#### Derivation of Feynman rules for higher order poles using cross-ratio identities in CHY construction

Received: May
Derivation of Feynman rules for higher order poles using cross-ratio identities in CHY construction
Kang Zhou 0 3
Junjie Rao 0 1
Bo Feng 0 1 2
Open Access 0
c The Authors. 0
0 No. 38, Zheda Road, Hangzhou, 310027 , P.R. China
1 Zhejiang Institute of Modern Physics, Department of Physics, Zhejiang University
2 Center of Mathematical Science, Zhejiang University
3 School of Mathematical Sciences, Zhejiang University
In order to generalize the integration rules to general CHY integrands which include higher order poles, algorithms are proposed in two directions. One is to conjecture new rules, and the other is to use the cross-ratio identity method. In this paper,we use the cross-ratio identity approach to re-derive the conjectured integration rules involving higher order poles for several special cases: the single double pole, single triple pole and duplexdouble pole. The equivalence between the present formulas and the previously conjectured ones is discussed for the rst two situations.
Gauge Symmetry; Scattering Amplitudes
1 Introduction
Rule I: single double pole
CHY con guration
Some examples
Analytic proof
Rule II: single triple pole
CHY con guration
Derivation of new Feynman rules
Comparison with conjectured formula
Rule for duplex-double pole
CHY con guration
Introduction
An =
Qin=1 dzi
of quantum
eld theories as multi-dimensional contour integrals over auxiliary variables
scattering equations
b2f1;2;:::;ngnfag
sab = 0 ;
a = 1; 2; : : : ; n ;
where sab
2ka kb is the Mandelstam variable and zab is de ned as zab
zb. The
also depends on the speci c eld theory.
in [20], and we will prove their equivalence for two cases.
poles, regarding a simplest special case. A brief conclusion is given in section 6.
Preparation
1To distinguish them from lines in Feynman diagrams, we will call the latter \Feynman lines".
node, where a solid line is counted as +1 and a dashed line
For a set
containing j j points of zi, we call a line connecting two points in
internal line of , and a line connecting at least one point inside
and the other outside
the external line of . The number of internal lines of
is denoted by L[ ], and that of
The order of poles corresponding to the set
is de ned as
[ ] = L[ ]
[ ] = 2
which will be useful later. Then, we have the following corollaries:
A simple set has 4 external lines.
A double set has 2 external lines.
A triple set has 0 external lines.
a simple set.
contribute poles, then C[ ] is de ned as3
may contain many subsets which correspond to di erent poles. Similar to the
ence is, if the full set
contributes a pole, C[ ] must include this pole. For example, for
However, if the full set contributes a pole, C[ ] now should be
C[ ] =
C[ ] =
2Again, the number of lines are counted as +1 for a solid line and
1 for a dashed line.
further modi cation is needed.
associated with a subset
and its complement , thus we have
C[ ] =
2 are two branches (two subsets with
2 =
) associated with the
endpoints of subset . The summation is over all correct divisions of 1
, 2, and a special
division is denoted as h 1 2i.
The major machinery we use in this paper is the cross ratio identity
s =
s =
i2 =fpg j2 =fqg
given in [22]. Let us give some explanations of (2.6):
sions, some choices will simplify the calculation.
(2) We have double sums over all subsets
and zpq ( xed for all terms in the sum) between subsets
increase E[ ] and
E[ ] by 2, thus from (2.2), [ ] and [ ] are reduced by 1. Similarly, two numerators,
i.e., zip in the subset
and zjq in , reduce L[ ] and L[ ] by 1, thus from (2.1), [ ]
and [ ] are reduced by 1.
gauge choice, as well as the corresponding pole.
Rule I: single double pole
conjectured formula in [20] is given as
RIule[pA; pB; pC ; pD] =
that all subsets A have [A]
0, except one subset
[ ] = 1. Furthermore, for
above, we can give some statements of CHY con gurations.
