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This is to certify that the
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"Using the Barbieri—Remiddi Lowest—
Order Equation for Calculating the
Decay Rate and the Energy Shift of
Positronium"

presented by

Alireza Abbasabadi

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USING THE BARBIERI-REMIDDI LOWEST-ORDER EQUATION
FOR CALCULATING
THE DECAY RATE AND THE ENERGY SHIFT

OF POSITRONIUM

BY

Alireza Abbasabadi

A THESIS

Submitted to
Michigan State University
in Partial fulfillment of the requirements

for the degree of

DOCTOR OF PHILOSOPHY

Department of Physics and Astronomy

1985

ACKNOWLEDGMENTS

I am most grateful to my advisor, Professor Wayne W. Repko, for
suggesting the topic of this dissertation and for his patient
guidance and support throughout this work. Without his constant

help and encouragement, this thesis would not have become a reality.

I also extend my thanks to Professor Julius Kovacs for his

support during my graduate study years at Michigan State.

Thanks also to Dr. Alberto Devoto, who gave valuable comments

and assisted me in the learning of the SCHOONSCHIP program.

I would like to thank the entire staff at Physics and Astronomy

Department for making this department a pleasant place to work.

Thanks also to Debra Robbins for the typing of this manuscript.

I thank the National Science Foundation and Michigan State

University for their financial support.

Many thanks to all the people who made my stay in East Lansing
an enjoyable experience, and to my fellow graduate students, and

especially to my friend Hossein Zoheidi.

Finally, I wish to thank my fiance, Patty, for run" love,

understanding and patience throughtout the writing of this thesis.

ii

TABLE OF CONTENTS

List of Tables

List of Figures

Chapter 1.

Chapter 2.

Chapter 3.

Chapter A.

Introduction

Solving the Bethe-Salpeter Equation

for Positronium

Contribution of One-Photon-Annihilation
Channel to the Energy-shift of the
Ground-State of Orthopositronium in the

First-Order Perturbation Theory

Contribution 6E1

Contribution 6E2

Contribution 6E3

Contribution 6E“

Contribution 6E5

Contribution 5E6

Contribution 6E7

Summary of the Calculation of the Total
Energy-Shift of the Orthopositronium

Ground-State due to the One-Photon-Annihilation

Channel

Decay Rate of the Ground-State of

Parapositronium in the First-Order

vi

vii

10

20

22
23
28
36
38
AA
56
58

61

Perturbation Theory

A. Contribution M

B. Contribution M

C. Contribution M

D. Contribution M

E. Contribution M

F. Contribution M6

G. Contribution M

H. Contribution M

8

I. Summary of the Calculation of the Total

Invariant Amplitude and the Decay Rate of

the Parapositronium Ground-State

Chapter 5. Summary and Conclusion

Appendices

Appendix

Appendix

Appendix

Appendix

Appendix

Appendix

Evaluation of an Integral related

to the Contribution 6E3

Evaluation of an Integral related

to the Contribution 6E1

Proof of an Identity

Evaluation of an Integral related

to the Contribution M1

Calculation of a Trace related

to Mu

Evaluation of some Integrals
related to the Contribution M2

iv

65
67
72
73
80
8A
90
95
95

99

102

107

110

113

116

122

Appendix G. Evaluation of some Integrals 130

related to the Contribution MA

List of References 138

Table 1.

Table 2.

LIST OF TABLES

Summary of the Calculation of the

Energy-Shift

Summary of the Calculations of the

Amplitudes and the Decay Rate

vi

6O

98

Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

1.

2.

3.

LIST OF FIGURES

Kernel 6K, where the Coulomb-like kernel 20
KC is given by (2.31). Lines with a dash

indicate inverse propagators.

Energy shifts 6E's, where GC and Kc 22

are given by (2.29) and (2.31), respectively.

(a) Definition of quantity Auv' (b) Graphical 2A

representation of "uv' (c) Equality which

is correct up to the 0(a) correction. Gc is

given in (2.29) and k = (2n, 8).

A

Contributions 6E3a and 6E3b, where R 28

.)
is given by (2.30), and k = (2K, 0).

Diagram which is related to 6E3b. R is 33
defined in (2.30).

. . a b
Contributions 6E“ and 6E“ . 36
The identity which is the direct result 37
of the equation (3D.2).
Contributions 6E 3and 6E b. Lines with 39

5 5

a dash through them indicate inverse propagators.

vii

Figure 9.

Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

10.

11.

12.

13.

1A.

15.

16.

17.

Contributions 6E6a and 6E6b.

Vertex correction for 6E6a and 6E6b.

Equalities which hold up to the 0(a)
corrections, where D00 is the instantaneous

Coulomb interaction (see (3C.2A).)

Equalities which are correct up to
the 0(a); they are the result of

equation (3F.25).

Equalities which are correct up to

leading order, i.e. 0(1).

Equalities which are correct up to

the 0(a) correction.

Equality which is the result of comparing

Figure 2(a) with Figure 2(g).

Graphical representation of the total
energy-shift, correct up to the order a

correction.

Kernel 6K, where the Coulomb-like kernel K0 is

given by (2.31). Lines with a dash indicate

viii

A5

H5

A7

53

54

55

56

59

61

Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

Figure

18.

19.

20.

21.

22.

23.

2M.

25.

26.

270

inverse propagators.

The only (GK') which contributes, up to

the 0(a) correction.

Quantity n. up to the 0(a) correction.

Contributions to the ivariant decay amplitude,
where G0, KC and C are given by (2.29),
(2.31) and (A.5) respectively. Lines with a

dash indicate inverse propagators.

Diagram related to M2, where k = (2K, 8),
It: (19;) and k2 =0.
Diagram related to M3, where k = (2K, 8)

and h= (191:).

Contributions Mua and Mub.

Diagram related to MA

a

Diagram related to Mub.

Contributions M5a and MSb.

The identity which is the direct result

ix

62

63

6A

68

72

7A

714

77

81

81

Figure 28.

Figure 29.

Figure 30.

of the relation (3D.2).

Contributions M6a and M6b, where R 85
and C are given by (2.30) and (A.5)
respectively. Line with a dash indicates
inverse propagator.

. a b
Contributions M7 and M7 . 90
Total invariant decay amplitude, correct up 96
to the 0(a) correction. Line with a dash
through it indicates inverse propagator, and

the normalization constant C is given by (A.5).

Chapter 1

Introduction

Positronium, which consists of one electron and one positron,
is a pure quantum electrodynamic (QED) bound system. It is
experimentally accessible and therefore its study is another
sensitive test of QED. It also can be used to test any two-body
relativistic equation for the case in which the interaction is
purely electromagnetic. Since positronium consists of a particle
and an antiparticle, which can annihilate and create photons, the
lifetime of positronium is very short, 10-.10 sec. If we want to get
a gross structure of positronium, we may compare it with hydrogen
atom. In this case, since electron and positron have equal masses,
the reduced mass will be-% me. Using this reduced mass, we find out

that the Bohr radius of positronium atom is twice that of hydrogen

atom, or the ionization energy is half of the hydrogen atom and etc.

The spin of positronium is sum of the spin of electron and spin
of positron. Therefore it can have spin 0 (singlet state) or spin 1
(triplet state). Positronium with spin 0 is called parapositronium

(p-Ps), and with spin 1 is called orthopositronium (O-Ps). We use

28+1

the notation L where J, L and S are total angular momentum,

J,
orbital angular momentum and spin of the system respectively; and

beside they are related by the relation 3 = E + S. By using this

notation, the states of positronium can be grouped as

ls 3s - 1P 3P 3P 3

1. 1. 0’ P and so on.

1’ 2

We can show [1] the charge parity (charge conjugate) of
positronium has the eigenvalues (_)2+s, and for photon it has the
eigenvalue (-). Therefore, for positronium decay into photons, we
must have the following selection rule (since C,1flmacharge

conjugation, is a good quantum number)
1+
(-> S = (-)“ (1.1)

where,riis thexummer of photons into which the positronium can
decay. Therefore in ground-state (l = 0), parapositronium (s = 0)
can decay into an even number of photons, and orthopositronium
(s = 1) can decay into an odd number of photons. Since from
conservation of energy-momentum, the decay of orthopositronium intc>
a single photon is forbidden, the minimum number of photons that
cmthopositronium, in ground-state (2 = 0), can decay into is three
photons. For parapositronium, in ground-state (2 = 0), the minimum

number of photons that it can decay into is two photons.

The calculated decay rate, in lowest-order approximation, for

p-Ps (parapositronium) is [2]

5 2 _
Fth(p-Ps -+ 2Y) = gr-(E¥%J = 8.0325 x 109 sec 1 , (1.2)

while the measured value is [3]

_ _ 9 -1
rexp(p Ps) - (7.99“ i .011) x 10 sec. . (1.3)

The calculated decay rate, in lowest-order approximatitni, for o-Ps

(orthopositronium) is [A]

2
2 2 6 mc
rth(o-Ps -+ 3Y) §;-(n - 9) a (-fr9

7.2112 x 106 sec-1 , (1.u)

while the measured value is [3]

6 -1
rexp(o Ps) - (7.051 i .005) x 10 sec. , (1.5)

(for a review of experimental advances in positronium see the review
by Rich [5] and for new measurement of decay rates see [3], and for

new development in QED see [6].)

The accuracy of the measurement of p-Ps, relation (1.3), is not
high enough to test the radiative corrections to the decay rate
(1.2). While for the o-Ps decay rate measurement, relation (1.5),
its accuracy is high enough to test the radiative corrections to the
decay rate (1.“). In fact the decay rates, including the radiative

corrections, are

2

I‘th(p-Ps) = I'O(p-Ps)[1 - (5 - 13‘) + In a'.‘ + 0(a2)]. (1.6a)

dIQ
uflk)
Q

2 1

_ g _ -2 -1 ‘
rth(o Ps) ro(o Ps)[1 " (10.266 i .011) 3 a In a .

+ O{ (g)2}] 9 (1.73)

wherelkxp-Ps), the lowest-order decay rate of p-Ps is given by
(1.2), and FO(o-Ps), the lowest-order decay rate of o-Ps is given by
(1.“); the 0(a) correction to p-Ps was done by Harris and Brown [7].
and the 0(a) correction to o-Ps (the value that we quoted in (1.7a))

and the order 02 In a 1 corrections, to both para-and

orthopositronium, are given by Caswell and Lepage [8].

In relations (1.6a) and (1.7a), if we assume that the

coefficients of 02 and (%)2 terms are unity, then we find

Fth(p-Ps) = (7.98M i .001) x 109 sec—1 , (1.6b)

(7.0386 1 .0002) x 106 sec“1 . (1.7b)

Fth(o-Ps)

For the decay rate of p-Ps, comparison of (1.3) with (1.6b)
shows that the theory and experiment are in agreement, but the
accuracy of measurement is not high enough to check the coefficients

of a and 0:2 In 01 1 terms. In fact if we write equation (1.6a), up

to the order a, in the following form

rth(p-Ps) = Fo(p~Ps)(1-ka) , (1.8)

experimental result (1.6b) gives the following possible range of

values for k

k 2:.66 i .2 . (1.9)

We note that the calculated value of k

2
k=l(5-1'——)z.806 (1.10)
1! ll .

is one of the possible values in (1.9). But since the range of the
values of k:h1(fl.9) is not narrow enough, we do not consider the
agreement between the measured and calculated values of the decay
rate of p-Ps as a sufficient criterion for the test of the radiative

corrections.

For the decay rate of o-Ps, comparison of (1.5) with (1.7b)
shows that the agreement between theory and experiment is not

satisfactory, and in fact there is a discrepancy.

To resolve the unsatisfactory agreement between the I‘th(o-Ps)
and Pexp(o-Ps), Gidley et al. [3] initiated more measurements of the

o-Ps decay rate.

Up to now all measured values of I‘(o-Ps) have been above the
calculated one, therefore one may think that, maybe, there are other
channels available for the o-Ps to decay to. I]: is suggested [9]
that the following process could be responsible (at least partially)

for the discrepancy between theory and experiment
0
o-Ps + a + Y, (1.11)

vfliere, a0 is a neutral particle of mass ma < 2me

O

Amaldi et al. [10] measured the decay rate of the above
process. For the mass ma varying from 100 to 900 kev, they found
the following range of upper limits for the ratio of decay rates of

the above mode to the 3X mode

6

6 - 1 x 10’ . (1.12)

P (o-Ps + aOY)
F (o-Ps + 3Y)

 

R = ~ 5 x 10”
Therefore, their result shows that the process (1.11) can not

resolve the discrepancy between rexp (o-Ps) and rth (o-Ps).

There is another quantity of interest, the ground-state
hyperfine splitting of positronium, AB = E (s = 1) - E (s = O), the
energy difference between the s = O and s = 1 levels with n = 1.

The measured value of this quantity is [11]
AEexp (hfs) = 203.3885 i .0010 GHZ , (1.13)

while the theoretical value is

chau 7 a 32
AEth(hfS) =‘——F—— [g - F'C‘g + 2 ln 2) +

own
9
5...;
3
Q
+
c:
,‘
Q
n)
L—J

a 203.u00 GHZ. (1.14)

The first two terms were calculated by Karplus and Klein [122], and
calculaticnicu'the 02 1n 0-1 term started by Fulton, Owen and Repko
[13], and completed by Caswell and Lepage [1N], and Bodwin and

Yennie [15].

Contribution of 02 term to the hyperfine splitting, if we
assume its coefficient is unity, is of order .005 GHZ. Therefore in
order to have a meaningful comparison of theoretical result,
relation (1.1“), with experimental result, relation (1.13), we need

to calculate the 0 (a2) correction to the hyperfine splitting.

As expressed by Buchmuller and Remiddi [16], most methods which
proposed and used in the past - for calculating the radiative shifts
of energy levels of positronium - work only1n31x>the order

02 In a 1 correction and in practice they can not find a2 correction

to the AE, relation (1.1M).

In Chapter 2, we discuss the method that Barbieri auui Remiddi
[17] introduced for solving the positronium problem, which in
principle can be used for finding decay rate or energy splitting of

positronium up to the any desired order in a.

Throughout our work, in subsequent Chapters, we use the Coulomb
gauge. The reason, beside the others, is that in covariant gauge,
for positronium, there are some spurious terms, such as o (013) and

3

o (a In 0-1) corrections to the energy levels, which appear in some
Feynman diagrams (for a discussion of these problems and the
cancellation of these contributions see [18]), while hi(knuomb
gauge we do not have such spurious terms.

In Chapter 3, after introducing a perturbative expansion for

«energy levels [16], we calculate the energy shift, GB, of the

(nethopositronium (6E = E - 2K, where E is the ground-state energy of

2
o-Ps and K = (1 - 9—H)1/2)' We consider those contributions which
come from the one-photon-annihilation channel and contribute up to

the first-order of perturbation theory (up to the o (a) correction).

The energy shift 6E that we find is

——-—(1-————-— (1.15)

which agrees with the result of Karplus and Klein [12]. The
interesting point is that we find the finite value for (SE, without
performing the wave-function and vertex renormalization subtractions
(for the regularization and renormalization of QED in Coulomb gauge

see [19] and [20]).

In Chapter N,an%er writing down the perturbed wave-function
for parapositronium, in terms of the zeroth-order wave-function, we
calculate the decay rate of p-Ps. We consider those decays which
contribute up to the first-order of perturbation theory (up to the

0(a) correction). They are the decays of p-Ps into two photons.

The decay rate that we obtain is

 

r (p-Ps + 2 1()= ”Sh“ [1 - § (5 - 17)] . (1.16)

which agrees with the result of Harris and Brown [7]. Here also we
find the result (1.16), without performing the wave-function and

vertex renormatization subtractions.

Throughoutcnu'work, we use the notation and conventions of
BJorken and Drell [21]. We also use the natural units 11 = c = 1,
and take the electron mass me = 1. For regularizatitni, we employ

the method of dimensional regularization [22].

10

Chapter 2
Solving the Bethe-Salpeter Equation

for Positronium [17], [23]

The formulation of the bound-state problem in QED, which is
given by the relativistic Bethe-Salpeter (BS) equation [2w],
although from theoretical point of view is complete, it lacks a
tractable and systematic computation procedure. In practice, in
order to solve BS equation, which is an integral equation, one
specifies a lowest-order equation, then by using this equation as a.

starting point, one does a perturbative calculation on it.

In general this lowest-order equation has a kinetic and an
interaction term. The interaction term is responsible for the
boundrstates and supposedly is the largest part of the full BS
kernel“ Ihn the limit of low-momenta these two terms should reduce
to the kinetic energy and the Coulomb potential of the Schrgdinger
equation, and also it is advisable that in the limit of high-momenta
the kinetic term describes the free propagation of the electron and

positron.

In the following, we discuss a lowest-order equation which has

the above properties and can be solved exactly.

