][24] h 31 ih 41 i = 4, I get 2 i [34] 2 h 12 i 2 st = 2 i h 12 i 4 h 12 ih 23 ih 34 ih 41 i (2{9) The second presen tation is holomorphic in the left handed spinors, whic h will turn out to b e in teresting later on. As a sligh tly more non-trivial example, I shall compute the amplitude M (+ ; + ; ; ; ) Fig. 2-1 I shall dene the p olarization v ectors to b e 1 = p 2 j 4] h 1 j [14] ; 2 = p 2 j 4] h 2 j [24] 3 = p 2 j 3] h 1 j h 13 i ; 4 = p 2 j 4] h 1 j h 14 i ; 5 = p 2 j 5] h 1 j h 15 i (2{10) 20

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Here I ha v e used p 1 and p 4 as reference legs. A general observ ation is that a v ertex that in v olv es b oth legs that are used as reference v anishes, so diagrams 2,3 and 4 are zero. A closer lo ok at the pro ducts b et w een a pair of p olarization v ectors sho ws that all the diagrams in v olving four p oin t v ertices are zero. W e are th us left with only three diagrams to compute. I will start with some small building blo c ks according to Fig. 2-2 : A = [54] h 41 i 4 [45] h 51 i 5 B = [35] h 51 i 5 [53] h 31 i 3 C = 1 2 [43] h 12 i ( p 2 p 3 ) + [43] h 32 i 3 [32] h 21 i 2 D = [42] h 21 i 2 [41] h 12 i 1 (2{11) With these building blo c ks, w e can calculate further larger blo c ks Fig. 2-3 : E = A 1 [ g (1 2 3) + g (2 + 3 4 5) + g (4 + 5 1) ] = 0 + A [4 j 2 + 3 j 1 i + 0 = A [4 j 5] h 5 j 1 i F = A 3 [ g (1 + 2 3) + g (3 4 5) + g (4 + 5 1 2) ] = 3 1 2 [54] h 41 i [4 j 2 3 j 1 i 1 2 [45] h 51 i [5 j 2 3 j 1 i + A [3 j 2 j 1 i = 3 [54] h 41 i [42] h 21 i [45] h 51 i [52] h 21 i A [3 j 2 j 1 i = 3 [54] h 21 i [23] h 31 i A [32] h 21 i G = B 4 [ g (3 + 5 4) + g (4 1 2) + g (1 + 2 3 5) ] = 0 + 4 1 2 [35] h 51 i [5 j 4 2 j 1 i 1 2 [53] h 31 i [3 j 4 2 j 1 i + 1 2 B [4 j 2 3 5 j 1 i = 4 [35] h 51 i [54] h 41 i [53] h 31 i [34] h 41 i + B [4 j 2 j 1 i = 4 [35] h 41 i [42] h 21 i + B [42] h 21 i (2{12) 21

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Finally E C = A [4 j 5] h 5 j 1 i 1 2 [43] h 12 i ( p 2 p 3 ) + [43] h 32 i 3 [32] h 21 i 2 = 1 2 [4 j 5] h 5 j 1 i [43] h 12 i [54] h 41 i [4 j 2 3 2 j 1 i [45] h 51 i [5 j 2 3 2 j 1 i 0 1 2 [4 j 5] h 5 j 1 i [32] h 21 i [4 j 5] h 5 j 1 i [54] h 21 i = 1 2 [4 j 5] h 5 j 1 i [43] h 12 i [54] h 41 i [4 j 2 3 2 j 1 i [45] h 51 i [5 j 2 3 2 j 1 i = 1 2 [4 j 5] h 5 j 1 ih 2 j 1 i 2 [54] [43][42] h 41 i [43][52] h 51 i + [32][45] h 51 i = 1 2 [45] 2 h 21 i 3 h 51 i [42][32] (2{13) F D = 3 [54] h 21 i [23] h 31 i A [32] h 21 i [42] h 21 i 2 [41] h 12 i 1 = 2 3 [54][42] h 21 i 2 [23] h 31 i 0 + 5 2 [45] h 51 i [32][42] h 21 i 2 + 0 = 1 2 [34] h 21 i [54][42] h 21 i 2 [23] h 31 i + 1 2 [54] h 21 i [45] h 51 i [32][42] h 21 i 2 = 1 2 [54] h 21 i 4 [42] 2 [23] (2{14) G D = 4 [35] h 41 i [42] h 21 i + B [42] h 21 i [42] h 21 i 2 [41] h 12 i 1 = 0 0 + [42] 2 h 21 i 2 5 2 [35] h 51 i 3 2 [53] h 31 i 0 = 1 2 [42] 2 h 21 i 2 [54] h 21 i [35] h 51 i [34] h 21 i [53] h 31 i = 1 2 [42] 3 h 21 i 4 [35] (2{15) Assem ble them together: 22

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G D h 21 i [12] h 35 i [53] + F D h 21 i [12] h 45 i [54] + E C h 23 i [32] h 45 i [54] = h 21 i 3 [42] [42] 2 h 54 ih 32 i [42][23] h 53 ih 32 i + [45][12] h 51 ih 53 i 2[12] h 53 ih 54 ih 32 i = h 21 i 3 [42] [21][41] h 51 ih 13 i 2[12] h 53 ih 54 ih 32 i (2{16) After putting bac k the normalization factor: ( i ) 2 p 2 5 [14][24] h 13 ih 14 ih 15 i (2{17) w e get p 2 5 h 12 i 4 2 h 12 ih 23 ih 35 ih 54 ih 41 i (2{18) T o go to large N c w e simply m ultiply it b y ( ig s p N c = 2) 3 2.2 The Ligh t Cone Setup The ligh t cone gauge F eynman rules are usually obtained b y the lagrangian metho d, namely w e rst set A = 0 and in tegrate out A Then the F eynman rules can b e read o from the lagrangian, whic h is a function only of A 1 and A 2 Here, I com bine the co v arian t v ertex functions with the ab o v e spinor helicit y metho d to obtain the F eynman rules in a more rexible w a y But this will require us to x a = a = [1 ; 0] T The gluon propagator in ligh t cone gauge is i ( g g + k k + g + k k + ) k 2 + i (2{19) Note the metric is diag (1 ; 1 ; 1 ; 1) throughout the dissertation. The n umerator can b e factored in to 23

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( g g + k k + g + k k + ) = ( ^ + ^ ) g + g + k +2 k 2 (2{20) Where ^ ; are ligh t cone gauge p olarization v ectors giv en b y = 1 p 2 ( k 1 ik 2 k 0 + k 3 ; 1 ; i; k 1 ik 2 k 0 + k 3 ) = 1 p 2 ( k ^ k + ; 1 ; i; k ^ k + ) ; ^ = = p 2 j ] h k j ; ^ = p 2 j k ] h j (2{21) They satisfy = 1 and k = 0. These p olarization v ectors are dened b oth on-shell and o-shell. Note from no w on, the denition of j k ], j k i [ k j and h k j will b e that of the app endix A whic h is sligh tly dieren t from the con v en tional ones. The F eynman rules are obtained b y dotting the p olarization v ectors in to the co v arian t three or four p oin t v ertices. The tri-gluon v ertex Fig. 2-4 for example, b ecomes V g g g = g f abc ( 1 )( 2 )( 3 )[ g ( p 1 p 2 ) + g ( p 2 p 3 ) + g ( p 3 p 1 ) ] Setting 1 ; 2 = ^ 3 = the ab o v e b ecomes g f abc [( p 2 p 3 ) + p ^1 p +1 ( p 2 p 3 ) ^ + ( p 3 p 1 ) + p ^2 p +2 ( p 3 p 1 ) ^ ] = 2 g f abc ( p 1 + p 2 ) + p +1 p +2 K ^ 21 (2{22) where K i;j := ( p +i p j p +j p i ). They are related to spinor pro ducts according to K i;j = p +i p +j [ p i j p j ] = p +i p +j p a i p j a The spinor notation here is also dieren t from the con v en tional one [ 20 ]. The reader can refer to the app endix for an explanation of the spinor notation. 24

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The gluon propagator Eq. 2{19 almost factorizes in to the pro duct of t w o p olarisation v ectors. While the third term on the rhs of Eq. 2{20 will mak e an extra con tribution to the four p oin t v ertex. F or example, consider the t-c hannel diagram Fig. 2-5 the rst t w o terms of Eq. 2{20 can b e asso ciated with the t w o tri-gluon v ertices, the third term, whic h describ es the mediation of A giv es 1 4 V i ( p 1 + p 4 ) 2 2 3 V + + ( p 1 + p 4 ) 2 ( p +1 + p +4 ) 2 = i 1 4 ( p +4 p +1 ) 2 3 ( p +2 p +3 ) ( p +1 + p +4 ) 2 (2{23) What's happ ening here is that the explicit factor of k 2 in the third term of Eq. 2{20 cancels the propagator, eectiv ely making a four p oin t con tact v ertex. The fermion-gluon v ertex Fig. 2-6 is the usual ig r t a W e can set = ^ or b y dotting ^ ; ; in to 1 Multiplying the spinors to the gamma matrix, w e get (assuming no w the fermion is left-handed) ig ( t a ) bc q 2 p +1 p +2 ( ^ ) 0 p 2 0 ( ) ( ) 0 p 1 0 = ig ( t a ) bc q 2 p +1 p +2 ( p 2)[ p 2 j ] h q j p 1 i = 2 ig ( t a ) bc q p +1 p +2 q ^ q + p ^1 p +1 = 2 ig t a p p +1 p +2 q + p +1 K ^ p 1 ;q 2 ig t a p +2 q + p +1 K ^ p 1 ;q (2{24) In Eq. 2{24 in order to a v oid dening what is p p + I c hose to asso ciate p +2 instead of p p +1 p +2 to a v ertex. This w on't cause an y problem, since a fermion line either alw a ys closes, or end up as an external particle (then the phase of the ro ot can b e dened arbitrarily). 1 Calling p olarizations b y or ^ is p oten tially confusing, esp ecially if y ou are lo oking at the diagram up-side-do wn. So,sometimes it is clearer to asso ciate ^ with 'in' and with 'out'. 25

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The fermion propagator is giv en b y ip r = ( p 2 + i ). W e can decomp ose p r according to p = p 2 p p ^ p p + = p 2 p + p ^ p + 1 h p p + 1 i + p 2 p 2 p + 0 0 0 # = p 2 p + j p i [ p j + p 2 p 2 p + j i [ j p = p 2 p + p ^ p p = p 2 p + 1 p p + h 1 p ^ p + i + 0 0 0 p 2 p 2 p + # = p 2 p + j p ] h p j + p 2 p 2 p + j ] h j (2{25) Since the fermion propagator almost factorizes to o, it is also p ossible to con tract a pair of v ertices here. F or example, the second term of Eq. 2{25 will con tribute to Fig. 2-7 according to g 2 ( t d ) ce ( t a ) eb q 2 p +1 p +2 [ p 2 j ( 3 ) i ( p 1 + p 4 ) ( p 1 + p 4 ) 2 ( 4 ) j p 1 i g 2 ( t d ) ce ( t a ) eb q 2 p +1 p +2 [ p 2 j ( p 2 j ] h p 3 j ) i ( p 1 + p 4 ) 2 ( ( p 1 + p 4 ) 2 p 2( p +1 + p +4 ) j i [ j ) ( p 2 j p 4 ] h j ) j p 1 i = 2 ig 2 ( t d ) ce ( t a ) eb p p +1 p +2 p +1 + p +4 2 ig 2 ( t d ) ce ( t a ) eb p +2 p +1 + p +4 (2{26) The scalar F eynman rules ha v e no susp ense in them at all, and can b e read o from an y eld theory b o ok. The F eynman rules that p ertain to our calculation will b e summarized in the app endix. The main prop ert y of the F eynman rules ab o v e is the absence of p 2.3 Brief Description of the Calculational Pro cedure In the Fig. 2-8 the k 's and q are the dual momen ta. They are related to the real momen ta coming in to the three legs according to p 1 = k 1 k 2 p 2 = k 4 k 1 and p 3 = k 2 k 4 The unregulated in tegrands ha v e a symmetry under k i k i + a whic h 26

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w ould ensure that eac h diagram dep ends only on the real momen ta. But here, as in [ 13 14 ], w e use a regulator exp ( 2 q ^ q ) that breaks this symmetry Hence, a regulated amplitude can dep end on the individual dual momen ta. This seemingly un wieldy regulator is designed for the w orld sheet description, but the result do esn't dier to o m uc h when a cut o regulator is used. The calculation roughly go es as follo ws, 1. Exp onen tiate all propagators according to i p 2 + i = Z 1 0 dT e iT p 2 F or the sc hematic diagram Fig. 2-8 W e ha v e = Z d 4 q (2 ) 4 i ( q k 2 ) 2 i ( q k 1 ) 2 i ( q k 4 ) 2 = Z d 4 q (2 ) 4 d 3 T i exp i X T i ( q k i ) 2 2. In tegrate out q lea ving a delta function relating q + to the F eynman parameters(this step requires the absence of q in all F eynman rules). = Z dq + d 2 q ? (2 ) 3 1 2 d 3 T i ( X T i k + i X T i q + ) exp i X T i ( q k i ) 2 3. In tegrate q 1 ; q 2 using exp [ q 2 ? ] as a damping factor. = Z dq + 16 2 dx i ( X x i 1) ( X x i k + i q + ) 1 x 2 x 4 ( k 2 k 4 ) 2 where x i := T i = P T i Here is the rub: w e cannot simply in tegrate o v er x 2 x 1 and x 4 b ecause the prefactor of this diagram will ha v e up to second order p oles at q + = k + i In order to sho w the 27

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cancellation of these p oles (gauge articial div ergences), w e pro ceed as follo ws: rst eliminate one F eynman parameter in fa v our of q + : k + 2 < q + < k + 1 : x 1 = x ( q + k + 2 ) k + 1 k + 2 ; x 4 = (1 x )( q + k + 2 ) k + 4 k + 2 ; x 2 = 1 x 1 x 4 ; x 2 [0 ; 1] k + 1 < q + < k + 4 : x 1 = (1 x )( q + k + 4 ) k + 1 k + 4 ; x 2 = x ( q + k + 4 ) k + 2 k + 4 ; x 4 = 1 x 1 x 2 ; x 2 [0 ; 1] No w dx 2 dx 1 dx 4 ( P x i 1) ( P x i k + i q + ) = dq + dxJ After in tegrating out x w e are left with a function of q + whic h is dened dieren tly in dieren t regions: k + 2 < q + < k + 1 and k + 1 < q + < k + 4 All these can b e visualized v ery clearly when w e represen t a F eynman diagram on the ligh t cone w orld sheet. The details can b e found in [ 13 14 ]. Our observ ation is that, in eac h region, all p oles cancel. 2 Hence w e can p erform the nal q + in tegral and obtain the results. T o summarize the computation pro cedure, p erform the q + inte gr al last PSfrag replacemen ts 1 2 3 4 5 = Figure 2-1. Fiv e-p oin t amplitude 2 in the case helicit y conserving amplitude, all p oles cancel up to infrared terms, but since infrared div ergence is alw a ys prop ortional to a tree, they are easy to recognize and deal with. 28

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PSfrag replacemen ts 1 2 2 3 3 4 5 5 A = B = C = D = Figure 2-2. Some building blo c ks that can b e computed separately PSfrag replacemen ts 1 2 3 3 4 4 5 5 5 E = F = G = Figure 2-3. Second lev el building blo c ks PSfrag replacemen ts ; b; p 2 ; a; p 1 ; c; p 3 Figure 2-4. T ri-gluon v ertex PSfrag replacemen ts ; p 1 ; p 2 ; p 3 ; p 4 Figure 2-5. Propagator in the t-c hannel is con tracted PSfrag replacemen ts ; a q c; p 1 b; p 2 Figure 2-6. F ermion-gluon v ertex 29

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PSfrag replacemen ts b; p 1 c; p 2 ; ^ ; p 3 ; ; p 4 Figure 2-7. F ermion propagator con tracted PSfrag replacemen ts q k 2 k 1 k 4 1 2 3 Figure 2-8. Here w e assume only ( k 2 k 4 ) 2 6 = 0, and k + 4 > k + 1 > k + 2 30

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CHAPTER 3 BO X REDUCTION This c hapter con tains a lot of tec hnical details, the reader ma y skip it for rst reading. Our general pro cedure for ev aluating b o x diagrams is 1. Ev aluate a b o x diagram that is free of articial, collinear and infrared div ergences using text b o ok metho d (com bining denominator with F eynman's tric k, shift the momen tum, p erform momen tum and F eynman parameter in tegrals, etc). 2. Ev aluate a b o x that con tains collinear or infrared div ergence but is free of articial div ergence b y subtracting those collinear or infrared div ergences in the form of triangle diagrams, and then use metho d 1. 3. A b o x that has articial div ergences has to b e reduced to triangle diagrams. In the next section, I will describ e ho w to ac hiev e 2 and 3. The dual momen ta assignmen t giv en b y Fig. 3-1 will b e used through out all the calculations. 3.1 Bo x with a Helicit y Violating Sub-diagram The F eynman rules for the gluon, scalar and fermion in the ligh t cone gauge are v ery similar. An y three p oin t v ertex will ha v e the follo wing form: R ( k + i ) K ^ ; ij where R ( k + ) is a rational function of the + comp onen t of external momen ta and K ^ ; ij = ( p +i p ^ ; j p +j p ^ ; i ) is in tro duced in the app endix. This allo ws us to write the n umerator of an y b o x diagram as R ( k + ) K ^ ; ij K ^ ; k l K ^ ; mn K ^ ; op (3{1) W e can also lab el a b o x b y the helicit y of the four corners, regardless of the details of the b o xes. F or example the follo wing Fig. 3-2 will b e abbreviated as ^ ^ __ b o x. R ( k + ) in general will con tain p oles of k + These p oles are usually in terpreted using Cauc h y principle v alue or some other prescriptions. W e c hose to sho w directly that these p oles are all fak e, namely when w e compute a ph ysical quan tit y these p oles will 31