First, we have
E[ ] = 2 ;
which means that there are two and only two lines connecting subset
and its complement
the two lines are given by a point a 2
connecting to , then
[ =fag] = L[ =fag]
1) = L[ ]
= L[ ]
1) = 1 ;
connecting point a and the subset
A 6=
, (3.3) contradicts with this assumption. This means that there are two points
and two points b1; b2 2
, such that there are one line connecting a1; b1 and
Second, we will show that there is no subset A
containing both a1; a2 satisfying
=A. Since [A] = 0,
and d0 belong to four di erent simple subsets, as shown in gure 1a.
Third, we will show that there is no subset
S , such that
0 = [ ] = L[ ]
1) = L[ ] + L[ ] + L[ ; ]
= (L[ ]
1)) + (L[ ]
1)) + (L[ ; ]
Since from our assumption [ ]
0, we have (L[ ; ]
0. With the
condition L[ ; ]
[ ] = 0 ;
[ ] = 0 ;
L[ ; ] = 2 =) a1; a2 2
which contradicts with the second observation in the previous paragraph.
to get the correct results.
factors in the denominator) connecting ; . To deal with this case, one can use, for example,
zabzdc =
(a) Original integrand
(b) Cross-ratio identity
red lines represent the term zziijaz0az0jcc00 provided by the cross-ratio identity.
contributions, the number of compatible subsets must be n
3. Since we have argued that there is no single pole = S
, all single poles are either in
can contain at most (j j
2) compatible poles, while
can contain at most
2) ones. Thus their combinations contribute only (n
4) compatible single poles,
and we need to include the double pole to get nonzero contributions.
Some examples
Feynman rules later although explicit examples will not be given.
I4;a =
z132z334z23z41
This example contains only one double pole s
= f1; 2g. There are two lines [14],
[23] connecting
di erent gauge choices, for the rst one [2; 3; ], we have
I4;a =
z132z334z23z41
s12 z122z324z223z421
For the second one [2; 4; ], we have
I4;a = z132z334z23z41
s12 z122z324z23z41z13z24
gauge choice, no new pole is introduced, so its calculation is simpler.
previous subsection, for a given Feynman diagram,
has been split to two subsets A1; A2
choice is [p 2 A2; q 2 B2; ], it is possible to nd two subsets
B2 satisfying
the following conditions
[ ] = 0 ;
[ ] = 0 ;
=) L[ ; ] = 1 :
Thus after multiplying the CHY integrand by sij zziijpzzpjqq , we have
] = L[ ] + L[ ] + L[ ; ] + 1
1) = 0 ;
p 2 A1; q 2 B2. Furthermore, since the splitting of
into A1; A2 is, in general, arbitrary,
a1; a2 (which connect to b1; b2 2
) from the second observation in the previous subsection,
choice for all Feynman diagrams, which can be either [a1; b2; ] or [a2; b1; ].6
The next example is the 5-point CHY integrand
I5;a =
z132z23z324z425z53z51
which, according to the Feynman rule (3.1), will lead to
RIule[f1g; f2g; f3; 4g; f5g] +
RIule[f1g; f2g; f3g; f4; 5g]
1 2p1p3 + 2p2p45 =
Now we derive this result using the cross ratio identity for the double pole s12 (so
= f1; 2g
have been canceled when summing all contributions together.
z132z23z324z425z53z51 s12
s12 z122z23z324z425z53z51z13z25
s12 z122z23z324z425z53z51z14z25
z132z23z324z425z53z51 s12
s12 z122z23z324z425z53z51z13z25
s12 z122z23z324z425z53z51z24z13
solution: summing up (3.14) and (3.15), we arrive
s13 + s25 + s24 =
which matches the Feynman rule (3.13).
\Feynman rules"; (2) Di erent Feynman rules come from di erent gauge choices.
analytic proof.
Analytic proof
and for all others [A]
0, we have the following statement:
diagrams contain the double pole s12 .
There are four special nodes a0; b0 2
connecting nodes a0; d0 and one connecting nodes b0; c0.8
, such that there is one line
8Again, this claim holds only if we assume the numerator is just 1.