The BS equation for the Green function G is [2“]

duk

(20)

 

G(P.p.q)=GO(P.p.q)+f “GO(P.p.k)

11

 

x f 'u x (p, k, k') 0 (p, k', q), (2.1)
(20)

and the lowest-order equation which is proposed by Barbieri and

Remiddi [17] is

GC (P. p. q) = GO (P. p) [(21r)Ll 0” (p-q)

A

+ fd p'u Kc (P, p, p') Gc (P, p', q)]. (2.2)
(20)

 

First we specify GO and Kc' In the C.M. frame where P = (2w, 8) (2w
is the total energy of positronium and 2w-2 is the binding energy),

GO and K0 are chosen as

1 1 (1)[ 1 1(2)

_ ' L— _ _ + ’
p + wYO 1 + 15 p wYO 1 18

 

 

GO (W, P) = [ v (203)

 

and
Kc (w. E. a) = - i) (E. 3) R (w. 5) v0 (5 - 3) R (w. 3). (2.u)

where

+ + 1 + + + + (1)
A (p. q) = 16 E E N N [(Np + p - Y) (1 + YO) (Nq + q - 1)] ’

p q p q ’
+ -> -> + (T)
x [(Nq + q - v) (1 - YO) (Np + p . 1)] , (2.5)

and

+
8(W.D)=——"—.E
/E + w p

and V0 is the scalar Coulomb

12

potential

 

Np = Ep + 1 , (2.6)
e2
‘3; ) (2.7)

In the above relations, (1) stands for electron line and (2) stands

for positron line.

By specifying Kc and GO, as we did, (nus finds the exact

solution to the equation (2.2) for the Green function Gc'

In C.M.

frame this solution has the following form

0C (w, p, q) = 00 (w, p) [(2101l 0

“(p-q)

+ 1 R (w. B) “c (w. E. a) R (w. a) A (S. a) 00 (w. 6)]. (2.8)

where, the scalar function HC

+ +
HC (w, p. q) =

x 115 - 3'2 p - [1/(uw2 — u)] (E: - we) (E: - we) (1 - p)2} .

1P - 3|

is

2

+ An a v f d p p-

1
v

0

1
(2.9)

13

where

(2.10)

c
u
.4
u
mun

We note that He has pole at w = K (from now on we assume w < K)

2
H0 (w, E, 5) + 32 1' Y 1 , (2.11)

K (52 + Y2) (a2 + Y2) w - K

 

W + K

where

K = ./1- Y , (2.12)
and it corresponds to the ground-state energy.

In equation (2.9), if we replace the real integral with contour
integral [25]

i

inv
2 Sin v e f0 d p (° - .) 9 (2.13)

f; d p (. . .) +

where the path 0 begins at p = 1 + oi and terminates at p = 1 - oi,
after encircling the origin within thexunite circle, then we find

that He has poles at

1H

|-<
NM

n = 1, 2, 3,... , (2.14)

which<xnrespond to all excited states of energy (including the
ground-state). But since we are interested in ground-state, the Hc
which is given by equation (2.9) is the one that we will use in our
calculations.

Now, since Hc has pole at w = K, therefore by equation (2.8) G0

has pole at w = K too. At the pole GC is

1 Eve (1)) we (D)

GC (14. p. q) -> 2 K w _ K . (2.15)

 

 

where summation is over all degenerate states (for ground-state
(2 = o) degeneracy is only due to spin). Using (2.15), one can find
wo's, inns zeroth-order Barbieri-Remiddi wave-functions. For

orthopositronium ground-state (s = 1, I = 0)

i n a
$0 (p) = +2

(P

(K - E ) {E + K
P P

¢
+ Y2)2 0 Ep Np

 

 

(Np-p-Y)(1+Yo)£

(m) + _+.->
Y (Np p Y)

X _ , _ _ . 9
(PO + K Ep + ls) (p0 K + Ep IE)

(2.16)

and its "conjugate" transpose

15

-T -i n a
111 (p)-—- 0

2 22 E
0 (15+)) 0 pr

 

 

-+.+ _ A(m).-+ _+.-)
(Np p 1) (1 1p) g 1 (Np p 1)

(p0 + K - Ep + is) (pO - K + Ep - 16)

 

X

. (2.17)

(m)

where ¢o and spin 1 polarization vectors g are

3
(9—'1/2. 5(0) = (O, O, 1) , 5(i) ='l~ (1, i i, 0) . (2.18)

¢o g 80 /§

For parapositronium ground-state (s = o, l = o)

_ 2 i w a
111 (p) - p2

(K - E ) VB + K
¢ P P
O
(P

+ Y2)2 o E

 

 

P

(E +Y -E~i¥)¥
p o o 5

X _. _ _ 9 (2019)
(p0 + K Ep + 15) (p0 K + ED is)

and its "conjugate" transpose

-T a 2 1 n a (K ' Ep) 'Ep + K

v p 0
o (p2 + Y2)2 0 Ep

 

 

-> +
(Ep-YO+p-YYO)Y5
(PO + K " Ep + 18) (p0 ' K + Ep - is)

X

(2.20)

In the above relations we defined 80 by the following relation

i=7 1) Y . (2.21)

One may use (2.15) for finding an integral equation for me.

This equation can be found by taking the residuum of (2.2) at w = K

A
_ d p' 1 1
WC (P) ' GO (K: P) f (2w)“ K0 (K: P, P ) $0 (P ) . (2022)

 

We can show that the following normalization condition for

these wave-functions is satisfied

 

f (211)}; (21:11“ 5° W 53%; [GO-1 (P' p' Q)
— Kc (P. p. 0)]1o (q) =1 , (2.23)
where
P=(2K,‘6),.<=(1-12)”2,1=% . (2.211)

In order to use the perturbation theory around w = K, one needs
the following quantity (in Chapters 3 and A we will use this

quantity)

. . Z w w
0p(1<)a[0p(w) -—1———‘-’-———9-]

2K w - K w + K ' (2°25)

which is finite (pole contribution is subtracted), and can be found

by using (2.8) and (2.9)

17

A

0p (K, p, q) = (211)ll a”

(p - q) 0p (K. p) + 1R (K, S) R (K. a)

x 00 (K. p) 1 (E. 6) 0p (n. q) fip (K. p. q) . (2.26)

where He is finite (pmfle contribution is subtracted). and is given

A

in [23]. Up to the first order of perturbation theoryy lap can be

written as

 

 

 

 

 

. 2 2
1 1 5 -2Y -2Y
H (K. p. q) = 1.. {-+ -[-+ ———————-+
c A B 2 +2 + Y2 52 + Y2
Y2 -Y2 Y2
p + K - E + is p - K + E - 16 q + K - E + 16
2 1
-Y 28 - A - B
+ - K + E - 1 + f dp p 2 ]} 9 (2.27)
QO 0 Ap + B (1 - p)
where
+ + 2 1 +2 2 +2 2
A=|p-q| ,B=——2-(p+Y)(q+Y). (2.28)

In the course of calculations of positronium decay rate or
energy splitting, in some Feynman diagrams we divide the
contribution of each diagram into two parts. One which comes from
the first term in CC, equation (2.26), and the other which comes

from the second term in 60' Let's write (2.26) in the following

form

18

A

Gc (K. p. q) = (21r)Ll 6" (p - 0) GO (K. p) + R (K. p. q) . (2.29)

A

where R is
a (K. p. q) = 1 R (K. E) R (K. 3) Go (K. p) 1 (E. E)

x GO (K, q) HC (K, p, q) . (2.30)
Beside R, there is another quantity that we always refer to, it

is KC (K, E, a), which is Kc (w, B, a) at w = K, or from (2.A)
Kp (K. 5. a1 = -1 x (S. 5) R (K. 5) v0 (6 - 3) R (K. E) . (2.31)

There are some remarks that we should mention. In 1952,
Salpeter [26] derived an equation which is the approximate version
of the BS equation and has the properties which we mentioned at the
beginning of the present Chapter; but the exact analytic solution of
this equation is not known. This equation and its improved version
which was given by Cung et a1. [27], before 1978 had been the
starting point of nearly all perturbative calculations for the
positronium and muonium. In 1978, the year that Barbieri and
Remiddi [17] proposed their lowest-order equation, Caswell and
Lepage [1A] also proposed an equation which is essentially

equivalent to the Barbieri-Remiddi equation.

It is also necessary to note that the Dirac equation, for a

system which consists of an electron in an external Coulomb

19

potential, can be solved exactly and it gives the correct energy
levels up to the order a“ (fine structure) [28]. One may think that
the Dirac equation is a good lowest-order equation to start with;
but this is not correct. The Dirac equation for an electron in an
external Coulomb potential can be obtained from the BS equation, by
considering the interaction kernel as a sum of the all irreducible
crossed ladder graphs in the limit of infinite mass of positmmi
[17]. In fact it is an equation for one fermion, not for two

fermions.

In any case, we do not need to demand that the lowest-order
equation, by itself, should give the correct order “A for energy
levels of positronium. As long as a lowest-order equation, by using
a perturbation expansion, gives the correct energy levels in terms
of the successive powers of a, it will be a good lowest-order
equation to start with (explicit calculation shows that beside

powers of a, we have also In a).

20

Chapter 3
Contribution of One-Photon-Annihilation Channel
to the Energy-Shift of the Ground-State of

Orthopositronium in the First-Order Perturbation Theory
The energy-shift of the orthopositronium ground-state, in terms
of the perturbation kernel 6K which is given by Barbieri and Remiddi
[17]. can be written [16], up to the first order in perturbation
theory as

1 as = (5K) + (6K 0p 6K) + (6K) (6K') + 0(6K3). (3.1)

where 6K = K - KC, and is given graphically in Figure 1.

2:. a.
(a)6K=- +>W<+§ T§++T

v
(1.1 E: ——

Figure 1. Kernel 6K, where the Coulomb-like kernel K0 is given

Kc

1---1

by (2.31).. Lines with a dash indicate inverse
propagators.
GC and K0 are given by equations (2.29) and (2.31),

respectively.

21

131 (3.1), SE is defined as 6E = E - 2K where E is the ground-
state energy, and prime stands for %:-%E , and (...) means the

expectation value with respect to $0. For example,

A A
(0K') f dp GL1)

1 a
mM——aummpflwwh Gm
(2w)” (2“)A 0 2K 3K 0

where (110 is the zeroth-order Barbieri-Remiddi wave-function [17],

which for the orthopositronium ground-state is given by the equation

(2.16).
Now using (3.1) and the graphical representation of 6K, Figure
1, one gets the energy-shift of the ground-state of

orthopositronium, up to the first-order of perturbation theory.

The contribution of one-photon-annihilation channel, up to the

0(a) correction, can be written as

7
6E = l 6Ei . (3.3)
i

where, GE's are represented graphically in Figure 2.

(a) 16E1= (M)
(b) 1032=(>-~@M<)

(c) 16E

22

(d) 108“ = -2 xi )

(e) 1625 = A :j:i1gg::>~n-<::)
(f) 16E6 A V)

7 2 <:::7“"7<:> :::i:f)>

(g) 16E

A

Figure 2. EMergy shifts GE's, where Gc and K0 are given by

(2.29) and (2.31), respectively.

In the following sections we calculate the contributions GE's.

A. Contribution (SE1

Using Figure 2(a), contribution 6E1 can be written as

 

_ uv
16E1 - I d pp < -ieY w (p) > —155—
(2“) AK
6“ - T
x I——9—u- < 11 (q) (-ieY ) > . (311.1)
0 V
(20)

where < > stands for trace. The double integral factorized, and it

is the case for all other diagrams.

23

Using (2.16) for wave-function $0 and performing pO integration
(by virtue of Cauchy's theorem), and then performing the angular

integrations, one gets

 

 

dup
f <-ieY w (p) >
(2w)“ u o
= ie¢ g“ 39-f” p d p / + K (2E + 1) . (3A.2)
0 3n 0 (52 + Y2).? Ep P P

The integral in the right-hand side of (3A.2), up to the 0(a)
correction, is performed in Appendix B; using that result, equation

(8.11), we find

 

 

u
f d 9,, < - 1er 110 (p) > = mo 5“ 5(1+%) . (311.3)
(2“)
and since
d“ -T d" *
f qp < 1pc (q) (—1evp) > = [1' “PT < -1er 1110 (p) >] , (311.11)
(2w) (2")

equation (3A.1) gives (up to the 0(a) correction)

) . (3A.5)

B. Contribution 6E2

For 6E Figure 2(b), let's first consider the<nuuudty App

2’
which is represented graphically in Figure 3(a).

1
1
P++k
(.1 A
UV K [4 1
l’vigk 2- 9"?“
P+Lk
(b) Hpv = K p. y K
P-fk

Figure 3. (a) Definition of quantity App. (b) Graphical
representation of pr. (c) Equality which is

correct up to the 0(a) correction. (a: is given in

(2.29) and k = (2x, 3).

Using the Feynman rules and equaticnl (2.29) fxn~ Go, the

quantity AW (u, v = 1, 2, 3) can be written as

 

d“ d" 1 1 2 1
A = -f p“ q+[< - ieY iS (p + Ek)(-ieY )iS (p - Ek)>
“V (2H) (2H) u V

x (2w)“ 6u(p - q) + iR(K, E)R(K, 3)AC(K, p, q)< - ieYuiS1(p +%k)
.. 11(‘15. 5)1sl(q + -;-k)(-ier)iSZ(q - gmzfifi. S)
x 182(p --%k)>] , (38.1)

where 1 stands for electron and 2 for positron. Let's write (38.1)

in the following form

A = H + H' , (38.2)
uv uv uv

25

 

 

where

dnp 1 1
pr = -f A <-1erIS(p + Ek)(-leYv)lS(p - 2k>> , (38.3)

(21!)

and

ion dup qu + + A -1
H' =-—— f R(K, p)R(K, q)H (K, p, q)(E E N N )
“V “ (2.)“ (2.)” ° p q p q

.{[(p + 3.12- 1][(p - gk>2— 1][<q + ék)2~ 1][<q - %k)2— 111“

X<Yu<v +-%k + 1)<Np + S - i)(1 + YO)(Nq + 3 - i)(q + 5x + 1))V

x(4 -‘%u + 1)(Nq + a - 1)‘l,‘ YO)(Np + S - v)

1
x (p " 5K ‘1’ 1)) , (38.14)
which means that pr is the contribution of the first term of Go and
“Liv is the contribution of the second term of Gc (see (2.29) for

A

Gc)'_

The quantity Huv is represented graphically in Figure 3 (b),
and obviously it is of order a. For 111'N we want to show that its
contribution is of order higher than 0, therefore we can ignore it.
In equation (3B.A), we note that the most contribution comes

from E, 25 ~ Y and p0, qO ~ Y2. In fact after performing po and q0

26

integrations (poles at p0 = K - ED and q0 = K - Eq give the most

contribution) and keeping the lowest powers of E and q, we get

H (K E 5)
n' ~ 0 fd3pd3q C . (38.5)

 

 

“V (52 + )2)(52 + Y2)
where E is
c
2 2

- + + 1 1 5 AY AY

H (...p, q)=A1ra{—+—[—- -

C A B 2 +2 + Y2 +2 + Y2

1 28 r A - B

+ I do p ]} . (38.6)

 

° Ap + B (1 - p)2

and quantities A and B are given by (2.28).

By scaling E + 13 and a + YE, we note that 80 is of order fi-and
therefore pr is of order (:2; so we can ignore 111'N and up to the

0(a) correction (38.2) gets the following form
A = n , (33.7)
which is represented graphically in Figure 3(0).

Now, by virtue of Figures 2(b) and 3(0), contribution 6E2 is

A .
153 = f-—9—EE < - ierwo(p) >[li n“V(k) :1}

2 (2n) k2 k2

26

integrations (poles at p0 = K - ED and q0 = K - Eq give the most

contribution) and keeping the lowest powers of E and 3, we get

fie (K, B! a)

 

 

3 3
"' “ 0 Id Pd 8 . (33.5)
uv (52 p Y2)(a2 + Y2)
where E is
c
2 2
- T * 1 1 5 AY AY
H (K, p! Q) = “WGL— +'— L— --——————— —._______
C A B 2 +2 + Y2 +2 + Y2
+ (699 28 - A - Bp ]} . (38.6)

Ap + B (1 - p)2

and quantities A and B are given by (2.28).

By scaling B + Y; and 5 + Y3, we note that He is of order g-and
therefore IIL'W is of order 012; so we can ignore Him and up to the

0(a) correction (38.2) gets the following form
A = 1T ’ (38.7)
which is represented graphically in Figure 3(c).

