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Solutions to the Problems 555
3.6. This is simply obtained by taking the absolute value squared of ( 3.3.41), and
restrict the p-integration as indicated in the region specified by /SOH, picking
up the spin/ESC, and dividing by the normalization factor N.
3.7. By using the projection given in ( I.21), we have
1
2X
/ESC/... |
.P˙.p//CR0//c142DP˙.p//CR0;. 2 m/2PC.p//CR0PC.p/D2p0./NUL/CRpCm/;
.2m/2P/NUL.p//CR0P/NUL.p/D/NUL2p0./CRpCm/; P˙.p0;p///CR0P/BEL.p0;/NULp/D0:
Using the anti-commutation relation f a.y/; /c142
b.y0/gDıabı3.y/NULy0/,f o r
y0Dy00.D0/,i n ( 3.5.11), we note that due to the last identity in the above
set of equations that th... |
556 Solutions to the Problems
3.10. The/CANintegral may be expressed in terms of the sine function, or equivalently
ID"/SYN
2Z.dQ/
.2/EM/4A/SYN.Q/eiQxZC1
/NUL1d/NAKei./SIQ=2//NAK
D/SI/SYNZ.dQ/
.2/EM/4A/SYN.Q/h
1/NUL1
3Š/DLE/SIQ
2/DC12
C1
5Š/DLE/SIQ
2/DC14
/NUL/SOH/SOH/SOHi
eiQx;
and is an odd function in /SI.
3.11. Fro... |
even function of s. Thus the above integral may be rewritten as
2ImW.e/
VTD/NUL1
8/EM2Z1
/NUL1ds
s3sin.sm2/h/NUL
sjeEjcoth sjeEj/SOH
/NUL.seE/2
3/NUL1i
;
D1
8/EM2ImZ1
/NUL1ds
s3e/NULism2h/NUL
sjeEjcoth sjeEj/SOH
/NUL.seE/2
3/NUL1i
:
It is precisely because of the .seE/2=3, term within the square brackets that
the point... |
Solutions to the Problems 557
Fors'/NULin/EM=jeEj, with nD1;2;::: , the integrand approaches:
e/NULism2=/STX
s2/NUL
sCin/EM
jeEj/SOH/ETX
, and the residue theorem gives
2ImW.e/
VTD1
8/EM2Im./NUL2/EMi/1X
nD1jeEj2
/NULn2/EM2e/NULn/EMm2=jeEjD˛E2
/EM21X
nD1e/NULn/EMm2=jeEj
n2;
coinciding with the one in ( 3.8.8 ).
3.13. Th... |
gU@/SYN@/ETBU/NUL1:
Accordingly,
@/SYNA0
/ETB/NUL@/ETBA0/SYNDU.@/SYNA/ETB/NUL@/ETBA/SYN/U/NUL1/NULh
U@/SYNU/NUL1;UA/ETBU/NUL1i
Ch
U@/ETBU/NUL1;UA/SYNU/NUL1i
/NULi
gh
U@/SYNU/NUL1;U@/ETBU/NUL1i
;
/NULigŒA0/SYN;A0/ETB/c141D/NULigUŒA/SYN;A/ETB/c141U/NUL1/NULh
U@/ETBU/NUL1;UA/SYNU/NUL1i
Ch
U@/SYNU/NUL1;UA/ETBU/NUL1i
Ci
gh
... |
558 Solutions to the Problems
which provides the gauge transformation of G/SYN/ETBD@/SYNA/ETB/NUL@/ETBA/SYN/NUL
igŒA/SYN;A/ETB/c141.