Now we consider the Feynman rule with di erent gauge choices. For [a0; c0; ], the
corresponding cross ratio identity is
1 =
i2 =fa0g j2 =fc0g
as shown in gure 1b. Now we consider new CHY integrands Iorg zijza0c0
A S B =
and C S D =
, a0 2 A, b0 2 B, c0 2 C and d0 2 D)9 from these CHY
be divided into the following four parts:
G(a0; c0)I =
G(a0; c0)II =
G(a0; c0)III =
G(a0; c0)IV =
i2A=fa0g j2C=fc0g
i2A=fa0g j2D
i2B j2C=fc0g
zia0 zjc0 will not contain the pole s1A . Similarly, the numerator zjc0 will lead
to the fact that there is no pole s1C . Altogether, we
1
contribute to the pole structure sAsBsCsD . The same argument tells that, we can exclude
nd the G(a0; c0)I part will not
1 due to the numerator zia0 . In other
is given by
Summing over all possible splittings A; B; C; D, we get
X sij =
X 2pB 2 pD C[A]C[B]C[C]C[D];
a0 2 A =
=B ; b0 2 B ; c0 2 C =
=D ; d0 2 D ;
9As we have proven, four special nodes must be in four di erent corners.
in [22]. The discussion of sign is too complicated for us to give a general, simple argument.
coe cient
thus the other possible Feynman rule is
X sij =
X 2pA 2 pC C[A]C[B]C[C]C[D] ;
a0 2 A ; b0 2 B =
=A ; c0 2 C ; d0 2 D =
=C :
X 2pA pC + 2pB pD C[A]C[B]C[C]C[D];
a0 2 A ; b0 2 B =
=A ; c0 2 C ; d0 2 D =
=C ;
Averaging over these two contributions, we get the Feynman rule
! C S D) in order to get the correct nal answer.
Rule II: single triple pole
CHY con guration
The assumption of only single triple pole requires that all subsets A have [A]
one subset
con gurations.
First, the triple set
and its
complement subset . In other words, subset
give weight-4 graphes by themselves
gives a simple pole, so is B =
gives a simple pole, so is D =
and node n 2
. Furthermore, when split at the two ending points of internal
with B =
=A and C =
nontrivial numerators.
(a) Original integrand
(b) Cross-ratio identity
red lines represent the term zzijjnzz11ni provided by the cross-ratio identity.
such that
0 and L[ ; ] = 0. Finally, for
contribution like
s3 C[A]C[B]C[C]C[D] ;
where the factor X is what we need to derive for the Feynman rule.
Derivation of new Feynman rules
chose the gauge [1; n; ] to get
1 =
sij zjnzi1 =
as can be seen in gure 2b. For given (i; j), the CHY integrand Iorg zijz1n
zjnz1i is nothing but
lines between
. One may worry about the numerator zjn in
and z1i in . But as
(or n and j in the subset ). Thus, one can use the open-relation in [23] (eq. (3.4))
the previous result to write down
for the gauge choice [1; j; ] for the second step. Or
X 2pB 2 pD C[A]C[B]C[C]C[D] ;
X 2pA 2 pC C[A]C[B]C[C]C[D] ;
for the gauge choice [1; j; ]. Or
X 2pB 2 pD C[A]C[B]C[C]C[D] ;
X 2pA 2 pC C[A]C[B]C[C]C[D] ;
X (2pB pC )(2pB pD)
C[A]C[B]C[C]C[D] ;
X (2pB pC )(2pA pC )
C[A]C[B]C[C]C[D] ;
for the gauge choice [1; j; ]. Or
sum, the P
is allowed because we have
j2 nfng sij produces (2pB pC ). Also, changing of summation ordering
xed the rst gauge choice 1 2
for all splittings
choice leading to the rule (4.8).
12Again, we need to stick to the same rule for all splittings of hABCDi to get the right result.
For simplicity, let us assume that
= f1; 2; : : : ; mg,
= fm + 1; m + 2; : : : ; ng. The
A =
s3 C[A]C[B]C[C]C[D] ;
by average two di erent gauge choices in previous sections.