Now, by virtue of Figures 2(b) and 3(c), contribution 6E2 is

u
__ d p _ -i uv -i
16E - f p < ievap(p) >[——-n (k)-—§}

2 (21) k2 k

 

27

u
x f-—9—3— < 1T (Q)(-ieY ) > . (38.8)
(2")A o v

where “uv' the vacuum polarization, can be found from (38.3), 1vhich

up to the 0(a) is

in a 8 2
Huv (k) = :7]; [(1 + '2‘) I'(-a) 1’ 3] (kn kv - k 81W) 9

k=(2.<,6), a=-’2—‘-2 (n+A), I‘(-a)+-1_-5-,a>o,(38.9)

(for finding “uv’ by using the cut-off method, see [29]).

Equation (3A.3) implies

u
_d_Il_ -1 a, =
f(2")“ < 1ervo(p) > 5p (50 o) . (38.10)

therefore (38.8) gets the following form

 

u
-121. a - £3. dp _
1682 - 3p [(1 + 2)r( a) + 3] f (2“)” < 1evap(p) >

. uv A

x '13 f d q~ < 0T (q)(-ieY ) > . (38.11)
k2 (2p)A o v

 

which by comparing it with (3A.1), we find

-o a 8 .
151:2 = -3—p- [(1 + -2-)1‘(-a)+ §](16E1) . (38.12)

28

Charge renormalization takes care of the first term, therefore by

virtue of (3A.5)

(1LI -8a
5E2 = '1"— (T1?) ’ (38.13)
which is correct up to the 0(a) correction.
C. Contribution 6E3
For 683, Figure 2(0). we decompose it as
a b
(SE = (SE + (SE C.1

where, the first term comes from the first term of Gp, equation

A

(2.29). and the second term comes from the second term cfi‘Chp; they

are represented in Figure A.

(a) 1083a= (W)

N

 

*
= —-e¢o a “ /2'(1 + %)
. 6
(b) 16E3 =2 .fl} )
*

A

Figure A. Contributions 6E3a and 683b, where R is given by

(2.30), and k = (2K, 3).

29

We first calculate 6E3a. From Figure A(a)

a =.:i *u .9
6E e105 ./2(1+p)

 

3 2
dup 1 1
x f (2")u < -1e Ap(p - 5k. 0 +-§k)vp(p) > . (30.2)

where A11 is due to the vertex correction and can be found by the

following relation

 

A(p-—k.p+%1<)=i°‘ Jane—L—Ji . (30.3)

where

D =62 02 [(q+ p -%k)2 -1][(q+ (ӣ102 - 1] . (30.11)

A = [_62890 _ qpqO + q0(quoo + qodpo)]pru-Y0 _ B“ ’ (3C.5)

CD
I

- [“32890 - qpq° + q°(qu°o + q°épo)]Ydedep . (3C 6)
1 1
c=(n+¢-§u+1)1p(q+v+§k+1) . (30.7)

Note that in the numerator of (3C.3). we added and subtracted the

term 8“, the term which gives ultraviolet divergence. So we find A“

A

(and J' dnq TEL) in 11 dimension (since it gives finite contribution)

and Bu in n dimension. Therefore

30

+2 0 0 0 +2
A = - Y 0 Y - 0 + 0 Y + Y 0 — A
p q p p d pd q d p 0 q 0 HA 0 qu

2 +2 2 2 2
+ q q Yu 0 q oYu 2 qoq qup . (30.8)
+2 2 +2 2
B = -2 2 - n + - n Y + Y + 2 Y . C.
p ( ) q qu q [(3 )q p q p p qpqp O] (3 ‘9)
From (3C.2) and equation (2.16) for we, we note that it is
necessary to evaluate the quantity <ApF>, where

+ + " -> + ->
F = (Np - p - Y)(1 + Yo)€ - Y(Np - p o Y) . (30.10)

By virtue of (3C.3), we obtain

 

10 u <A F) ia n <8 F>
<A F> =-——— f d q u +-——— I 0 q-—E- . (30.11)
H 3 D 3 D 1
An An
where
2 u
<ApF> = -32(Ep + 1) q a . (30.12)

2

<30F> = [32 (2 - n) 620 p + 16 (3 - n) 32002 + 16 (n - 3) 5”

+ 16 quOZ] 5“ (30.13)

In finding (3C.12) and (30.13) we used these arguments: from the
structure of integrals in (3C.11) we notice that for the first

integral the small values of q(q ~ Y) gives the 0(1) and 0(0)

31

contributions to (SE a, while the non-small values of q gives only

3
the 0(a) contribution. For the second integral only non-small
values of q contribute and give the 0(a) contribution. Therefore in

(30.12) we dropped terms such as‘Y2q2

+ + 2 -> 2
. P ’ q q . (p ° 5) and etc.
In (3C.13) we put p = O, Y = O. Beside these approximations we also
used the relation po -- K -ED and p ~ Y, which follow from the fact

that the main contribution, up to the 0(a), comes fvwxn the pole in

wave-function.

USing (30.12) and (3C.13) hi C3C.11), we notice that we need
the following integrals, which can be found by the table of

integrals which is provided in [19]

2

qug—-=iir I131 1

__ I 0y . (30.1u)
0 /x o k2y2 _ k2y + E: _ 52x

 

 

 

 

 

 

+2 2
q Q _.

f 0“q D“ = 13 (1 - 3a)r(-a) (u = 1. 2. 3) . (30.15)
929 2

f dnq p° = -21112 - 102F(-a) , (3C.16)
+1 .

x dnq-flfi = ‘1: (3 - 7a)r(-a) . (30.17)
azq 2

f 0“q ° = ’1“ (1 + 3a)F(-a) , (3C.18)

D A

32

In relations (3C.15) - (30.18), we dropped 3 and p0 and Y, since we
are interested in finding the values of these integrals up to the

0(1), which corresponds to the 0(a) correction to GEa3.

The integral in (3C.1A) is performed in Appendix A and is given

by equation (A.7). the result is

f 0 qq—=l‘p—sin. (————P--—) - 21112 . (30.19)

D
/32 p Y2

Using these results in equation (3C.11), we find

u -
(A F) =-51§— [31 (E + 1) sin 1 (————Jl—-—) - 8 + r(-a)] . (30.20)
u T p D '35—““‘§
/p + Y

where the first term will give the 0(1) and the rest the 0(a)

corrections to 68:.

From equations (3C.2), (3C.10), (30.20) and equation (2.16) for

 

 

 

wo' we find

a a6 /- a 3 Ep + K
55 = 2 (1 +-—) f d p

3 16113 ” 6 (52 + Y2)2E (E + 1)

p p
21103 +1) _1
x [ 2 sin (————11———) - 8 + r(-a)] . (30.21)
/32 + Y2

Since in the square brackets the first term gives the:<>(1) and the
second and third terms give the 0(a) corrections to 68a, for the

second and third terms we use the approximation Ep = 1, K = 1.

33

Therefore, using these approximate relations and then performing p

integration on these two terms we find

 

5 5
6E: =-%; /2 (1 + %- [-/§ +-—§ r(—a)
m p {E + K _
+ a fodp p2 p 2 2 sin 1 (———4B—-)] . (3C.22)
(p + Y ) E +2 2
p /p + Y

The ithegra1.i11 (3C.22) is performed in the Appendix A and is
given by equation (A.20). Using that result we find, up to the 0(a)
correction

u
a

01
6E3 ='_E [2 -

52..

(I C!
p -§ +-§p r(-a)] . (30.23)

To find 68;, we note that in the diagram of the Figure 5. up to

the 0(a) correction, only the instantaneous Coulomb interaction DOO

0f the photon propagator D contributes, where

iv

1 _ v v A
D)v(p) =.;§ [ giv + +2 ] ' (30.2A)

    

P—iu

Figure 5. Diagram which is related to dish

3. R is defined in

(2.30).

3A

Dominant contribution comes from small p and q1 (but it is not
the case for q), more specifically, p and q1 are of the order a, so

we keep the lowest powers of p and Q1 and we set Ep = Eq = 1,
1

where ever it is legitimate (for more details see Section D of

Chapter A for calculation of MB which is similar to the GED.)

3

After performing PO and q0 integrations we find (poles are

9 Q10

at p0 = K - E = K - E and q0 = K - Eq)

p. 010 Q1

 

6
oEb = a /2 (1 + 9) f d3pd3qd3q (2E + 1) WE + K
3 3811117 1 6 1 q q

1

. 80(K. 6. 6,)[(32 + Y2>2<3 “ 51>2<31 * *2)‘52 * 12)Eq]— ' (3C725)

or after performing p integration

 

 

5 (2E +1)./E +1<fi(1<,3,5)
683:: °‘ 5 /2(1 + %)f(13q<13q1 9 p2 ‘212 +2 0 2 1 . (30.26)
19211 (01+Y)(q +Y)Eq

A

where EC comes from Hc’ equation (2.27). after performing p0, qO and

Q10 integrations, or

 

 

(30.27)

35
The integral in (3C.27) does not contribute. It is so because

by direct integration we can show (see Appendix C)

I 03.11: (62) f 0361 [1 2B ’ A ‘ B" 2] -- 0 . (30.20)
Ap + B (1 - p)

 

where F(82) is any function of a2 which satisfies the conditnni

(C.13) of the Appendix C

q. +2
p2 2 F(q ) < w . (3C.29)
which in our calculations it is the case.

Therefore, using (3C.27) in (3C-26) and performing Q1

integration

 

 

 

p 5 a 3 (2E +1)./E +1:
003=°2/'2‘(1+—6-)qu 32 23
2An (q + Y ) Eq
2
Y (2 + K + E )
i- a
" [2 +2 2 ] '. (30'30)

q + Y

The second term in square brackets only for q ~ Y contributes. For

this term we use the approximations Eq = 1 and K = 1. Therefore

ash-i/E(1+9-)Id3q 1
3 802 6 (52 + Y2)2

 

36

2
1 A /2'Y
x [312:1- (ZEq + 1) VEq + K ‘ W] . (3C.31)

The integraticnicni the first term in square brackets is
performed in Appendix B and is given by the equation (8.11).. Using
that result and performing the integration on the second term in the

square brackets we obtain

0'
Q
J:

20
0E =-—E (1 +-§—) . (30.32)

UL)

So, by virtue of equations (30.1), (30.23), auui (30.32), the

contribution 683 is

[3 + 0 - —— + —2—p— r(-a)]. (30.33)
D. Contribution 68”

For GE“, Figure 2(d), we use the same decomposition that we had

for 6E3, equation (30.1), or

11 + 6E“ . (30.1)

where dEpa and dEub are represented in Figure 6.

(a) iGEpa -2 KS >

u
1
ml—e
(0
e.
o
m
a:
1:
'81
A
_)
4.
<3le
V
V

(b) 16E" =~2 Ki 0’

 

a b
A and 6E“ .

Figure 6. Contributions 6E

Using the following identity, which we gave it by the equation

(2.22) (momentum integrations are implicit)

we = GOchO . (30 2)

we obtain the identity which is respresented in Figure 7.

1(1 =

(l
I

Figure 7. The identity which is the direct result of the

equation (3D.2).

earid from it we find the contribution 6E a

u
083—-2 (SE (30 3)
11 ' 1 ' '
<>r~ by virtue of (3A.5)
a an 20
6E” =-H (-2 --§—) . (38.A)

38

Calculation of dEub is similar to 6E3b. Ihi fact, by applying

the same approximation that we used for (SE b, at the very beginning

3
we note the following relation which is correct up to the 0(a)

correction
68 = —68 . (30.5)

(for the details of calculation see Section E of Chapter A for the

calculation of M5b which is similar to GEub).

Equations (3D.5) and (30.32) give

u
60 b ='EE (—1 —-§1) , (3D.6)

annd equations (3D.1), (3D.A) and (3D.6) give

68 = 25 (-3 --51) (3D 7)
A A 3 ' i

E. Contribution 0E5

To find (SE25, Figure 2(e), we decompose its contribution into
‘tvvo parts, as we did in (30.1) for 683
a b
E = E + E E.
6 5 6 5 6 5 . (3 p1)

"P1€3re GESa and 6E5b are represented in Figure 8.

39

(a) 1685a 11 >

e¢oa*“/2 (1 + g)

a.
(b) idES 6?

II
J:

e¢o€*“/2'(1 + 9)

 

Figure 8. Contributions GESa and 6E5b. Lines with a dash

through them indicate inverse propagators.
For the electron self-energy we use the following form [19]
19 +

2(9)='%Fr(‘a)(¢‘1)+%;{~gp°i-%0Y

-f1-9£{(1 _ x)5 . ? + 1]1n x + 2f1 dx[(1 - x)p - 1]ln Y
Opf‘ O
x

+2-p-Yfgdx/xfzpduln2} , (3E.2)

undere, the mass-renormalization is already performed and

X
ll

1 + 52(1 - x) , y = 1 - 02(1 - x)-ie .

1 - p02(1 - u) + 32(1 - xu) -ie . (32.3)

N
II

For the contribution 6858, Figure 8(a), we write

A0

a _ _ *u pg
6E5 — A0000: 1/2 (1 + 6)

11
d p 1 1
x f (210” <YpS(p + Ek)2(p + 5101;0(0) > . (38.11)

 

Using (3E.2) for £(p +-%k), by replacing p by p + %k III (38.2)

and (3E.3). we find

 

6E5a = 620 5*“/2’r(-a)f dupu < Y 0 (p) >
O (21;) u 0

A
-a2¢og*u/§ f _d__P__

1
< Y S(p + —k)
(2")A p 2

x [0

+ +
1 p - Y + 02(0O + K)Yp + 0 10(0) > . (30.5)

3]

innere
C2 = 12--f1-9£ (1 - x)ln X -2f1 dx(1 - x)ln Y + 2f1 dx f1 du ln 2 ,

1 6 of _ o 0 o

x

C: = --l + 2f1 dx (1 - x)ln Y

22 2 o 0 ’
03 = 41—“ In x -2 11 dx 1n Y . (38.6)

0 /—' O
x

The first integral in (38.5) up to the order that we are

~1111:erested is (see equation (3A.3))

A1

A
I”

(20)

 

- - H'-
p < Ypup(p) > - /2'5 /2 . (38.7)

For the second integreal in (3E.5), we can show that it is of
higher orders so we can ignore it. In fact, after performing the po
integration (dominant contribution comes from the pole hi the wave-
function, or for p0 = K - Ep; we close the contour in the upper half
pflane of p0). then by scaling E + Y3 we note that in order this

integral contributes, the following trace should be, at most, of the

order 02

<Y (p + K Y + 1)[0 E - i + 0 (p + K)Y + 0 ]
u o 1 2 o O 3

x (Np - 0 ° Y)(1 + YO)€ - Y(Np - p - Y) > . (38.8)

where p0 = K - Ep.

It is not difficult to see that we need to find 01 up to the

o (1), but for C2 and 03 up to the order 012. From the relations in

(38.3) and (38.6). with the replacement p by p + %k, we find the

fwallowing approximate relations

10
C1 ,3 ,

0 = -2 - 4(52 + Y2)1n[2(52 + Y2)]-2(f52 + Y2)

2

03 = 2 + A(E2 + Y2)1n[2(32 + Y2)]"% 32 . (38.9)

A2

where we have used the following approximations

S , 1c =1 - —Y . (38.10)

After finding the trace in (38.8) and using (38.9), we note
that it is of the order higher than 02. Therefore the second

integral in (38.5) does not contribute.

Using (38.7) in (38.5) we obtain

.fi'

685a = “—p [- % r(-a)] . (38.11)

For the contribution 685b, Figure 8(b), we write

6E

 

A A + +
b 100 ¢OE*u/§(1 +.E)f d p d q, R(K, p)R(K, q)

5 A (2")A (20)“ EquNqu

X

" + + + +
HC(K. q. p)<(Np + p - Y)(1 - YO)(Nq + q ° Y)

1 1 + +
$01 3k)YpS(q + §k)(Nq + q - Y)(1 + YO)

X

x (Np +5 . Y’)S(p + gkmb + %k)wo(p)> , (38.12)

Where we have used (2.30) for R.

Using relations (AD.9) for simplifying the trace in (38.12),

and using (38.2) for )3, we note that (by scaling B + Y5 and a 4 Y3)

A3

the most contribution comes from E, 8 ~ Y and the poles at

p0 = K - 8p and q0 = K - Eq (there is only one pole for qo).
After performing the p0 and q0 integrations, we obtain

5 I; (K, a! E)
b - a "a _ 3 3 0
GE - [ r( a)]fd pd q ,2 2 2 .2 2

5 16115 u" (p + Y ) (q + Y )

+ [contribution from the second term of 2]. (38.13)
where AC is given by (38.6).