3.15./NUL
/CR/SYN.@=i@z/SYN/Cm/SOH
SC.z/NULy/Dı.4/.z/NULy/. Upon multiplying the latter, from the
right, by S/NUL1
C.y/NULx/, and integrating with respect to y, we obtain : S/NUL1
C.z/NUL... |
DRijmsinh˛ıj3h0/ESCjexpŒ/NULi˛J03/c141expŒ/NULi/DC2n/SOHJ/c141:
With ngiven in ( 4.2.24), we may refer to ( 2.2.11) to infer that R33Dcos/DC2,
R23Dsin/RSsin/DC2,R13Dcos/RSsin/DC2, and use the fact that sinh ˛D
jpj=m, to conclude that the state in ( 4.2.23) becomes multiplied by pby
the action of P,w h e r e pis given i... |
prove. Accordingly suppose that W/SYN¤0,t h a t i s ,jW0jDj Wj>0.F r o m
the orthogonality of the two vectors: P0W0DjPjjWjcos/DC2orjW0jjP0jD
jPjjWjjcos/DC2jfrom whichjcos/DC2jD1.T h a t i s , WD/NAKP. On the other
hand, the relation W0P0DW/SOHPD/NAKjPj2also implies that we also have
W0D/NAKP0,s i n c e P0DjPj>0, which ... |
Solutions to the Problems 559
4.4. W0D.1=2/"ij kPiJjk.B u t.1=2/"ij kJjkDJi, hence W0DP:J.O n t h e
other hand, WiD.1=2/"i/ETB/ESC/NAKP/ETBJ/ESC/NAK. The latter may be rewritten as P0JiC
"0ij kPjJ0k.B u t J0kD/NULNk, from which WDP0J/NULP/STXN.
4.5. Upon writing A/SYN.x/DtcAc/SYN.x/, and using the commutation rule Œta;... |
@/ETBF˛ˇD@˛F/ETBˇ/NUL@ˇF/ETB˛, and the asymmetry of F˛ˇthe conservation law
follows@/SYNT/SYN/ETBD/NUL1
2F˛ˇ@/ETBF˛ˇCF/SYN˛@/SYNF/ETB˛D0.T/SYN/ETBis traceless.
4.7./NUL
@L=@.@/SYN //SOH
rDi /CR/SYN. Recall from ( I.7)t h a t S/ETB/NAKD.i=4/Œ/CR/ETB;/CR/NAK/c141. Hence
from the definition of ˝/SYN/ETB/NAKin (4.4.6 ), tak... |
560 Solutions to the Problems
4.8. The general expression for the energy-momentum tensor, given in the
previous problem is T/SYN/ETBD.1=4 i/ /DLE
/CR/SYN !@/ETBC/CR/ETB !@/SYN/DC1
. In particular,
T00D1
2i /c142 !@0 ; T0kD1
4i /NUL
/CR0 !@kC/CRk !@0/SOH
:
The Dirac equations give @0 D/CR0./CR/SOH/NUL !rCim/SOH
,@0 D... |
b.x0/gDı3.x/NULx0/,f a.x/; b.x0/gD0.
To prove the identity ./ETX/, note that the following ones, in turn
/c142
a.x/ b.x/ /c142
c.x0/ d.x0/D /c142
c.x0/ d.x0/ /c142
a.x/ b.x/
/NULıadı3.x/NULx0/ /c142
c.x0/ b.x/Cıbcı3.x/NULx0/ /c142
a.x/ d.x0/;
/c142
a.x/ d.x0/ /c142
b.x/ c.x0/D/NUL d.x0/ /c142
c.x0/ /c142
a.x/ b.x/
/N... |
Solutions to the Problems 561
4.11. By multiplying the given equation, in turn, by /CR/SYNand@/SYNand by considering
the two resulting equations simultaneously , we are led to the following
equivalent two equations ./CR@/DC1/CR/SYN@/SYN,@KD@/SYNK/SYN,/CRKD/CR/SYNK/SYN/
m@ D./CR@/CR K/C@K;/CR D/NUL2./CR@/CR KC@K/=.im2/... |
description, while the coefficient of ./NUL/ETXCm2/of the second term is the non-
propagating, non-singular term, mentioned in the statement of the problem.
The equations in ( 4.7.137), (4.7.138) are also satisfied for K/SYND0.