[1; n; 1; ], we get
which is given by
RIuIle[pA; pB; pC ; pD]
and (4.8).
hABCDi
(2pA pC )(2pA pD)+(2pB pC )(2pB pD)+(2pC pA)(2pC pB)+(2pD pA)(2pD pB)
where the rule (4.7) has been used. For the second gauge choice [1; n; n; ], we get
we need to consider two di erent cases. When m is also in the subset A, we get
(2pB pC )(2pB pD)
(2pB pC )(2pA pC )
C[A]C[B]C[C]C[D] ;
C[A]C[B]C[C]C[D] ;
(2pB pC )(2pB pD)
C[A]C[B]C[C]C[D] :
When m is not in the subset A, we get
(2pA pC )(2pA pD)
C[A]C[B]C[C]C[D] :
When (m + 1) 62 D, we get
n;(m+1)2D
n2D ;(m+1)62D
(2pD pA)(2pD pB)
When (m + 1) 2 D, we get
(2pA pC )(2pA pD)
C[A]C[B]C[C]C[D]
(2pB pC )(2pB pD) (2pA pC )(2pA pD)
C[A]C[B]C[C]C[D] :
Summing over (4.16) and (4.17), the expression of the fourth gauge choice is
C[A]C[B]C[C]C[D]
(2pB pC )(2pA pC ) (2pD pA)(2pD pB)
C[A]C[B]C[C]C[D] :
(2pB pC )(2pA pC )
C[A]C[B]C[C]C[D] :
(2pD pA)(2pD pB)
C[A]C[B]C[C]C[D] :
n;(m+1)2D
in (4.10) as
A =
X =
(2pA pC )(2pA pD) + (2pB pC )(2pB pD) + (2pC pA)(2pC pB) + (2pD pA)(2pD pB) :
(2pB pC )(2pA pC ) (2pD pA)(2pD pB)
C[A]C[B]C[C]C[D]
(2pB pC )(2pB pD) (2pA pC )(2pA pD)
C[A]C[B]C[C]C[D] ;
= s (sA sB) + K1 ;
and similarly
We will prove that
as follows
(2pB pC)(2pB pD) (2pA pC)(2pA pD)
(2pB pC)(2pB pD)
sD(2pC pD + 4pB pC)
= (2pC pD)(sA
sB) 4sCsD sC(2pC pD + 4pB pD) sD(2pC pD + 4pB pC)
Thus to reproduce the rule (4.9), we need to show that
m+1;n2D
s (sA sB)C[A]C[B] =
s (sD sC)C[C]C[D] =
(p2A p2B)2 + (p2C p2 )2 + 29(p2A + p2B)(p2C + p2D)
D
C[A]C[B]C[C]C[D]
X (2pB pC)(2pB pD) (2pA pC)(2pA pD)C[A]C[B]C[C]C[D] : (4.21)
n;(m+1)2D
Putting (4.26), (4.27) and (4.28) back to (4.21), we see the identity is proved.
other words, we will have
sB)C[A]C[B] =
C[A1]C[A2]C[B]
C[A]C[B1]C[B2] ;
If B contains only one point, we have
sB)C[A]C[B] =
C[A1]C[A2]C[B] :
of several C[ 1]C[ 2]C[ 3]'s, where 1
, 2 and 3 are simple sets satisfy 1 S
2 S
3 = .
under a special division h 1 2 3i. There are several cases:
fA1 =
1; A2 =
3; B =
2g and fA =
1; B1 =
2; B2 =
3g, thus, after summing
over all correct divisions, the coe cient of C[ 1]C[ 2]C[ 3] is 1 + 1
1 = 1. Notice
1; B1 =
3; B2 =
2g gives the same
situation B =
2 S
3 as fA =
1; B1 =
2; B2 =
3g, so we chose only one of them.
An alternative way is to add them together and divide by the symmetry factor 2.
fA1 =
1; A2 =
2; B =
3g, thus the coe cient is 1.