We can show that contribution of the second term of )3 is of
liigher orders, so we may neglect it. In fact, for the contribution
of’the second term of Z we have a similar situation that we had for

61358; here we need to find the following trace, at most, up to the

or~der012
<(N ++°Y1~Y N-IoiYN-I-ier N-I-Y’
p p )( 0)( q q ) u( q q )( O)( p p )

+ -) + -> A +
x[c1 p - Y + 02(2r - 8p)Yp+C ](N -p - Y)(1+Y0)€ ° 1

3 P

x (Np - E ~i)> . (3E.1A)

It is not difficult to show that we need to know 0 up to the

1

QC 1 ) , but 02 and 03 up to the order 012. Therefore we may use (38.9)

Fol" values of 01, C2 and 03. After finding the trace in (38.1A) and

HM

using (3E.9), we note that it is of order higher than a2. Therefore

in (3E.13) only the first term contributes.

Using (38.6) for go and then (3C.28), relation (3E.13) gets the

following form

7

 

 

 

b -a 3 3 1
6E = r(-a)fd pd q
5 16fl5 (32 + Y2)2(52 + Y2)
2 2
1 1 5 --m -Lw
x[-+4-+ + H . (£49
A B 2 +2 + Y2 +2 + Y2

where A and B are given by (2.28).

Performing p and q integrations, we obtain

6E b = 274-[—9-r(—a)] . (3E.16)
and by virtue of (3E.1), (3E.11) and (3E.16) we find

, u
6E5 =-§E L:%% r(-a)] . (3E.17)

F. Contribution 686

For 6E6, Figure 2(f), we decompose its contribution in the same

waY‘that we did in (3C.1) for 6E3

6 + 6E6 , (3F.1)

85

where GEéa and 6E6b are represented in Figure 9.

(a) 16E6

 

‘
(b) 16E6b 14y)

 

Figure 9. Contributions 6E6a and 6E6 .

First we consider the vertex correction in these diagranms (see

Figure 10.)

m

Figure 10. Vertex correction for 6E6a and 6E6b.

Chantributions 6E6a and 6E6b can be written as

N H
*
536a = 4611120121105 “[271 + 3M d qu d p" <b“(p - q)
‘ (2H) (2H)

 

1 1 1 1
x szm - Ekwusm + Emlm + 5k. p + 5k)vo(p)> . (3F.2)

N6

duq1 h

1.1
(2H) (2H) (2H)

m. 23mm 31) -
x . HC(K. q. Q1)<D

 

lv( _ )
E E N N p q1
q q1 q Q1

X

1 + + -> ->
sz(q1 3k)(Nq + q1 -1)<1 mm} + q - 1r)

1

1 1 -) -> -> +
x S(q --§k)Yus(q +-§k)(Nq + q - Y)(1 + YO)(Nq1 + q1 - Y)
x S(q1 + 13mm, + g1. p + g-moe» . (31.7.3)

where DAV is given by (3C.2N).

Using relations (HD.9) for simplifying the trace ill (3F.3) we

4'

9

. -> -> -> +
note that (by scaling p + Yp, q_+-Yq and Q1 + Yq1) the most

and 61 ~ Y and the poles at

.041

contribution comes from B,
p = K -13 , qo = K - Eq and q1O = K - Eq1 (for (3F.2) we have the
same situation with the exception that there are only p and q

variables.)

Performing the po, qo and Q10 integrations (for (3F.2) cnmly p0

and q0 integrations) we see that up to the 0(a) correction only

0 contributes. This means that A = 0 (since DOi = o, for

V

i

1, 2, 3).. Therefore only D00, the instantaneous Coulomb

interaction contributes. We summarize these results in Figure 11.

47

h

(a)

(b) .6, ::

Figme‘fl. Equalities which hold up to the 0(a)

 

corrections, where D00 is the instantaneous

Coulomb interaction (see (3C.2u).)

Therefore, contributions 6E6a and 6E6b get the following form

 

6
a a 5*“fd3 3 [(p2 2 2 +2 + Y2)(E _ + 2] 1

6E pd q + Y ) (q Q)
32w

1 1 ‘ +
<vu(1 + YO)AO(q +-§k, p + 5k)(1 + YO)5 - Y> . (3F.4)

X

6 * _
6b -—41<7 a “Id3pd3qd3q1 “C(K' 6,. a)
256w

6E

[(132 + 12)2<Ef + 12)<32 + 12><B - 51>21“

X

X

1 1 ‘ +
<Yu(1 + YO)AO(q1 +-§k, p +-§k)(1 + YO)§ - Y> . (3F.5)

where EC is given by (3C.27).

Relations (3F.N) and (3F.5) show that we need to find the
leading order of A0. In the following we find A0, but we keep in

mind that AC is sandwiched between two (1 + YO)'s.

N8

Using Feynman rules the quantity A0 (p', p) (where p stands for

p +-%k and p' stands for q + %k or q1 +-%k) can be written as

. n
Ao(p'. p) = ifig-IEES [Fuva(d + p' + 1)10<q + p + 111“] . <3F.6>
11’

where

D = q2 52[(q + p)2 -1][(q + D')2 '1] 1

2
Fuv= -gW a - qu qv + qo(qu6: + qug) . (3F.7)

Since p0 and po' are of the 0(1), and B and 5' are of the 0(a),

(3F.6) gets the following form

n
, = id d q uv
A0(p . p) ”"3 f—B- [F Yv(q + 1 + YO)YO(d + 1 + v0)vu] , (3F.8)

 

or if we use the fact that in (3F.N) and (3F.5).1¥)is sandwiched

between two (1 + YO)'s

. n
AO(P'. p) = -39-r959 [qiqz + 2<n - 2>q§ 62 + (3 - n>q2 62

“W3

+ u qoq2 + u q2] . (3F.9)

Using the table of integrals, which are provided in [19], we

obtain the following relations

N9

 

 

 

 

2 2
Q q
n o 2 1 dx_ 1 .1 , _ , 2
qu p 111 f0 _fody{A[p +y(po po )1
/x 1
-1 xaA aI‘(-a)}
2 1 ’
q q
n o _ _. 2 1 Ex. 1 .1 , _ ,
qu p - 111 fo'rfodyA [p0 +y(pO p0 >] .
x 1
2 a
. 2 1 d 1 1
fdnq-g— = In f0 -£-fody-K- .
.5 1
n 62q2 2 1 a
fd q D = in I‘(-a)fodx A2 , (3F.10)
and also the following relation
<12 32 2
n o _ 2 1 1 x , _ , 2
Id q D 111 f0 dx f0 dy{‘A-;[po + y<po p0 )]
- l xa+1Aa r(-a)} (3F 11)
2 3 ' °
where

2 2 2 2 2
AF [p0'+ y<po - PO')] -X[3' + 116-3)] +1 - p' + y(p' - p)

2 2 2 2
2 2 2 2
AB‘ x[p' + y<p - p')] +1 - p' + y(p' - p ) (3F~‘2)

50

It is not difficult to note that for the integrals in (3F.10).
it is legitimate to use the approximate values of p and p', before

performing these integrals.
The approximate relations that we use in these integrals are
+ +
p = P ' = 1 . p = p' = o . A1 = A2 = 1 . (3F.13)

Using (3F.13) in (3F.10) and then performing integrals we

obtain

 

 

 

q 2q2
fdnq OD = -in2 F(-a) ,
q q2
fdnq O = -2112 ,
p
2
fdnq-g— = 2112 ,
D
n +2 2 2
Id q qu = in r(-a) . (3F.1u)

For the integral in (3F.11) it is not clear that we can use the

approximate relations in (3F.13); so first we perform the x

 

 

integration
q 232 2
n o . 2 1 C 1 B A + B
fd q p - in IQ dy-—§[§A B +‘K— 1n ( B )

51

N

2
-EE ln B + (£5 - 1)ln(A + 8)]
A A

in2 1
u f0 dy[T(-a) + 1 -

+

 

. (3F.15)

>1m

where

2
A=[p'+y(p-p')] .B=1-p'2+y(p'2-p2).

C = [po' + y(pO - po’)]2 . (3F.16)

Since the approximate values of A, B and C are (for all values

of y, 0.5 y_: 1)
A=1,B=o,C=1 , (3F.17)

we can use these values in (3F.15) before performing the

integration.

Using (3F.17) in (3F.15) and then performing the y hategration

we obtain

. 2
fdnq OD -—l—Z—[l‘(-a)-1] . (3F.18)

 

Now, by virtue of (3F.9). (3F.1H) and (3F.18) we find

1 _ E_, _
AO(P . p) - H“ F( a) . (3F.19)

52

Relation (3F.19) is derived under the assumptknithat A0 is
going to be sandwiched between two (1 + YO)'s. Therefore the proper

way of writing (3F.19) is

a
1 z __ _

(1 + YO)Ao(p . p)(1 + 10) (1 + 10>[ufln a)]<1 + 10) . (3F.20)

'We note that, we have proved the legitimacy of using the

following approximation in the integral in (3F.9)
..)
p' = p = (1. o) . (3F.21)

However, if we use (3F.21) at the very beginning, 1.63., using
(3F.21) 1J1 (3F.6) where A0 is not yet sandwiched between two
(1 + Yb)'s, we find a term which is proportional to (1 - YO) but
undefined. Still there is a fast and simple way of fhufing the

result (3f.20).

By considering the following approximation (instead of (3F.21))

212 2
p'=p=(21<-Ep.p)=(1-Y -35.3) . (3F.22)

we can write (at this point we only use the equality p' = p)

1 1
Ao(q1+§k,p+§k)

1 1
Ao(q +31“ p +—24<)

1 1
110(1) + 5k. p + 5k) . (3F.23)

53

1k, p + %k), which can be

Therefore, we need to find Ao(p + 2

found by the Ward identity

A (p. p) = - —§- 2 (p) . (3F.2u)
0 3p ,
0
Using equation (3E.2) for Z, and approximate relations (3F.22),
and keeping only the leading orders we obtain

1
Ao(p +-§k. p +

Ni.“

01
) ='EF {YOT(-a) + u(1 -Yo)
x [1 + 1n(232 + 212)]} , (3F.25)

where if we sandwich this AO between two (1 + YO)'s we find the same

form that we had in (3F.20).

Since the second term in (3F.25) does not<xnuxfibute, we can
represent the result in (3F.25) effectively by the Figure 12.

(a) D A0 7 [€77- 11.3)] I?"

90

O

(b) 2:» 4-35,; 1121)] D“

Figure 12. Equalities which are correcttu>1x>the 0(a);

they are the result of equation (3F.25).

Now, using the result (3F.20) or (3F.25) in the relations

(3F.u) and (3F.5). we find

5H

 

 

7 + + +9—
6363 = a 5 F(-a)fd3pd3q[(p2 + Y2)2(q2 + Y2)(p - q)2] 1 . (3F.26)
16w
b a7 333- +~>
6E6 = 8 T(-a)fd pd qd q1 Hc(K. Q1. q)
128w .

x [(32 + Y2)2(3f + 12)(32 + 12)(3 - 51)2]-1 . (3F.27)

or by virtue of the equation (3C.27) for go and equation (3C.28) and

then performing the integrals

u
6E6a = E74-[%-r(-a)] , (3F.28)
as b - 35 [—°1 r(- )] (3F 29)
6 - u 2n a '

Finally, relations (3F.1), (3F.28) and (3F.29) give

u
a 3a
6136 = 7 [31; r(-a)] . (311.30)

We note that, there is a close connection between 6136, Figure

2(f), and GE Figure 2(c). In fact, since we are interested in the

3!
leading order, we can derive the results (3F.28) and (3F.29) by
observing that in the diagrams in the right-hand sides of the

equalities in Figure 12,I%K)can be replaced by Duv' We represent

this approximation in Figure 13.

(a) D =

55

(p) p". 2:

Figure 13. Equalities which are correct up to leading

order, i.e. 0(1).

Comparing the results in the Figures 9, 11, 12 and 13 we obtain

equalities which are represented in Figure 1“.

*
(a) 16E6a= e¢og “/E'fig-P(-a) :::Z:::>fifi-

*
(b) 15E b= ed) 5 “5—9- I‘(-a) .6,
6 0 Mn
Figure 1”. Equalities which are correct up to the 0(a)

correction.

Now, the comparison of the Figure 11 with Figure 114 gives the

following relation which is correct up to the 0(a) correction

SE = —9-r(-a) GE

6 2“ (3F.31)

3 9

and by virtue of the equation (3C.33) for GE we recover the

3!
equation (3F.30) for 656.

There is also a relationship between 6E6 and 685, which CENT be

found, simply, by comparing (3F.30) with (3f.17)

6E = -6E . (3F.32)

56

G. Contribution 6E7

For calculating 6E Figure 2(g), by virtue of the Figure 15

7’
__LC:5__’
16E = 216E

7 1
———7l-——-——

Figure Hi Equality which is the result of comparing

Figure 2(a) with Figure 2(g).

we write

 

1:
_ d p _1 _§_ _ ‘1 _ 1 'T
”7 ‘ 2531f 112 3k < S (p Emi’o‘p)
(2v) 0
x 2(p + J2-k)wo(p) > . (30.1)
k0 = 2K

Using relations (2.16) and (2.17) for wave-functions and

relation (3E.2) for 2, we obtain

-ia
6E72 11

8Kw

 

u _ 3
6E1P(-a)fd p(k + Ep)(K Ep)

x [(52 + Y2)u(pO + K - Ep + ie)2(po - K + Ep - 102]-1

~ia6
u

 

+

6E1fdup(K + E )(K - Ep)2
8Kn p

+2 2 H 2 _ _ 2 -1
x [Ep(p + Y ) (po + K Ep + 16) (p0 K + Ep is) ]

1 %k . 32)] . (3C.2)

 

where

-—k ) - f X In X

‘11?

X
+ 2x; dx{(1 - x)[Ep(pO +-%ko) - 32] - 1} ln y

+ 232f; dx /§'f; du ln 2 , (30.3)

and X, Y and Z are given by (3E.3) (with the obvious replacement of

1
p by p +-§k).

After performing pO integration (dominant contribution

corresponds to the pole at p0 = K - Ep) we find

-a6 3 K + E
6E = (SE I‘(-a)fd p T“
7 16113 1 (p2 + 12)“

 

2

6 (K + E )
— a 3 6E1fd3p 92 p
321 Ep(p

 

 

+ Y2)5

3F -.§§_
x [F + (K - Ep)(2-§E; apo)] . (30.1)

The second term in square brackets of (3C.11), because of the

following identity, is zero

58

1 BF
8k 2 3p . (30.5)

0 0

Using relation (3C.3), we note that F is of order higher than
a2, i.e., the first term in square brackets also does not
contribute. Therefore, only the first integral in (BG.N) survives.

Using the approximation Ep = K = 1 in this integral, we find

_.:2 _
6E7 - 2" F( a) 6E1 , (36.6)

or by virtue of relation (3A.5) for 6E1

.1:

a -a
5157 — 7 [3“— r( a)] . (3G.7)
H. Summary of the Calculation of the Total Energy-Shift

of the Orthopositronium Ground-State due to the One-

Photon-Annihilation Channel

Using relations (3.3), (3C.1), (3D.1), (3D.3), (3D.5) and
(3F.32), we obtain the total energy-shift of the orthopositronium
ground-state

6E = -6E + 6E2 + GB + 6E , (3H.1)

which we represent it graphically in Figure 16.

59
16E = 2 “

+ >.o~< +4: >“<

Figure 16. Graphical representation of the total energy-

shift, correct up to the order a correction.

Relatnn1(3H.1), or its graphical representation Figure 16,

shows that, up to the 0(a) correction the second term of 60' R, does

not give any contribution to the energy-shift 6E.

Using relation (3H.1), (3A.5), (38.13), (3C.23) and (30.7), we

obtain

11

ds=9—u(1-ia—-

11

212°

) . (311.2)

whcih agrees with the result of Karplus and Klein [12].

We present all results of calculations of the present Chapter

in the Table 1.

60

Table 1

Summary of the Calculation of the Energy-Shift

 

 

 

 

 

 

 

 

 

G“ a
5E1 1 T (1 + *)
11
a -8a
532 "H (7.11?)
53 Ef-[3 + a - BE-+pg— P(-a)
3 H N 2w
1.1
a ”a
6El4 '—E (‘3 "§-)
L1
a -3a _
5135 7 [—2—11- I'( a)]
u
a 3a _
636 7 ['27: l"( a)]
a" -a
6E7 71:2? F(‘3)]
u
a “a 8a
“3 =1:1 6E1 ‘11“ ’7-51?)

 

 

 

61

Chapter u
Decay Rate of the Ground-State of Parapositronium

in the First-Order Perturbation Theory

For the decay rate we use the perturbed wave-function 111 which

can be written [30] as

_ " 2

= 1
0 up 1110+ (6K )11;O+Gc 6K1po+o(6K) , (11.1)
where GC is given by (2.29) and C is the normalization constant auui
6K = K - K0 is represented graphically in Figure 17.

.21.— ~71...
2.1:. .1 ..
_+—-

E
w 1,

_‘C:7—

Figure 17. Kernel 6K, where the Coulomb-like kernel Kc is
given by (2.31). Lines with a dash indicate

inverse propagators.