4.12. Consider the equation for 0in (4.7.137)./CR@= iCm/ 0D0forK/SYND0.
It may be rewritten a... |
in (2.3.3 ),u.˙1/may be taken (see ( I.25)) and conveniently normalized, as
u.C1/Dp
jpj/DC2/CANC
0/DC3
;u./NUL1/Dp
jpj/DC20
/CAN/NUL/DC3
;X
/ESCu./ESC/u./ESC/D/CRp; |
562 Solutions to the Problems
where/CAN˙are defined in ( I.13). We may then write
PCijD/NUL1
2X
/NAK;/NAK0ei
/NAKek/ETX
/NAK/CR`/DLE
u.C/u.C/Cu./NUL/u./NUL//DC1
/CRke`
/NAK0ej/ETX
/NAK0:./ETX/
By considering, for example, the vector pto be along the 3-axis, and eCD
.1;/NULi;0/=p
2,e/NULD./NUL1;/NULi;0/=p
2, the followin... |
m2/DC10/SYNı.4/.x0/NULx/;or equivalently
•
•K/SYN.x/V/ETB.x0/D1
m2ı/ETB0/DC10/SYNı.4/.x0/NULx/:
On the other hand, ./NULi/•=•K/ETB.x0/h0Cj0/NULiDh0CjV/ETB.x0/j0/NULi, and hence
from ( 4.6.38 ),
./NULi/•
•K/SYN.x/./NULi/•
•K/ETB.x0/h0Cj0/NULiDh0Cj/NUL
V/SYN.x/V/ETB.x0//SOH
C0/NULi
/NULih0Cj•
•K/SYN.x/V/ETB.x0/j0/NULi
Dh... |
Solutions to the Problems 563
from which the statement in the pr oblem follows upon multiplying by i, and
is a consequence of the presence of a dependent field.
4.16. Although V0.x/is a dependent field we note that
/NAK./NULi/•
•K/SYN.x/.i/•
•/DC1.x//CR/SYN./NULi/•
•/DC1.x/h0Cj0/NULi
D/NAK./NULi/•
•K/SYN.x/h0Cj/NUL
.x//... |
.dx0/D/SYN/ETB.x/NULx0/J/ETB.x0/D/NAKR
.dx0/DC.x/NULx0/@0
/ETBJ/ETB.x0/D/NAK/US.x/:
5.3. Going through these various steps, we have: En.K;R/expŒ/NULiEn.K;R/T/c141
Did
dTexpŒ/NULiEn.K;R/T/c141D/NULd2
dT2/DLEexpŒ/NULiEn.K;R/T/c141
En.K;R//DC1
;
expŒ/NULiEn.K;R/T/c141
En.K;R/D2i
T@
@K2exph
/NULir
K2Cn2/EM2
R2Ti
:
Therefor... |
564 Solutions to the Problems
independently of f.jkj/where
CDZ
d˝1Z
d˝2ˇˇˇe/NULijkjn1/SOHR1e/NULijkjn2/SOHR2Ce/NULijkjn1/SOHR2e/NULijkjn2/SOHR1ˇˇˇ2
:
Upon using the elementary integrals,
Z1
/NUL1dxxe/NULikaxD2ihcoska
ka/NULsinka
k2a2i
;CD32/EMh
1Csin2ka
k2a2i
;
we obtain the following expression
hciD/DLEcoska
ka/NULsin... |
FŒ•=•J/c141h0Cj0/NULi0/CRin (5.7.27). Then
@/SYNh0CjA/SYN.x/j0/NULiD/NAKFŒ•=•J/c141Z
.dx0/DC.x/NULx0/@0/ETBJ/ETB.x0/h0Cj0/NULi0/CR:
Upon using the identity: FŒ•=•J/c141J/ETB.x/D/STX
J/ETB.x/C•FŒT/c141=•T/ETB.x//ETXˇˇ
T/ETBD•=•J/ETB,t h e
expansion
SC.y;y0Ie0T/DSC.x/NULx0//NULie0Z
.dy/SC.y/NULy1//CR/SYNT/SYN.y1/SC.y1/NU... |
Solutions to the Problems 565
and the identity
Z
.dx0/.dy1/DC.x/NULx0/@x0
/SYN•
•T/SYN.x0/SC.y/NULy1//CR/SYNT/SYN.y1/SC.y1/NULy0/
D/NULiŒDC.x/NULy/SC.y/NULy0//NULDC.x/NULy0/SC.y/NULy0//c141;
which applied to the exp Œi/DC1SC.:Ie0OA//DC1/c141 factor leads to the stated result.