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b e cancelled. T o sho w this w ould en tail us to express the amplitude as a holomorphic functions of k + and analyze the p oles one b y one. But the b o x in tegrand in general will ha v e a denominator that is quadratic in the F eynman parameters and k + In tegrating it b y force will in tro duce logarithm and di-logarithms with v arious argumen ts. Although w e kno w of certain relations b et w een (com binations of ) di-logarithms and double logarithms, the application of these relations is hard to automate. F ortunately for the case at hand, all b o x diagrams can b e reduced to triangle-lik e in tegrands, whose manipulations are considerably easier. Here I describ e the b o x reduction tec hnique in detail. The lo w er half of the b o x Fig. 3-2 is the 't c hannel' of (+ + + ) scattering pro cess, whic h is zero. So w e can replace this 't c hannel' with min us the 's c hannel', whic h will result in a triangle diagram. But since the amplitude is only truly zero on-shell, while the t w o in ternal lines are certainly o-shell, I will get some more terms that are prop ortional to the virtualit y of the t w o in ternal lines ( p 25 and p 26 ). Sc hematically I ha v e Fig. 3-3 : More concretely Fig. 3-3 is written as 1 ( p 1 + p 6 ) 2 K ^ 16 K ^ 52 = 1 ( p 1 + p 2 ) 2 K ^ 21 K ^ 65 + p 26 ( p 1 + p 6 ) 2 ( p 1 + p 2 ) 2 K ^ 52 K ^ 21 + p 25 ( p 1 + p 6 ) 2 ( p 1 + p 2 ) 2 K ^ 21 K ^ 16 (3{2) Clearly in this case, the b o x is reduced in to three triangle-lik e diagrams. Here I list its con tribution: First, dene some functions that o ccur ubiquitously: 32

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H s : = ( k + 4 q + )( k + 2 + q + ) s e r ( k + 2 k + 4 ) 2 ; k + 2 < q + < k + 4 H t : = ( k + 3 q + )( q + k + 1 ) t e r ( k + 1 + k + 3 ) 2 ; k + 3 < q + < k + 1 H d : = ( k + 3 q + )( k + 2 + q + ) s e r ( k + 3 k + 2 )( k + 4 k + 2 ) ; k + 2 < q + < k + 3 = ( q + k + 3 )( k + 4 q + ) s e r ( k + 4 k + 2 )( k + 4 k + 3 ) ; k + 3 < q + < k + 4 H l : = ( k + 4 + q + )( k + 3 q + ) t e r ( k + 3 k + 4 )( k + 1 + k + 3 ) ; k + 3 < q + < k + 1 = ( k + 4 + q + ) 2 t e r ( k + 4 k + 1 )( k + 4 k + 3 ) ; k + 1 < q + < k + 4 H u : = ( k + 1 q + )( q + k + 2 ) s e r ( k + 1 k + 2 )( k + 4 k + 2 ) ; k + 2 < q + < k + 1 = ( q + k + 1 )( k + 4 + q + ) s e r ( k + 4 k + 1 )( k + 2 k + 4 ) ; k + 1 < q + < k + 4 H r : = ( k + 2 + q + ) 2 t e r ( k + 1 k + 2 )( k + 3 k + 2 ) ; k + 2 < q + < k + 3 = ( k + 2 q + )( q + k + 1 ) t e r ( k + 1 + k + 2 )( k + 1 + k + 3 ) ; k + 3 < q + < k + 1 (3{3) Next, I list the results for the t w o mo del b o xes ^ ^_ and ^ _^ ^ ^_ : i 8 2 T r t a t b t c t d k + 2 < q + < k + 3 : + 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H s ) + 1 = 8 ( q + k + 2 ) 2 ( k + 1 k + 4 )( k + 3 k + 4 ) k + 2 k + 4 log ( H r ) 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H r ) 33

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k + 3 < q + < k + 1 : + 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H s ) + 1 = 8 ( k + 3 k + 4 )( k + 2 k + 1 ) ( k + 3 k + 1 )( k + 2 k + 4 ) ( k + 3 k + 2 + k + 4 k + 3 k + 4 k + 2 + 2 q + k + 2 q + 2 ) k + 1 + k + 3 k + 4 k + 2 + q + 2 k + 3 2 q + k + 4 k + 3 q + 2 k + 2 + q + 2 k + 4 log ( H t ) + 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H t ) 1 = 8 ( q + k + 4 ) 2 ( k + 2 k + 1 )( k + 2 + k + 3 ) k + 2 k + 4 log ( H l ) 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H l ) + 1 = 8 ( q + k + 2 ) 2 ( k + 1 k + 4 )( k + 3 k + 4 ) k + 2 k + 4 log ( H r ) 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H r ) k + 1 < q + < k + 4 : + 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H s ) 1 = 8 ( q + k + 4 ) 2 ( k + 2 k + 1 )( k + 3 k + 2 ) k + 2 k + 4 log ( H l ) 1 = 4 ( q + k + 4 )( q + k + 2 ) ( k + 2 k + 4 ) s K ^ 43 K 21 log ( H l ) No w assume that its co ecien t is C + A= ( q + k + 4 ) 2 1 = ( q + k + 4 ) 1 what to do next is to com bine this with the co ecien ts of the log H terms, then extract the p olynomial part through partial fraction and p erform the in tegrals to get its con tribution. 1 = 36 ( k + 4 k + 3 )( k + 2 k + 3 )( k + 1 k + 4 )( k + 1 k + 2 ) C + 1 = 8 k + 1 k + 2 k + 4 + k + 3 s K ^ 43 K 21 1 = 24 ( k + 1 2 + k + 3 2 k + 1 k + 2 k + 4 k + 1 k + 3 k + 2 k + 4 k + 3 + 2 k + 4 k + 2 ) C s K ^ 43 K 21 + 1 = 8 log( s ) k + 4 + k + 2 + 2 A s K ^ 43 K 21 1 = 24 log( s ) ( k + 2 k + 4 ) 2 C s K ^ 43 K 21 + 1 = 24 log( t ) ( k + 4 k + 3 )( k + 2 k + 3 )( k + 1 k + 4 )( k + 1 k + 2 ) C 1 = 8 log( t ) k + 4 + k + 2 + 2 A s K ^ 43 K 21 + 1 = 24 log( t ) ( k + 2 k + 4 ) 2 C s K ^ 43 K 21 (3{4) 34

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If its co ecien t w ere C + A= ( q + k + 4 ) 2 1 = ( q + k + 4 ) 1 then the result can b e obtained from the ab o v e b y replacing k + 2 $ k + 4 k + 1 ( k + 3 ) $ k + 1 ( k + 3 ). What is not included in Eq. 3{4 is of the form: ::: ( q k i ) 2 + ::: q k i log H (3{5) where ::: only dep ends on the external momen ta. The second order p oles will ha v e to cancel ev en tually the rst order p ole alw a ys com bines to b ecome prop ortional to a tree amplitude with a univ ersal structure. ^ _^ k + 2 < q + < k + 3 : 1 = 8 ( q + k + 4 ) 2 ( k + 1 + k + 2 )( k + 3 k + 2 ) k + 2 k + 4 log ( H s ) + 1 = 8 ( q + k + 3 ) 2 ( k + 1 k + 4 )( k + 1 + k + 2 ) k + 3 k + 1 log ( H d ) 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H d ) + 1 = 8 ( k + 1 + q + ) 2 ( k + 3 k + 4 )( k + 3 k + 2 ) k + 3 k + 1 log ( H u ) + 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H u ) k + 3 < q + < k + 1 : 35

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1 = 8 ( k + 3 k + 4 )( k + 2 k + 1 ) ( k + 2 k + 4 )( k + 3 k + 1 ) ( k + 3 k + 2 + 2 q + k + 2 k + 4 k + 2 + k + 3 k + 4 q + 2 ) k + 1 + k + 3 k + 4 k + 2 + k + 3 q + 2 2 k + 3 q + k + 4 + q + 2 k + 4 q + 2 k + 2 log ( H s ) 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H s ) 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H t ) 1 = 8 ( q + k + 3 ) 2 ( k + 1 k + 4 )( k + 2 k + 1 ) k + 3 k + 1 log ( H d ) + 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H d ) + 1 = 8 ( k + 1 + q + ) 2 ( k + 3 k + 4 )( k + 3 k + 2 ) k + 3 k + 1 log ( H u ) + 1 = 4 ( k + 1 + q + )( q + k + 3 ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H u ) k + 1 < q + < k + 4 : + 1 = 8 ( q + k + 2 ) 2 ( k + 1 k + 4 )( k + 3 k + 4 ) k + 2 k + 4 log ( H s ) 1 = 8 ( k + 3 + q + ) 2 ( k + 1 k + 4 )( k + 2 k + 1 ) k + 3 k + 1 log ( H d ) + 1 = 4 ( k + 3 + q + )( k + 1 + q + ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H d ) 1 = 8 ( k + 1 + q + ) 2 ( k + 3 k + 4 )( k + 3 k + 2 ) k + 3 k + 1 log ( H u ) 1 = 4 ( k + 3 + q + )( k + 1 + q + ) ( k + 3 k + 1 )( t ) K ^ 14 K 32 log ( H u ) No w assume that its co ecien t is C + A= ( q + k + 1 ) 2 1 = ( q + k + 1 ) 1 its con tribution w ould b e 36

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1 = 36 ( k + 4 k + 3 )( k + 2 k + 3 )( k + 1 k + 4 )( k + 1 k + 2 ) C + 1 = 8 k + 4 + k + 3 + k + 1 k + 2 ( t ) K ^ 14 K 32 1 = 24 ( k + 2 k + 1 2 k + 3 k + 1 + k + 4 k + 3 + k + 1 k + 4 + k + 2 k + 3 k + 4 2 k + 2 2 ) C ( t ) K ^ 14 K 32 + 1 = 24 log( s ) ( k + 4 k + 3 )( k + 2 k + 3 )( k + 1 k + 4 )( k + 1 k + 2 ) C + 1 = 8 log( s ) k + 1 + 2 A + k + 3 ( t ) K ^ 14 K 32 1 = 24 log( s ) ( k + 3 k + 1 ) 2 C ( t ) K ^ 14 K 32 1 = 8 log( t ) k + 1 + 2 A + k + 3 ( t ) K ^ 14 K 32 + 1 = 24 log ( t ) ( k + 3 k + 1 ) 2 C ( t ) K ^ 14 K 32 (3{6) Just as in Eq. 3{4 the p ole terms are dropp ed here. 3.2 Bo x without a Helicit y Violating Sub-diagram But of course, not all b o xes are going to ha v e a helicit y violating subtree, I no w turn to a second, m uc h more complicated case Fig. 3-4 Assuming this b o x diagram can b e written as A ( q + k + 1 ) 2 + B ( q + k + 1 ) K 16 K ^ 52 K 3 ; 5 K ^ 6 ; 4 (3{7) Here I ha v e used the notation of dual momen ta to mak e the formalism more symmetric lo oking. W rite 37

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A ( q + k + 1 ) 2 + B ( q + k + 1 ) K 16 K ^ 52 = A ( q + k + 1 ) 2 ( K 16 K ^ 52 + K ^ 16 K 52 K ^ 16 K 52 ) + B ( q + k + 1 ) ( K 16 K ^ 52 K ^ 16 K 52 + K ^ 16 K 52 ) = A ( q + k + 1 ) 2 ( K 16 K ^ 52 + c:c ) + B ( q + k + 1 ) ( K 16 K ^ 52 c:c ) ( A ( q + k + 1 ) 2 B ( q + k + 1 ) ) K ^ 16 K 52 = A ( q + k + 1 ) 2 1 2 ( k + 1 q + )( k + 1 + q + ) s + A ( q + k + 1 ) 2 1 2 ( k + 4 k + 1 )( k + 2 q + ) + ( k + 1 k + 2 )( q + k + 4 ) ( q k 1 ) 2 + A ( q + k + 1 ) 2 1 2 ( k + 1 + k + 2 )( k + 1 + q + )( q k 4 ) 2 + A ( q + k + 1 ) 2 1 2 ( k + 4 + k + 1 )( k + 1 q + )( q k 2 ) 2 + B ( q + k + 1 ) ( K 21 K ^ 65 c:c ) ( q + k + 1 ) ( k + 4 k + 2 ) ( A ( q + k + 1 ) 2 B ( q + k + 1 ) ) K ^ 16 K 52 (3{8) Notice that the rst four terms in Eq. 3{8 con tain co v arian t pro ducts of momen ta: s ( q k 1 ) 2 ( q k 4 ) 2 ( q k 2 ) 2 The co ecien t of s is simply sA = 2 (the anno ying 1 = ( q + k + 1 ) 2 has disapp eared, the fth term also has this prop ert y). T erms lik e ( q k i ) 2 will simply cancel one of the four propagators, eectiv ely making a triangle-lik e terms. No w Let us lo ok at the sixth term of Eq. 3{8 Putting bac k the factor of K 3 ; 5 K ^ 6 ; 4 giv es [ A= ( q + k + 1 ) 2 B = ( q + k + 1 )] K ^ 16 K 52 K 3 ; 5 K ^ 6 ; 4 No w that t w o ^ 's (or ) are next to eac h other, this term is in fact nothing but the previous mo del diagram. In summary the ev aluation of the second, fth and sixth terms of Eq. 3{8 go es through without further t wist. Ho w ev er, the rst(hence forth called half scalar b o x), third and fourth term ha v e collinear div ergence individually (on top of the infrared div ergence), some more manipulations are needed. 38

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Subtraction of Collinear Div ergence: A collinear div ergence in a virtual pro cess happ ens when the momen ta ro wing through t w o in ternal propagators connected to a common massless external leg b ecome parallel. F or a b o x diagram, there can b e a collinear div ergence at eac h massless corner. When only transv erse comp onen ts propagates in the lo op, the v ertex function will v anish due to helicit y conserv ation. So, a b ona de b o x in ligh t cone gauge will ha v e no collinear problem. Ho w ev er, in the manipulations of Section 3.2 w e ha v e tamp ered with the v ertex structures of the lo w er sub-tree, so there will b e collinear div ergences in some of the resulting pieces. But they will cancel when all the pieces are collected. The collinear div ergence in a b o x diagram can b e lo cated in to t w o triangle-lik e diagrams. This pro cess is rather lik e doing a partial fraction. Consider K 3 ; 5 K ^ 6 ; 4 i ( q k 1 ) 2 i ( q k 4 ) 2 i ( q k 3 ) 2 i ( q k 2 ) 2 (3{9) T ak e limit q k 4 (so the righ t propagator is soft). Eq. 3{9 b ecomes K 3 ; 5 K ^ 4 ; 3 i ( q k 1 ) 2 i ( q k 4 ) 2 i ( q k 3 ) 2 i s (3{10) T ak e limit q k 1 (so the b ottom propagator is soft). Eq. 3{9 b ecomes K 3 ; 5 K ^ 1 ; 4 i ( q k 1 ) 2 i ( q k 4 ) 2 i t i ( q k 2 ) 2 (3{11) Naturally if these t w o 'p oles' are subtracted from Eq. 3{9 w e should ha v e a term free of collinear div ergence, whic h will b e demonstrated next. According to Chapter 2 the Sc h winger represen tation of Eq. 3{9 is (with H giv en b y x 1 x 3 t + x 2 x 4 s + 2 where is a small mass used here to regulate temp orarily the p ossible collinear div ergences) 39

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Z T 3 dT ( x i 1) d 4 x i 2 T (2 ) 3 exp iT q 2 ? + iT H ( k + 4 k + 3 ) q + x 2 K 43 + x 1 K 32 ( k + 3 k + 2 ) q ^ + x 4 K ^ 43 + x 1 K ^ 14 = Z T 3 dT ( x i 1) d 4 x i 2 T (2 ) 3 exp iT q 2 ? + iT H 1 2 ( k + 4 k + 3 )( k + 3 k + 2 ) q 2 ? + x 2 x 4 K 43 K ^ 43 + x 1 x 4 K 32 K ^ 43 + x 2 x 1 K 43 K ^ 14 + x 21 K 32 K ^ 14 = Z T 3 dT ( x i 1) d 4 x i 16 2 T exp iT H 1 2 ( k + 4 k + 3 )( k + 3 k + 2 ) ( iT ) 2 + x 2 x 4 iT K 43 K ^ 43 + x 1 x 4 iT K 32 K ^ 43 + x 2 x 1 iT K 43 K ^ 14 + x 21 iT K 32 K ^ 14 In tegrate o v er T : Z 1 8 2 ( x i 1) dx i [ 1 4 H ( k + 4 k + 3 )( k + 3 k + 2 ) + x 2 x 4 2 H 2 K 43 K ^ 43 + x 1 x 4 2 H 2 K 32 K ^ 43 + x 2 x 1 2 H 2 K 43 K ^ 14 + x 21 2 H 2 K 32 K ^ 14 ] The rst t w o terms w on't ha v e collinear div ergence, the can b e set to zero, but the rest will ha v e to b e ev aluated with a small to regulate the collinear div ergences. There are basically t w o t yp es of F eynman parameter in tegrals in this case. The others are either nite or could b e obtained from these t w o. Z x i 1 x 21 ( x 1 x 3 t + x 2 x 4 s + 2 ) 2 = 1 2 st log 2 2 s s 2 t ( s + t ) 2 log 2 s t + 2 st log 2 s + 1 t ( s + t ) log ( s t ) + 4 st 2 s 2 t ( s + t ) 2 Z x i 1 x 1 x 2 ( x 1 x 3 t + x 2 x 4 s + 2 ) 2 = 1 2( s + t ) 2 log 2 s t 2 st log 2 s 1 t ( s + t ) log ( s t ) 2 st 2 2( s + t ) 2 (3{12) Collecting all the div ergen t terms from Eq. 3{9 : 40

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( 3{9 ) div = iA 8 2 1 8( t ) log 2 2 + 1 2( t ) log 2 1 4( t ) log s log 2 K 32 K ^ 14 iA 8 2 1 4( t ) log 2 ( K 43 K ^ 14 + K ^ 43 K 32 ) (3{13) Eq. 3{10 and Eq. 3{11 simply giv e K 3 ; 5 K ^ 4 ; 3 i ( q k 1 ) 2 i ( q k 4 ) 2 i ( q k 3 ) 2 i s = Z T 2 dT dx 1 dx 3 dx 4 ( P x i 1) 2 T (2 ) 3 k + 4 k + 3 ( t ) q x 1 ( t ) K 32 0 ( t ) K 43 K ^ 43 exp ( iT q ? 2 + iT H ) = 1 8 2 [log 2 + 2 log ( t ) ] 1 2 s ( t ) K ^ 43 K 32 K 3 ; 5 K ^ 1 ; 4 i ( q k 1 ) 2 i ( q k 4 ) 2 i t i ( q k 2 ) 2 = Z T 2 dT dx 1 dx 2 dx 4 ( P x i 1) 2 T (2 ) 3 k + 4 k + 3 ( t ) q x 1 ( t ) K 32 x 2 ( t ) K 43 K ^ 14 exp ( iq ? 2 + iH ) = 1 8 2 [(log s 2) log 2 1 2 log 2 2 ( 1 2 log 2 s + 4 2 log s )] 1 2 s ( t ) K 32 K ^ 14 + 1 8 2 [log 2 + 2 log s ] 1 2 s ( t ) K 43 K ^ 14 This, m ultiplied b y sA= 2, is the same as the half scalar b o x con tribution Eq. 3{13 The complete con tribution of Eq. 3{9 with Eq. 3{10 and 3{11 subtracted is (p ole terms omitted) 41

