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Now using the results of section A.7.3 with

;

= p

(2.99)

x1p0 + x2p

(2.100)

= (x1 + x2)2m2 x1x2 q2 + (1 x1 x2) 2

(2.101)

where

we get,

q = p0

(2.102)

1 x1

i u(p0) loopu(p)

(3) Z0

dx1

dx2

(u(p0) u(p) " ( 2 + d)(x12m2 + x22m2 + 2x1x2p0 p) 4p0 p

+4(p + p0)

p0 + x

p) +

(2 d)2

2 !#

+u(p)u(p)m 4(x1p0 + x2p) 4(p0 + p) (x1 + x2)

2(2 d)(x1 + x2)(x1p0 + x2p) )

(2.103)

i u(p) hG(q2) + H(q2) (p + p0)i u(p)

(2.104)

where we have de ned9 ,

G(q2) 4 " 2 2 Z0

dx1

1 x1

dx2 ln

) m

)

(x1 + x

x2q + (1

1 x1

+ 2q

2(x1 + x2)

x1x2q

1 Z0

x1x2q2 + (1 x1

x2) 2

2 (x1 + x2)2m2

x1x2q2

(x1 + x2)2m2

+ (1 x1 x2) 2

2(x1

+ x2)(4m2

q2)

(2.105)

H(q2) 4 "

dx1

1 x1

dx2 (x + x )2m2

2x q2 + (1

)

(2.106)

x ) 2

2m (x1

+ x ) + 2m (x

+ x

1 2

Now, using the de nition of Eq. (2.85), we get for the renormalized vertex,

u(p0) R(p0; p)u(p) = u(p0) h G(q2) + Z1

+ H(q2) (p + p0) i u(p)

(2.107)

1 is calculated in the0

limit of q = p0

0 it is convenient to

use the Gordon

+ p)

identity to get rid of the (p

term. We have

u(p0) (p0 + p) u(p) = u(p0) 2m i q u(p)

(2.108)

To obtain Eq. (2.106) one has to show

that the symmetry of the integrals in x1

implies that the

coe cient of p is equal to the coe cient of

and therefore,

u(p0) R(p0; p)u(p) = u(p0) G(q2) + 2mH(q2) + Z1 i H(q2) q u(p)

= F1(q2) +	i	q F2(q2)	(2.109)

	2m
where we have introduced the usual notation for the vertex form factors,
F1(q2) G(q2) + 2mH(q2) + Z1			(2.110)
F2(q2) 2mH(q2)			(2.111)
The normalization condition of Eq. (2.86) implies F1(0) = 0, that is,
Z1 = G(0) 2m H(0)			(2.112)

We have therefore to calculate G(0) and H(0). In this limit the integrals of Eqs. (2.105) and (2.106) are much simpler. We get (we change variables x1 + x2 ! y),

" 2

y2m2

+ (1

y) 2

G(0)

2 Z0

dx1

Zx1

dy ln

+ 8ym

(2.113)

y) 2

y2m2 + (1

2m y + 2m y

H(0)

dx1

Zx1

(2.114)

y2m2 + (1 y) 2

Now using

dy ln

= 2

dx1 Zx1

2 1!

y m + (1

+ 8ym

1 Zx1

= 7 + 2 ln

y2m2 + (1 y) 2

2m y + 2m y2

dx1 Zx1 dy

y2m2 + (1 y) 2

(where we took the limit ! 0 if possible) we get,

G(0)

+ 6 ln

+ 2 ln

H(0)

4 m

Substituting the previous expressions in Eq. (2.112) we get nally,

" 4 + ln

Z1 =

2 ln

(2.115)

(2.116)

(2.117)

(2.118)

(2.119)

(2.120)

in agreement with Eq. (2.83) and Eq. (2.93). The general form of the form factors Fi(q2), for q2 6= 0, is quite complicated. We give here only the result for q2 < 0 (in section 4.3 we will give a general formula for numerical evaluation of these functions),

F1(q2) =

(

2 ln

+ 4! ( coth 1) tanh

8 coth Z0

tanh d )

F2(q2) =

(2.121)

2 sinh

where

sinh2

(2.122)

4m2

In the limit of zero transfered momenta (q = p0 p = 0) we get

F1(0) = 0

(2.123)

F2(0) =

a result that we will use in section 5.1 while discussing the anomalous magnetic moment of the electron.