By the same method just developed, the othe... |
order they cannot connect together thre e different expressions of the form
Œ/DC1S/CR/SYNS/DC1/c141.B2leads to.1=2/Œ./NUL i/D/SYN1/SYN2CD/SYN1/ETB1J/ETB1D/SYN2/ETB2J/ETB2/c141K/SYN1/SYN2,a n d
the first term in the latter .1=2/./NUL i/D/SYN1/SYN2K/SYN1/SYN2involves no external lines
and should be omitted due to th e nor... |
/DC1S/CR/SYN3S/DC1/c141K/SYN1/SYN2h
D/SYN1/SYN3D/SYN2/ETB2J/ETB2CD/SYN2/SYN3D/SYN1/ETB1J/ETB1i
;
retaining only connected parts. |
566 Solutions to the Problems
5.8. This involves four terms. The term obtained by multiplying the first term
in (5.9.23) by its complex conjugate is given by
1
.p2/NULp0
2/4Œu.p0
2;/ESC02//CR/SYNu.p2;/ESC2/u.p01;/ESC01//CR/SYNu.p1;/ESC1/
/STXu.p1;/ESC1//CR/ESCu.p0
1;/ESC01/u.p2;/ESC2//CR/ESCu.p02;/ESC02//c141
D1
.2m/41
... |
1/NULˇ2/. The four spinors of the outgoing
electrons may be written as
u.p0
1;/ESC01/Ds
p0Cm
2m
/CAN1
/ESC/SOHp0
1
p0Cm/CAN1!
;u.p0
2;/ESC02/Ds
p0Cm
2m
/CAN2
/NUL/ESC/SOHp0
1
p0Cm/CAN2!
;
where
n1D.0;sin/US1;cos/US1/;n2D.0;sin/US2;cos/US2/;p0
1D/CRmˇ.1;0;0/D/NULp02;
/CANjD.e/NULi/EM=4cos./USj=2/ ei/EM=4sin./USj=2/>;j... |
Solutions to the Problems 567
This leads tojAj2=2DPŒ/US1;/US2/c141as given in the problem, satisfying the
completeness relation:
PŒ/US1;/US2/c141CPŒ/US1C/EM;/US2/c141CPŒ/US1;/US2C/EM/c141CPŒ/US1C/EM;/US2C/EM/c141D1:
If only one spin is measured: P Œ/US1;/NUL/c141DPŒ/US1;/US2/c141CPŒ/US1;/US2C/EM/c141D1=2,
and similarly... |
the identity follows.
5.11. From the method of the Feynman parameter representation, as given
in (II.34), in Appendix Iat the end of the book,
Z
.dk/1
.k2Cm2/2/ETX2
.k2C/ETX2/D2/ETX2Z1
0xdxZ.dk/
/STX
k2C/ETX2C.m2/NUL/ETX2/x/ETX3
Di/EM2Z1
0/ETX2xdx/NUL/ETX2C.m2/NUL/ETX2/x/SOHD/ETX2
m2/NUL/ETX2Z1
0dxh
1/NUL/ETX2
/ETX2C.m... |
over x by parts , and using the elementary integral ( II.8), we obtain ID |
568 Solutions to the Problems
i/EM2ŒCuvCI2/c141,w h e r e Cuvis given in ( 5.10.21), and
I2D.p2Cm2/Z1
0x.1/NUL2x/dx
Œ.p2Cm2/x.1/NULx/Cm2x2C/SYN2.1/NULx//c141
CZ1
0x.2m2x/NUL/SYN2/dx
Œ.p2Cm2/x.1/NULx/Cm2x2C/SYN2.1/NULx//c141:
In the first integral I.1/
2,w em a ys e t .p2Cm2/D0, in the denominator,
obtaining, I.1/2DŒ.p2C... |
. Hence I2D2CŒ.p2Cm2/=m2/c141ŒCir/NUL2C3/NUL2Cir/c141,
Z.dk/
Œ.p/NULk/2Cm2/c141Œk2C/SYN2/c141Di/EM2h/NUL
CuvC2/SOH
/NUL2/CRpCm
m/NUL
Cir/NUL1/SOH
CO/DLE
./CRpCm/2/DC1i
;
where we have used the identity in Problem 5.10. The statement of the
problem follows upon multiplying this integral by /NULie2Œ/NUL./NAKC1/./CRpCm//N... |
Solutions to the Problems 569
behavior
I1D2Z1
2mjxjdte/NULt
tD/NUL2Ei./NUL2mjxj/'/NUL2/CR EC2ln/DLE1
2mjxj/DC1
CO.mjxj/;
where/CRED0:5772157::: denotes Euler’s constant. On the other hand,
it is justifiable to take the limit mjxj!0inside the second integral. This
amounts to evaluate the integral
I0
2/DC1Z1
.2m/2dM2
M2/D... |
problem then follows from ( 5.10.44), (5.10.49), by taking the limit k2!0
in this order, where ˘.0/D1/NULZ3.