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1 8 2 A log s t 1 8 s ( k + 3 + k + 4 )( k + 2 + k + 3 ) ( t ) s 1 4 s (( t ) s )( t ) K ^ 14 K 32 + 1 4 s (( t ) s )( t ) K ^ 14 K 43 + 1 4(( t ) s ) K ^ 43 K 32 + 1 8 2 A [log 2 s t + 2 ] 1 16 s ( k + 3 + k + 4 )( k + 2 + k + 3 )( t ) (( t ) s ) 2 1 8 s 2 (( t ) s ) 2 ( t ) K ^ 14 K 32 + 1 8 s (( t ) s ) 2 K ^ 14 K 43 + 1 8 s (( t ) s ) 2 K ^ 43 K 32 (3{14) So, when ev aluating Eq. 3{9 whic h is collinearly div ergen t b y itself, I shall subtract Eq. 3{10 and 3{11 from it, to mak e 3{9 nite. Next I will sho w that Eq. 3{10 and Eq. 3{11 also cancel the collinear div ergence in the third and fourth term of Eq. 3{8 Putting the triangle-lik e terms in Eq. 3{8 together with Eq. 3{10 3{11 : ( k + 1 + k + 2 )( k + 1 + q + )( q k 4 ) 2 + ( k + 4 + k + 1 )( k + 1 q + )( q k 2 ) 2 1 ( q + k + 1 ) 2 i ( q k 1 ) 2 i ( q k 2 ) 2 i ( q k 3 ) 2 i ( q k 4 ) 2 K 3 ; 5 K ^ 6 ; 4 + K 3 ; 5 K ^ 4 ; 3 i ( q k 1 ) 2 i ( q k 4 ) 2 i ( q k 3 ) 2 i s s + K 3 ; 5 K ^ 1 ; 4 i ( q k 1 ) 2 i ( q k 4 ) 2 i t i ( q k 2 ) 2 s (3{15) Supp ose q k 1 = ( p 1 p 2 ), q k 2 = (1 + )( p 1 p 2 ). The ab o v e expression b ecomes 1 ( q k 1 ) 2 1 ( q k 2 ) 2 K 3 ; 5 ( k + 1 + k + 2 ) ( k + 1 k + 2 ) 1 (1 + ) t (1 + ) K ^ 14 + 0 + 0 + 1 s s 1 t K ^ 14 = 0 Supp ose q k 1 = ( p 4 p 1 ), q k 4 = ( 1)( p 4 p 1 ). The ab o v e expression b ecomes 42

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1 ( q k 1 ) 2 1 ( q k 4 ) 2 K 3 ; 5 0 + ( k + 4 + k + 1 ) ( k + 4 k + 1 ) 1 ( 1) t ( K ^ 14 + K ^ 24 ) + K ^ 4 ; 3 1 (1 ) t 1 s s + K ^ 1 ; 4 1 t 1 s s = 0 Th us all the collinear div ergences cancel. The complete con tribution of 3{10 and 3{11 and third fourth term of Eq. 3{8 are quite complicated. I observ ed that the rational part of their con tribution is zero, so if there is a w a y to get the logarithmic parts through some other metho ds (suc h as unitarit y), w e can b e spared all these ordeals. T o summarize, I ha v e subtracted 3{10 and 3{11 from Eq. 3{9 and then added them bac k to the triangle-lik e terms in Eq. 3{8 to mak e b oth parties collinearly nite. I w an t to add that the in tro duction of will not mess up gauge co v ariance as p eople w ould normally think. PSfrag replacemen ts 1 k + 2 2 k + 1 3 k + 4 4 k + 3 q + Figure 3-1. Dual momen tum assignmen t with k + 2 < k + 3 < k + 1 < k + 4 PSfrag replacemen ts ^ ^ p 6 # p 5 # % p 1 p 2 p 3 & p 4 Figure 3-2. Mo del b o x with a helicit y violating subtree 43

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PSfrag replacemen ts = + + Figure 3-3. Graphical represen tation of b o x reduction PSfrag replacemen ts ^ ^ p 6 # p 5 # % p 1 p 2 p 3 & p 4 Figure 3-4. Mo del b o x with alternating helicit y 44

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CHAPTER 4 MASSLESS AMPLITUDES I will b e follo wing [ 21 ] and decomp ose an n-particle amplitude in to M n = X per m 0 T r ( t a 1 t a 2 :::t a n ) M ( p 1 ; 1 ; p 2 ; 2 ; ::: ; p n ; n ) (4{1) where per m 0 is o v er non cyclic p erm utations for complex represen tations, non cyclic and non rerexiv e p erm utations for real represen tations. In the follo wing results, the represen tation is assumed to b e the adjoin t represen tation. 4.1 Tw o P oin t F unctions A factor of ig 2 = (16 2 ) f cad f dbc = ig 2 = (16 2 )T r t a t b will b e omitted in the follo wing list.4.1.1 Gluon Self-Energy Refer to Fig. 4-1 ( g + ; g +; s ) = 1 3 k ^ 2 1 + k ^ 1 k ^ 3 + k ^ 2 3 ( g + ; g +; q ) = 4 3 k ^ 2 1 + k ^ 1 k ^ 3 + k ^ 2 3 ( g + ; g +; g ) = 2 3 k ^ 2 1 + k ^ 1 k ^ 3 + k ^ 2 3 (4{2) where the rst t w o argumen ts in tell the sp ecies of particle and the helicit y and the third denotes the particle in the lo op. F or the ab o v e assignmen t of helicit y the con tribution should b e zero due to Loren tz co v ariance. They are only nonzero b ecause the regulator used here do esn't resp ect Loren tz co v ariance. This is purely an artifact, and it has to b e cancelled b y a coun ter term. Also, it can b e observ ed that ( g + ; g + ; s ) N s + ( g + ; g +; q ) F N f + ( g + ; g +; g ) N g = 0 in the N = 4 SYM case. 45

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( g + ; g + ; s ) = 1 6 p + p ^ ( g + ; g + ; q ) = 2 3 p + p ^ ( g + ; g + ; g ) = 7 3 p + p ^ + IR terms (4{3) The notation g + simply means that I am using j ] h j or g + as p olarization v ector. The rst t w o terms should also b e zero due to Loren tz co v ariance, so they ha v e to b e set to zero with coun ter terms. Here are the helicit y conserving 2 p oin t functions: ( g + ; g ; s ) = Z 1 0 dxx (1 x ) p 2 log x (1 x ) p 2 e r = p 2 5 18 1 6 log p 2 e r ( g + ; g ; q ) = 2 Z 1 0 dx x 2 + (1 x ) 2 p 2 log x (1 x ) p 2 e r = p 2 26 9 4 3 log p 2 e r ( g + ; g ; g ) = 2 Z 1 0 dx x 1 x + 1 x x + x (1 x ) p 2 log x (1 x ) p 2 e r = p 2 67 9 + 11 3 log p 2 e r + Z 1 0 dx 2 1 x + 2 x log x (1 x ) p 2 e r (4{4) Here x is in fact ( q + k + 1 ) = ( k + 3 k + 1 ). Sev eral commen ts are in order: In the last line of Eq. 4{4 the x in tegral is certainly div ergen t, this is due to the articial div ergence. This div ergen t in tegral is what w e call the infrared term (with x in terpreted as ( q + k + 1 ) = ( k + 3 k + 1 ). T emp orarily forgetting ab out these infrared terms, it is easily observ ed that the sum of these diagrams is zero with the N = 4 eld con ten t. I shall commen t on this later. 46

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In the ab o v e list, all quadratic div ergence 1 = are omitted. Due to the w a y the ligh t cone w orld sheet computation is setup, all the tadp oles are b eing dropp ed. These tadp oles will only con tribute to the quadratic 1 = div ergence. As a result, the 1 = terms can actually ha v e non-trivial p + dep endence. When the p + dep endence is of the t yp e 1 =p + it can b e in terpreted as a w orld sheet cosmological constan t as in [ 13 ]. While in other cases, the p + dep endence could b e of the form (1 =p + ) log p + whose in terpretation has not b een fully understo o d y et. There is one small p oin t ab out n umerics that I w an t to p oin t out here. The 2 p oin t function naturally comes with a color structure of f cad f dbc = T r t a t b when w e try to t the 2 p oin t functions in to the big picture as in Fig. 4-2 the trace factor b ecomes (assuming that t a is in a real represen tation): f abe T r [ t e t f ] f f cd = T r[( i )[ t a ; t b ]( i )[ t c ; t d ]] = T r [ t a t b t c t d ] + T r [ t a t b t d t c ] + T r[ t b t a t c t d ] T r [ t b t a t d t c ] 2T r [ t a t b t c t d ] (4{5) Also, there is alw a ys another factor of i 2 coming from the t w o propagators connecting to the 2 p oin t function. As a comparison, I shall giv e the t w o p oin t function that describ es the 'propagation' of A The argumen t g + means that I am taking the p olarization v ector to b e g + ( g + ; g + ; s ) = p +2 4 9 1 6 log p 2 e r ( g + ; g + ; q ) = p +2 20 9 4 3 log p 2 e r ( g + ; g + ; g ) = p +2 8 9 1 3 log p 2 e r W e observ e that the logarithmic piece that will giv e a cut matc hes b et w een ( g + ; g ) and ( g + ; g + ) when the lo op particle is fermion or scalar. This is a hin t that w e shall use coun ter terms to enforce the total agreemen t b et w een ( g + ; g ) and ( g + ; g + ), due to the 47

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consideration of Loren tz co v ariance. This observ ation shall b e a guide to x coun ter terms later on. Clearly there is a mismatc h when the gluon is in the lo op. This mismatc h is due to the presence of infrared div ergences in ( g + ; g ; g ), while ( g + ; g + ; g ) is free of infrared div ergence. 4.1.2 F ermion and Scalar Self-Energy ( q + ; q +) = Z 1 0 dx 2 + 3 x 3 x 2 (1 x ) x p 2 log x (1 x ) p 2 e r = p 2 6 + 3 log p 2 e r + Z 1 0 dx 2 1 x + 2 x log x (1 x ) p 2 e r The div ergen t x in tegral is lik ewise in terpreted as infrared term. I shall also giv e the result when I use j i or j ] as spinors instead of j p i or j p ]. ( q + ; q + ) = p 2 p + 2 log p 2 e r where p ro ws with the direction of the fermion line. W e again observ e a mismatc h of log p 2 term, due to the infrared div ergence. The scalar self-energy diagram is giv en b y ( s; s ) = Z 1 0 dx 2 + 4 x 4 x 2 (1 x ) x p 2 log x (1 x ) p 2 e r = p 2 8 + 4 log p 2 e r + Z 1 0 dx 2 1 x + 2 x log x (1 x ) p 2 e r 4.2 Three P oin t F unctions The general structure of 3 p oin t function Fig. 4-3 is const + log p 2o tree + term + const ( k 2 ? + k 1 ? + k 4 ? ) 48

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The term will arise if the o-shell leg has unlik e helicit y A factor of g 3 = 8 2 f dae f ebf f f cd = ig 3 = 8 2 T r[ t a t b t c ] is omitted. 4.2.1 Gluon V ertex Correction ( g + ; g + ; g ; s ) = 2 p +3 p +1 p +2 K ^ 2 ; 1 1 6 log p 2o e r 1 9 1 6 p +1 p +2 p +23 ( g + ; g + ; g ; q ) = 2 p +3 p +1 p +2 K ^ 2 ; 1 4 3 log p 2o e r 32 9 + 2 3 p +1 p +2 p +23 ( g + ; g + ; g ; g ) = 2 p +3 p +1 p +2 K ^ 2 ; 1 11 3 log p 2o e r + 70 9 1 3 p +1 p +2 p +23 + IR terms (4{6) where = 1 if leg 3 is o shell and 0 otherwise, and p o is the momen tum of the only o-shell leg. The infrared terms presen t in the gluon triangle diagram will ev en tually b e com bined with infrared terms from other diagrams to b ecome prop ortional to a tree. They are giv en in [ 14 ] and are not rep eated here. All of the ab o v e also con tain an anomalous term: ( g + ; g + ; g ; s ) ano = 1 3 ( k ^ 2 + k ^ 1 + k ^ 4 ) ( g + ; g + ; g ; q ) ano = 4 3 ( k ^ 2 + k ^ 1 + k ^ 4 ) ( g + ; g + ; g ; g ) ano = 2 3 ( k ^ 2 + k ^ 1 + k ^ 4 ) here the k 's are dual momen ta, they arise b ecause is an exp onen tial damping factor of the transv erse dual momen ta. Had w e used a cut o regulator (( q ? k 1 ? ) 2 < for example) instead of the anomalous terms w ould c hange according to ( k 2 ? + k 1 ? + k 4 ? ) ( k 2 ? k 1 ? ) + ( k 4 ? k 1 ? ) These p olynomial terms m ust b e canceled b y coun ter terms. The MHV triangles (with all three legs ha ving the same helicit y) giv e 49

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( g + ; g + ; g +; s ) = ( K ^ 21 ) 3 p +1 p +2 p +3 2 3 p 2o ( g + ; g + ; g +; q ) = ( K ^ 21 ) 3 p +1 p +2 p +3 8 3 p 2o ( g + ; g + ; g +; g ) = ( K ^ 21 ) 3 p +1 p +2 p +3 4 3 p 2o (4{7) Here, w e again observ e that with N = 4 SYM eld con ten t, the gluon v ertex correction v anishes. As a comparison, I shall giv e the result when the o shell leg tak es on + comp onen t. ( g + ; g ; g + ; s ) = ( p +1 p +2 ) 1 6 log p 2o e r 5 18 ( g + ; g ; g + ; q ) = ( p +1 p +2 ) 4 3 log p 2o e r 26 9 ( g + ; g ; g + ; g ) = ( p +1 p +2 ) 5 3 log p 2o e r + 31 9 + IR terms ( g + ; g + ; g + ) = 0 (4{8) Although ( g + ; g ; g + ) are not directly used in the computation of the scattering amplitudes, they are a guide to x the coun ter terms 4.2.2 F ermion V ertex Correction The fermion v ertex correction Fig. 4-4 is giv en b y ( q + ; q + ; g +) = 2 q + K ^ p 2 ;q 3 log p 2o e r + 6 + IR terms (4{9) The notation here is that: q means an incoming fermion line while q means an outgoing fermion line; q + means that the fermion is righ t handed; g + means that the gluon is righ t handed whic h corresp onds to ^ The reader can refer to Fig. 4-4 The factor ig 3 = (8 2 )T r[ t a t b t c ] is omitted as usual. The other com bination is 50

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( q + ; q + ; g ) = 2 p +2 q + p 1 K p 1 ;q 3 log p 2o e r + 6 + IR terms (4{10) While ( q ; q ; g +), ( q + ; q + ; g ); and ( q ; q ; g ), ( q + ; q + ; g +) are related b y c harge conjugation 1 These results up to the IR terms are the same with an y one of the legs b eing o shell. W e can also ha v e the case when a gluon is o shell but taking on + comp onen t. ( q + ; q + ; g + ) = 2 p +2 log p 2o e r + 2 + IR terms 4.2.3 Scalar V ertex Correction Scalar v ertex Fig. 4-5 is giv en b y ( s; s; g +) = 2 q + K ^ 2 ; 1 4 log p 2o e r + 17 2 + ( k ^ 2 + k ^ 1 + k ^ 4 ) + IR terms (4{11) with ( s; s; g ) equal to ( s; s; g +) with the ob vious c hange of ^ to This result is also the same up to IR terms with an y one of the legs b eing o-shell. 4.2.4 F our-P oin t F unctions The b o x diagrams are to o cum b ersome to presen t here, but I do w an t to p oin t out that, up to the infrared terms, the total con tribution of the b o x diagrams in a sp ecic amplitude is surprisingly simple. I shall list the rational part of the con tribution of some b o xes in conjunction with the scattering pro cesses in whic h they o ccur. 4.3 Scattering Amplitudes I shall start with gluon scattering, with a factor of ig 4 = (8 2 )T r[ t a t b t c t d ] omitted. 1 up to the external line factors whic h are dened asymmetrically see the explanation b elo w Eq. 2{24 51

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4.3.1 Helicit y Violating Amplitudes The tree lev el amplitude for four gluons with the same helicit y is zero. A t one lo op, the amplitude is A ( g + ; g + ; g + ; g +; s ) = 4 3 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st A ( g + ; g + ; g + ; g +; q ) = 16 3 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st A ( g + ; g + ; g + ; g +; g ) = 8 3 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st (4{12) The tree lev el amplitude with only one unlik e helicit y is zero to o. A t one lo op lev el, the amplitude is A ( g + ; g + ; g + ; g ; s ) = 1 6 ( s + t ) K ^ 2 13 p +2 p +4 K ^ 43 K 32 K 21 K ^ 14 A ( g + ; g + ; g + ; g ; q ) = 2 3 ( s + t ) K ^ 2 13 p +2 p +4 K ^ 43 K 32 K 21 K ^ 14 A ( g + ; g + ; g + ; g ; g ) = 1 3 ( s + t ) K ^ 2 13 p +2 p +4 K ^ 43 K 32 K 21 K ^ 14 (4{13) It is easily observ able that the helicit y violating amplitude is zero if there is an y amoun t of sup ersymmetry [ 17 ]. F or example: N = 1 : 16 3 1 2 + 8 3 1 = 0 2 3 1 2 + 1 3 1 = 0 N = 2 : 16 3 1 + 8 3 1 + 4 3 2 = 0 2 3 1 + 1 3 + 1 6 2 = 0 N = 3(4) : 16 3 2 + 8 3 1 + 4 3 6 = 0 2 3 2 + 1 3 1 + 1 6 6 = 0 (4{14) Since for gluon scattering up to one lo op, adding sup ersymmetry is simply taking in to accoun t the m ultiplicit y of eac h sp ecies, hence N = 3 is the same as N = 4. In the list of amplitudes, the fermions are assumed to b e dirac, this explains the strange lo oking 1/2 m ultiplicit y for the fermions in the rst line of Eq. 4{14 52

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4.3.2 Helicit y Conserving Amplitudes The helicit y conserving amplitude is non-zero at tree lev el. They are giv en b y [ 20 ] A ( g + ; g + ; g ; g ) = ig 2 f abe f ecd 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 A ( g + ; g ; g + ; g ) = ig 2 f abe f ecd 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 (4{15) The factor f abe f ecd can b e con v erted to 1 =C ( G )T r [ t a ; t b ][ t c ; t d ] 2 =C ( G )T r t a t b t c t d Here I ha v e c hosen t a to b e the structure constan ts, while most literature pic ks t a to b e in the fundamen tal represen tation. So, instead of the one lo op amplitude gaining a factor of N c here the tree amplitude is down b y a factor of N c A t one lo op lev el, the amputated Green's function is (with external leg corrections omitted) 53