In relation (N.1), $0 is the zeroth-order Barbieri-Remiddi
wave-function [17]. which for the ground-state of parapositronium is
given by (2.19), and Prime stands for éz'gzy and (..Q)nwans the

expectation value with respect to $0 (see (3.2)).

The normalization constant C, up to the order a, is [31]

c" = 1 + (6K') . (u.2)

The only (6K') which contributes (up to cfla)) is the one that we

represent it in Figure 18. We can show that its
__4C::h_.’

(GK') = 2
___.f—-—-—'

Figure 18. The only h“(') which contributes, up to the

0(a) correction.

value is the same as the (5K') for orthopositronium. So by

comparing relation (3C.6) with Figure 15, we obtain

1 _._E -
(6K ) - 2“ T( a) . (4.3)

Therefore, from relations (N.1)-(N.3). the perturbed wave-

function, up to the 0(a) correction, is
= c
p c[1po + (6K )10 + 0c 6K p0] , (u.u)
where C, the normalization constant, is
C =1+ 74- I‘(-a) . (11.5)

To find the decay rate up to the 0(a) correction, we need to

consider only the decays to two photons. It is so because for

63

parapositronium, decays to odd number of photons are forbidden and

decays to four photons are of the higher orders.

Let's express the decay amplitude as [31]

1
(2w 2m'K)

 

T(p-Ps + 2Y) = M (p-Ps + 2Y) . (4.6)

1/2

where, w = “1|, (0' = |1’| and 1: and 11' are wave vectors of the two

photons; M the invariant decay amplitude is

u

M(p-Ps . 2v) = f d 9“ <nT w(p)> . (1.7)
' (2w)

 

The quantity'11 is represented graphically by irreducible graphs in
the Figure 19, and we assume the mass and charge renormalizations

are already performed.

 

l 5,11 .~a
P+zk
+
n = Y I ’ ‘t —+
52K ,,___L~u

 

 

Figure 19. Quantity n, up to the 0(a) correction.

In C. M. frame, M can be expressed as

M = f h - (e x e') , (4.8)

A A
where f is a constant and k.is the direction of the decay line and e

A

and e' are polarization vectors of the two back to back photons.

64

The decay rate is

3 3

d d ' (211)" 5% - k- 11') z 1112 , (11.9)

(2103 (2103 8"

 

1
F=fif

where, k = (2K, 6).
By virtue of (4.6), (4.8) and (4.9) we obtain
r = 811? |r|2 , (14.10)
or, up to the order a correction
r=%17|f|2 . (11.11)
The invariant amplitude M, according to the Figures 17-19 and

relations (4.5) and (4.7), can be expressed up to the 0(a)

correction as

8
M=.§ M, , (11.12)

where M 's are represented graphically in Figure 20.

 

1
11,:
P+ K 1
(a) M1 = 2c "
1""i” l k—K,a
(b) M = 110 1

 

(c)

(d)

(e)

(f)

(g)

(h)

M3 = 20
Mu = 20
M5 =
M6 =

65

 

QVJ

) _

3

7".—

 

M = 4C(

Figure 20.

/~n.
M=4C 1

(
( )1:
(

 

)3...
3541

Contributions to the invariant decay
amplitude, where Gc’ Kc and C are given by
(2.29), (2.31) and (4.5) respectively. Lines

with a dash indicate inverse propagators.

In the following Sections we calculate the contributions Mi's.

For M

M = 2Cf d p
(2“)

1

 

19

q
4

A. Contribution M1

Figure 20(a), we write

< -ie 2' i S(p + %k -k)(—ie ¢)¢O(p) > , (4A.1)

and using (2.19) for wo' after performing pO integration, we obtain

66

a 20‘2 3 Ep + K

M -—— Co K 0 e x E'fd p A
1 1' 0 E13652 + Y2)2(Ep -5 ° 1*)

P

(E2+1<2-213-R’)1/2 ‘

(1111.2)

In the square brackets of (4A.2), the first term is due to the pole
of wave-function and it gives the 0(1) and also gives contribution
to the 0(a) correction; the second term is due to the pole in
propagator and it gives contributLon to the 0(a) correction.
Therefore, for the second term we may use the approximation K = 1

(since for this term only non-small values of E contributes).

After performing angular integrations, (4A.2) gets the

following form

 

 

 

 

2 °° p
M = 8C a p k - e x e' {f dp in (E + p)
1 o o (32 + Y2)2E p
p
m E+1 (E+p)(E-1+/62+2-2p)
-fdp—p—-——ln[ p p ]} . (11111.3)
p Ep-1+/p+2+2p

The first integral is performed in Appendix D. For the second

integral, we found it by numerical integration. The results are

2
first integral = /2 (:3 + G - 2 ln 2 --§%g) , (4A.4)

67

second integral = /2 A , (4A.5)

where, G = .91596559... is the Catalan Constant and A =

.49172096... .

Therefore using (4.5) for C, contribution M

1 up to the 0(a)

correction is

2
M1 = M011 + T4g-[M-a) + 80 - 8A - 16 ln 2 —-3%—]} . (4A.6)

where

3
ll

4 /2lw a ¢o K ° 6 x e' , (4A.7)
is the lowest order invariant amplitude.

B. Contribution M

 

2
For M2, Figure 20(b), we write
d” 1
M2 = 4Cf p“ < -ie t' i S(p + 5k -'K)(-ie)
‘ (2n)

n 1 1
x Ai(p + E3 -'k. p + 5k)eiwo(p)> (1 = 1. 2. 3) . (13.1)

where momenta p, k and.k, and polarization vectors eanuie' are

shown in Figure 21.

68

 

 

p+~1§k 11,5:
,, :VVVvVv?”“M~
Lkgxk
1 V
p -Ԥk k -'k, e'
< M

 

Figure 21. Diagrwun related to M where k = (2K, 8),

2’
k: (K, 1:) and k2 = 0.

After performing p0 integration (dominant contribution comes
from the pole in the wave-function, or for p0 = K - Ep) and scaling
S + YE we note that the dominant contribution comes from E of order
Y} therefbre we neglect p wherever it is possible. We find up to

the 0(a) correction

 

3
-12 dp
M‘na‘bofif-e 22

<d'(1+v ~16)
(p +Y) ' O

1 1
x 111(1) + 5k -k. p + 24081” + YO)Y5> . (1113.2)

where we have used (4.5) for C.

Since we need only the leading order of A1, we neglect p in A1

and perform the rest of the integration in (4B.2). We find

M2 = nape/'2’ ei<¢'[(1 + YO) + K o1(1 — 1(0)]

1 1
x 111 (3k -k. 540(1 + YO)Y5> . (48.3)

69

1where we introduced another factor of (1 + YO) in order to simplify

the calculation of the trace.

Ai can be found by the following relation

n
_ in d g uv + . +
Ai -~;;§ f D F Yv(q + R Y + 1)Yi(q + YO + 1)Yu , (48.4)

where
p = 32 q2[(q + -;—k)2-1][(q + -%k 402-1] , (1113.5)

Fuv = _8uv a2 _ quqv + q°(qu6: + qvég) . (48.6)

Let's separate the term which gives the ultraviolet divergence

from those which give finite contribution

n
ia G q uv
111 «;;§ fi—D— F quYquD

 

. u
1a -g_g uv
+ f D [F vqui(1 + Yonu
+ F“VY (1 + fi - ?)Y (4 + 1 + Y )Y ] (48 7)
v . i o u ’ °

where only the first integral gives ultraviolet divergence.

Using (48.7) in (48.3) we obtain

7O

./2 2 a“ d“
”2 = ‘ I? 0‘ ¢oei[f—I3g—<>n * Fag-(>111 ' (“B-8)

where, effectively

+ + 2 2 +2 2
g o —
on... <2n Y[qo q 11 + (3 n>q q Y1

+

+ 2(2 - n)qi 62 a - v]1015¢'> . (13.9)

__ +2+ +_ +.+->.—>
<>D__<4Yi(qoq R Y qoq kq Y

.. 2 + . 9 1
do q k 1010152: > . (1113.10)

To find M2, relation (48.8), we need to perform the q

integration. This can be done by using the table which is provided

in [19]. We find

+ term proportional to ki kj;

so gives zero contribution to

 

the trace of (48.9) , (48.11)
n +2 2 2
Id q 359— = in I‘(-a) , (1113.12)
q 32
fduq O = in2(1 - ln 2) . (48.13)

71

The evaluation of these three integrals was easy, so we gave

only the final results. However, the evaluation of the rest of the

integrals that we need is not straight forward. In Appendix F, we

have shown how to perform these integrals. Theruxnflts are (see

relations (F.11), (F.14) and (F.39))

 

 

 

 

2 2
C10 q 2 112 — —
D =i1r[-r(-a)—8+—-2-+u./21n(1+./2)
-11 ln2(1+ 5)] . (1113.111)
2
q q 2
OD = 1112[- 1,7 + 2 ln2(1+ 5)] , (1113.15)
11 90‘1th 2 11 3112 95
qu——D——-—-111 hjkD[—§- 16+21n2-—2—-ln (1+5)
+ %~ln2(1 + f2)]
.2 3 112 35'
+ in 6j1[--§ +'T6 - ln 2 + 2 ln (1 + /2)

--l-1n2(1 + 5)] . (118.16)
by virtue of (48.8) - (48.16) we find

Mofi[2r(-a)+12-u-81n2+1151mm+5)

72

-2ln2(1 + /2)] . (48.17)
where Mo is given by (4A.7).

C. Contribution M

3
For M3, Figure 20(0), we write
d" 1 1
M = ch—i— <-ie 2' 1 S(p + —k - k)(-i)):(p + —+< -k)
3 11 2 2
(2“)
x i S(p + —k12 -k)(-1e ¢)po(p)> , (no.1)

where the momenta p, k and)(, and polarization vectors 5 and e' are

shown in Figure 22.

 

 

 

p + 51‘ k, E
7 A.“ .’,
V
< v‘v—W
1
p --§k k -k, 6'

Figure 22. Diagram related to M3, where k = (23:, 8) and
h = (K. K).

After performing pO integration (the dominant contribution
comes from the pole in the wave-function, or for p0 = K - Ep) and

scaling B + YB, we notice that the leading contribution to M3 cxmnes

73

from E of the order Y. Therefore we neglect B and replace K by 1

everywhere except for the term-:E—l——§. We find
P + Y ‘

1 2 ' +.-> 1_ +o+
”3331751’0“ <95 (1+1: Y)E(-2-k k)(1+k Y)

3

d p
x 2(1 + Y )Y >f . (4C.2)
- o 5 (32 + Y2)2

 

By performing p integration and using relation (38.2) for Z and

finding the trace, we obtain

M=M—°i

1dx 1
3 o 1111 [-r(—a) + fO—:1n x + 2 Io dx In x] , (110.3)

/x

where MO is given in (4A.7). X and Y are given in (38.3), which in

present case they are
X = Y = 2 - x . (4C.4)

Using (4C.4) in (4C.3), we obtain (up to the order a

correction)

M3=M0fir~[-r(-a)—6+111n2+u51n(1+5)] . (40.5)

D. Contribution “4

For MD, Figure 20(d), we use the same decomposition that we

used for 683, relation (3C.1), or

(4D.1)

where MDa and MDb are represented in Figure 23.

a
(a) MD = 20 . (b) MD = 20

 

1' 1

Figure 23. Contributions MD3 and Mub.

Contribution MDa can be written as

4 4
a = 32n2a2Cf d p d q, Duv

M u (q - p)
(21) (21)

u
1 , 1 _ 1
x <YvS(q - Efi)é S(q + 5k k)tS(q +-§k)YDwo(p)> . (4D.2)

where momenta p, q, k and k, and polarization vectors 2 and e'.are

shown in Figure 24, and p0, Duv and C are given by the relations

(2.19), (3C.24) and (4.5) respectively.

 

 

p+—k q+-12'k kve
> ‘7 'W
a?
p-';_k (1";‘k k-klt 5’
‘4 4 A-‘W

 

Figure 24. Diagram related to MDa.

After penfinmung pO integration (up to the 0(a) correction,

only the pole in wave-function contributes, or for po = K - Ep) and

75
scaling E + Yp, we note that the 0(1) and 0(a) corrections both come

from the small values of B, or S of order Y.

By neglecting 52, Y2 and higher orders in B and Y, wherever it

is possible, we obtain (see Appendix E)

 

 

 

a 1d d p d q 4
M =M C I f [ +1
.4 0 4n5 (52 + Y2)2 D (a _ +)2
2 4 2 2
0 q0 q1 q0 q1
“32 " 2 ‘7' +2 ] ’ (”'3)
q q i q k

where MO is given in (4A.7) and i = 1, 2, 3 (no summation on i), and
p = [(q -%1<)2 -1][(q + %k)2 -1][(q + 7‘2) -k)2 - 1] . (4D.4)

we shouldtmnndon that in deriving (4D.3) we also neglected
terms of trm~form B - a. We note that in square brackets of (4D.3)
the first term gives 0(1) and 0(a) corrections and the other terms
give only 0(a) correction which comes from non-small values of q;
the 0(1) comes from small q (of order Y). Therefore 3 - (3 in
numerators of all terms, gives the 0(a2) correction which we are not
interested, so we can neglect it.

Da from (4D.3), we do not use any more

For finding M
approximation with respect to p and q. The only approximation that

we use, wherever possible, is related to Y and K.

76

Performing the p integration in (4D.3), we obtain

 

 

u 2 u 2
a is d qr 4 q0 Q0
”11 =—_3CMopr+2 +1.72” 2
2n + Y q q
q q2q
-i—i-—:C23——i—] . (110.5)
1 q k1

These integrals are performed in Appendix C. Using Uneresults

(0.2), (G.10). (0.26) and (6.38) of the Appendix G, we obtain

 

 

2
a 2
MD =MO{1+FS-T‘-[I‘(-a)+4ln2-1§+2ln (1+5)
-8/21n(1+/2)]} . (ups)
For MDb, Figure 23(b), we write
11 duq L1
MDb = -2iCn2a2f d DD 14 d Q4 R(K. a)R(K, a1)
(21) (21) (21)
8 (K.q.q)
c 1 uv _
x BE NN D (p q,)<> , (110.7)
qq1qq1

where

1 + + + ->
<>a <sz(q1 —k.2 )(NC11 + q1 Y)(1 YOHNq C) Y)

77

.. S(q -%1<)¢' S(q + 13k «)2 S(q + ékmaq +3 ~1’)

(1 + 1 )(N + " ?)s< +1k) u
x 0 Q1 q1 q1 2 Yuwo(p)> . ( D.8)
In relation (4D.7), the momenta and polarizathon vectors are shown

+ A
in Figure 24; Duv, R(K, q), 11.10 and Hc are given by the relations

(3C.24), (2.6). (2.19) and (2.27) respectively.

 

 

 

p +-§k q1 + 5k q + —k k, e
7 7 7 V‘W
12
\/
-lk -lk -11< k-h '
D 2 Q1 2 q 2 . e
b

Figure 25. Diagram related to M4 .

Calculation of M b is similar to the GE b (Section C of Chapter

4 3

3). The only difference (apart from the wave-function and presence

cfl‘polarization vectors) is the presence of the propagator

S(q + %k - k) which its pole at q0 = -(Eq2 + K2 - 2a . 1:).1/2 gives

contribution of the 0(a).
In fact in (4D.7) by using the following relations

-(1 - YO)(Nq - q ° Y)

, 4D.
qo _ K + Eq _ ie ( 9a)

 

+ -> 1
(1 - 10m)q + q - Y)S(q - 5k) =

77

x S(q —%k)¢v S(q + 121. 40.: S(q + 32—1<)<Nq + a - i)

x (1 + YO)(N + 31 - 7)s(q1 + gmuwomw . (110.8)

q1
In relation (40.7), the momenta and polarizathon vectors are shown

in Figure 24; Duv, R(1<, a), $0 and Hc are given by the relations

(3C.24), (2.6). (2.19) and (2.27) respectively.

 

 

 

1 1 1
phi“ q1+§k ‘1’“? “'5
7 r 7 ‘ww""'
ii
\/
-—11< -—4< ~11< k-R '
p 2 <11 2 q 2 .6

Figure 25. Diagram related to Mub.