5.16. Using the normalization condition of /SUB.x/, the change in potential energy is
/SOHU.x/D/NUL˛/DC4Z
d3x0/SUB.x0/
jx/NULx0j/NUL1
jxji
D/NUL˛Z
d3x0/SUB.x0//DC41
jx/NULx0j/NUL1
jxj/NAK
:
leading to an approxi... |
570 Solutions to the Problems
The x0- integrand, multiplying /SUB.x0/, is easily worked out, for example, by
a Legendre polynomial expansion,7to be
Z
d3x/DC41
jx/NULx0j/NUL1
jxj/NAK
D2/EMZ
jxj<jx0jdjxjjxj2/DC42
jx0j/NUL2
jxj/NAK
C0D/NUL2/EMjx0j2=3;
andR
d3x0/SUB.x0/jx0j2D12/CR2.T h i s g i v e s
•ED8˛4m
n3m2/CR2:
5.17.... |
and the above two equations in turn, lead to
ŒAa.x0;x0/;@/NUL1
3@b@0A3.x0;x//c141/NULŒAa.x0;x0/;@0Ab.x0;x//c141Diıabı.3/.x0/NULx/;
ŒAa.x0;x0/;@/NUL1
3@0A3.x0;x//c141Di@a
r2ı.3/.x0/NULx/;
7Recall:.jx/NULx0j//NUL1D.1=r>/P1
nD0.r<=r>/nPn.cos/DC2/, in a standard notation, P0.cos/DC2/D
1;P1.cos/DC2/Dcos/DC2,R1
/NUL1dc o s/D... |
Solutions to the Problems 571
from which the following key equal-time commutation relations emerges
ŒAa.x0;x0/;@0Ab.x0;x//c141Di/DLE
ıab/NUL@a@b
r2/DC1
ı.3/.x0/NULx/; a;bD1;2:
The expression A3D/NUL.@a=@3/Aathen leads to the equal time commutation
relations in question.
5.19. The Fourier transform of the left-hand side... |
./NULi/•
•/DC1.x/K
DexpŒie0/ETX.z//c141/DC1. z/K:
But
expŒie0/ETX.z//c141ŒS/NUL1
C.z/NULx/Cı.4/.z/NULx/e0/CR/SYN/NUL
@/SYN/ETX.x//c141
DS/NUL1
C.z/NULx/expŒie0/ETX.x//c141;
(see Problem 3.15). Hence upon multiplying the former equation by:
expŒ/NULie0/ETX.y//c141SC.y/NULz/, and integrating over z, we obtain
./NULi/•
•/... |
572 Solutions to the Problems
5.21. Upon setting e 0/NUL
•=•/SUB. z//SOH
/CR/SYN/NUL
•=•/SUB.z//SOH
DOf/SYN.z/, with/DC1!/SUB,/DC1!/SUB,
J/SYN!K/SYN, w eh a v ef r o m( 5.15.1 ),
exphZ
.dx//STX/NUL
/DC1/SYN/ETB/NUL@/SYN@/ETB
/ETX/SOH
J/ETB.x//ETX•
•K/SYN.x/i
FŒ/SUB;/SUB;K/SYN;/NAK/c141ˇˇˇ
K/SYND0
Dexphi
2.Of/SYNCJ/SYN/... |
5.22. Set e 0/NUL
•=•/SUB. z//SOH
/CR/SYN/NUL
•=•/SUB.z//SOH
DOf/SYN.z/. Then from (5.15.20),
./NULi/•
•/DC1.x/.i/•
•/DC1.y/FŒ/DC1;/DC1;J/SYN;/NAKD0/c141ˇˇˇ
/DC1D0;/DC1D0
D./NULi/•
•/SUB.x/.i/•
•/SUB.y/expŒOQ/c141FŒ/SUB;/SUB;K/SYN;/NAK/c141ˇˇˇ
/SUBD0;/SUBD0;K/SYND0;./ETX/
OQDe0h
Qa/SYN
y•
•K/SYN.y//NULQa/SYNx•
•K/SYN.x... |
Solutions to the Problems 573
With expŒOQ/c141generating translations in K/ETB, the right-hand side of the former
equation./ETX/is given, in matrix multiplication notation in spacetime, by
expŒi/RS.J/SYN//c141./NULi/•
•/DC1.x/.i/•
•/DC1.y/exphi
2.Of/SYNCJ/SYN/D/SYN/ETB.Of/ETBCJ/ETB/i
/STXexp/STX
/NULiOf/SYN@/SYN/ETX/c1... |
•/DC1.y/FŒ/DC1;/DC1;J/SYN;/NAKD0/c141ˇˇˇ
/DC1D0;/DC1D0;
Dei/TABŒJ/SYN/c141e/NULie2
0/STX