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A ( g + ; g + ; g ; g ; s ) = 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 1 18 + 1 6 log e r t 1 6 + 1 3 A ( g + ; g + ; g ; g ; q ) = 2 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 19 9 2 3 log e r t 1 3 + 2 3 A ( g + ; g + ; g ; g ; g ) = 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 log 2 s t + 2 11 3 log e r t + 73 9 1 3 + 2 3 (4{16) A ( g + ; g ; g + ; g ; s ) = 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 s 2 t 2 2( s + t ) 4 log 2 s t + 2 + s (2 t 2 5 st s 2 ) 6( s + t ) 3 log s t + 1 6 log e r s + ts 2( s + t ) 2 + 1 18 1 6 + 1 3 A ( g + ; g ; g + ; g ; q ) = 2 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 st ( t 2 + s 2 ) 2( s + t ) 4 log 2 s t + 2 + s (5 t 2 + st + 2 s 2 ) 3( s + t ) 3 log s t 2 3 log e r s + ts ( s + t ) 2 + 19 9 1 3 + 2 3 A ( g + ; g ; g + ; g ; g ) = 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 ( s 2 + st + t 2 ) 2 ( t + s ) 4 log 2 s t + 2 + s 3 (14 t 2 + 19 st + 11 s 2 ) ( s + t ) 3 log s t 11 3 log e r s + ts ( s + t ) 2 + 73 9 1 3 + 2 3 (4{17) The sym b ol ab o v e is the relev an t four p oin t v ertex: 2( p +1 p +3 + p +2 p +4 ) = [( p +1 + p +4 )( p +2 + p +3 )] or 2( p +2 p +3 + p +1 p +4 )[( p +1 + p +2 )( p +3 + p +4 )] + 2( p +1 p +2 + p +3 p +4 ) = [( p +1 + p +4 )( p +2 + p +3 )]. Here I also giv e the rational con tribution of the corresp onding b o x diagrams: 54

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B ( g + ; g + ; g ; g ; s ) = 4 9 all terms B ( g + ; g + ; g ; g ; q ) = 16 9 all terms B ( g + ; g + ; g ; g ; g ) = 8 9 all terms B ( g + ; g ; g + ; g ; s ) = 1 9 + st 2( s + t ) 2 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 B ( g + ; g ; g + ; g ; q ) = 4 9 2 st ( s + t ) 2 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 B ( g + ; g ; g + ; g ; g ) = 2 9 + st ( s + t ) 2 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 (4{18) So, the rational part of a b o x diagram v anishes in N = 4 SYM together with the earlier observ ation that t w o and three p oin t functions v anish in N = 4 SYM This agrees with the 'no-triangle' assertion and some other tec hnical observ ations that are commonly used no w ada ys to simplify the computation of N = 4 SYM amplitudes. Using a b o x reduction pro cedure [ 18 ], an in tegrand with d p o w ers of momen ta in the n umerator and n propagators can b e reduced to tensor b o x in tegrals of degree up to d + 4 n F or gauge in teractions, all three p oin t v ertices ha v e one p o w er of momen ta, so there is n p o w ers of momen ta in the n umerator of an n-gon diagram. Then the result of reduction is a com bination of degree four b o x in tegrals, and some of the degree four b o x in tegrals can b e further reduced to triangle and bubble in tegrals. In N = 4 SYM due to the ultra violet cancelation b et w een dieren t sp ecies, the degree of an n-gon in tegrand will b e n-4, so the result of reduction is th us scalar b o x in tegrals. This essen tially is the 'no triangle' assertion, whic h w e did observ e in Section 4.2 of v ertex corrections. While it can also b e sho wn that the scalar b o x in tegrals will not in an y w a y pro duce rational terms, this I ha v e explicitly sho wn in the ab o v e list. The gluon scattering in N = 4 SYM is v ery simple (up to infrared terms): 55

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A ( g + ; g + ; g ; g ; S Y M ) = 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 log 2 s t + 2 A ( g + ; g ; g + ; g ; S Y M ) = 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 log 2 s t + 2 (4{19) without any ne e d for c ounter terms 4.3.3 Restoring Gauge Co v ariance So far, w e ha v e only studied the pure gluon scattering, and w e ha v e encoun tered some non-gauge-co v ariance (non-Loren tz-Co v ariance) suc h as the hanging four p oin t v ertex. W e ha v e to complete ev erything to a tree in order to main tain Loren tz co v ariance. A t four p oin t lev el, the spinor structure of the leading order amplitude is the unique one that agrees with all the helicit y assignmen t. So for the result to b e Loren tz co v arian t, it has to b e prop ortional to the leading order. This remains true to all orders if sup ersymmetry is presen t according to [ 17 ], whic h sa ys MHV amplitudes are prop ortional to tree amplitudes. W e cannot just simply use a coun ter-term to cancel the hanging four p oin t v ertex, b ecause they are not p olynomials in external momen ta. Ho w ev er, w e can adjust the relativ e strength of the exc hange diagram to the con tact diagram b y adding a term prop ortional to p 2 to the self-energy term Eq. 4{4 This mo dication only c hanges the eld strength renormalization b y a constan t, hence is p erfectly allo w ed. With this term, the co ecien t of s and t c hannel exc hange diagram is shifted. So if w e pic k the n umerical factor in fron t of p 2 to b e 1 = 6 ; 1 = 3 ; 1 = 3 for scalar, fermion and gluon resp ectiv ely then they will matc h the co ecien t of the lone four p oin t v ertex, completing it to a full tree. This brings ab out a c hange in the n umerical factor: 1 = 18 5 = 18, 19 = 9 13 = 9, 73 = 9 67 = 9 [ 14 ]. I ha v e also done the computation with dimension regulation. The pro cedure w as to use dimension regulation to regulate the transv erse momen tum in tegral, and as so on as this is done, will b e set to zero. Hence the infrared regulator is still k + The computation sho ws that the hanging four p oin t v ertex and the pure n um b er will v anish, 56

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this is exp ected since I ha v e used a gauge in v arian t regulator. The n umerical factor in this sc heme will come out in agreemen t with [ 23 ], who used dimension regulation through and through, (not that the n umerical factor is an y thing imp ortan t, as it can b e altered b y a redenition of coupling constan t). This t yp e of coun ter-term will b e put to a more sev ere test later on when w e consider the scattering of not just gluons but quarks and scalars. These coun ter terms had b etter b e univ ersal in the sense that they are only the prop ert y of the self-energy bubble, and should not dep end up on what pro cess it is em b edded in to. 4.3.4 All 2-2 Pro cesses A ( s; s; g + ; g ) = 8 K ^ 2 13 K 2 14 sp +3 p +4 p +21 9 2( s + 2 t ) ( s + t ) log e r s 2 s ( s + t ) log e r t s 2 + st + t 2 ( s + t ) 2 log 2 s t + 2 1 2 1 + ( p +2 p +1 )( p +4 p +3 ) ( p +1 + p +2 ) 2 + 1 A ( g + ; s; g ; s ) = 8 K ^ 2 21 K 2 32 stp +1 p +3 p +22 9 2 log e r s 2 log e r t log 2 s t + 2 1 ( 2) A ( s; s; s; s ) = 59 s 2 + 77 st + 59 t 2 3 st 3(2 s + 3 t ) s log e r s 3(2 t + 3 s ) t log e r t 2( s 2 + t 2 ) st log 2 s t + 2 1 2 ( p +1 p +2 )( p +3 p +4 ) ( p +1 + p +2 ) 2 + ( p +4 p +1 )( p +2 p +3 ) ( p +2 + p +3 ) 2 while A ( s; s; s; s ) 0 = 2( s 2 + st + t 2 ) = ( st ).The ab o v e are the pro cesses only in v olving b osons, the sc heme for pic king coun ter term describ ed in Section 4.3.3 remains v alid, namely w e can nd a univ ersal set of n umerical factors that will complete all the ab o v e amplitudes in to trees. This set can b e c hosen as -1/3 for ( g + ; g ; g ), -1/3 for ( g + ; g ; q ), -1/6 for ( g + ; g ; s ) and -1/2 for ( s; s ). 57

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Ho w ev er, when fermions are in v olv ed, I failed to nd a univ ersal set of coun ter terms that will x the problem. I shall list the result of the computation rst. A ( q ; q ; q ; q ) = 4 K ^ 21 K 43 tp +1 p +2 67 9 11 3 log e r t log 2 s t + 2 1 3 4 p +3 p +4 ( p +1 + p +4 ) 2 (4{20) A ( q ; q ; q ; q ) = 4 K ^ 13 K 24 ( s + t ) stp +1 p +3 67 9 11 3 t s + t log e r s 11 3 s s + t log e r t 2 s 2 + st + 2 t 2 2( s + t ) 2 log 2 s t + 2 1 3 4 p +2 p +4 ( p +1 + p +4 ) 2 4 p +2 p +3 ( p +1 + p +2 ) 2 (4{21) The notation here is that a q represen ts an incoming fermion line, a q represen ts an outgoing one, '+' corresp onds to righ t handedness while corresp onds to left handedness. So the pro cess corresp onding to Eq. 4{20 and Eq. 4{21 is Fig. 4-6 : A ( q ; q ; q + ; q +) = 4 K ^ 14 K 32 sp +1 p +3 67 9 11 3 log e r s log 2 s t + 2 1 3 4 p +2 p +4 ( p +1 + p +2 ) 2 In fact, b y c harge conjugation, A ( q ; q ; q + q +) can b e related to A ( q ; q ; q ; q ), whic h then can b e obtained from rotating A ( q ; q ; q ; q ) clo c kwise b y one notc h (up to the external line factors). As w as explained in Section 2.2 instead of asso ciating p p + to eac h fermion line, I only asso ciate a factor of p + to an outgoing line, but nothing to an incoming line. This is simply b ecause a square ro ot alw a ys causes troubles in automated computation. F or example, 4 K ^ 14 K 32 = ( sp +1 p +3 ) should really b e 4 K ^ 14 K 32 = ( s p p +1 p +2 p +3 p +4 ), whic h is quite clear as the latter is Loren tz co v arian t (can b e written as spinor pro ducts), but the former is not. The phase of the square ro ot here can b e giv en arbitrarily 58

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A ( s; s; q ; q ) = 4 K ^ 13 K 14 sp +1 p +3 92 9 29 6 log e r s 2 t + s 2( s + t ) log 2 s t + 2 2 3 2 p +4 ( p +2 p +1 ) ( p +1 + p +2 ) 2 So far, the eect of using a non-co v arian t regulator is the mismatc h b et w een the exc hange v ertex and 4 p oin t v ertex in the nal result. Ho w ev er, in the case of gluon fermion scattering, the mismatc h is m uc h w orse. A ( g + ; g ; q ; q ) = 8 K ^ 2 13 K 32 K 24 stp +1 p +2 p +23 6 3 log e r s log 2 s t + 2 1 3 s c hannel tree A ( g + ; g ; q + ; q +) = 8 K ^ 14 K ^ 13 K 2 32 stp +1 p +2 p +3 p +4 6 3 log e r s log 2 s t + 2 1 3 s c hannel tree Here w e see that the mismatc h is b et w een s and t c hannel. By s-c hannel tree, I mean the s-c hannel exc hange diagram and also its descenden t four p oin t v ertex. A ( g + ; q ; g ; q ) = 8 K ^ 2 21 K 43 K 32 stp +1 p +3 p +22 6 3 log e r t log 2 s t + 2 Here, it is mere coincidence that ev erything matc hes (the planar condition eliminates the u-c hannel tree). The problem of xing coun ter terms will b e revisited in the future w ork discussion. PSfrag replacemen ts a b q p k 3 k 1 Figure 4-1. Tw o-p oin t function and the dual momen tum assignmen t 59

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PSfrag replacemen ts a b c d e f Figure 4-2. A self-energy diagram em b edded in a scattering pro cess PSfrag replacemen ts ^ ; a; p 1 ^ ; b; p 2 ; c; p 3 k 1 k 4 k 2 q Figure 4-3. T riangle diagram PSfrag replacemen ts ^ ; a r ; b; p 1 r ; c; p 2 k 1 k 4 k 2 q Figure 4-4. F ermion v ertex correction corresp onding to Eq. 4{9 PSfrag replacemen ts ^ ; a q b; p 1 c; p 2 k 1 k 4 k 2 Figure 4-5. Scalar v ertex correction corresp onding to Eq. 4{11 PSfrag replacemen ts p 1 ; l p 1 ; l p 2 ; l p 2 ; l p 3 ; l p 3 ; l p 4 ; l p 4 ; l Figure 4-6. F our-fermion scattering corresp onding to Eq. 4{20 and Eq. 4{21 60

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CHAPTER 5 BREMSSTRAHLUNG IN LIGHT CONE In this c hapter, I shall deal with the infrared terms. 5.1 A List of Infrared T erms All the amplitudes giv en in Section 4.3.4 are amputated Green's function, the external leg corrections are not included. As w e can see that the self-energy diagram alw a ys con tains a term log e r p 2 x (1 x ), whic h giv es a m ulti-particle branc h cut on the p ositiv e real axis, stopping us from doing w a v e function renormalization. This can b e cured b y summing o v er collinear emissions or absorptions from the external legs. The analysis of [ 14 ] sho w ed that doing so is equiv alen t to replacing log e r p 2 x (1 x ) with log e r 2 x (1 x ), b eing the jet resolution. F or ( g ; g ; g ; g ; s ) or ( g ; g ; g ; g ; q ), this substitution alone is enough to regulate the infrared div ergence (the triangle or b o x diagrams in v olving fermions or scalars are dev oid of further infrared div ergences). While all the other pro cesses (whenev er there is a gluon propagator juxtap osed b et w een t w o massless external legs), the computation is fraugh t with IR terms. But these IR terms when com bined nally are univ ersal, whic h sho ws that they are ph ysical infrared div ergences, instead of ligh t cone gauge artices. Here I list all of them rst. All momen ta app earing b elo w are dual momen ta, with the real momen ta giv en b y p +1 = k + 1 k + 2 p +2 = k + 4 k + 1 p +3 = k + 3 k + 4 p +4 = k + 2 k + 3 and k + 2 < k + 3 < k + 1 < k + 4 see Fig. 5-1 61

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ig 4 8 2 T r[ t a t b t c t d ] k + 2 < q + < k + 3 log ( k + 4 q + )( k + 2 + q + ) s e r ( k + 2 k + 4 ) 2 1 q + k + 3 + 1 q + k + 1 + log ( k + 1 q + )( q + k + 2 ) s e r ( k + 1 k + 2 )( k + 4 k + 2 ) 2 q + k + 1 + log ( k + 3 q + )( k + 2 + q + ) s e r ( k + 3 k + 2 )( k + 4 k + 2 ) 2 q + k + 3 + log ( k + 2 + q + ) 2 t e r ( k + 1 k + 2 )( k + 3 k + 2 ) 2 q + k + 2 k + 3 < q + < k + 1 log ( k + 4 q + )( k + 2 + q + ) s e r ( k + 2 k + 4 ) 2 1 q + k + 3 + 1 q + k + 1 + log ( k + 3 q + )( q + k + 1 ) t e r ( k + 1 + k + 3 ) 2 1 q + k + 2 + 1 q + k + 4 + log ( k + 2 q + )( q + k + 1 ) s e r ( k + 1 + k + 2 )( k + 2 k + 4 ) 2 q + k + 1 + log ( q + k + 3 )( k + 4 q + ) s e r ( k + 4 k + 2 )( k + 4 k + 3 ) 2 q + k + 3 + log ( k + 4 + q + )( k + 3 q + ) t e r ( k + 3 k + 4 )( k + 1 + k + 3 ) 2 q + k + 4 + log ( k + 2 q + )( q + k + 1 ) t e r ( k + 1 + k + 2 )( k + 1 + k + 3 ) 2 q + k + 2 k + 1 < q + < k + 4 log ( k + 4 q + )( k + 2 + q + ) s e r ( k + 2 k + 4 ) 2 1 q + k + 3 1 q + k + 1 + log ( q + k + 1 )( k + 4 + q + ) s e r ( k + 4 k + 1 )( k + 2 k + 4 ) 2 q + k + 1 + log ( q + k + 3 )( k + 4 q + ) s e r ( k + 4 k + 2 )( k + 4 k + 3 ) 2 q + k + 3 + log ( k + 4 + q + ) 2 t e r ( k + 4 k + 1 )( k + 4 k + 3 ) 2 q + k + 4 62

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The ab o v e is alw a ys m ultiplied b y the corresp onding tree amplitude (b y a tree, I mean whatev er I wrote in fron t of the square brac k et in the list of amplitudes). As long as the pro cess has infrared div ergence, the infrared div ergence will b e of the ab o v e form. The infrared terms ab o v e will b e com bined with soft Bremsstrahlung and collinear emission (absorptions) along with the self-energy insertions on the external legs to giv e a nite result. The basic idea is the same as the standard treatmen t of infrared div ergence. That is, w e insist on measuring jets (within a certain resolution ) rather than gluons or quarks. 5.2 Bremsstrahlung Pro cess I will b e fo cusing on nal state bremsstrahlung radiation (region b et w een leg 3 and 4) Fig. 5-2 and study the four gluon core scattering pro cess as an example. p i ; i = 1 : : : 4 and k are assumed to b e incoming, and k will b e called the extra one for no w. The momen ta of the 2-2 core pro cess will b e denoted as j i i = 1 : : : 4 (j for jet). P ossible sources of div ergence asso ciated with region 34 are 1. k is collinear with p 3 or p 4 2. k is soft. W e can lo ok at roughly ho w eac h t yp e of div ergences get canceled. F rom Fig. 5-3 it is quite natural to guess that the collinear div ergence arising from k b eing parallel with p 3 is going to b e cancelled b y the corresp onding self-energy bubble on leg 3. The cancellation will w ork as long as k is not to o soft. W e can mak e sure of this b y setting k + a w a y from zero as a cut o. What happ ens when k 0 is that w e lose coherence. More concretely when k is not to o soft, w e only need to w orry ab out the case when it is attac hed to p 3 When k 0, it can b e attac hed to either p 3 or p 4 So w e will b e considering the follo wing cancelation of Fig. 5-4 There is no natural b oundary as to when k is soft enough. In fact, w e can rst imp ose an articial b oundary A suc h that when k + > A w e use the sc heme in Fig. 5-3 when k + < A w e use the sc heme in Fig. 5-4 When the dust settles the A dep endence disapp ears. 63