3Passarino-Veltman Integrals

In this section we present the Passarino-Veltman decomposition of the one-loop tensor integrals.

3.1The general de nition

The description of the previous sections works well if one just wants to calculate the divergent part of a diagram or to show the cancellation of divergences in a set of diagrams. In this last case one just uses the results of section A.8. If one actually wants to numerically calculate the integrals the task is normally quite complicated. Except for the self-energy type of diagrams, the integration over the Feynman parameters is normally quite di cult.

To overcome this problem a scheme was rst proposed by Passarino and Veltman [1]. This scheme with the conventions of [6, 7] was latter implemented in the Mathematica package FeynCalc [2] and, for numerical evaluation, in the LoopTools package [3, 8]. The numerical evaluation follows the code developed earlier by van Oldenborgh [5]. We will now describe this scheme. We will write the generic one-loop tensor integral as

	(2 )4 d			k 1	k p
Tn 1 p		Z	ddk			Dn 1	(3.1)
Tn 1 p	i 2	Z	ddk	D0D1D2		Dn 1	(3.1)

where we follow for the momenta the conventions of Fig. 10, with

Di = (k + ri)2 mi2 + i

(3.2)

p1 k pn

p2 k+r1 pn-1

p3 k+r3pi

Figure 10:

and the momenta ri are related with the external momenta (all taken to be incoming) through the relations,

		j
		X	; j = 1; : : : ; n 1
rj	=	pi	; j = 1; : : : ; n 1
		i=1
		n
		X
r0	=	pi	= 0	(3.3)

i=1

Notice that a factor of i=16 2 is taken out. This is because, as we have seen in section 2, these integrals always give that pre-factor (see also A.3). So with our new convention that pre-factor has to included in the end. Factoring out the i has also the convenience of dealing with real functions in many cases.10 From all those integrals in Eq. (3.1) the scalar integrals are, has we have seen, of particular importance and deserve a special notation. It can be shown that there are only four independent such integrals, namely (4 d = )

A0(m02) =

(2 )

(3.4)

i 2

k2 m02

ddk

(2 )

B0(r102 ; m02; m12) =

ddk i=0

(3.5)

i 2

[(k + ri)2

mi2

]

(2 )

C0(r102 ; r122 ; r202 ; m02; m12; m22) =

ddk i=0

(3.6)

i 2

[(k + ri)2

mi2

]

D0(r102 ; r122 ; r232 ; r302 ; r202 ; r132 ; m02; : : : ; m32) =

(2 )

ddk i=0

(3.7)

i 2

[(k + ri)2

mi2

]

where

rij2 = (ri rj )2

;

8 i; j = (0; n 1)

(3.8)

Remember that with our conventions r0

= 0 so ri20

= ri2. For simplicity, in all these

expressions the i part of the denominator factors is suppressed. The general one-loop tensor integrals are not independent and therefore their decomposition is not unique. We

10The one loop functions are in general complex, but in some cases they can be real. These cases correspond to the situation where cutting the diagram does not corresponding to a kinematically allowed process.

follow the conventions of [2, 3] to write

B	(2	)4 d	Z
		i 2
B	(2	)4 d	Z
		i 2
C	(2	)4 d	Z
		i 2
C	(2	)4 d	Z
		i 2
C	(2	)4 d	Z
		i 2
D	(2	)4 d	Z
		i 2
D	(2	)4 d	Z
		i 2
D	(2	)4 d	Z
		i 2
D	(2	)4 d	Z
		i 2

ddk k

[(k + ri)2

m2]

i=0

ddk k k

[(k + ri)2

mi2]

i=0

ddk k

[(k + ri)2

]

i=0

ddk k k

[(k + ri)2

]

i=0

ddk k k k

[(k + ri)2

]

i=0

ddk k

[(k + ri)2

]

i=0

ddk k k

[(k + ri)2

mi2]

i=0

ddk k k k

[(k + ri)2

]

i=0

ddk k k k k

[(k + ri)2

]

i=0

(3.9)

(3.10)

(3.11)

(3.12)

(3.13)

(3.14)

(3.15)

(3.16)

(3.17)

3.2The tensor integrals decomposition

These integrals can be decomposed in terms of (reducible) functions in the following way:

B	=	r1 B1					(3.18)
B	=	g B00 + r1 r1 B11					(3.19)
C	=	r1 C1 + r2 C2					(3.20)
C	=	g C00 +	2				(3.21)
C	=	g C00 +	ri rj Cij				(3.21)
			X
			i=1
C	=	2	+ g ri + g ri ) C00i +		2	Cijk	(3.22)
C	=	(g ri	+ g ri + g ri ) C00i +		ri rj rk	Cijk	(3.22)
		X			X
		i=1			i;j;k=1
		3
		X
D	=	ri Di					(3.23)
		i=1
D	=	g D00 +	3	Dij			(3.24)
D	=	g D00 +	ri rj	Dij			(3.24)
			X
			i=1
D	=	3	+ g ri + g ri ) D00i +		2		(3.25)
D	=	(g ri	+ g ri + g ri ) D00i +		ri rj rk Dijk		(3.25)
		X			X
		i=1			i;j;k=1

D = (g g	+ g g + g g ) D0000	3
		+ i;j=1 g ri rj + g ri rj + g ri rj
		X
+g ri rj + g ri rj + g ri rj D00ij			3	Cijkl :	(3.26)
			+ i;j;k;l=1 ri rj rk rl
			X

All coe cient functions have the same arguments as the corresponding scalar functions and are totally symmetric in their indices (not in their arguments). In the FeynCalc [2] package one generic notation is used,

PaVe hi; j; : : : ; fr102 ; r122 ; : : :g; fm02; m12; : : :gi	;	(3.27)
for instance
B11(r102 ; m02; m12) = PaVe h1; 1; fr102 g; fm02; m12gi :		(3.28)

All these coe cient functions are not independent and can be reduced to the scalar functions. FeynCalc provides the command PaVeReduce[...] to accomplish that. This is very useful if one wants to check the cancellation of divergences or gauge invariance, where a number of diagrams have to cancel. We emphasize that the decomposition of the tensor integrals is not unique. Our choice is tied to the conventions of Fig. 10.

4QED Renormalization with PV functions

In this section we will work out in detail a few examples of one-loop calculations using the FeynCalc package and the Passarino-Veltman scheme.

4.1Vacuum Polarization in QED

We have done this example in section 2.1 using the techniques described in sections A.3, A.4 and A.5. Now we will use FeynCalc. The rst step is to write the Mathematica program. We list it below:

(*********************** Program VacPol.m **************************)

(* First input FeynCalc *)

<< FeynCalc.m

(* These are some shorthands for the FeynCalc notation *)

dm[mu_]:=DiracMatrix[mu,Dimension->D] dm[5]:=DiracMatrix[5] ds[p_]:=DiracSlash[p] mt[mu_,nu_]:=MetricTensor[mu,nu] fv[p_,mu_]:=FourVector[p,mu]

epsilon[a_,b_,c_,d_]:=LeviCivita[a,b,c,d] id[n_]:=IdentityMatrix[n] sp[p_,q_]:=ScalarProduct[p,q] li[mu_]:=LorentzIndex[mu]

L:=dm[7]

R:=dm[6]

(* Now write the numerator of the Feynman diagram. We define the constant

C=alpha/(4 pi)

num:= - C Tr[dm[mu] . (ds[q] + m) . dm[nu] . (ds[q]+ds[k]+m)] (* Tell FeynCalc to evaluate the integral in dimension D *)

SetOptions[OneLoop,Dimension->D]

(* Define the amplitude *)

amp:=num * FeynAmpDenominator[PropagatorDenominator[q+k,m], \ PropagatorDenominator[q,m]]

(* Calculate the result *)

res:=(-I / Pi^2) OneLoop[q,amp] ans=Simplify[res]

(******************** End of Program VacPol.m **********************)

The output from Mathematica is:

	2	2	2	2	2	2	2	2	2
	Out[4]= (4 C (k	+ 6 m	B0[0, m , m ] - 3 (k			+ 2 m ) B0[k , m , m ])
	2			2
	(k g[mu, nu] - k[mu] k[nu])) / (9 k )

Now remembering that,
			C =						(4.1)
			C =	4					(4.1)
				4
and		i (k; ") = i k2P T (k; ")
		i (k; ") = i k2P T (k; ")							(4.2)

we get

(k; ") =

8 m2

B0(0; m2; m2) +

1 +

2m2

! B0(k2; m2; m2)#

(4.3)