G.0//NUL G.x/NULy//ETX
.i/./NULi/•
•T.x/.i/•
•T.y/FŒT;T;J/SYN;/NAK/c141ˇˇˇ
TD0;TD0;;./ETX/ETX/
/TABŒJ/SYN/c141D/NUL1
2.@/SYNJ/SYN/G.@/ETBJ/ETB//NULe0Z
.dz/J/SYN.z/@z
/SYNŒG.z/NULx//NULG.z/NULy//c141:
Upon dividing (**) by F Œ0;0; J... |
574 Solutions to the Problems
6.2. We explicitly have
Œr/SYN;r/ETB/c141cbDgofcab.@/SYNA/ETB
a/NUL@/ETBA/SYNa/Cg2o/NUL
fcdafaeb/NULfceafadb/SOH
A/SYN
dA/ETB
e:
Using the identity fcdafaeb/NULfceafadbD/NULfcbafadeDfcabfade, the result follows
upon factoring out gofcab, and using the definition of G/SYN/ETB
a.
6.3. From th... |
/STXexpi/NAK
2ŒZ
.dx//RSa.x//RSa.x//c141exp/NULi
2/NAKŒZ
.dx/@/SYNA/SYN
a.x/@/ETBA/ETB
a.x//c141;
upon completing the squares in the exponential, and shifting the variable /RSa.
The result follows after integration over the latter variable.
6.5. From Problem 6.1:A./DC2//SYN
a'A/SYNaC.ıab@/SYNCgofacbA/SYNc//DC2b. The co... |
4MpQ˛/DC4F2.Q2/i
u.p;/ESC/; |
Solutions to the Problems 575
as readily follows by invariance argum ents and application of the Dirac
equation./CRpCMp/u.p;/ESC/D0.Mpdenotes the mass of the proton. This
gives
d/ESC
d˝ˇˇˇ
TFD˛2
4E2sin4#
2E0
E"/DLE
F2
1CQ2
4M2
p/DC42F2
2/DC1
cos2#
2CQ2
2M2
p/DLE
F1C/DC4F2/DC12
sin2/DC2
2#
:
The result follows upon sett... |
the momentum of an emerging quark qrelative to that of the electron. Hence
d/ESC=d˝/3e4X
f.e2
f=e2/.1Ccos2#/;
where e fis the charge of the quark of a given flavor, and the factor 3 is
for the three different colors. The cross section then works out to be /ESC/
3e4P
f.e2
f=e2/.16/EM/=3 . Upon comparison of this expressi... |
576 Solutions to the Problems
which by using the integral representation over kin (II.7) in Appendix II, at
the end of the book, gives
i
.4/EM/D0=2/DLE1
Q2/DC1.1/NULı=2//NUL/NUL
1/NULı
2/SOH
/NUL.3/2Z1
0dx.1/NULx/./NUL1Cı=2/Zx
0dzz./NUL1Cı=2/:
Finally carrying out the z-integral, followed by the use of the integral ( I... |
after setting an odd integral in kequal to zero. The integral involving the k/SYNk/ETB
part may be ultraviolet-regularized using the integral in ( III.8), while the
integral involving .p/SYN
1.1/NULx/Cp/SYN2z/.p/ETB
1.1/NULx/Cp/ETB2z/may be infrared-
regularized as in Problem 6.8. Finally, the.x;z/– integrations yield ... |
these terms cancel out again the 1=Q2factor just mentioned. That is, in all the
terms contributing to the fermion-gluon vertex, the 1=Q2factor multiplying
./SYN2
D0=Q2//NULı=2is canceled out in the infra-red regularized part. The most infra-
red singular part in evaluating the vertex function comes from the first integr... |
Solutions to the Problems 577
6.11.