Calculation of Mub is similar to the 6E3b (Section C of Chapter

3). The only difference (apart from the wave-function and presence

ofxxflarization vectors) is the presence of the propagator

S(q + %k - k) which its pole at q0 = -(Eq2 + K2 - 2E3 - {)1/2

contribution of the 0(a).

gives

In fact in (4D.7) by using the following relations

-> +
-(1— 110)(Nq - q - Y)
qO - K + Eq - is

(1 - mm + 6 . 'v’)s(q - %k) =

 

, (4D.9a)

78

(Nq - q - Y)(1 + YO)

 

 

 

 

 

1 ~> ->
S(q +-§k)(Nq + q - Y)(1 + YO) - qo + K _ Eq + is . (4D.9b)
, + 1 (1 + Yo)(Nq — 3 . 7)
(1 + YO)(Nq + q . Y)S(q +.2k) = q0 + K - Eq + is ' (4D.9c)
-> +
-(N - q ' Y)(1 - Y )
_ 1 + . + _ a L O
S(q 310mg + q Y)(1 Yo) qo _ K + Eq _ 1.: . (110.901)
and after performing p0, q10 and q0 integrations, we
+ + ‘
obtain (p. Q1 ~ Y)
b a3 ‘ ‘ " 3 3 3
M = C¢k°exc'fdpdqdq/E +K
.4 8 6 o 1 q
11 .
2 " -
x [(52 + 12>2<E - E1) (5?. 12m? + Y2)(Eq - a .maq] 1
- + + — + + K - Eq
x [HC(K' Q9 Q1) + HC'(K’ q: Q1) 2 2 + + 1/2] . (“0.10)
(E + K ‘ 2q “'R) '

q

In deriving (4D.10), we used these observations: for po and

C110 only poles at p0 = K - ED and Q10 = K - Eq contribute, which

1

means that only D00, the instantaneous Coulomb interaction of the

photon propagator Duv contributes; for do, both poles at q0 = K - E

2 2
and q0 — (Eq + K

q
- 2 a 0 [0.1/2 contribute.

In relation (4D.10), EC' is

79

 

(1413.11)

where fic is given by (3C.27). fic is related to the pole at

q = K - Eq, and ac. is related to the pole at qo= -(E 2 + K2

0 q
-223 - {If/2.
In (4D.10), the first term in the second square brackets gives

..)
the 0(1) and 0(a) corrections. The 0(1) comes from c1-‘Y and 0(a)
comes from all values of a.' The second term gives only
contributions to the 0(a) correction which comes from non-small

values of 3. Therefore in (4D.1CD we may use the following

approximate relation

9 ) (110.12)
which is correct up to the 0(a) correction.

Relation (4D.10), after performing p integration and using

(4D.12), gets the following form

Mb=a_2_
4 4

C¢ k - e x e'fd3q d3q “E + K a (K. E. E )
4n 0 7 q c 1

+2 2 2 + 2 + A -1
“[(q1+Y)(q+Y)(Eq-q-k)Eq]

[ K - E(1
x 1+
(E:+K2-2E-Ii’)

 

1Q] (1413.13)

80

By virtue of (3C.27) for Hc’ and using (3C.28) and then

performing q1 integration, we obtain

 

 

b 2a 3 1 3 q
M =-—-— C¢ K 0 e x e'fd q L— A
A" " ° <52+12)2 Zequs -21’-k)
K - E 2
q a V2
x [1 + ] --———-—-—} (up.1u)
(Eq2 + K2 _ 26"?)1/2 62 + Y2 ,

Performing the q integration on the last term and comparing the
integral of other terms with equation (4A.2) (therefore using the

result (4A.6)). we find

 

2
b-1211- __ -11..
MM - MO{2 + 8" [3 r( a) + 80 8A 16 In 2 2 ]} . (4D.15)
Finally, (4D.1), (4D.6) and (4D.15) give
3 a 3 11n2
Mu = M015 +‘EF 15 r(-a) + 120 - 12A - 20 In 2 - u

-8 .6 ln(1+ /'.2‘) + 2 ln2(1+ /‘2‘)]} . (113.16)

E. Contribution M5

For M Figure 20(e), we use the same decomposition that we

5’

used for 6E3, relation (3C.1), or

5 5 5 , (1115.1)

where Msa and M5b are represented in Figure 26.

 

 

 

1 1 1
91'2““ ‘31:?k q+§k “'5
r I ' 7 v‘vAv‘vM
(b) M5” = -2c K, F)
- 1k - 1. - 1. I??? .

Figure 26. Contributions M5a and MSb.

As for the case of dEua, by virtue of the relatkn1(3D.2) we

find the identity which is given in Figure 27.

:x
11

Figure 27. The identity which is the direct result of the

relation (3D.2).

Therefore, we can write

5 ‘1 , (1112.2)

or from (4A.6) for M1, we obtain

82

2
a _ _ _g_ _ _ _ _ 3“
M5 - MO{1+ufl[I‘(a)+8G 8A 16an ———2]} . (1413.3)
For Msb, Figure 26(b), calculation is very similar to the Mub.
In fact, in Mub by replacing iDuv(p - Q1)(-ieYu)(1)(-ieYv)(2)In!

+ + . . b . .
KC(K, q1, p), we obtain an expre331on for (-)M ; where Kc is given

5
by (2.31), and (1) stands for electron line and (2) for positron
line. The (-) comes from the definations of Nut) and M5b (compare

Figure 23(b) with Figure 26(b).)

In the following, by using this replacement, we find Msb from

As we noticed in the course of calculation of Mub, 3 and 51

were small (3, 31 ~ Y) and only v = u = o contributed. Therefore,
ifideriving (4D.10):fium1(4D.8) we used the following approximate
relation

uv _ _. (1) _. (2) _
iD (p q1)( 1eYu) _ ( ieYV) -

iDOO(p - Q1)(-ieY )(1)(-ieY )(2) , (ua.u)
o o _
which by virtue of (3C.24) for Duv’ is

iD”"(p - Q1)(-ieYu)(.1)(-ieYv)(2) = {31515—3 YO(1)YO(2) . (113.5)
' (p - <11)

Relation (2.31) for Kc’ for small 3 and 51 gives

 

 

4111a (

K( 1+ +)
CK, q]! p _) 9 2
(p-q1)

) . (4E.6)

Relation (4D.8), for small 5 and 3 is (see relation (4D.7) for

b

”4 )

<> b = <1 (...)1 (1)(1 + v )1 > , (4E.7)
u o 5 1

where (1 4-‘n3YY5 comes from wo’ and we use subscript Mub for <> to
indicate that this trace is related to the Mub (note that only

v = u = o contribute).

Let's write (4E.7) in the following form

<>M b (1)
u

(2)
<(...)1rLl (1+YO)Y5Yv > . (1113.8)

To find <>M b, according to the results (4E.5) and (4E.6), in.
5

1+YO(1)1-Y0(2)

(1) (2) )
2

(4E.8) we replace Yo Yo by (-)0-73--) (

 

Now, using this replacement and following relations

(1)

Y (1 + YO)Y5(YO)(2)

= 110(1 + YO)Y5YO = -(1 + YO)Y5 , (113.9)

1+Y(1) 1-Y(2) 1+1! 1-Y

O 1 O O O
(—)( ) (1 + YO)Y5( 2 > = -< 2 )(1 + 10)15( 2

 

 

 

 

)

= —(1 + y0)y5 , (4E.10)

84

we infer the value of OM b
5

<> b = <> b . (4E.11)

 

M5 = -Mu , (43.12)
or by virtue of the relation (4D.15) for Mub
b 1 a 1
M5 - M015 +‘4F [5 r( a) + 120 12A
932
- 24 in 2 "'4’]} . (43.13)
Finally, (43.1), (43.3) and (43.13) give
_ _ i. 2.. -.§ - _
M5 - MO{ 2 + u" [ 2 r( a) 200 + 20x
15n2
+40 ln 2 + 4 ]} . (43.14)

F. Contribution M6

For M6, Figure 20(f), we use the following decomposition

(4F.1)

85

where M6a and M6b are represented in Figure 28.

 

 

 

 

 

a ; (=77 ’
1 ‘1'
p - —k k -|¢, 2'
21
x W
P + 1k q +‘lk k E
> 2A 2* - ’
(b) Mb-4C r "A
6 " A
1 R 1 \/
p--2-k q-—2+< k-k.e'
#1 ll ‘1 W

 

 

Figure 28. Contributions M68 and M6b, where R and C are
given by (2.30) and (4.5) respectively. Line
with a dash indicates inverse propagator.

Calculations of these diagrams are similar to 6E (see Section

 

5
E of Chapter 3).
For M6a, Figure 28(a), we write
a d“ 1 1
M =- 16 ime p <9.“ S(p+—k -k)¢ S(p+-k)
6 7 4 2 2
(24)
x 2(p + ék)1po(p) > , (43.2)

and using (3E.2) for Z, we find

M6

 

2 d” 1
a = 4 101 c I‘(-a)f p” < 213(1) + 5k ~11)“: (p) >
(211) °

86

4
- 4ia2C fi—94E— < d' S(p + l1<z--k)¢S(p + 1k)
(211)” 2 2

x[C S - 1 + 02(1)O + K)YO + c 10(p) > , (43.3)

1 3]

where C1, C2 and C3 are given in (3E.6).
The first integral in (4F.3) can be found, simply, by comparing
(4A.1) with (4A.6). the integral, up to the order that we are

interested, is

u
I—d—RE < d'S(p + %k 42).: 110(1)) > = 3.1“; MO . (43.4)
(24) '

For the second integral of (4F.3), after performing Unapo
integration (dominant contribution comes from the pole in wave-
function, or for po = K - Ep, and 3 ~ Y; we close the countor in the
upper half-plane of po) and then by scaling E + YB, we note that in
order this integral contributes the following trace should be, at

most, of the order a2
<¢'(p + i.- 7 + 1)¢(p + KY0 + 1)[c13 - 7 + C2(pO + K)Yo + C3]
+ +
x(Ep + YO - p - Y vow5 > , (4F.5)

where p0 = K - E .

87

It is not difficult to show that in order to find this trace,
we need to find C1 up to the 0(1) and 02 and C3lnito the 0032).

These coefficients, up to these desired orders, are given by (3E.9).

After finding the trace in (4F.5) and using (3E.9), we note
that this trace is of the order higher than a2. Therefore the
second integral in (4F.3) does not contribute and only the first

integral gives contribution.

Using (4F.4). (4.5) and (4F.3), we obtain

68 = Mop}. Hm] , (43.6)

M 2n

For M6b, Figure 28(b), we write

4

b d p duq R(K a)R(K E) A
= - naCf E E N N HC(K, q, p)<> , (4F.7)

M
(211)“ (2101I q p q p

6

where

<> a <(Np +5 -1’)(1 - 10m:q +5.1)s(q -%k)¢'

1 1 + +
xS(p + 5k - k)¢S(q + -2-k)(Nq + q - Y)(1+ 10)
+ -> 1
mp + p - Y)S(p + %k)2(p + Emwom) > , (4F.8)

and we have used (2.30) for R.

88

After using (3E.2) for Z and (2.19) for $0 and also the
relations (4D.9) for simplifying the trace in (4F.8), and then
+ + —) +

performing the p0 and q0 integrations (by scaling p + Yp and q + Yq
we note that the dominant contribution comes from S, 6 ~ Y and the
poles at p0 - K - ED and q0 = K - E ; for Q0 there is only one

q
pole), and then using the following approximate relations

a , K = 1 - —N , (4F.9)
we find

11 (K. a. E)
b aC -a 3 3 c
M =-——— M L—— r(-a)]f d pd q +

 

+ contribution from the second term in Z, (4F.10)
where RC is given by (38.6).

We can show that contribution of the second temnzhlilis of
higher orders, so we can neglect it. In fact, for contribution of
the second term in Z we have a similar situation that we had for
11:; here we need to find the following trace, at most, up to the

order a2

<(Np + B - 1)(1 - 10m:q - E - 1)¢'(p + i - 1 + 1)¢

+ + + + -> +
x(Nq - q - Y)(1 + YO)(Np - p - Y)[c1 p- Y + C2(2K - 3p)vO + c3]

89

+ ~>
x(Ep + YO - p - Y YO)Y5 > . (43.11)

It is not difficult to show that we need to know C1 up to the

0(1) and C and C up to the 0(a2). Therefore we may use relations

2 3
(38.9) for these coefficients.

After finding the trace in (4F.11) and using (313.9), we note
that this trace is of the order higher than a2; so we neglect

contribution of the second term in 2.

Therefore from (4F.10) we have

_ 11 (K. E. E)
116 = — M [74-3 I‘(-a)]fd3pd3q °

(4F.12)
815 o

 

2 .

+2 )

(p + 12)2(32 + Y

This integral is the one that we had in (3F.13). By comparing

(3F.13) with (3E.16) we infer the value of this integral

R (K E E) 5
fd3pd3q +2 0 2 2 +2 2 = 8: , (43.13)
(p + Y ) (q + Y )

 

and by using (4.5) for C, we find, up to the 0(a) correction

b

'01
M6 = MOLE; r(-a)] . (43.14)

Now, relations (4F.1), (4F.6) and (4F.16) giveiusthe final

result for contribution M6

9O

‘30
M6 = MOLEF- T(-a)] . (43.15)

G. Contribution M7

For M Figure 20(g), we use the following decomposition

7!

M =M +M , (4G.1)

where, M a and M b are represented in Figure 29.

 

 

 

 

 

 

7 7
p + 1k q + 1k k e
l ; 2 A 2 2 '
(a) M7a = 4C Y
‘1 ‘1 74W
-4 -4 k—k '
P 2 q 2 . 8
p +-lk q + 1k q + 1k k e
.32 fr”‘\ 1 ‘2 2 ’
r ‘ 7 v... ,
(b) M7b = 4c J 3 1r
- 1.. 31. S4. “1%. 1. .
. . . a b
Figure 29. Contributions M7 and M7 .

Calculation of these diagrams are very similar to 6E6 (see
Section F of Chapter 3), so we try to give only the summary of the

calculation.

91

a .
For M7 we write

4 4
M73 = 64 c1232; G q“ d P“ <>a , (4C.2a)
‘ (24) (23)

 

where
0 a <p“"(p - (m S(q - lk>¢'s(q + 14. 42).:
a v 2 2
1 1 1
X S(q 1' -2—k)AA(q + —k9 P + EkNJOUD > . (“G-21>)

2

For M7b, using (2.30) for R, we write

 

H + +
b . 2 2 a"; d q1 dup R(K’ q1)R("' q)
(211) (211) (211) (1 Q1 q <11
x HC(K, q1, q)<)b , (4G.3a)

where

:5 UV — —1 + o + -
<>b.. <1) (p Q1)YVS(q1 21<)(Nq1+q1 Y)(1 YO)

xmq +6-1)S(q --2-k)ef'S(q + 124 -k)2s(q + 2—4)

+ + + + 1
><(Nq + q - Y)(1+ Yo)(Nq1+ q1- Y)S(q1+ 2k)

1 1
xAA(q1 +-2k, p + 2k)wo(p) > . (4G.3b)

92

In the above formula, A is the vertex correction and DI“ is given

A
by (3C.24).

We note that relations (4C.2) and (4C.3),zumuw from a
multiplicative factor and the wave-functhon, are the same as
relations (3F.2) and (3F.3) respectively, if in the latter relations
we use the following replacement

Y“ + é'S(q + 2k - 10¢ . (40.4)

Since for M7a and M7b we are interested in the leading order

(which corresponds to the 0(a) correction), (4C.4) can be written as
Y +-—2'(1+‘1E-1’)¢ , (40.5)

where, we used the same approximation that we applied for 6E6,
namely 3 ~ Y and q0 ~ Y2. In fact, (4G.5) shows that we can use all
approximations that we employed for 61336, namely, 3, 31, 6~ Y, and

A = v = o, and that for p0, and q0 integrations the dominant

q1o

poles correspond to p0 = K - E K - E and q0 = K - E .

p’ q10 = Q1 q

Therefore (4G.2a), after peforming po and q0 integrations, gets

the following form which is very similar to (3F.4)

_ 3 2 -1
M78 = 1‘; /2¢Ofd3pa3q[(p’2 + Y2)2(q2 + 12)<S - 2132]
' 411

 

93

, . + 1 1
mi (1 + 1. mm + Yonom + 24. p + 2"”? + vows > , (46.6)

and (4G.3a) after performing p0, q1O and q0 integraticnus, gets the

following from which is very similar to (3F.5)

b ~ia3
M7 = 6
323

3 3 3 ‘ + +
V5'10fd pd qd Q1 HC(K. Q1. q)

 

'1
x[(p2 + 12)2(qf + Y2)(q2 + Y2)(p - q1)2] ~

+ + 1 1
x<¢'(1 + k - Y)¢(1 + YO)AO(q1 + 2k. p + 2k)(1 + YO)Y5> . (40.7)

where RC is given by (3C.27), and we have used (4.5) for C.