3 k2

To obtain the renormalized vacuum polarization one needs to know the value of (0; "). To do that one has to take the limit k ! 0 in Eq. (4.3). For that one uses the derivative of the B0 function

B00

(p2; m12; m22)

B0(p2; m12; m22)

@p2

to obtain

(0; ") =

B0(0; m2; m2) +

m2B00 (0; m2; m2)

Using

B00 (0; m2; m2) =

6m2

we nally get

(0; ") = Z3 =

B0(0; m2; m2)

and the nal result for the renormalized vertex is:

R(k) =

+ 1 +

2m2

B0(k2; m2; m2) B0(0; m2; m2)

" 3

If we want to compare with our earlier analytical results we need to know that

B0(0; m2; m2) = " ln m2

(4.4)

(4.5)

(4.6)

(4.7)

(4.8)

(4.9)

Then Eq. (4.8) reproduces the result of Eq. (2.54). The comparison between Eq. (4.8) and Eq. (2.56) can be done numerically using the package LoopTools [3].

4.2Electron Self-Energy in QED

In this section we repeat the calculation of section 2.2 using the Passarino-Veltman scheme. We start with the Mathematica program,

(********************** Program SelfEnergy.m ***********************)

(* First input FeynCalc *)

<< FeynCalc.m

(* These are some shorthands for the FeynCalc notation *) dm[mu_]:=DiracMatrix[mu,Dimension->D] dm[5]:=DiracMatrix[5]

ds[p_]:=DiracSlash[p] mt[mu_,nu_]:=MetricTensor[mu,nu] fv[p_,mu_]:=FourVector[p,mu]

epsilon[a_,b_,c_,d_]:=LeviCivita[a,b,c,d] id[n_]:=IdentityMatrix[n] sp[p_,q_]:=ScalarProduct[p,q] li[mu_]:=LorentzIndex[mu]

L:=dm[7]

R:=dm[6]

(* Tell FeynCalc to reduce the result to scalar functions *)

SetOptions[{B0,B1,B00,B11},BReduce->True]

(* Now write the numerator of the Feynman diagram. We define the constant

C= - alpha/(4 pi)

The minus sign comes from the photon propagator. The factor i/(16 pi^2) is already included in this definition.

num:= C dm[mu] . (ds[p]+ds[k]+m) . dm[mu]

(* Tell FeynCalc to evaluate the one-loop integral in dimension D *)

SetOptions[OneLoop,Dimension->D]

(* Define the amplitude *)

amp:= num \ FeynAmpDenominator[PropagatorDenominator[p+k,m], \

PropagatorDenominator[k]]

(* Calculate the result *)

res:=(-I / Pi^2) OneLoop[k,amp] ans=-res;

The minus sign in ans comes from the fact that -i \Sigma = diagram *)

(* Calculate the functions A(p^2) and B(p^2) *)

A=Coefficient[ans,DiracSlash[p],0];

B=Coefficient[ans,DiracSlash[p],1];

(* Calculate deltm *)

delm=A + m B /. p->m

(* Calculate delZ2 *)

Ap2 = A /. ScalarProduct[p,p]->p2

Bp2 = B /. ScalarProduct[p,p]->p2

aux=2 m D[Ap2,p2] + Bp2 \

+ 2 m^2 D[Bp2,p2] /. D[B0[p2,0,m^2],p2]->DB0[p2,0,m^2]

aux2= aux /. p2->m^2

aux3= aux2 /. A0[m^2]->m^2 (B0[m^2,0,m^2] -1)

delZ2=Simplify[aux3]

(******** ********* End of Program SelfEnergy.m ********************)

The output from Mathematica is:

			2		2
A = -(C (-2 m + 4 m B0[p , 0, m ]))
2		2		2	2	2	2
C (p	+ A0[m ] - (m			+ p ) B0[p , 0, m ])
B = -(-----------------------------------------							)
				2
			p
2		2	2	2		2
C (m - A0[m ] - 2 m			B0[m , 0, m ])
delm = ------------------------------------
		m
	2	2		2		2	2
delZ2= C (-2 + B0[m , 0, m ] - 4 m					DB0[m , 0, m ])

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