X
nPn.2/EM/4ı4.pCQ/NULPn/hP;/ESCjj/SYN.0/jnPnihnP njj/SYN.0/jP;/ESCi
DX
nPnZ
.dy/ei.pn/NULP/NULQ/yhP;/ESCjj/SYN.0/jnPnihnP njj/SYN.0/jP;/ESCi
DZ
.dy/e/NULiQyhP;/ESCjj/SYN.y=2/j/SYN./NULy=2/jP;/ESCi
where we have used the fact hP;/ESCjj/SYN.0/jnPni
Dei.y=2/Pe/NULi.y=2/PnhP;/ESCjeŒ/NUL... |
which establishes ( 6.9.13). Equations ( 6.9.14), (6.9.16) follow upon multiply-
ing ( 6.9.13)b y2xM, with xDQ2=2M/ETB, and finally using, in the process,
the definitions in ( 6.9.15).
6.13. We explicitly have (see also ( 6.9.7 ))
W/SYN/ETB
iD1
2/EM/CANMe2
i
e2Zd3p0
2p00.2/EM/3Œ:/c141/SYN/ETB.2/EM/4ı.4/./CANPCQ/NULp0/;
w... |
578 Solutions to the Problems
6.14. The integral on the left-hand side is equal to
Z
.dp0//STX.p00/ı.p02C/CAN2M2/ı.4/./CANPCQ/NULp0/
D/STX./CANP0CQ0/ı/NUL
./CANPCQ/2C/CAN2M2/SOH
Dı.2/CANPQCQ2/:
The result in question then follows upon taking 2QPoutside the argument
ofı.2/CANPQCQ2/.
6.15. The vertex function V/SYNfor a ... |
involving no./DC1/SYN/ETB/NULQ/SYNQ/ETB=Q2/t e r m .T h i sl e a d st o W1D0and the results
stated in the problem follow.
6.16. We may write
X
ie2
ixfi.x/Dxf.x/X
ie2iD.2=3/ xf.x/D.2=3/.1=3/X
ixfi.x/;
where we have used the fact thatP
ie2
iD.4=9/C.1=9/C.1=9/D2=3.
Upon integration over x, this gives the relation stated i... |
Solutions to the Problems 579
we have used the relation ˛s.Q2/D1=.ˇ0t/, to lowest order. From the chain
rule d=dtD.d/FS=dt/.d=d/FS/, the relation follows.
6.19. (i) This directly follows by noting that .AnDn/NULBnCn/D/NAKC
n/NAK/NULn,a n d t h a t
AnCDnD/NAKCnC/NAK/NULn, upon carrying out the multipli cation of the thr... |
nf<43 . On the other hand .An/NULDn/>4 BnCn,a n d.AnCDn/>p
2.AnCDn/
with the latter being negative. Hence
/NAK/NUL
nD.1=2/ΠAnCDn/NULp
.An/NULDn/2C4BnCn/c141
>.p
2=2/Œ2D n/c141>/NULp
2Œ.11=2/C.nf=3/C6#n
4/c141;
as follows from ( 6.11.18). (iv) Finally
./NAKC
n/NULAn/D.1=2/Œ/NUL.An/NULDn/Cp
.An/NULDn/2C4BnCn/c141>0;
./N... |
580 Solutions to the Problems
with condition Q/NUL.0/DV/NUL1.0;a/. Using the fact that
dh/NUL.t;a;0;a/=dtD/NULigh/NUL.t;a;0;a/A0.t;a/;
we obtain d Q/NUL.t/=dt
D/NUL igh/NUL.t;aI0;a/V/NUL1.t;a/V.t;a//STX
A0.t;a//NUL.1=ig/.d=dt//ETX
V/NUL1.t;a/;
which is just/NULigQ/NUL.t/AV
0.t;a/. Upon integration from 0toT,g i v e s :... |
/RS0/DC3
:
Upon considering the expression /RS0=/RSC, and writing the latter as
./RS0=/RSC/D.aCib/, with aand breal, it is easily checked, by equating
the resulting upper entry to zero, i.e., by setting
/NUL
cos.'=2/Cin3sin.'=2//SOH
C.in1Cn2/sin.'=2/.aCib/D0;
that this equation has always a solution for all real aand b... |
Œ/CR/SUB.1/NUL/CR5//c141aBŒ/CR/SUB.1/NUL/CR5//c141cD/ETX
: |
Solutions to the Problems 581
Multiplying the Fierz identity8
./CR/SUB/ab./CR/SUB/cdD/NULıadıcb/NUL1
2./CR/SUB/ad./CR/SUB/cb/NUL1
2./CR5/CR/SUB/ad./CR5/CR/SUB/cbC./CR5/ad./CR5/cb;
by.1/NUL/CR5/bB.1/NUL/CR5/dDand using the identities f/CR5;/CR/SYNgD0,./CR5/2D1,
give/NULŒ/CR/SUB.1/NUL/CR5//c141aDŒ/CR/SUB.1/NUL/CR5//c141c... |
conservation of energy and momentum that the maximum value of the energy
Eeattained by the electron corresponds to the neutrinos moving in the same
direction leading to EejmaxDM/SYN=2, we readily obtain, upon carrying the
k-integration, the decay rate stated in the problem.