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The metho d describ ed ab o v e w as used in [ 14 ], it in v olv es a careful analyzing of phase space to a v oid double coun ting. Ho w ev er the b eautiful P ark e-T a ylor expression for helicit y amplitudes greatly simplies the computation: the rst diagram in Fig. 5-2 is giv en b y (in the large N c limit) ( g p N c ) 3 h p 1 j p 2 i 4 h p 1 j p 2 ih p 2 j p 3 ih p 3 j k ih k j p 4 ih p 4 j p 1 i 2 = g 6 N 3 c 2 ( p 1 p 2 ) 4 ( p 1 p 2 )( p 2 p 3 )( p 3 k )( k p 4 )( p 4 p 1 ) (5{1) It's quite clear from Eq. 5{1 that when k is collinear with p 3 ; 4 or soft, there will b e div ergences. As stated b efore, w e ha v e to set k + a w a y from zero in order to cut o infrared div ergence, but this do es not regulate collinear div ergence, so a temp orary cut-o will b e used ten tativ ely to cut o collinear div ergence. In order to parameterize the phase space, dene (refer to Fig. 5-5 ) x = 2 Q ( p 3 ) Q 2 ; y = 2 Q ( p 4 ) Q 2 ; Q = p 1 + p 2 (5{2) The in tegration region of x and y is depicted in Fig. 5-5 w e will only b e in tegrating x and y in the 'L' shap ed region. R is related to detector resolution according to R := 2 =s The ph ysical meaning of these cuts is quite clear: when x 1, k p 4 = 1 = 2( k + p 4 ) 2 = 1 = 2( Q + p 3 ) 2 = 1 = 2 Q 2 (1 x ) = 0; when x; y 1, k Q 0 (since Q is time-lik e, k 0). Here and after, s and t will denote the Mandelstam in v arian ts for the 2-2 core pro cess. A tric k used in [ 24 ] turned out useful. W rite 1 ( p 3 k )( k p 4 ) = 1 ( p 3 k )( k ( p 3 + p 4 )) + 1 ( p 4 k )( k ( p 3 + p 4 )) = 2 s (1 y ) 2 s (2 x y ) + 2 s (1 x ) 2 s (2 x y ) (5{3) 64

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The rst term will only div erge when p 3 and k are collinear and(or) k is soft. The second term will only div erge when p 4 and k are collinear and(or) k is soft. F or the second term, the in tegration of x and y in area A: 1 R < x < 1 ; 1 x < y < 1 is div ergen t, but nite in the area B : 1 R < y < 1; 1 y < x < 1 R These t w o areas will b e treated sligh tly dieren tly Region A: Since x 1, p 4 and k are either collinear or k is soft. All x at inno cuous places can b e set to b e 1. F urther dene L to b e a reference (ligh t-lik e, L 0 > 0) 4-v ector, suc h that L p = p + No w the 3 b o dy phase space in tegral is giv en b y sx 64(2 ) 5 d n 3 dxd n 4 k d n 3 is the angular in tegral of p 3 d n 4 k includes the angular in tegral of p 4 and k dx represen ts the in tegral of the norm of ~ p 3 Note that if the phase space in tegral is isotropic, it can b e written as the standard form sdx dy = 128 3 for a 3-b o dy nal state. W e can factorize the phase space in tegral in to [1 = (8(2 ) 2 ) d n 3 ] [ sx= (8(2 ) 3 ) dxd n 4 k ]. Here p 3 is pic k ed out as 'the sp ecial one', since in region A, p 3 is alw a ys hard, and is almost equal to j 3 of the 2-2 core pro cess. I will in tegrate out [ sx= (8(2 ) 3 ) d n 4 k ], while the [1 = (8(2 ) 2 ) d n 3 ] part is what should b e compared to the 2-2 pro cess. The second term of Eq. 5{3 is parameterized as g 6 N 3 c 2 ( p 1 p 2 ) 4 ( p 1 p 2 )( p 2 p 3 )( p 4 k )( k ( p 4 + p 3 ))( p 4 p 1 ) (5{4) = g 6 N 3 c s 4 4 s ( p 2 + p 3 ) 2 ( p 4 + p 1 ) 2 1 ( p 4 k )( k ( p 4 + p 3 )) = g 6 N 3 c s 4 s ( p 2 + p 3 ) 2 ( p 4 + p 1 ) 2 1 s (1 x ) s (2 x y ) 65

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As x 1, ( p 2 + p 3 ) 2 t ( p 1 + p 4 ) 2 y t The rst appro ximation is alw a ys go o d in region A, the second is go o d as long as p 4 is not to o soft. Plug in the phase space in tegral, th us arriving at Z sx 64 3 g 6 N 3 c s 2 st ( y t ) 1 (1 x )(2 x y ) dxd n 4 k = Z x 64 3 g 2 N c A 2cor e 1 (1 x )(2 x ) ( 1 y + 1 2 x y ) dxd n 4 k (5{5) where A cor e = g 4 N 2 c ( s=t ) 2 let us drop the factor g 2 N c A 2cor e hereafter. d n 4 k can b e expressed as d n 4 k := 2 dudv p ( u 1 u )( u u 0 ) u := 1 2(1 y ) x v := 1 2 j k + j j Q + + p +3 j 1 2 j k + j j j + 4 j ; ( j k + j < j j + 4 j ) w := 1 2 j p +3 j (1 x ) j Q + + p +3 j x := 1 2 b (1 x ) x ; b = j j + 3 j j j + 4 j (5{6) W e can rst pic k a particular v alue of p 3 then go to the CM frame of p 1 + p 2 + p 3 to ev aluate the in v arian t expression ( k + p 4 + p 1 + p 2 + p 3 ) ( k 2 ) ( p 24 ) d 4 k d 4 p 4 Notice in this frame, u is in fact cos ( ^ p 3 k ), v is cos ( ^ Lk ) and w is cos ^ p 3 L Then after some exercises in Euclidean geometry in Fig. 5-6 w e can obtain the expression for d n 4 k u 0 ; 1 := v w p (1 v 2 )(1 w 2 ) are the lo w er/upp er limit of u ( y ) in tegral. The limit placed up on k + is for breaking the Bose symmetry b et w een p 4 and k and to mak e sure that the appro ximation ( p 1 + p 4 ) 2 y t w orks (the dep endence will drop out). P erforming the u in tegral using 66

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Z 1 1 1 a + bx 1 p 1 x 2 = p a 2 b 2 ( a > b > 0) = p a 2 b 2 ( a > b > 0) Z 1 1 2 dv Z 1 1 R dx sx 64 3 A 2cor e 1 (1 x )(2 x ) 2 x ( 2 q ( 2 x 1 + u 1 )( 2 x 1 + u 0 ) + 2 q ( 2 x 1 u 1 )( 2 x 1 u 0 ) ) (5{7) The rst square ro ot can b e obtained from the second one b y replacing v b y v So I will only compute the second square-ro ot. The follo wing indenite in tegral will p erhaps b e useful: Z dx 1 p x 2 + 1 1 a + bx = 1 p a 2 + b 2 log b + ax + p ( a 2 + b 2 )( x 2 + 1) a + bx The second term of Eq. 5{7 b ecomes 1 16 2 Z 1 1 2 dv 1 v log 1 v (1 + v ) + log [(1 v ) + (1 + v 2 b ) R ] + q [(1 v ) + (1 + v 2 b ) R ] 2 + 8 bR 2 (1 + v ) 4 bR (5{8) here w e ha v e treated the t w o cut-o 's R and dieren tly: is set to zero with xed v (xed k + ) while R is tak en to b e small but non-zero. This treatmen t agrees with the strategy giv en in the paragraph ab o v e Eq. 5{1 Notice the second term in the square brac k et is nite (when k + is k ept a w a y from zero), while the rst will b e com bined with self mass con tributions to cancel the dep endence. W e can mak e a c hange of v ariable (the y b elo w is not the previous y ): 67

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[(1 v ) + (1 + v 2 b ) R ] + q [(1 v ) + (1 + v 2 b ) R ] 2 + 8 bR 2 (1 + v ) = 4 bR y 1 v = 2 R 1 by 2 y (1 b ) y (1 R ) + R ; dv dy = 2 R b (1 R ) y 2 + 2 bR y + 1 bR [ R + y (1 R )] 2 (5{9) then the in tegral b ecomes 1 16 2 Z y 1 y 0 b (1 R ) y 2 + 2 bR y + 1 bR [ R + y (1 R )][1 by 2 y (1 b )] log y + 1 16 2 Z 1 1 2 dv 1 v log 1 v (1 + v ) = 1 16 2 Z y 1 y 0 1 R [ R + y (1 R )] + 1 1 y 1 y + 1 =b log y + 1 16 2 Z 1 1 2 dv 1 v log 1 v (1 + v ) (5{10) where y 0 R (1 ) = ; y 1 = 1. The ab o v e in tegral b ecomes 1 16 2 Z y 1 y 0 1 R [ R + y (1 R )] log y 2 6 dilog(1 + b ) + 1 16 2 Z 1 1 2 dv 1 v log 1 v (1 + v ) = 1 16 2 1 2 log 2 R dilog 1 + log R 1 log 2 3 dilog(1 + b ) + Z 1 1 2 dv 1 v log 1 v (1 + v ) (5{11) W e can do one more thing to mak e it more symmetric: as b alw a ys comes in pairs with 1 =b w e ma y replace dilog (1 + j j + 3 =j + 4 j ) with 1 = 4 log 2 j j + 3 =j + 4 j 2 = 12 b y using the form ula dilog [1 = (1 x )] + dilog [1 =x ] = 2 = 6 1 = 2 log 2 (1 =x 1) The rst term of Eq. 5{7 is obtained from the ab o v e b y substituting v to v and the limit of y in tegral b ecomes y 2 [0 ; R = (1 )]. W e get 68

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1 16 2 dilog 1 1 + log R 1 log 1 1 + Z 1 1 2 dv 1 + v log 1 + v (1 v ) = 1 16 2 dilog 1 2 6 log R log (1 ) 1 2 log 2 + 1 2 log 2 (1 ) + 1 2 2dilog log 2 (1 ) + 2 3 + 2 log log (1 ) = log R log (1 ) (5{12) In fact, b y noticing that du= p ( u 1 u )( u u 0 ) b eha v es lik e ( times) function as x 1, I can set y = 1 x (1 u ) = 2 to b e 1 x (1 v ) = 2 (1 + v ) = 2 in the rst term of Eq. 5{5 whic h will lead to the same result in a m uc h quic k er w a y Region B No w p 4 is the hard momen tum, ( p 4 + p 1 ) 2 is equal to t while ( p 3 + p 2 ) 2 is appro ximated as xt The three b o dy phase space in tegral should lik ewise b e factored as d n 4 dy d n 3 k Since there is no need here to hold k + xed, and the restriction j k + j < 0 j Q + + p +4 j can b e lifted, sy = (8(2 ) 3 ) dy d n 3 k is simply sdxdy = (16 2 ). Here I ha v e used a dieren t partition: 0 b ecause the partition asso ciated to leg 3 need not b e the same as leg 4. Z g 6 N 3 c sy dy d n 3 k (8(2 ) 3 ) s 4 s ( p 2 + p 3 ) 2 ( p 4 + p 1 ) 2 1 s (1 x ) s (2 x y ) = Z g 6 N 3 c sdxdy 16 2 s 4 s ( xt ) t 1 s (1 x ) s (2 x y ) = 1 16 2 2 12 g 2 N c A 2cor e (5{13) I ha v e dealt with the rst term of Eq. 5{3 the second is similar: 1 16 2 1 2 log 2 R dilog 1 0 + log R 1 0 0 log 0 2 3 dilog(1 + j j + 4 =j + 3 j ) + Z 0 j j + 3 j 0 dk + k + log k + ( j j + 3 j k + ) + 2 12 + dilog 1 1 0 + log R 0 1 0 log 1 1 0 + Z 0 j j + 3 j 0 dk + j j + 3 j k + log j j + 3 j k + k + (5{14) 69

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So far, I ha v e co v ered the rst diagram of Fig. 5-2 when k do es not dominate o v er neither p 3 nor p 4 ( k + < j j + 4 j k + < 0 j j + 3 j ). T o complete the region k + > j j + 4 j w e need to do a computation in region 14. g 6 N 3 c s 4 4 s ( p 2 + p 3 ) 2 ( p 4 + p 1 ) 2 1 ( p 4 k )( k ( p 4 + p 3 )) = g 6 N 3 c s 3 8( p 2 + p 3 ) 2 ( k ( p 4 + p 3 )) 1 p 4 k 1 p 4 ( k p 1 ) + 1 p 4 p 1 1 p 4 ( k p 1 ) (5{15) with ( k + p 4 ) 2 < R s and j k + j > j j + 4 j One p oin t that I should ha v e emphasized earlier is that in the presence of 1 = ( p 4 k ), w e can only mak e appro ximations con trolled b y O ( p 4 k ), and similarly in the presence of 1 = ( p 4 p 1 ), w e can only mak e appro ximations con trolled b y O ( p 4 p 1 ), otherwise there will b e errors of the t yp e R log This rule mak es the ev aluation of the rst term m uc h easier: w e can set k to b e parallel to p 4 in all the irrelev an t terms. g 6 N 3 c s 3 8( p 2 + p 3 ) 2 ( k ( p 4 + p 3 )) 1 p 4 k 1 p 4 ( k p 1 ) = g 6 N 3 c s 3 8 t ( k ( p 4 + p 3 )) 1 p 4 k 1 p 4 ( k p 1 ) No w set up a parametrization for region 14: ~ x = 2 p 1 ( p 1 p 4 k ) ( p 1 p 4 k ) 2 ~ y = 2 k ( p 1 p 4 k ) ( p 1 p 4 k ) 2 Abbreviate ( p 1 p 4 k ) as T Then k p 4 can b e w ork ed out as 1 = 2(1 ~ x ) T 2 With this parametrization: g 6 N 3 c s 3 8 t ( k ( p 4 + p 3 )) 1 p 4 k 1 p 4 ( k p 1 ) = g 6 N 3 c s 3 t ~ y s 1 (1 ~ x ) T 2 1 (2 ~ x ~ y ) T 2 In the limit ~ x = 1, T 2 = j t j In fact, the ab o v e expression can b e calculated without eort, simply b y iden tifying as 1 and b as p +1 = j p +4 j ( p 1 and p 4 here are the core v alues), w e can b orro w the result from Eq. 5{11 70

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Z g 6 N 3 c T 2 d ~ x d n 4 k 64 3 s 3 t ~ y s 1 (1 ~ x ) T 2 1 (2 ~ x ~ y ) T 2 = g 2 N c A 2cor e 16 2 1 2 log 2 ~ R dilog 1 1 + log ~ R 1 log (1 ) 2 3 dilog(1 + j + 1 j j + 4 j ) + Z j j + 4 j j j + 4 j dk + j j + 4 j k + log j j + 4 j k + ~ k + + 1 16 2 dilog 1 + log ~ R 1 log 1 + Z j j + 4 j j j + 4 j dk + k + log k + ~ ( j j + 4 j k + ) # (5{16) Comparing this to Eq. 5{11 5{12 w e see that the dep endence cancels. No w w e lo ok at the second term of Eq. 5{15 it in fact can b e in terpreted as a false jet, since it will mak e a con tribution when p 4 is parallel with p 1 th us the nal state will b e a three w ell separated jets unless p 4 is soft. Our task is to compute this term in the region 2 k p 4 < R s with j j + 4 j > j k + j > j j + 4 j This region can b e dissem bled in to a part f 1 ~ x < ~ R g T f ~ y < 1 ~ R g whic h pro duces a 1 = (16 2 ) 2 = 12 (the restriction on k + mak es no dierence). There is the other part: f 1 ~ x < ~ R g T f 1 ~ y < ~ R g T fj k + j > j j + 4 jg Here I only la y out the strategy if one w ere to compute it honestly First compute in region f 1 ~ y < ~ R g with no restriction on k + then subtract the en tire region of f ~ x < 1 ~ R g T f 1 ~ y < ~ R g again with no restriction on k + The v alidit y of this strategy lies in the fact that there exists an upp er limit of order O ( ~ R ) for j p +4 j b ey ond whic h there is no in tersection with the region f 1 ~ x < ~ R g T f 1 ~ y < ~ R g while the limit j k + j > j j + 4 j translated to j p +4 j < (1 ) j j + 4 j is w ell b ey ond the said upp er limit. Y et in practice, the condition that w e can only mak e appro ximation of order p 4 p 1 = 0 mak es the ev aluation (and in terpretation) quite hard. So w e mak e a compromise, and allo w for errors of R log trusting that it will go a w a y with a complete calculation. I shall set not only p 4 p 1 = 0 but also p 4 k = 0. Then, this term will cancel against some disconnected diagrams whic h will b e explained later. So the completed result for Eq. 5{4 is 71

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1 16 2 1 2 log 2 ~ R 2 4 dilog(1 + j + 1 j j + 4 j ) + Z j j + 4 j 0 dk + j j + 4 j k + log j j + 4 j k + ~ k + + 1 16 2 1 2 log 2 R 2 4 dilog(1 + j j + 3 j j j + 4 j ) + Z j j + 4 j 0 dk + k + log k + ( j j + 4 j k + ) (5{17) It mak es a similar con tribution asso ciated to leg 3, with some suitable substitutions. Let us tak e a lo ok at the self-energy bubble on leg 4 Fig. 5-7 calculated with a cut o K i 4 ;k K i 4 ;k =p +4 k + > 2 to regulate the on shell div ergence. should b e in this case iden tied with 2 =s and ~ with 2 = j t j The w a v e function renormalization of these diagrams are (with the 1/2 from p Z and another 2 since it en ters as a cross term): left: Z 1 0 1 16 2 1 x + 1 1 x log x (1 x ) 2 middle: Z 1 0 1 16 2 (1 x ) 3 x log x (1 x ) 2 righ t: Z 1 0 1 16 2 x 3 1 x log x (1 x ) 2 (5{18) where x is j k + j = j j + 4 j Not all of Eq. 5{18 will con tribute to region 34. F or no w, I only tak e the rst line. Com bine Eq. 5{18 to the expression ab o v e: 1 16 2 1 2 log 2 R 1 2 log 2 ~ R 2 2 dilog(1 + j j + 3 j j j + 4 j ) dilog(1 + j + 1 j j + 4 j ) + Z j j + 4 j 0 dk + k + log k +2 s j +2 4 + Z j j + 4 j 0 dk + j j + 4 j k + log ( j j + 4 j k + ) 2 j t j j +2 4 Th us, the eect of a self-mass insertion is to replace b y 1 =sx (1 x ) or 1 = j t j x (1 x ). Fig. 5-8 represen ts all the diagrams that con tribute to the Bremsstrahlung pro cess. The second diagram of Fig. 5-2 or Fig. 5-8 when com bined with the second line of Eq. 5{18 will con tribute to leg 4: 72