Now by vnflnmaof the relation (3F.20) (that we found in the

Section F of Chapter 3), which in the present case can be written as
(1 +Y)A (q+-1—k p+—‘4<)(1+Y)=
. o o 2 ’ 2 . o
a
(1 + 10mm r( a)](1 + YO) , (4G.8a)
U+Ymm-J4p+4m+Y)=
o o 1 2 ’ 2 - o

(1 + Yo)[fi%-F(-a)](1 + YO) . (4G.8b)

relations (4G.6) and (46.7) get the following forms

94

 

3 -> -> + + -1
M7a =~JE§ MOI‘(-a)fd3pd3q[(p2 + Y2)2(q2 + Y2)(p - q)2] ' , (4G.9)
' 811
M b = “3 M r(-a)fd3 d3 03 3 ( + +)
7 6448 o p q <11 0 K. Q1: q
-1
x[(E2 + Y2)2(3? + Y2)(q2 + Y2)(S - 31)2] ' , (40.10)

where MO is given by (4A.6).

The integrals 1J1 (46.9) and (46.10) are those that we had in

(3F.26) and (3F.27), respectively. Therefore

a (1
M7 = M012; r(-a)] , (40.11)
M b = M [—E-f(-a)] (46 12)
7 0411 ’ '

and by virtue of (46.1)
- % -
M7 - Mo[—_41r I‘( a)] . (46.13)

By comparing (46.11) with (4F.6), and (46.12) with (4F.14) we

obtain the following relations
(40.14)

(46.15)

95

and consequently
(40.16)
H. Contribution M8

For M8, Figure 20(h), by comparing Figure 20(h) with Figures

20(a) and 18, we find
a 1
M8 (6K ) M1 , (4H.1)
and by virtue of the relations (4.3) and (4A.6), we obtain

M8 = MOL2— r(-a)] . (43.2)
I. Summary of the Calculations of the Total Invariant
Amplitude and the Decay Rate of the Parapositroniunl

Ground-State

By virtue of the relations (4.12), (4E.2), (46.12) and (46.16)

we find the total invariant decay amplitude

. (4I.1)

A

*which means that, up to the 0(a) correction, the second term in Gc’

R (see relation (2.29)), does not contribute.

96

We represent M, relation (4I.1), graphically in Figure 30.

W

M = 4 + 2 {: + 2C
.3... w-

-_J::L—l

 

 

H,

Figure 30. Total invariant decay amplitude, correct up to
the 0(a) correction. Line with a dash through
it indicates inverse propagator, and the

normalization constant C is given by (4.5).

In (4I.1), by virtue of (43.17), (40.5). (40.6) and (43.2) we

obtain

2
C! 11’
M = MO[1 + 45—2 -10)] , (41.2)

where M0, the lowest order invariant amplitude, is given by (4A.7).

New that we know the total invariant decay amplitude M, by
virtue of the relations (2.18), (4.8), (4.11), (4A.7) and (41.2) we

can find the decay rate (up to the order a correction)

5 2
I‘(p-Ps + 2Y) = 12— [1 - — (5 - 74)] . (41.3)

which agrees with the result of Harris and Brown [7].

97

We present all results of calculations of the present Chapter

in the Table 2.

98

Table 2
Summary of the Calculations of the

Amplitudes and the Decay Rate

 

2
Mo{1 + T4%{r(-a) + 86 - 8) - 16 ln 2 --3%-]}

 

Mofi[2r(-a)+112-4-8ln2+4/21n(1+/2)
- 2 ln2 (1 + /2)]

 

MO ifi-[-T(-a) - 6 + 4 in 2 + 4 /2'ln (1 + l2)]

 

 

11112

u
-8./21n (1+ .6) + 2 ln2 (1 +/§)]}

 

2.2;- __ _
MO{2 + u“ [2 r( a) + 120 12) 20 ln 2

 

1215])

“3.3.1-2--
M { 2 + u“ [ 2 r( a) 200 + 20) + 40 In 2 + 4

 

 

 

 

 

 

 

 

 

Chapter 5

Summary and Conclusion

USng the Barbieri-Remiddi lowest-order equation, equation
(2.2) which we discussed it in Chapter 2, and working in Coulomb
gauge we calculated the energy shift of orthopositronium and the

decay rate of parapositronium.

For the contribution of one-photon-annihilation channel to the
energy shift of the ground state of orthopositronium, using relation

(3.1), up to the order a correction we found (see Chapter 3)

i;

63=9T(1-i°‘—-

11'

312

(3H.2)

which is in agreement with the result of Karplus and Klein [12].

The interesting:point is that we derived relation (3H.2)
without performing the wavefunction and the vertex renormalization)
subtractions. It means that relation (3.1) which gives the energy
shift, is free of divergences and gives the finite result without
needing to perform the wavefunction and the vertex renormalization

subtractions (at least in the first-order perturbation theory.)

For the decay rate of the ground state of parapositronium,
using relation (4.9), up to the order a correction we found (see

Chapter 4)

100

5 2
I‘(p-Ps + 2Y) = 9—2- [1 - $5 - L1] , (41.3)

which is in agreement with the result of Harris and Brown [7].

Here also we derived the result (41.3) without performing the
wavefunction and the vertex renormalization subtractions. It
implies that the relation (4.9) which gives the decay rate, is free
of divergences and gives the finite result without needing to
perform the wavefunction and the vertex renormalization subtractions

(at least in the first-order perturbation theory).

In Chapters 3 and 4 we noticed that in derivations of the
results (3H.2) and (4I.3) we encountered some diagrams which "rule
of thumb" for finding their leading-order contributions would give
higher orders in a than their actual orders. Inn'instance, rule of
thumb for the leading-order contributions of the diagrams of Figures
2(c) and 20(d) gives order a, and for the diagrams of Figures 2(f)
and 20(8) gives order 02. However, the explicit calculation showed
that the leading-order contributions of the diagrams in Figures 2(c)
and 20(d) were of the order 1 and for the diagrams in Figures 2(ffi

and 20(8) were of the order a.

The reason that the rule of thumb for these diagrams does not
vunfl< is due to the kernel 6K, which was introduced in Chapter 2; it

can generate inverse powers of a.

101

We should also mention that in covariant gauge there are an
infinite number of diagrams which contribute to a given order in 0;
whereas i11<moulomb gauge we deal with a finite number of diagrams.
This is due to the photon propagator Duv' In the diagrams which one
or more virtual photons are exchanged between electron and positron
lines, the D00 part of photon propagator, in covariant gauge,
creates pole; whereas in Coulomb gauge D00 does not give a pole (see

relation (3C.24)).

In fact it was one of the reasons that we chose to work hi

Coulomb gauge; namely, dealing with a finite number of diagrams.

Our calculations showed that the method of Barbieri-Remiddi,
for solving the positronium problem, is straight-forward and
practical. Also in order to make our results useful for derivation
of higher order corrections to the quantities that we calculated up
to the order a correction, we presented all calculations of the

related diagrams separately.

APPENDICES

102

Appendix A
Evaluation of an Integral
related to the Contribution 663

Consider the following integral

 

(A.1)

where

k 4(1-Y) , E=p +1 . (11.2)

After performing the y integration, we find

 

I = f1 95- 2 {I-
O 51.13211— x) + 1211/2 2
-tan“[§ (62(1 - x) + 12)”2]} . (4.3)

We should mention that we are not interested in the quantity I
by itself; it is the following quantity which we need to find up to

the order $-

fd3p "E + K I , (A.4)

(32 + Y2)2E

 

where K = (1 - Y2)1/2.

103

By scaling B + Yp, we note that in (AJI) the term involving

tan 1 is of the order $3 therefore, for this term we may say 3 is of

the order Y and use the following approximation

 

tan”1[§(52<1 - x) + Y2)1/2] = $3211 - x) + ref/2 . (A.5)
Using (A.5) in (A.3) we obtain
n 1 dx 1
I‘Efo‘: 1/2'2 ’ (AT6)

/x [52(1-2) + Y2]

which by change of variable x + x2 and performing the integral and

using the approximate relation k = 2, we find

I = %sin-1(-—-p——-) —2 . (A.7)
/32 + Y2

As we mentioned before, we are interested in the quantity which

is given in (A.N).

First we find the contribution of the first term of I in (A.7).

which we define it as J
J = fmdp E + K sin-1 G——-Jl-—4 . (A.8)
0 +2 2 2 ———-——-— .
(P+Y)E /I+)2 2

Relation (A.8) can be written as

J=J +J , (A.9)

1014

 

 

where
J1 = Ide-—Eé 1 +2K2 sin—1 ¢-—-—J1-’-) . (A.10)
(p + Y ) / $2 + Y2
J = -f°°dp p (/1 + K - M)sin-1(—'—E‘—) . (A.11)
2 0 (+2 + Y2)2 E -:§————§ ,
p / p + Y

1 .
Integration by parts gives the J1, which up to the order-V is

(A.12)

 

For J2, we show that it is of the order higher than %, so it

can be ignored.

First we note

 

/1+.<-—————'E;J"30 , (A.13)
therefore
|J2| = Ide 2 p 2 2(/1+ K -————’I'3;.")sin'1( p ) , (11.11:)
D + ) / B2 + Y2

and since

105

sin (——‘i—> _<_

from (A.1H) we have

 

E-1

 

71 m p

+2 2 2 E
(p

+ Y )

Using the following relations

 

from (A.16) we obtain

+2
_lL.

.2

 

“ p

2/2 (52 + Y2

O!“

 

)2 E

g., VETT7E'3;/1 + K , /1 + K < /E' , (A.15)

(A.16)

(A.17)

(A.18)

 

= 1'__[lln(1“2Y )+1] , (A.19)

which means that J2 is not of order-%, so we neglect it.

Therefore, using (A.8), (A.9) and (A.12), we obtain the

following relation which is correct up to the order-1

 

 

Y
Ide +21 E E Z sin-1C--E--) = /2 g . (A.20)

107

Appendix B
Evaluation of an Integral
related to the Contribution 6E1
The following integral is the one that we need for finding 681
(its evaluathnmaflso serves as a sample for finding similar

integrals)

 

+2
I = I: +g dpz 2 /E + K (2E + 1) , (3.1)
(p + Y ) E
where
E = / 52 + 1 , K = (1 - Y2)1/2 . (8.2)

Scaling the momentum E by Y
5+ YE . (8.3)
we find that the leading order of I is
I ~ 0(1) . (8.”)
Y 1

We need to find the integral I up to the 0(1). Consider the

following relation

 

/E + K = /E + 1 (1 - 1 " ")1/2 , (8.5)

1 + E

108

which by (8.2) it is

/E + K = /8-+‘1[fi + 0012)]1/2 ~ /E + 1 , (8.6)

where we ignored the second term in square brackets, since it gives

contribution of order higher than 0(1).

Therefore, (8.1) gives

 

 

 

+2
I = f0 +2p d2 2 "E g 1 (2E + 1) , (8.7)
(p + Y )
or, using (8.2) for changing the variable p to E
on dE ___..
I =f (E+ 1)(2E+ 1) /E-1 (B.8)
1 2 2 2
(E - K )
By changing the integration variable
E = 1 +-1 x2 , (8.9)
2
we obtain
2 2 2
I=—_8-_fo dxxff 1:)” 23) . (8.10)
/2 (x + 4x + NY )

which by using table of integrals, up to the 0(1) is

109

(8.11)

110

Appendix C

Proof of an Identity

We want to show that the following integral is identically zero

I . fd3q F<EZ> fd3p +2 1 2 2-% ’A 1 28 ‘ 892 . (C.1)
(p + Y ) A9 + 8(1 - p)

 

 

where

2 1

 

) ’ Dips-1 ’(C02)

and F(32) is any function of 32 which satisfies the condition

(C.13).

We write (C.1) as

I = I + I , (c.3)

where

 

I1 =-Zl fd3q 8(62) fd3p (C.H)
p (p +Y)

1 1

+2 2 2 2
(8 +1!) Ap+B(1-p)

I = fd3q 8(62) fd3p

l
2 p

 

(C.5)

After performing the p integration in (C.U), we obtain

111

 

f dq T— F(Y q ) . ((3.6)

For 12, we write (x = cos 8, where o is the angle between E

 

9
and q)
I = 3"- f “3" I" dq 62 8(62)
2 p (52 + Y2)2 o
x f:: dx [(52 + a2 - 2pqx)p + 8(1 - p)2]-1 . (0.7)

or after performing the x integration

 

_ -Nn a +2 w p
12 - 2 f_co dq qF(q ) f0 dp-:§——-—§
p p + Y
x in [(p — q)2p + 8(1 - p)2] . ((2.8)

By scaling p + Yp, q + Yq and integration by parts, we obtain

2

 

-un w 2 +2 w dp
2 2 f_m dq QF(Y q )fo +2
9 p + 1

p - q,+ 8(52 + 1)p
(p - q)2 + 8(62 +1)(S2 +1)

X

. (C.9)

where

_ (1 - p)
B - Up . (C.10)

112

In (C.9), Unaintegration on variable p is not difficult to
perform. After performing this integral and retaining only those

terms which are an odd function of q, we obtain

“1232) . ((2.11)

 

I = o . (C.12)

We note that this result is based on the following assumption

f: dq ——‘l-—— F(Y2 32) = finite . (C.13)

113

Appendix D
Evaluation of an Integral
related to the Contribution M

1

Consider the following integral

 

I - I" dp (52 + 72)2E ln(E + p) , (D.1)
where
E = (p2 + 1)“2 , K = (1 - Y2)”2 . Y = %- . (D.2)
Scaling the momentum p by Y
D * YD 9 (D.3)

We notice the leading order of the integral in (D.1) is
I 3 0(1) 0 (Don)
Y

‘We need to find the integral I up to the 0(1). Using (8.6) in

(D.1), we obtain

 

I = f: dp g/E +212 ln(E + p) . (D.5)
(p + Y ) E

Let's write (D.5) in the following form

I=I +I , (0.6)

where

H
II

 

i/‘2fon dp p ln(E + p) . (D.7)
0 (D2 + Y2)2 .

 

I = I” dp p (/E + 1 - 12./2‘) ln(E + p) . (D.8)
o (p2 + Y2)2E

The integral I1, using the integration by parts, up to the 0(1)

is

I = V/E(_ - _) 0 (D09)

For I let's write it in the following form

2'

I2 = 1'de J— (/E + 1 - E/2)ln(E + p)

 

p313
2 2 2
f (1 13(23 " I; g (/——_E + 1 - 8/2)ln(E + p) . (0.10)
p (p + Y )

Scaling p by Y shows that the leading order of the second integral

is o(Y), therefore we may ignore it and write (D.10) as

= I: dp —%—-(/E + 1 - E/2)ln(E + p) . (D.11)

I
2 p E

115

Changing the variable p to E (see (D.2)), (D.11) gives

I = I” dE———1——-——- (/E +1 - E/2)ln(E + /E22 - 1) , (D.12)
2 1 2 2
(E - 1)
1 + x2
or by changing the variable E = ——§;——

I =-W2f°dx—§—L—-§(/§-1)2(x+/§+1)Inx , (0.13)
(x - 1)

and if we change x + x2, we obtain
I =—16/2f°°dx—-—’£————(x-1)2(x +x+1)lnx . (8.111)

The nuithod of partial fraction gives I in a form of integrals

2

which can be found in the standard tables of integrals. The result

is

2

_ l _ -1;
12 —/2(2+ G 2 ln 2 16) , (D.15)

where G = .915965594... is the Catalan's Constant.

Using (D.6), (D.9) and (D.15), we find up to the 0(1)

2
I=/2(§1-T-+G-2ln2-—3-1T—— . (8.16)
a 16

116

Appendix E

Calculation of a Trace related to Mu

In this appendix we calculate a trace which is related to the:
decay amplitude Mua, equation (“8.2). This calculation serves as a
sample for the method of approximation that we implemented for
finding nearly all other traces that we had hicnn'work. Traces

were found, mostly, by the SCHOONSCHIP program.
The quantity that we want to evaluate is
TE 8“”(p — q)<Yv(4 - KYO +1)é'(a[ + 12 ~1+1>¢
x(q + 1 + KYO)Yu(Ep + YO - 3 - ? YO)Y5> , (E.1)

where iDuV, the photon propagator in Coulomb gauge is given by

(3C.2u), and <> stands for trace.

As it can be seen from (ND.2), we are interested in the

following quantity (up to the order a)

 

3 u
IEa2f+2dp22fgfigT , (8.2)
(p +Y)

which is related to the Mua, relation (HD.2) after performing the pO
integration (we note that, up to the order a correction, only the
pole at po ==K:- Ep contributes (3 ~ Y)). The quantity D in (8.2)

is given by (48.4).

117

Relation (8.2) with scaling B + YE shows that I is of the order

4

I ~ a f 253 . (8.3)

The most contribution to I is of the 0(1), and it comes from the

pole at q0 = K - Eq ~ Y2 (if 6 ~ Y and if we assume <> ~ 1 in (8.1))

4 3
2.3 - 9.5.1. - .. l. . -
f D f 2 (I 9 T 2 a q Y 9 (E014)
a a
which means
I ~ 1 (only for 6 ~ Y) . (E.5)
If we assume a is not small (not of the order Y), and <> ~ 1,
then
d” d“
f-—Eg ~ I —Tfl-~ 1 and T ~ 1 (for non-small q) , (E.6)
which means
I ~ o(a) , (for non-small q) . ((E.7)

In short, the o(1) contribution to I comes from small values
of q, while the o(a) contribution to I comes from all values of q

(we always keep in mind that 3 is of the order Y).