8For many details on Fierz identity and some ... |
Index
Z2in QED, 275
Z3in QED, meaning of, 280, 327
PCAC, 118
SU.2/,379
SU.3/,380
SU.N/,369, 378
U.1/,223, 369
sin2/DC2W, experimental determination, 481
˘Cerenkov radiation and Apollo missions,
241
AdS/CFT correspondence, 30
anomalies, 74,111, 402
/EM0!/CR/CRdecay, 117
experimental verification, 120
non-abelian, 120
abe... |
charge of the electron, 80
charge renormalization, 282, 289, 332, 338
charge space, 371
chirality, 520
closed path, 375
color, QCD, 405
colors of quarks, 405, 406, 494
conditional probability, 137
connection, 373
connect
ion, non-abelian gauge theories, 373
Coulomb gauge formulation of QED, 304
Coulomb gauge, non-abeli... |
584 Index
dimensional regularization, 275, 530
dimensional transmutation, 423
dimensional-regularized integral, 276
Dirac equation
with external electromagnetic field, 80
Dirac propagator, 77
in external electromagnetic field, 80
Dirac quantum field, 79,80,85
donkey electron, 88
double valued representation, 146
Duffin-Kem... |
and wavefunction renormalization, 137
fi
ne-structure constant, definition of, 95
Fock vacuum, 98
Foldy-Wouthuysen-Tani transformation, 457
form factors, 405
Fourier Transform
Grassmann variables, 61
fractional charges of quarks, 496
functional derivatives with respect to
anti-commuting functions, 63
functional determina... |
Gell-Mann and Low function, 351, 352
Gell-Mann matrices, 380
generating functional, non-abelian gauge
theories in covariant gauges, 393
generations of quarks, 405
ghost fields, 394
ghosts and gauge invariance, 417
gluon splitting, 440
Goldstone bosons, 359
Goldstone Theorem, 359
Gordon decomposition, 82
Grassmann variab... |
Index 585
inhomogeneous Lorentz transformations
group properties of, 51
integration over commuting complex variables,
60
integration over complex Grassmann variables,
60
integrations over links, 466
Jacobian of transformation for Grassmann
variables, 58
jets, QCD, 406, 408, 444
Källen-Lehmann representation, 323
Lagran... |
parton model, 14,407parton splitting, 437, 447
partons, 408
path integral expression for QED, 246
Pauli-Lubanski (pseudo-) vector, 141
componens of, 141
photon propagator in covariant gauges, 230
photon spectral representation, 281, 323
plaquette, 464
Poincaré algebra, 49,141
Poincaré Transformations, 51
polarization c... |
nger parametric representation, 109
Schwinger’s constructive approach, 166
Schwinger’s poi nt splittin g method, 73,102,
131, 533
Schwinger-Feynm an boundary c ondition, 77,
78
self-mass of the electron, 275
sites, lattice, 464
skeleton expansion, 322
soft bremsstahlung, 293
solution of QED in Coulomb gauge, 307
soluti... |
586 Index
Spin & Statistics Connection, 32,55,162, 166
splitting factor, quark, 440
splitting factors of partons, 443, 448
spontaneous symmetry breaking, 357, 359,
360, 371, 478
structure functions, 407, 431
substitution law, 264
Super-Poincaré algebra, 49
survival probability of neutrino, 488
time anti-ordering, 459
t... |
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