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1 16 2 1 2 log 2 R 2 4 dilog(1 + j j + 3 j j j + 4 j ) + Z j j + 4 j 0 dk + k + log k +2 s j +2 4 + Z 1 0 ( 3 + 3 x x 2 ) log x (1 x ) R s (5{19) Note its con tribution here should b e attributed to region 34. This diagram w on't con tribute to region 14, b ecause it remains nite when p 4 0. The third is similar to the second: 1 16 2 1 2 log 2 ~ R 2 4 dilog (1 + j + 1 j j + 4 j ) + Z j j + 4 j 0 dk + k + log k +2 j t j j +2 4 + Z 1 0 ( 3 + 3 x x 2 ) log x (1 x ) R j t j (5{20) Note that it is credited to region 14. In summary the con tribution from Bremsstrahlung plus self-mass insertion to region 34 is 1 16 2 2 log 2 2 s 2 2 2 6 1 2 log 2 j j + 3 j j j + 4 j +2 Z j j + 4 j 0 dk + k + log k +2 s j +2 4 + 2 Z j j + 3 j 0 dk + k + log k +2 s j +2 3 + 67 9 11 3 log 2 And a symmetric con tribution to region 14: 1 16 2 2 log 2 2 j t j 2 2 2 6 1 2 log 2 p +1 j p +4 j +2 Z j j + 4 j 0 dk + k + log k +2 j t j j +2 4 + 2 Z j + 1 0 dk + k + log k +2 j t j j +2 1 + 67 9 11 3 log 2 As a summary although the kinematics in 14 and 34 region are v ery dieren t, the results are almost symmetric up to the false jets that I ha v en't included. Next, w e study ho w to com bine these results to the virtual pro cess, and defer the discussion of the false jets. 73

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5.3 Com bining with the Infrared T erms from the Virtual Pro cess The list of infrared terms in Section 5.1 can b e rewritten in a more symmetric form that is indep enden t of the relativ e size of the dual momen ta: Region 34: 1 8 2 Z j p +4 j 0 dk + k + log k +2 s j p +4 jj p +3 j + Z j p +3 j 0 dk + k + log k +2 s j p +3 jj p +4 j + Z j p +3 j j p +4 j dk + k + log ( j p +3 j k + ) j p +4 j ( j p +4 j + k + ) j p +3 j (5{21) Region 12: 1 8 2 Z j p +1 j 0 dk + k + log k +2 s j p +1 jj p +2 j + Z j p +2 j 0 dk + k + log k +2 s j p +1 jj p +2 j + Z j p +2 j j p +1 j dk + k + log ( j p +2 j k + ) j p +1 j ( j p +1 j + k + ) j p +2 j Region 41: 1 8 2 Z j p +1 j 0 dk + k + log k +2 ( t ) j p +1 jj p +4 j + Z j p +4 j 0 dk + k + log k +2 ( t ) j p +1 jj p +4 j + Z j p +1 j j p +4 j dk + k + log ( j p +1 j k + ) j p +4 j ( k + j p +4 ) j p +1 j Region 23: 1 8 2 Z j p +2 j 0 dk + k + log k +2 ( t ) j p +2 jj p +3 j + Z j p +3 j 0 dk + k + log k +2 ( t ) j p +2 jj p +3 j + Z j p +2 j j p +3 j dk + k + log ( j p +2 j k + ) j p +3 j ( k + j p +3 ) j p +2 j 74

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The rst t w o terms of Eq. 5{21 cancels the div ergence of Eq. 5{19 while the last term in tegrates to b e 1 = (16 2 ) 2 log 2 ( p +3 =p +4 ) Finally all the non-co v ariance pieces cancel and ev erything falls together nicely: 1 8 2 [ log 2 2 s + 2 3 + 67 18 11 6 log 2 ] (5{22) is the total con tribution of virtual and Bremsstrahlung pro cesses to region 34. While for region 14 w e ha v e similarly: 1 8 2 [ log 2 2 j t j 2 2 3 + 67 18 11 6 log 2 ] (5{23) Th us, w e can write do wn the total scattering probabilit y: P ( g + ; g + ; g ; g ; g ) 1 = P ( g + ; g + ; g ; g ; g ) 0 1 + g 2 s N c 8 2 2 log 2 2 s 2 log 2 2 j t j 2 3 + 67 9 11 3 log 4 e r j t j + log 2 s j t j P ( g + ; g ; g + ; g ; g ) 1 = P ( g + ; g ; g + ; g ; g ) 0 1 + g 2 s N c 8 2 2 log 2 2 s 2 log 2 2 j t j 2 3 + 67 9 11 3 log 4 e r s ( s 2 + st + t 2 ) 2 ( t + s ) 4 log 2 s j t j s 3 (14 t 2 + 19 st + 11 s 2 ) ( t + s ) 3 log s j t j ts ( t + s ) 2 And the probabilit y with N = 4 SYM is 75

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P ( g + ; g + ; g ; g ;S Y M) 1 = P ( g + ; g + ; g ; g ) 0 1 + g 2 N c 8 2 2 log 2 2 s 2 log 2 2 j t j 2 3 + [ 67 9 5 18 N s 26 9 N f ] + log e r 4 j t j [ 11 3 + 1 6 N s + 4 3 N f ] + log 2 s j t j P ( g + ; g ; g + ; g ;S Y M) 1 = P ( g + ; g + ; g ; g ) 0 1 + g 2 N c 8 2 2 log 2 2 s 2 log 2 2 j t j 2 3 + [ 67 9 5 18 N s 26 9 N f ] + log e r 4 s [ 11 3 + 1 6 N s + 4 3 N f ] + 1 2( s + t ) 4 log 2 s j t j [ s 2 t 2 N s + 2 st ( s 2 + t 2 ) N f + 2( s 2 + st + t 2 ) 2 ] + 1 6( s + t ) 3 log s j t j [ s (2 t 2 5 st s 2 ) N s + 4 s (5 t 2 + st + 2 s 2 ) N f 2 s (14 t 2 + 19 st + 11 s 2 )] + st 2( s + t ) 2 [ N s + 4 N f 2] (5{24) 5.4 The Inclusion of Disconnected Diagrams In this section I will discuss certain disconnected diagrams Fig. 5-9 whose imp ortance w as explained in [ 16 ]. Only the cross term b et w een the rst and others in Fig. 5-9 has the correct p o w er in coupling constan t, and the factor 2 E k (2 ) 3 3 ( ~ k ~ k 0 ) = (2 ) 4 4 ( k k 0 ) corresp onding to the forw ard particle line will force the t w o extra legs to ha v e the same momen tum. The fourth and fth diagram of Fig. 5-9 lo ok rather lik e self-energy diagrams, and indeed they will receiv e the same factor of 1 = 2 discoun t just as self-energy diagrams on the external legs do. The fourth (resp. fth) diagram will ha v e to b e ev aluated rst with p 4 (resp. p 1 ) o shell. When the dust settles, an y term that do not con tain p 21 or p 24 will b e dropp ed just the same as w e are w on t to drop the tadp ole diagrams. W e ha v e insisted on ha ving t w o partons scattering in to t w o jets, so the t w o extra gluons ha v e to b e soft. The F eynman rules giv e (with the appro ximation k 0) 76

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ig r N c 2 2 A 2cor e 8 p 1 p 4 ( k + p 1 ) 2 ( k p 4 ) 2 g 6 N 3 c s 2 t 2 t 2( k p 1 )( k p 4 ) (5{25) to b e ev aluated in the region ( k p 3 ) 2 < 2 and ( k p 4 ) 2 < 2 Next I sho w that it will cancel the false jet terms. Setting p 4 p 1 = 0 and p 4 k = 0 in the second term of Eq. 5{15 : g 6 N 3 c s 3 8( p 2 + p 3 ) 2 ( k ( p 4 + p 3 )) 1 p 4 p 1 1 p 4 ( k p 1 ) g 6 N 3 c s 3 4 ts 1 p 4 p 1 1 p 4 ( k p 1 ) The third diagram of Fig. 5-8 will double the ab o v e. T o mak e comparison with the disconnected diagrams, w e need to iden tify k with p 4 and p 4 with k : 2 g 6 N 3 c s 3 4 ts 1 p 4 p 1 1 p 4 ( k p 1 ) g 6 N 3 c s 3 2 ts 1 ( k p 1 )( k ( p 4 p 1 )) (5{26) The fth and sev en th diagram will giv e a similar con tribution: 2 g 6 N 3 c s 3 8( p 2 + p 3 ) 2 ( k ( p 1 + p 2 )) 1 p 1 p 4 1 p 1 ( k p 4 ) g 6 N 3 c s 3 2 ts 1 p 1 p 4 1 p 1 ( k p 4 ) No w iden tify k with p 1 and p 1 with k : g 6 N 3 c s 3 2 ts 1 p 1 p 4 1 p 1 ( k p 4 ) g 6 N 3 c s 3 2 ts 1 ( k p 4 )( k ( p 1 p 4 )) (5{27) The sum of Eq. 5{26 and Eq. 5{27 cancels Eq. 5{25 The cancelation ab o v e is ad ho c and certainly not the prettiest. F or one thing, w e ha v e made appro ximations of order O ( R log ), for another, the in terpretation of the disconnected diagrams as false jets is not in tuitiv e. Ho w ev er w e can see the similarit y b et w een Fig. 5-9 and the false jet terms: 5{26 (resp. 5{27 ) w ould b e a lot more natural had p 4 (resp. p 1 ) b e an incoming(resp. outgoing) particle. While if w e could c hange the sign of k 0 in Fig. 5-9 they w ould all b ecome true 77

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lo op diagrams. So b oth the false jets and Fig. 5-9 ha v e the 'incoming-outgoing-rev ersed' problem. I b eliev e a b etter w a y of dealing with them is p ossible. In [ 14 ], the false jets come more naturally They come from the diagrams with leg 4 absorbing a gluon collinear with itself, or leg 1 emitting a gluon collinear with itself. These pro cesses suer collinear div ergences, and they are cured b y the fourth and fth diagram in Fig. 5-9 while the cross term b et w een them is canceled b y the second and third in Fig. 5-9 PSfrag replacemen ts 1 k + 2 2 k + 1 3 k + 4 4 k + 3 q + Figure 5-1. Dual momen tum assignmen t PSfrag replacemen ts p 1 p 1 p 2 p 2 p 3 p 3 p 4 p 4 k k Figure 5-2. Tw o diagrams with an extra 'unseen' gluon the arro ws indicate helicities,all momen ta are incoming Figure 5-3. Cancelation of collinear div ergence 78

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Figure 5-4. Cancelation of the soft bremsstrahlung radiation against a virtual pro cess PSfrag replacemen ts x = 1 y = 1 0R Figure 5-5. Phase space in tegration region of x and y PSfrag replacemen ts ~ L ~ k ~ p 3 u v w Figure 5-6. Conguration of ~ k ~ p 3 and ~ L in the CM frame of p 1 + p 2 + p 3 Figure 5-7. Self-energy bubble on leg 4, they will cancel the dep endence in the Bremsstrahlung PSfrag replacemen ts 34 ; 14 ; 23 34 14 23 k 12 ; 14 ; 23 12 14 23 Figure 5-8. All non-v anishing Bremsstrahlung pro cesses, the n um b ers underneath them are the regions they con tribute to 79

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PSfrag replacemen ts p 1 p 1 p 1 p 1 p 1 p 2 p 2 p 2 p 2 p 2 p 3 p 3 p 3 p 3 p 3 p 4 p 4 p 4 p 4 p 4 k k 0 k 0 k 0 k k k Figure 5-9. Disconnected Bremsstrahlung with t w o extra 'unseen' gluons 80

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CHAPTER 6 GLUON SCA TTERING WITH MASSIVE MA TTER FIELDS I only computed gluon scattering amplitude with massiv e matter in the lo op, for these are the only pro cesses without an y in ternal gluon propagators. If it w ere not so, w e ha v e to devise a w a y to regulate the infrared div ergences. Dimension regulation certainly w orks, and the en tire problem is reduced to calculating the 'b o x co ecien ts'. But this is simply bringing coals to New castle, as there are p eople who ha v e already dev elop ed the tec hnique and can do it a h undred times faster than me. The new thing ab out our calculation is the ph ysical infrared cut o that do es ha v e a meaning, y et I ha v en't w ork ed out ho w to incorp orate this cut o in to massiv e amplitudes. 6.1 Computation T ec hnique The main dicult y for the massiv e eld calculation is the F eynman parameter in tegrals. Indeed, I can only reduce all F eynman parameter in tegrals in to a set of three denitiv e in tegrals. They are I ( s ) := Z 1 0 dx s sx (1 x ) + M = 4 s p s 2 + 4 M s sinh 1 r s 4 M J ( s ) := Z 1 0 dx 1 x log sx (1 x ) + M M = 2 sinh 1 r s 4 M 2 K ( s; t ) := Z 1 0 dx st stx (1 x ) + M ( s + t ) log ( sx (1 x ) + M )( tx (1 x ) + M ) M 2 (6{1) Where M is in fact m 2 + i As M 0: I ( s ) 2 log s M J ( s ) 1 2 log 2 s M K ( s; t ) 2 + 2 log s M log t M K ( s; t ) 2 J ( s ) 2 J ( t ) 2 log 2 s t (6{2) F or example, a simple in tegral: 81

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Z 1 0 ( x 1 + x 2 + x 3 + x 4 1) dx 1 dx 2 dx 3 dx 4 x 1 x 2 (1 x 1 x 2 ) ( x 1 x 3 t + x 2 x 4 s + M ) 2 can b e reduced to M s 2 ( s + t ) J ( s ) M t 2 ( s + t ) J ( t ) M 2 st ( s + t ) K ( s; t ) + 1 2 st The denition of Eq. 6{1 is in accord with [ 27 31 ]. The last in tegral of Eq. 6{1 can b e expressed as dilogarithm and logarithms. In [ 30 ], the author ga v e t w o equiv alen t expressions of the b o x in tegrals, one in terms of dilogarithms the other in terms of h yp ergeometric functions. But for the purp ose of this pap er, I nd it most con v enien t to use Eq. 6{1 to sp ell out the results. The F eynman rules for scalar elds remain go o d, but the decomp osition Eq. 2{25 do es not, since it in tro duces 1 =q + factors in to the F eynman rules. This will complicate the already complicated F eynman parameter in tegral. Also q + and q will b e treated the same in con trast to the massless case, where q is in tegrated out and q + is giv en b y P x i k + i = P x i In order to organize the gamma matrix algebra in the fermion part of the calculation, w e mak e use of the factorizabilit y of gluon p olarisation v ectors Eq. 2{21 to reduce pro ducts of gamma matrices to pro ducts of K ij = p +i p j p +j p i whic h had b een pro v ed to b e quite handy F or example, if w e are to calculate the diagram Fig. 4-1 W e w ould write do wn: ( g + ; g ; q ) = ( ig ) 2 T r( t a t b ) T r ([( q k 1 ) r + m ] r [( q k 3 ) r + m ] r ) ( )( ^ ) ( i ) [( q k 1 ) 2 m 2 ] ( i ) [ ( q k 3 ) 2 m 2 ] (6{3) The n umerator can b e written as 82

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T r m ( q k 1 ) ( q k 1 ) m 0 p 2 j i [ k 3 k 1 j p 2 j k 1 k 3 ] h j 0 m ( q k 3 ) ( q k 3 ) m 0 p 2 j k 1 k 3 i [ j p 2 j ] h k 3 k 1 j 0 = 2[ j q k 1 j i [ k 3 k 1 j q k 3 j k 1 k 3 i + 2 h k 1 k 3 j q k 1 j k 3 k 1 ] h j q k 3 j ] + h k 1 k 3 j m j i [ k 3 k 1 j m j ] + [ j m j k 3 k 1 ] h j m j k 1 k 3 i (6{4) F or the notation of spinor, see the app endix. The standard pro cedure of momen tum in tegration tell us to shift q q + xk 1 + (1 x ) k 3 and to replace q q b y q 2 g = 4, etc: [ j q k 1 j i [ k 3 k 1 j q k 3 j k 1 k 3 i = [ j (1 x )( k 3 k 1 ) j i [ k 3 k 1 j x ( k 1 k 3 ) j k 1 k 3 i + [ j a a j i [ k 3 k 1 j b b j k 1 k 3 i g q 2 4 = (1 x ) p 2 ( k 3 k 1 ) + p 2 x ( k 1 k 3 ) 2 2( k 1 k 3 ) + + [ j k 3 k 1 ] h k 1 k 3 j i q 2 2 = x (1 x )( k 1 k 3 ) 2 + q 2 2 (6{5) here w e ha v e used an iden tit y: a a b b g = 2 a b ab ; a a b b g = 2 ab a b (6{6) F or a more complicated example, a string of spinor pro ducts b ecomes (with q shifted to q + x 2 k 2 + x 1 k 1 + x 4 k 4 + x 3 k 3 ) [ p 1 j q k 2 j p 4 i [ j q k 3 j p 3 i = [ p 1 j q + x 1 ( k 1 k 2 ) + x 4 ( k 4 k 2 ) + x 3 ( k 3 k 2 ) j p 4 i [ j q + x 2 ( k 2 k 3 ) + x 1 ( k 1 k 3 ) + x 4 ( k 4 k 3 ) j p 3 i = [ p 1 j ] h p 3 j p 4 i q 2 2 + [ p 1 j x 4 p 2 j p 4 i [ j x 2 p 4 x 1 p 2 j p 3 i = ( 1) 1 p +3 p +4 K ^ 43 q 2 2 + p 2 x 4 p +1 p +2 p +4 K 12 K ^ 42 ( p 2 x 2 p +3 K ^ 34 p 2 x 1 p +3 K ^ 32 ) (6{7) 83

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Basically all of the spinor pro ducts can b e reduced to one of the K ij 's, and the reader can nd in the app endix some practical details as to ho w to organize the pro ducts of K ij 's. After the momen tum in tegral is done, w e can p erform the F eynman parameter in tegrals using Eq. 6{1 Ho w ev er, there is one more complication due to the regulator: the in tegration o v er q ^ ( ) is dieren t from q +( ) Since the momen tum in tegration is no longer homogeneous, the replacemen t of q q b y q 2 g = 4 is problematic. But fortunately when doing the con traction of q q g q 2 = 4, q ^ ( ) and q +( ) will nev er co exist. F or example, in the rst line of Eq. 6{5 only q + will app ear in the rst brac k et while only q ^ ; q ; q app ears in the second brac k et. So, when w e mak e the replacemen t q q g q 2 = 4, w e should remem b er that q 2 actually means q 2 k Another example, if w e ha v e a spinor pro duct suc h as in the rst line of Eq. 6{7 q ^ ; q ; q will app ear in b oth brac k ets, then q 2 means q 2 ? this time. 6.2 Self-Energy Diagrams A factor of ig 2 = (16 2 ) f cad f dbc = ig 2 = (16 2 )T r t a t b will b e omitted. F or Fig. 4-1 the results are ( g + ; g ; s ) m = 1 6 p 2 log e r m 2 + 1 12 ( p 2 4 m 2 ) 2 p 2 I ( p 2 ) 5 18 p 2 + 4 3 m 2 ( g + ; g ; q ) m = 4 3 p 2 log e r m 2 2 3 ( p 2 4 m 2 )( p 2 + 2 m 2 ) p 2 I ( p 2 ) + 26 9 p 2 + 16 3 m 2 (6{8) The gluon mass coun ter term in the expressions ab o v e has b een remo v ed already 6.3 T riangle Diagrams A factor of g 3 = (8 2 ) f dae f ebf f f cd = ig 3 = (8 2 )T r[ t a t b t c ] is omitted. F or Fig. 4-3 : 84