118

The trace <> in (8.1) has the following orders

<> ~ 1 + 0(a) for small q (5 ~ Y) , (8.8)

<> ~ 1 for non-small q , (8.9)

and-using the above considerations for I-all higher orders of 0
give rise to the o(a2) and higher order contributions to I (relation

(8.2)).

Therefore, for finding T, relation (8.1), we will use relations
(8.8) and (8.9) in order to simplify the calculation. One of the

implications of (8.8) and (8.9) is that we may use the following

approximations
1+2 12
8 ~ 1 +-§ p 1 , K ~ 1 -E Y 1 , (8.10)
+ + 2+ + -> ++ +
P‘Q"0: QQ'D”O. Q°Kp°q~o ,... (8.11)

(for any term which contains p).

Using (8.10) and (8.11) in relation (8.1), we obtain

2 2 2 2
q - 2Q + u 2Q - “q + +
T=[ + ]<d'¢k-YYOY5>

 

 

(8.12)

119

or since in numerators we can add any term of the forms 32, E . 3,
p2, etc. (any term which contains p gives higher orders so its
introduction does not change the result), we may write relation

(8.12) in the following form

q2 l1q2
U o o + +
T = L———————— + 1 --———————- --—-——————]<¢'¢ w - Y Y Y >
(E - E)2 (a - 5)2 (q - p)2 O 5
Q2
0 + +
+ [- 1 - + + 2]<¢'¢ q - Y Y0 Y5) . (8.13)
(q ~ 8)
Theifirst term in (8.13), —:ele:7§, gives the 0(1) and o(a)
' (q - 8)

contributions tx>lI. They come from all values of a; small 3 gives
both o(1)aumio(a), non-small 6 gives only o(a). We leave this

term in its present form.

For all other terms in (8.13), their contributions are of the
0(a) and they come from non-small values of a and q0 (for small a,
with either the pole at qO ~ 1 or qo ~ Y2, we get contributions of

orders higher then a for I).

Therefore we may use the following approximate relations

(12 Q2 (12 Q2
0 O O 0
(q - p) q (q - p) q

and write relation (8.13) in the form

 

q2 uqz
T=1u +1-—9-- °]<¢'d§-Y’YY>
+ 2 +2 2 o 5
(q-p) q q
2
+(-1-——-_)(2))<z"¢q--Y)YOY5> . (Ii-15)
' q

or sintna integration on q, relation (8.2), for the a . Y term gives

a quantity proportional to K - Y, (8.15) may be written as

2 u 2
T=[ ‘1 .1__9____39_-Ei_-0q1]
+ +2 . +2 2 +2
(q-p) q q ki Ci"1
+ +
1 o

where i = 1, 2, 3 (no summation on i).

For performing the integrations in (8.2), the easiest way is

.first tn) perform the p integration, which transforms the first term

8
of (8.16), -——————-—, to the form -—————-;
(2’1 - E)2 32 + Y2

the q integration, by the Feynman parameterization method. For

then we proceed to perform

doing this, for the first term, , we use the exact form of D,

62 + Y2
relation (MD.U); but for all other terms we use the approximation K

a 1.

This method of approximation, in practice, facilitates

calculations; it allows us, before performing the integrations, to

121

get rid of unnecessary terms (terms which give those higher order

corrections that we are not interested).

122

Appendix F
Evaluation of some Integrals

related to the Contribution M2

In this Appendix we evaluate three integrals that we need in
evaluation of M2.

In these integrals D is defined by (48.5) and q
integrations are performed by using the Tables provided in [19].

One of the integrals that we need is

 

 

2 2
q q
. 2
f dnq 0D = In [I1 + 12 - r(-a)] , (8.1)
where
1dx1 (1-)2
I1 = fo'f: f0 dy y , (F.2)
/x 1 + (1 - x)y
1 1 dx 1 2
12 =-§ f0 5::f0 dy ln[x + x(1 - x)y ] . (F.3)
x
For I1,

after changing x to x

and performing the x
integration, we obtain

1 (1- )2 2
I = 2fo dy-—————lL—-In(y + / 1 + y )
y/ 1 + y2

(F.u)

and by changing the variable y + / 1 + y2 = t

123

— — 2 — 1+5 lnt

I1 = -2 + 2 J2 ln(1 + J2) -2 ln (1 + /2) + h f dt 2 , (F.5)
1 t - 1
or
n2 2
I1=-2+—u-+2/21n(1+/2)-3ln(1+/2) . (8.6)
For 12, after performing y integration and changing x to x2, we

find

12 = I; dx[2 In x + ln(2 - x2) -2 + 2 tan-1/ 1 - x2], (F.7)
2
/ 1 - x
or
-— —- 1 1 -1 “2
12 = -6 + 2/2 ln (1 + /2) + 2fo dx tan / 1 - x , (8.8)
/ 2
1 - x
2
and by changing / 1 - x = t
I2 = -6 + 2/2 ln(1 + f2) + 21':) dt —-—1———tan-.1t , (F.9)
/1-t2
or [31]
2 _ 2 _
I --6+"—+2/21n(1+/2)-1n (1+/2) . (8.10)

2 N

124

Therefore from (F.1) by virtue of (F.6) and (F.10) we obtain

Q Q 2

fdnq —-9—— = i112[-I‘(-a) - 8 + “-5 + 11./‘2‘ ln(1+ /2)

 

 

- t1 ln2(1 + /'2)] . (F.11)
The next integral is
Q Q2
fduq OD = in2 I; 9%:f; dy —§ 2 . (F.12)
/x 1 + (1 - y) -x(1 - y) '

Changing x + x2, y + 1 - y and performing the x integration we find

2
H q q

 

= 182 I; dy-—aQL;;ll-ln(y + / 1 + y2) . (F.13)

W1+y2

and by change of variable y + / 1 + y2 = t and then performing the t

integration we obtain

2
,qq 2

Id q OD = i112 [- 1‘7 + 2 ln2(1+ /2)] . (p.121)

 

The last integral is

q Q- q
Iduq—‘P—Si—i = 1112(1" 11311, + g 511.) (i. 9. =1. 2. 3) . (8.15)

125

where

2
_ 1 dx x y z(y - z)
x xA
a l. 1 95 x Y ._E
2 f0 [0 dy f0 dz xA , (F.17)
22 2
A =-;— -(y - z) + 2(y - z) . (F.18)

From (F.15) we infer the following relation (since we are

. +2 2
interested in the leading order: n. = K = 1)

u q q
fd q D

 

= in2(f + 3g) . (F.19)

Comparing (”8.13) with (F.19) we find

f + 3g = 1 - 1n 2 , (F.20)

therefore we need to calculate f or g, but not‘boflucfi‘them. We

choose to calculate g.

From (F.17) and (F.18) we have

1 dx

0—

Z

 

J. x y
= 2 f f0 dy f0 dz , (F.21)

22 - x(y - z - 1)2 + x

or by using the following change of variables

126

 

 

z = a , y - z - 1 = B , x = x , (F.22)
x-1 x-1-8
1 1 dx
8 _ .5 _: f as f da 2 C12 . (F.23)
/x -1 o a - x8 + x
Performing the a integration, we find
1 1dx x 2 2 x“ 2
g = filo —: {fodu 1n[(1- x)u + x ]-f d8 In (x - x8 )} . (153214)
/x -1
where in the first integral we changed 8 to u-1. By performing the
B and u integrations we obtain
3 = 7:12” flag'é [ln(2 - x) -ln 2 + -——£——tan-1 /1 - x] , (F.25)
/x V1 - x
or by changing x to x
- 1 x2 -1 2
g=-2-ln2+2/2ln(1+/2) +fodx tan /1—x .(F.26)
2
/ 1-x

Let's represent the remaining integral by J

1 x2 -1 2
J = fodx tan ./ 1 - x , (F.27)
——-——7§ .
/ 1 - x

where by changing 1 - x

= y gets the following form

127
J = fgdy'/ 1 - y2 tan-1y 5 (F.28)
We write J in the following form
J = J + J , (F.29)
where

J1 = fgdy————— tan 1y . (F.30)
/ 1 - y2

and it is the same integral that we had in (F.9), so

N

ln2(1+ 5) , (F.31)

1r1
J‘TE

1

and J is

J2 = -f;dy ———l—-— (y tan—1y) . (9.32)
/ 2
1 - y

Let's try to find J by integration by parts

2

_ _ 1 _ 2 -1 __JL___
J2 - fody / 1 y (tan .y + 2) . (F.33)

1 + y

which by (F.28) we can write it as

128

 

 

-__1 _ 2 y
J2- Jfody/1 y 2 . (F.34)
1 + y

and by changing 1 - y2 = x2 and then performing x integration we get

J2 = -J +1 -/21n(1+/2) . (F.35)

Using (F.29), (F.31) and (F.35) we find

2
J= +%-[§ln(1+/2)-1iln2(1+/2) . (F.36)

l
2

Therefore (F.25). (F.28) and (F.36) give

 

g=-%+lg-ln2+3221n(1+/2)-%ln2(1+/2) , (F.37)

and by virtue of (F.20), f is

 

2
r=1-3" +2ln2-filn(1+/2_)-§ln2(1+/_2_) .(F.38)
2 16 2 H _
Finally, the integral in (F.15) by virtue of (F.37) and (F.38)
is

2
u qoqjqfl. 2 11 31r 9/5
qu——D-——-—=i1rhjk£—§ 16+21n2 —2—-1n(1+/2)

 

+ % ln2(1 + 5)]

129

 

2 .—
,2 __3_ 1'..- 31/2
+ 1,, 6j9.[ 2 + 16 1n 2 + 2 ln(1 + 15)

~151n2<1 + m] (1.3 =1. 2. 3) - <F-39)

130

Appendix G
Evaluation of some Integrals
related to the Contribution MN
In this Appendix we evaluate four integrals that we need hi
evaluation of Mu. In these integrals D is defined by (ND.N), avid q
integrations (except for the integral in (G.3)) are performed by

using the Tables given in [19].

The first integral is (in all integrals, except (G.11), we put

 

Y = o)
14 q
f 913—9“ - r) = 4112;; an: dy 2 x
1 My - ny + 1
= ‘11f2f; dX _______X tan-1(—-—-x-' ) 9 (0°1)
/ 1 - x2 / 1 - x2
and by integration by parts we obtain
14 q
G q ___i _ _ 2
f D (1 R.) — in . (0.2)
1
The next integral is
d“ -uq02
I = f 153' 2 2 (6.3)

 

1

which using the Feynman parametrization method, we find

131

 

 

I1 = -21n2f; dxf: dy I: dzL% +-3£5;§1—11] , (G.u)
where
A = (2y - z - x)2 + 1 - x2 (G.5)
By changing the variables
2y - z - x = v , z = u , x = x , (0.6)
the integral gets the following form
I1 = -2in2f; dxf: duf:-:v[ 2 1 2 + 2(x2 - 1) ] , (G.7)

v + 1 - x (v2 + 1 - x2)2

or after v integration

 

I = 21w2f1 dxfX du x ' u . (0.8)
1 o o 2 2
(x - u) + 1 - x

Changing the variables x - u = t, x = x and performing the t

integration, we find

I1 = -in2f; dx ln(1 - x2) , (0.9)

which after x integration and using (G.3). we obtain

132

d“ ~quo2 2
f —53 2 = -in (2 In 2-2) . (0.10)
q .

 

Now we consider the following integral (wherever possible we

put Y = O)

 

u
I = f d-9 u , (0.11)

which by using Tables of [19] it gives

1 dx x y 1
f0 __f0 dy f0 dz 2 , (C.12)

1r2
/x A2

12 = -N1

where approximately (we ignore higher orders, since they give 0012)

corrections)
A = i-(Zz - y)2 + (x - y)(2 + y - x) + Y2 . (C.13)
By following transformations
22 - y = u , x - y = v , x = x , (0.1”)

I2 gets the following form

x-v 1
dv f du-—— (G.15)

X

I = -Nifl f f

O_s
$115?
0
O
N

where

133

2

A =-§— + v(2 - v) + 12 . (0.16)

After performing u integration, we find

12 = -2in2 x; dx/x- I: dv X '2"
a[(x - v) + ax]

- 2112 I; dv-—l— fl dx tan“<x ’ v)

a £5

 

. (6.17)

where

a = v(2 - v) + Y . (G.18)

jhi (G.17). after using integration by parts on the x variable

of the second integral, we obtain

 

12 = —21112f; dv —1— tan-.1 (1 ' V)
a/E /§

-i112f1 dx/x fx dv 1- x - 3v , (G.19)
o o a 2
ax + (x - v)

and by changing the variable 1:; = t in the first integral, and
/a

performing v integration in the second integral, we find (up to the

o(a) correction)

1
.2? -1
I2 21w f0 dt tan t

1311

2

 

 

 

—1121; 9§;[%-ln x - 1n Y - x ' 5x +22 1n(2 ; x)]
/x 2(2 - x)

-in2f;-g£ (x - 6) 12- x tan-1 1 - x . (G.20)
/§' (x - 2)

After performing the first and second integrals, and changing

the variable / - x = t in the third integral, we obtain

12 = -112[$-- 1 - g-/2 ln(1 + /§)]

1

+ 2112f; dt tan- t

/ 1 - t2

2
1 1 _ 3t tan 1t . (0.21)
2 (1 + t2)2 ‘
/ 1 - t

 

2 1
21w f0 dt

Relation (F.31) gives the value of the first integral, and for

the second integral we use the following relation

 

2
tan 1t = g-cos 1 1 t , (G.22)
1 + t ‘
1 - t2
and the change of variable-——-——— = s, or
1+t
[1 dt 1 1 ' 3t2 tan-1t = {E f1 ds-—3§—:—l— 003-13 (0 23)
o -————-— 2 2 8 o -————-— '

/ 1 - t2 (1 + t ) / s - s

135

/2 1 s
= - —Efo ds-—--- (G.2N)

/ s + 32
= - %-+ %-/2'1n (1 + /2) , (0.25)
where, using integration by parts in (0.23) we found (6.2“).

Therefore, using (0.11). (0.21), (F.31) and (0.25) we obtain

(up to the o(a) correction)

 

u
qu 11 =—i112[$-2-2/21n(1+/2)

2

- "—4 + ln2(1 + ./'2')] . (0.26)

Next, we consider the following integral

H
II
H
L?
A
1
O
I
O
V
A
H
II

 

1, 2, 3) , (6.27)

which using Tables of [19]. gives

. 2 1
I - in f0

[(22 - y)2 _
3

.221. x y - _l
__10 (1ny dz (x y+1> 2 2A] . (0.28)

x/x xA

where

A =-%(22 - y)2 + (x - y)(2 + y - x) . (0.29)

136

Using following transformation
22 - y = u , x - y = v , x = x , (6.30)

I3 gets the following form

x ( fx-v { u2
f v + 1)dv du
1/; O 0 [u2 + v(2 - v)x]2

. 2 1 dx
I3 — in IQ

 

 

2 1 1 . (0.31)
u + v(2 - v)x '

nfl—s

and after performing u integration

 

 

 

 

. 2
I3 = “2‘ 1293‘; f:dy(:+1)(v x; (0.32)
/x v (1 - x) + x '
Performing v integration in (G.32) gives
in2 1 dx 1 x 1 ' -1
I = — f _— [-ln(2-x) + - tan F1 T 111 , (0.33)
3 2 052 1-x (1_x)3/2

and after performing the x integration on the first term, and

changing the variable x = 1 - t2 in other terms, we obtain

13 = 1112[-1 + ./2 ln(1 + ./2)]

2 _ _
[1.2.1. .. __1.._ tan-1, - 1.1.3.. 1.... 1t] . (0.311)

2 1
+ in fodt t 2

137

Using the following relation (it can be found by integration by

parts)

 

 

—— 2
f1dt—1-(1/1-t2tan-1t)=1+f1dt(-——1-—-tan1t+i—1—-:—t——
0 t2 0 ——3 t

./1- t
2
“/1 2t) . (0.35)
1+ t -
(G.3M) gets the following form
2 21 t/1- t2
I =i11[-2+/21n(1+/2)] +111f dt—-—-———- , (0.36)
3 o 1 + t2 .
0!“
I3 = -i112[3 - 2/2 ln(1+ /2')] . (0.37)
and finally by (G.27)
‘4 q 2 q 2 q
f dD‘1(- +3 - 32 l) = -1112[3 - 2/2 ln(1 + 5)] . (0.38)
q q k

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