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( g + ; g + ; g ; s ) m = 2 p +3 p +1 p +2 K ^ 21 1 6 log m 2 e r + ( p 2o 4 m 2 )( ( p 2o 4 m 2 ) 12 p 4o + p +1 p +2 p +23 m 2 p 4o ) I ( p 2o ) p +1 p +2 p +23 m 2 p 2o J ( p 2o ) 1 9 + 4 m 2 3 p 2o 1 6 p +1 p +2 p +23 (1 + 24 m 2 p 2o ) ( g + ; g + ; g ; q ) m = 2 2 p +3 p +1 p +2 K ^ 21 2 3 log m 2 e r +( p 2o 4 m 2 )( ( p 2o + 2 m 2 ) 3 p 4o + 2 p +1 p +2 p +23 m 2 p 4o ) I ( p 2o ) 2 p +1 p +2 p +23 m 2 p 2o J ( p 2o ) + 16 9 + 8 m 2 3 p 2o 1 3 p +1 p +2 p +23 (1 + 24 m 2 p 2o ) (6{9) where = 1 if leg 3 is o shell and 0 otherwise, and p o is the o shell momen tum. Their anomalous terms are iden tical with the massless result, whic h agrees with the fact that anomalous terms are UV eects. F or an MHV triangle: ( g + ; g + ; g +; s ) m = ( K ^ 21 ) 3 p +1 p +2 p +3 4 m 2 ( p 2o 4 m 2 ) p 6o I ( p 2o ) + 4 m 2 p 4o J ( p 2o ) + 2( p 2o + 24 m 2 ) 3 p 4o ( g + ; g + ; g +; q ) m = 2 ( K ^ 21 ) 3 p +1 p +2 p +3 8 m 2 ( p 2o 4 m 2 ) p 6o I ( p 2o ) + 8 m 2 p 4o J ( p 2o ) + 4( p 2o + 24 m 2 ) 3 p 4o (6{10) 6.4 Scattering Amplitudes A factor of ig 4 = (8 2 )T r[ t a t b t c t d ] is omitted. A ( g + ; g + ; g + ; g +; s ) m = 8 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st 1 6 m 4 st K ( s; t ) A ( g + ; g + ; g + ; g +; q ) m = 16 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st 1 3 2 m 4 st K ( s; t ) (6{11) 85

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A ( g + ; g + ; g + ; g ; s ) m = K ^ 2 13 p +2 p +4 K ^ 43 K 32 K 21 K ^ 14 (2 t + s ) tm 2 2( s + t ) s (1 4 m 2 s ) I ( s ) (2 s + t ) sm 2 2( s + t ) t (1 4 m 2 t ) I ( t ) + (2 s + t ) t 2 m 2 ( s + t ) 2 s J ( s ) + (2 t + s ) s 2 m 2 ( s + t ) 2 t J ( t ) + stm 2 2( s + t ) 2 (1 2 m 2 ( s + t ) st ) K ( s; t ) + 2(2 s 2 st + 2 t 2 ) m 2 st + ( s + t ) 6 A ( g + ; g + ; g + ; g ; q ) m = 2 K ^ 2 13 p +2 p +4 K ^ 43 K 32 K 21 K ^ 14 (2 t + s ) tm 2 ( s + t ) s (1 4 m 2 s ) I ( s ) (2 s + t ) sm 2 ( s + t ) t (1 4 m 2 t ) I ( t ) + 2(2 s + t ) t 2 m 2 ( s + t ) 2 s J ( s ) + 2(2 t + s ) s 2 m 2 ( s + t ) 2 t J ( t ) + stm 2 ( s + t ) 2 (1 2 m 2 ( s + t ) st ) K ( s; t ) + 4(2 s 2 st + 2 t 2 ) m 2 st + ( s + t ) 3 (6{12) In con trast to the massless case, the external leg factors are included in the results b elo w for the helicit y conserving amplitudes, since w e can p erform w a v e function renormalization no w. A ( g + ; g + ; g ; g ; s ) m = 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 ( m 2 2 s 2 m 2 3 t 2 m 4 st + 4 m 4 3 t 2 + 1 12 ) I ( t ) + m 4 s 2 K ( s; t ) 2 m 2 s + 4 m 2 3 t + 1 18 1 6 log e r m 2 1 3 1 6 + 1 3 A ( g + ; g + ; g ; g ; q ) m = 2 2 K ^ 4 12 p +3 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 ( m 2 s + 2 m 2 3 t 4 m 4 st + 8 m 4 3 t 2 1 3 ) I ( t ) + ( m 2 s + 2 m 4 s 2 ) K ( s; t ) 4 m 2 s + 8 m 2 3 t + 19 9 + 2 3 log e r m 2 2 3 + 2 3 4 3 (6{13) 86

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A ( g + ; g ; g + ; g ; s ) m = 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 ( tm 2 (17 st + s 2 + 4 t 2 ) 6( s + t ) 3 s + 2 tm 4 (2 t + 5 s ) 3( s + t ) 2 s 2 + t (5 st 2 s 2 + t 2 ) 12( s + t ) 3 ) I ( s ) +( sm 2 (17 st + t 2 + 4 s 2 ) 6( s + t ) 3 t + 2 sm 4 (5 t + 2 s ) 3( s + t ) 2 t 2 + s (5 st 2 t 2 + s 2 ) 12( s + t ) 3 ) I ( t ) +( 2 stm 2 ( s + t ) 3 s 2 t 2 ( s + t ) 4 )( J ( s ) + J ( t )) + ( 2 stm 2 ( s + t ) 3 + m 4 ( s + t ) 2 + s 2 t 2 2( s + t ) 4 ) K ( s; t ) + 2 m 2 (2 t 2 + 2 s 2 + 3 st ) 3( s + t ) st + s 2 + t 2 + 11 st 18( s + t ) 2 1 6 log e r m 2 1 3 1 6 + 1 3 A ( g + ; g ; g + ; g ; q ) m = 2 2 K ^ 4 13 p +2 p +4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 ( tm 2 (5 s 2 + 2 t 2 5 st ) 3( s + t ) 3 s + 4 tm 4 (2 t + 5 s ) 3( s + t ) 2 s 2 t (5 s 2 + 2 t 2 + st ) 6( s + t ) 3 ) I ( s ) +( tm 2 (5 t 2 + 2 s 2 5 st ) 3( s + t ) 3 t + 4 sm 4 (5 t + 2 s ) 3( s + t ) 2 t 2 s (5 t 2 + 2 s 2 + st ) 6( s + t ) 3 ) I ( t ) +( 4 stm 2 ( s + t ) 3 + st ( s 2 + t 2 ) ( s + t ) 4 )( J ( s ) + J ( t )) + ( m 2 ( s t ) 2 ( s + t ) 3 + 2 m 4 ( s + t ) 2 st ( s 2 + t 2 ) 2( s + t ) 4 ) K ( s; t ) + 4 m 2 (2 t 2 + 2 s 2 + 3 st ) 3( s + t ) st + 19 s 2 + 19 t 2 + 47 st 9( s + t ) 2 + 2 3 log e r m 2 2 3 + 2 3 4 3 (6{14) The last n umerical factors 1 = 3 and 2 = 3 inside eac h square brac k et is due to the amputation of external legs: [lim p 2 0 ( p 2 ) =p 2 ] 1 = 2 6.5 Photon Photon Scattering The amplitude of photon-photon scattering can b e obtained from the ab o v e results fairly easily all w e need to do is to replace g 4 T r[ t a t b t c t d ] with e 4 and sum all the crossings. The reader migh t think that w e should also remo v e the triangle diagrams from the amplitudes since these diagrams in v olv e tri-gluon v ertices whic h are absen t in an ab elian gauge theory But, these diagrams will automatically cancel eac h other when w e sum o v er all crossings. This cancellation go es b y the name of U (1) decoupling. The amplitudes are listed in the app endix. When w e sum o v er all the crossings, the coun ter term that is prop ortional to the four p oin t v ertex will v anish, while the pure n um b er 2 = 3 will b ecome 2, whic h has to 87

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b e subtracted to restore gauge co v ariance. W e obtain the follo wing photon scattering amplitude: A (+ ; + ; + ; +) = ie 4 4 2 4 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 p +3 p +4 st 2 4 m 4 st K ( s; t ) 4 m 4 su K ( s; u ) 4 m 4 tu K ( t; u ) A (+ ; + ; + ; ) = ie 4 4 2 stK ^ 2 13 p +2 p +4 2 uK ^ 43 K 32 K 21 K ^ 14 2 + 4 m 2 s + 4 m 2 t + 4 m 2 u [ J ( s ) + J ( t ) + J ( u )] + 2 m 2 u 4 m 4 st K ( s; t ) + 2 m 2 t 4 m 4 su K ( s; u ) + 2 m 2 s 4 m 4 tu K ( t; u ) A (+ ; + ; ; ) = ie 4 4 2 tK ^ 4 12 p +3 p +4 sK ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +2 8 m 2 s + 4 m 2 t 2 t s 1 I ( t ) + 8 m 2 s + 4 m 2 u 2 u s 1 I ( u ) + 8 m 2 s 4 t 2 + 4 st + 2 s 2 s 2 J ( t ) + J ( u ) + 2 m 2 t 4 m 4 st K ( s; t ) + 2 m 2 u 4 m 4 su K ( s; u ) + 2(4 t 2 + 4 st + s 2 ) m 2 stu 4 m 4 tu + 2 t 2 + 2 st + s 2 s 2 K ( t; u ) 2 A (+ ; ; + ; ) = ie 4 4 2 stK ^ 4 13 p +2 p +4 u 2 K ^ 43 K ^ 32 K ^ 21 K ^ 14 p +1 p +3 8 m 2 u + 4 m 2 t 2 t u 1 I ( t ) + 8 m 2 u + 4 m 2 s 2 s u 1 I ( s ) + 8 m 2 u 4 t 2 + 4 ut + 2 u 2 u 2 J ( t ) + J ( s ) + 2 m 2 t 4 m 4 ut K ( u; t ) + 2 m 2 s 4 m 4 su K ( s; u ) + 2(4 t 2 + 4 ut + u 2 ) m 2 stu 4 m 4 ts + 2 t 2 + 2 ut + u 2 u 2 K ( t; s ) 2 (6{15) The spinor structure ab o v e has b een set up to b e uni-mo dular and in v arian t under crossings of t w o legs. The results here agree with [ 27 ] and [ 31 ]. Note that in the rst term of Eq.(127.18) of [ 31 ], the authors seemed to ha v e left out terms of 4 =s + 2 =t and 4 =s + 2 =u in the co ecien t of B ( t ) and B ( u ) (their B function is eectiv ely our I function) resp ectiv ely as Eq.(127.18) will not lead to Eq.(127.20) without those t w o terms. 88

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CHAPTER 7 CONCLUSIONS AND FUTURE W ORK 7.1 Conclusion T o conclude, I ha v e studied the renormalization of gauge theory on the ligh t cone w orld sheet. In order to do so, I computed all the four p oin t amplitudes in gauge theory The b o x reduction tec hnique w as dev elop ed to extract the articial div ergences and IR div ergences. The articial div ergences are the rational functions con taining 1 =q + p oles, they come from the ligh t cone gauge propagators and w ere sho wn to cancel in a gauge in v arian t quan tit y The IR div ergences, regulated b y setting q + a w a y from zero, w ere com bined with the Bremsstrahlung con tributions to giv e a Loren tz co v arian t scattering cross section. The calculation of Bremsstrahlung con tributions w as also done in the ligh t cone fashion to facilitate the comparison with the virtual pro cesses. A mismatc h in the rational parts of the amplitudes w ere prev alen t, the restoration of gauge co v ariance w as only partially addressed. Finally the scattering of gluon b y gluon with general massiv e matter w as computed for completeness, and the ligh t b y ligh t scattering amplitudes w ere obtained along the w a y Next I giv e some discussion on some unresolv ed issues and an outlo ok for the future w ork. 7.2 Restoring Gauge Co v ariance in the Ligh t Cone In [ 13 14 15 ], w e insisted on only allo wing coun ters terms that are p olynomials in the target space. Th us, when w e sa w that there w as a hanging four-p oin t v ertex in an amplitude, w e could not put in a four p oin t v ertex as a coun ter term, but instead w e mo died the self-energy b y a term const p2to adjust the strength of the exc hange diagrams. But this sc heme do es not restore the gauge co v ariance for all the amplitudes. Here I suggest a new system of putting in coun ter terms. Diagrams suc h as ( g+; g+) en ter the amplitudes as 'double quartic' graphs, similarly ( g + ; g ; g+) en ters as 'quartic sw ordsh' diagrams, b oth of whic h are treated as 1PIR graphs in the canonical ligh t cone formalism.89

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T reating them as 1PIR graphs forbids us to adjust their strength, for they usually con tribute a m ultiple of four p oin t v ertices, whic h is not a p olynomial in the momen ta. In Chapter 4 I compared ( g + ; g ; s ) with ( g+; g+; s ) and ( g + ; g ; q ) with ( g+; g+; q ). Here I giv e the result again: ( g + ; g ; s ) = p2 5 18 1 6 log p2 er ( g+; g+; s ) = p+2 4 9 1 6 log p2 er ( g + ; g ; q ) = p2 26 9 4 3 log p2 er ( g+; g+; q ) = p+2 20 9 4 3 log p2 er The non-rational parts of the self-energy con tributions are the same regardless of the index on W e ha v e also reason to b eliev e that, when there is no infrared div ergence, the rational part should also matc h due to Loren tz co v ariance. More sp ecically v acuum p olarization should b e of the form = gp2 pp ( p2). This tells us that ( g + ; g ) = ^ _p2( p2) = p2( p2) ( g + ; g+) = 0 ( g+; g+) = p+2( p2) Hence, apart from the factor of p2and p+2, ( g + ; g ) should b e equal to ( g+; g+). Therefore, w e ha v e to in v ok e coun ter terms to force this equalit y I c hose to asso ciate to eac h ( g+; g+; s ) a term 1 = 6, and to ( g+; g+; q ) a term 2 = 3, I cannot quite nd what is the correct v alue for ( g+; g+; g ) b ecause of the infrared div ergence. So I simply dened it to b e 1 = 3, c hosen suc h that these coun ter terms v anish with N = 4 SYM eld con ten t. These coun ter terms are going to aect the four p oin t v ertex that is deriv ed from the exc hange diagrams. The gluon v ertex correction diagram has a similar problem:90

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( g + ; g ; g +; s ) = 2 p+3 p+1p+2K^ 2 ; 1 1 6 log p2o er 1 9 ( g + ; g ; g+; s ) = ( p+1 p+2) 1 6 log p2o er 5 18 ( g + ; g ; g +; q ) = 2 p+3 p+1p+2K^ 2 ; 1 4 3 log p2o er 32 9 ( g + ; g ; g+; q ) = ( p+1 p+2) 4 3 log p2o er 26 9 The logarithmic piece matc hes, so I will asso ciate to ( g ; g ; g ; s ) a term 1 = 6, to ( g ; g ; g ; q ) a term 2 = 3 and to ( g ; g ; g ; g ) a term 1 = 3 to enforce the total agreemen t b et w een ( g + ; g ; g +) and ( g + ; g ; g+). Again, the n um b er for ( g ; g ; g ; g ) is hand pic k ed so that there is no need for coun ter terms in N = 4 SYM. These coun ter terms will aect the strength of the exc hange v ertex. There are also mismatc hes b et w een ( s; s; g +) and ( s; s; g+), whic h w e cannot easily determine due to the IR div ergence, so w e thro w in t w o arbitrary n um b ers and adjust these t w o n um b ers to mak e all the amplitudes w ork. The n um b ers are determined to b e 1 = 2 for ( s; s; g +) and 0 for ( s; s; g+) and ( q ; q ; g ). Note that all of the mo dications ab o v e can b e ac hiev ed b y p olynomials in the external momen ta 1 I shall rep ort ho w this coun ter term system is w orking out. First, I list all amplitudes in T able. 7-1 note that b y s-pt I mean the 4 p oin t v ertex that is deriv ed from the s-c hannel exc hange diagram. The eect of the old regularization sc heme is listed in T able. 7-2 W e see b y comparing 7-1 and 7-2 that all the b osonic amplitudes are xed as long as there is only one sp ecies of scalar, while fermionic amplitudes generally ha v e problems. The new sc heme giv es T able. 7-3 By comparing 7-1 and 7-3 all the mismatc hes are xed. But of course, this sc heme answ ers as man y questions as it raises. The lo ose threads include ho w to determine the coun ter term 1 the structure 2 p +3 =p +1 p +2 K ^ 2 ; 1 do esn't lo ok lik e a p olynomial, but actually its p + dep endence comes from the p olarization v ectors 91

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for those diagrams that ha v e IR div ergences, and whether they v anish in N = 4 SYM and ho w to realize them on the ligh t cone w orld sheet. Finally I w an t to p oin t out an in teresting observ ation that whenev er a gen uine four p oin t v ertex (meaning not deriv ed from an exc hange diagram) exists, the amplitude will ha v e a constan t mismatc h.7.3 T riangle AnomalyAlthough I ha v e stuc k to the adjoin t represen tation, in whic h the cubic in v arian t dabcis zero, it is still necessary to lo ok at ho w anomaly calculation turn out in the ligh t cone. The rst PSfrag replacemen ts k 1 k 1 k 2 k 2 k 3 l k 3 l a; p 1 ,^a; p 1 ,^b; p 2 ,^b; p 2 ,^c; k c; k Figure 7-1. T riangle anomalydiagram of Fig. 7-1 is giv en b y T r h itbitaitci T r [ 2 ( q k2) 2 ( q k1) ( k3 k1) ( q k3) ] ( i )( q k2)2( i )( q k1)2( i )( q k3)2= i T r h tbtatci 2 h j ( q k2) j k1 k2] h j ( q k1)( k3 k1)( q k3) j k2 k3] ( i )( q k2)2( i )( q k1)2( i )( q k3)2= 2T r h tbtatci h j ( q k2) j k1 k2] h j ( q k3) j k2 k3] ( i )( q k2)2( i )( q k3)2 h j ( q k1) j k2 k3] ( i )( q k2)2( i )( q k1)2 (7{1) After the usual steps of momen tum in tegrals: Z idxi 16 2 h j x3( k3 k2) j k1 k2] h j x2k2 ?+ x3k3 ?j k2 k3] + h j x2( k2 k1) j k2 k3] h j x1k1 ?+ x2k2 ?j k2 k3] = i 16 22 K_ 12 3 p+1p+2 p+2( k2+ 2 k3)_+ p+1( k1+ 2 k2)_ (7{2) The second diagram of Fig. 7-1 is similar:92

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T r h itbitaitci T r [ 2 ( k3 q ) ( k3 k1) ( k1 q ) 1 ( k2 q ) ] ( i )( q k2)2( i )( q k1)2( i )( q k3)2= i T r h tbtatci 2 h j ( k3 q )( k3 k1)( k1 q ) j k1 k2] h j ( k2 q ) j k2 k3] ( i )( q k2)2( i )( q k1)2( i )( q k3)2= 2T r h tbtatci h j ( k2 q ) j k2 k3] h j ( k1 q ) j k1 k2] ( i )( q k2)2( i )( q k1)2 h j ( k3 q ) j k1 k2] ( i )( q k2)2( i )( q k3)2 (7{3) After the momen tum in tegral: Z idxi 16 2 h j x1( k1 k2) j k2 k3] h j x1k1 ?+ x2k2 ?j k1 k2] + h j x2( k2 k3) j k1 k2] h j x2k2 ?+ x3k3 ?j k2 k3] = i 16 22 K_ 12 3 p+1p+2 p+1( k2+ 2 k1)_+ p+2( k3+ 2 k2)_ (7{4) So the sum of these t w o diagrams giv es i 8 2T r h f ta; tbg tci 2 K_ 12 p+1p+2 p+1( k2+ k1)_+ p+2( k3+ k2)_ (7{5) The righ t handed fermion giv es the same con tribution as ab o v e (the F eynman diagram itself giv es a negativ e sign relativ e to the left handed con tribution, while the trace factor giv es a second negativ e sign). But if w e put tcto b e 1 and study the axial U (1) curren t (commonly kno wn as the ab elian anomaly), the div ergence of this curren t will b e t wice the ab o v e result. Or w e can k eep the theory c hiral, and lo ok at the div ergence of the c hiral curren t (kno w as the non-ab elian anomaly) only Note that the dierence b et w een these t w o cases is of order A3, whic h will not sho w up here. F ^ F is of the form K_ 12K_ 12= ( p+1p+2) and 0 if the t w o gluons ha v e dieren t helicit y So some subtractions ha v e to b e made for Eq. 7{5 to b e of the correct form (again, due to the regulator w e used, Eq. 7{5 dep ends on eac h dual momen ta). I ha v en't sho wn the calculation for the case when the t w o gluons ha ving dieren t helicit y the result is non-zero, so it has to b e subtracted b y coun ter terms to o. Here, w e see that the coun ter terms asso ciated to three-p oin t93

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function are p opulating fast, there ma y or ma y not b e an economical c hoice of coun ter term(s) that will tak e care of all these problems.7.4 Tw o-Lo op and n-P oin t AmplitudesThe commonly adopted metho d for calculating one lo op n-p oin t amplitude is to use (generalized) unitarit y [ 25 26 ] to calculate b o x-co ecien ts, and to use recursion relations to recycle old results. And the results can b e c hec k ed partially b y lo oking at its collinear limit soft limit and m ulti-particle factorization prop erties. W e ha v e basically calculated the four particle amplitude in the ligh t cone gauge b y brute force, should there b e a need to obtain amplitudes with more legs or ev en higher lo op amplitudes, to w a y to pro ceed is certainly not brute force. Let us fo cus on the N = 4 SYM theory as a rst step. W e kno w that in this theory all in tegrands can b e reduced to scalar b o xes. This is v ery useful if w e are using dimension regulation since the IR (or collinear) div ergences are regulated. But w e'd lik e to stic k to our IR regulator, so scalar b o xes ha v e to go through some more subtractions to b ecome infrared safe. I don't kno w y et ho w is this going to tell on the pro cedure of computing b o x co ecien ts. The IR div ergence is lo cal in the sense that it is presen t whenev er there is a four-p oin t MHV subtree in the lo op diagram, so the one lo op IR structure w e found in Chapter 5 will p ersist in to a higher p oin t amplitude. This feature p erhaps can help us to dene an IR safe part in an amplitude and 'b o otstrap' it to a larger amplitudes.94

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T able 7-1. List of mismatc hes in all the amplitudes Amplitude s-exc h t-exc h s-pt t-pt const A ( g + ; g + ; g ; g ; s ) 0 0 -1/6 1/2 A ( g + ; g + ; g ; g ; q ) 0 0 2/3 -2 A ( g + ; g + ; g ; g ; g ) 0 0 -1/3 1 A ( g + ; g ; g + ; g ; s ) 0 0 -1/6 -1/6 0 A ( g + ; g ; g + ; g ; q ) 0 0 2/3 2/3 0 A ( g + ; g ; g + ; g ; g ) 0 0 -1/3 -1/3 0 A ( s; s; g + ; g ) 0 0 -1/2 1/2 A ( g + ; s; g ; s ) 0 0 2 A ( s; s; s; s ) 0 0 -1/2 -1/2 0 A ( q ; q ; q ; q ) 0 -1/3 0 A ( q ; q ; q ; q ) 0 0 -1/3 -1/3 0 A ( s; s; q ; q ) 0 -2/3 0 A ( g + ; g ; q ; q ) -1/3 0 -1/3 0 0 A ( g + ; g ; q + ; q +) -1/3 0 -1/3 0 0 A ( g + ; q ; g ; q ) 0 0 0 0 T able 7-2. List of the eect of the old coun ter terms sc hemes Amplitude s-exc h t-exc h s-pt t-pt const A ( g + ; g + ; g ; g ; s ) -1/6 -1/6 0 0 A ( g + ; g + ; g ; g ; q ) 2/3 2/3 0 0 A ( g + ; g + ; g ; g ; g ) -1/3 -1/3 0 0 A ( g + ; g ; g + ; g ; s ) -1/6 -1/6 0 0 0 A ( g + ; g ; g + ; g ; q ) 2/3 2/3 0 0 0 A ( g + ; g ; g + ; g ; g ) -1/3 -1/3 0 0 0 A ( s; s; g + ; g ) -1/2 -1/2 0 0 A ( g + ; s; g ; s ) -1/2 -1/2 0 A ( s; s; s; s ) -1/2 -1/2 0 0 0 A ( q ; q ; q ; q ) 1/3 0 0 A ( q ; q ; q ; q ) 1/3 1/3 0 0 0 A ( s; s; q ; q ) 1/6 0 0 A ( g + ; g ; q ; q ) 1/3 0 0 0 0 A ( g + ; g ; q + ; q +) 1/3 0 0 0 0 A ( g + ; q ; g ; q ) 0 0 0 0 95

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T able 7-3. List of the eect of the new coun ter terms sc heme Amplitude s-exc h t-exc h s-pt t-pt const A ( g + ; g + ; g ; g ; s ) -1/3 -1/3 -1/6 0 A ( g + ; g + ; g ; g ; q ) 4/3 4/3 2/3 0 A ( g + ; g + ; g ; g ; g ) -2/3 -2/3 -1/3 0 A ( g + ; g ; g + ; g ; s ) -1/3 -1/3 -1/6 -1/6 0 A ( g + ; g ; g + ; g ; q ) 4/3 4/3 2/3 2/3 0 A ( g + ; g ; g + ; g ; g ) -2/3 -2/3 -1/3 -1/3 0 A ( s; s; g + ; g ) -1 -1 -1/2 0 A ( g + ; s; g ; s ) -1 -1 0 A ( s; s; s; s ) -1 -1 -1/2 -1/2 0 A ( q ; q ; q ; q ) 0 1/3 0 A ( q ; q ; q ; q ) 0 0 1/3 1/3 0 A ( s; s; q ; q ) -1/2 1/6 0 A ( g + ; g ; q ; q ) 1/3 0 1/3 0 0 A ( g + ; g ; q + ; q +) 1/3 0 1/3 0 0 A ( g + ; q ; g ; q ) 0 0 0 0 96

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APPENDIX A SPINOR NOT A TION IN THE LIGHT CONE r = 0 0 := (I ; ~ ) := (I ; ~ ) ab = ab = a b = a b = i 2 p a = ab p b p a = p b ba j p ] := p a j p i := p a [ p j := p a h p j := p a (A{1) So far the spinor notations are common to all, and in the last line of Eq. A{1 I ha v e conformed to the 'hep-ph' notation: j p ] is assigned a lo w er index while j p i an upp er index. In ligh t cone, unlik e what w e had in Section 2.1 the reference spinor is xed to b e = = 10 = = 0 1 (A{2) W e can dene the ligh t cone v ersion spinor as p = p ^ p + 1 p = 1 p p + (A{3) The spinors satisfy the Dirac equation if p is ligh t lik e. Note that they don't ha v e the correct normalization, namely p 6 = p p but they ha v e the merit that p = ( p ) The p olarization v ectors of gluon can b e written as ^ a a = p 2 j p ] h j h j p i = p 2 j p ] h j ; a a = p 2 j ] h p j [ p j ] = p 2 j ] h p j (A{4) K ^ ij = p +j p +i h p j j p i i ; K ij = p +i p +j [ p i j p j ] [ j p i j p j i = h p j j p i j ] = p 2 p +j K ^ j i ; h j p i j p j ] = [ p j j p i j i = p 2 p +j K j i (A{5) The K ij 's satisfy 97

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X j K ij = 0 p +i K j k + p +k K ij + p +j K k i = 0 K ^ l i K ^ j k + K ^ l k K ^ ij + K ^ l j K ^ k i = 0 X j K ^ ij K j k p +j = p +i p +k X j p 2j 2 p +j (A{6) The third line of Eq. A{6 is called the Sc houten iden tit y: [ ij ][ k l ] + [ j k ][ il ] + [ k i ][ j l ] = 0. In the curren t case w e are dealing with, i; j run from 1 to 4, but only t w o of the six K ij 's are indep enden t, sa y K 43 and K 32 And an y pro duct of K ij 's with total helicit y 4 can b e reduced to either ( K ^ 43 ) 4 or ( K ^ 43 ) 3 K ^ 32 Pro duct of helicit y 2 can b e reduced to ( K ^ 43 ) 2 and K ^ 43 K ^ 32 Pro duct of helicit y 0 can b e reduced to 1 and K ^ 32 K 43 The reduction is in general a formidable task for h uman, but quite a piece of cak e for computers, as all our calculations are done with computers. 98

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APPENDIX B FEYNMAN R ULES W e remind the reader that g + ( ) corresp onds to a ^ ( ) on the gluon line, q ( q ) corresp onds to an incoming (outgoing) fermion line, q + ( ) corresp onds to righ t(left) handedness, s corresp onds to a scalar. The fermion gluon v ertex Fig. B-1 is giv en b y V ( q ; q ; g +) = 2 ig ( t a ) bc p +2 q + p +1 K ^ p 1 ;q ; V ( q + ; q + ; g +) = 2 ig ( t a ) bc 1 q + K ^ p 2 ;q V ( q ; q ; g ) = 2 ig ( t a ) bc 1 q + K p 2 ;q ; V ( q + ; q + ; g ) = 2 ig ( t a ) bc p +2 q + p +1 K p 1 ;q (B{1) The gluon scalar v ertex Fig. B-2 is giv en b y V ( g + ( )) = 2 ig ( t a ) bc 1 p +b + p +c K ^ ( ) p c ;p b (B{2) Here w e ha v e used real scalar elds and hence it transforms in a real represen tation. The gluon fermion 4 p oin t v ertex Fig. B-3 is giv en b y V ( q ; q ; g + ; g ) = 2 ig 2 ( t d ) ce ( t a ) eb p +2 p +1 + p +4 + 2 g 2 f dae ( t e ) cb ( p +3 p +4 ) p +2 ( p +3 + p +4 ) 2 V ( q + ; q + ; g + ; g ) = 2 g 2 f dae ( t e ) cb ( p +3 p +4 ) p +2 ( p +3 + p +4 ) 2 V ( q ; q ; g ; g +) = 2 g 2 f dae ( t e ) cb ( p +3 p +4 ) p +2 ( p +3 + p +4 ) 2 V ( q + ; q + ; g ; g +) = 2 ig 2 ( t d ) ce ( t a ) eb p +2 p +1 + p +4 + 2 g 2 f dae ( t e ) cb ( p +3 p +4 ) p +2 ( p +3 + p +4 ) 2 (B{3) The gluon scalar four p oin t v ertex Fig. B-4 is giv en b y V ( g + ; g ; s; s ) = V ( g ; g + ; s; s ) = ig 2 ( t a ) ce ( t b ) ed + ( t b ) ce ( t a ) ed + g 2 f abe ( t e ) cd ( p +3 p +4 )( p +2 p +1 ) ( p +1 + p +2 ) 2 V ( g + ; g + ; s; s ) = V ( g ; g ; s; s ) = 0 (B{4) 99

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F or tri-gluon v ertex Fig. 4-3 V ( g ; g ; g ) = g f abc [ 1 2 ( p 1 p 2 ) 3 + 2 3 ( p 2 p 3 ) 1 + 3 1 ( p 3 p 1 ) 2 ] Setting ; = ^ = the ab o v e b ecomes V ( g + ; g + ; g ) = g f abc [( p 2 p 3 ) + p ^1 p +1 ( p 2 p 3 ) ^ + ( p 3 p 1 ) + p ^2 p +2 ( p 3 p 1 ) ^ ] = 2 g f abc ( p 1 + p 2 ) + p +1 p +2 K ^ 21 (B{5) The gluon four p oin t v ertex receiv es con tribution from t w o sources: the left diagram in Fig. B-6 is simply the co v arian t four p oin t v ertex: V 1 = ig 2 f abe f ecd [ g g g g ] + f dae f ebc [ g g g g ] + f cae f ebd [ g g g g ] (B{6) The second is obtained b y shrinking a propagator. V 2 = g 2 f dae f ebc [ 1 4 ( p 4 p 1 ) ] ig + g + ( p 1 + p 4 ) 2 ( p +1 + p +4 ) 2 ( p 1 + p 4 ) 2 [ 2 3 ( p 2 p 3 ) ] = ig 2 f dae f ebc ( 1 4 )( 2 3 )( p 4 p 1 ) + ( p 2 p 3 ) + ( p +1 + p +4 ) 2 There are t w o cases in whic h the gluon four p oin t v ertex is nonzero: V ( g + ; g + ; g ; g ) = 2 ig 2 f dae f ebc p +1 p +3 + p +2 p +4 ( p +1 + p +4 )( p +2 + p +3 ) + f ace f ebd p +3 p +2 + p +1 p +4 ( p +1 + p +3 )( p +2 + p +4 ) V ( g + ; g ; g + ; g ) = 2 ig 2 f abe f ecd p +2 p +3 + p +1 p +4 ( p +1 + p +2 )( p +3 + p +4 ) + f dae f ebc p +1 p +2 + p +3 p +4 ( p +1 + p +4 )( p +2 + p +3 ) (B{7) When using these v ertices, w e need to w atc h the indices of structure constan ts closely: not all terms are going to mak e con tributions to T r[ t a t b t c t d ]. 100

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The fermion four p oin t v ertex comes from con tracting a pair of three p oin t v ertices connected b y a gluon propagator. It is giv en b y (for conguration of Fig. B-7 ) V ( q ; q ; q ; q ) = 4 ig 2 ( t e ) da ( t e ) cb p +3 p +4 ( p +1 p +4 ) 2 (B{8) with the ob vious restriction that fermion line 1 and 4 ha ving the same handedness while 2 and 3 ha ving the same handedness. The scalar four p oin t v ertex Fig. B-8 comes from con tracting a pair of three p oin t v ertices connected b y a gluon propagator. It is giv en b y V ( s; s; s; s ) = ig 2 ( t e ) ab ( t e ) cd ( p +3 p +4 )( p +1 p +2 ) ( p +1 + p +2 ) 2 (B{9) The scalar fermion four p oin t v ertex Fig. B-9 also comes from con tracting a gluon propagator: V ( s; s; q ; q ) = ig 2 ( t e ) ab ( t e ) cd 2 p +3 ( p +2 p +1 ) ( p +1 + p +2 ) 2 (B{10) with the restriction that the fermion lines ha ving the same handedness. PSfrag replacemen ts ^ ; a ; a q c; p 1 c; p 1 b; p 2 b; p 2 Figure B-1. Gluon-fermion-fermion 3 p oin t v ertex 101

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PSfrag replacemen ts ; a; q c; p c b; p b Figure B-2. Gluon-scalar-scalar 3 p oin t v ertex PSfrag replacemen ts e e b; p 1 b; p 1 c; p 2 c; p 2 a; p 4 a; p 4 d; p 3 d; p 3 Figure B-3. Tw o diagrams con tribute to the fermion-gluon 4 p oin t v ertex. Note these are not exc hange diagrams but rather four p oin t v ertices gotten through cancelling a 1 =p 2 factor in a fermion propagator or a gluon propagator, see Eq. 2{25 PSfrag replacemen ts ^ ; b; p 4 ^ ; b; p 4 ; a; p 3 ; a; p 3 c; p 1 d; p 2 Figure B-4. Scalar-gluon 4 p oin t v ertex. Again, the second diagram is not an exc hange diagram but rather a four p oin t v ertex obtained through shrinking a gluon propagator PSfrag replacemen ts ; b; p 2 ; a; p 1 ; c; p 3 Figure B-5. T ri-gluon v ertex 102

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PSfrag replacemen ts ; b; p 2 ; a; p 1 ; c; p 3 ; d; p 4 ; b; p 2 ; a; p 1 ; c; p 3 ; d; p 4 Figure B-6. Gluon 4 p oin t v ertex PSfrag replacemen ts a; p 1 c; p 3 b; p 2 d; p 4 Figure B-7. F ermion 4 p oin t v ertex PSfrag replacemen ts a; p 1 c; p 3 b; p 2 d; p 4 Figure B-8. Scalar 4 p oin t v ertex PSfrag replacemen ts a; p 1 c; p 3 b; p 2 d; p 4 Figure B-9. Scalar F ermion 4 p oin t v ertex 103

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BIOGRAPHICAL SKETCH Jian Qiu w as b orn in 11.2.1979 in Shanghai, China. He w as raised up in a small to wn Le-Ping in the Jiang-Xi pro vince, where he nished the rst t w o y ears of elemen tary sc ho ol. He mo v ed bac k to Shanghai in 1988 and w en t to elemen tary sc ho ol, middle sc ho ol and high sc ho ol there. F rom the sixth to the t w elfth grade, he also to ok part in comp etitions in remote con trolled race car. After graduating from Shi-Er high sc ho ol in 1998, he w as admitted to F udan Univ ersit y to study ph ysics. He w as rst in terested in Nuclear ph ysics, then slo wly shifted his in terest to theoretical ph ysics and mathematics. He then graduated from F udan Univ ersit y in 7.2002 with the bac helor's degree. In the same mon th, he left China and came to the Univ ersit y of Florida to pursue his graduate study in ph ysics. And he has b een in the Ph ysics Departmen t ev er since. 106