Thus, the coe cient of any C[ 1]C[ 2]C[ 3] is 1, we get
sB)C[A]C[B] = s
One may observe that (4.30) can be ignored when counting the coe cient of
the structure of B is not important. The set
can always be divided into three simple
two simple subsets) will be neglected automatically.
side, i.e., the (sA
sB)2C[A]C[B] part. By similar argument, we can write it as
(sA sB)2C[A]C[B] = (sA sB)@
C[A1]C[A2]C[B]
C[A]C[B1]C[B2]A:
If one of A and B is a single point, one term in the bracket vanishes.
sB)2C[A]C[B] also provides several C[ 1]C[ 2]C[ 3]'s. Let us count the coe cient
of C[ 1]C[ 2]C[ 3] with 1 2
1, under the summation P(sA
sB)2C[A]C[B] which is over
fA1 =
1; A2 =
then the coe cient is
s 2 3 ) = s 1 2 3 = s :
hABi, We get
which has proven the relation (4.26).
sB)C[A]C[B] =
sB)2C[A]C[B] = s
Exactly similar argument shows
m+1;n2D
s (sD sC )C[C]C[D] =
(sC sD)2C[C]C[D] = s
Then we turn to the most di cult relation (4.28). We divide P K1C[A]C[B]C[C]C[D]
relation (4.27).
into following three parts:
Y1 =
Y2 =
Y3 =
Y1 =
and will treat them one by one.
First, using (4.31) we get
4sC sDC[A]C[B]C[C]C[D] ;
(sC + sD)C[C]C[D]
(sC + sD)(sB
sA)C[A]C[B]C[C]C[D] ;
sC (2pC pD +4pB pD)+sD(2pC pD +4pB pC ) C[A]C[B]C[C]C[D] ;
1; D2 = 2; C = 3g,
Y1 = 3
Secondly, considering Y2 gives
Y2 =
(2pC pD + 4pB pD)
C[C1]C[C2]C[D]
+ (2pC pD + 4pB pC )
C[C]C[D1]C[D2] C[A]C[B] :
C[C1]C[C2]C[D] +
C[C]C[D1]C[D2]A
Now we x A, B, and count the coe cient of C[ 1]C[ 2]C[ 3] with n 2
1 under the
sum+ 4p 3 pB + 2p 1 3 p 2
+ 4p 2 pB + 2p 2 3 p 1
2(sA sB)+2s 1 +2s 2 +2s 3 C[A]C[B] C[ 1]C[ 2]C[ 3]
C[ 1]C[ 2]C[ 3] C[ 1]C[ 2]C[ 3]
C[A]C[B]
C[ 1]C[ 21]C[ 22]C[ 3] +
C[ 1]C[ 2]C[ 31]C[ 32] ;
Y1, thus the coe cient is
Substituting it into (4.41) we get
Y2 =
[ 1]C[ 2]C[ 3]C[ 4] with n 2 1 and
coe cient under the summation over divisions
1 S 2 S 3 S 4 =
1 2 3 can be counted by following six
f 11 =
1; 12 =
4; 2 =
2; 3 =
3g, f 1 =
; 21 =
; 22 =
3; 3 =
C[ 1]C[ 2]C[ 3] C[ 1]C[ 2]C[ 3]
C[A]C[B]
Y2 =
as a new quintic vertex.
Finally, we consider Y3 to obtain
Y3 =
C[A]C[B]
C[C1]C[C2]
C[D1]C[D2] ; (4.45)
Considering the coe cient of C[ 1]C[ 2]C[ 3]C[ 4] with n 2
under the summation over divisions hCDi, one can
nd three correct con gurations:
1; D2 =
3; C1 =
2; C2 =
Y3 =
C[A]C[B]
and the result is given as
Similar calculation gives
n;m+12D
K1C[A]C[B]C[C]C[D]
C[ 1]C[ 2]C[ 3] C[ 1]C[ 2]C[ 3] :
K2C[A]C[B]C[C]C[D]
C[ 1]C[ 2]C[ 3] C[ 1]C[ 2]C[ 3] :
To nish our proof, let us we consider the last term of relation (4.28)
(sA + sB)(sC + sD)C[A]C[B]C[C]C[D]
C[A1]C[A2]C[B] +
C[A]C[B1]C[B2]
C[C1]C[C2]C[D] +
C[C]C[D1]C[D2] ;
which leads
n;(m+1)2D
Thus, we have
This ends the proof of relation (4.28).
(sA + sB)(sC + sD)C[A]C[B]C[C]C[D]
have shown that the nal result of the integration can be expressed as
XC[A]C[B]C[C]C[D]
gration rule. Now instead of just cubic vertexes like h
! ABi and
new quartic vertexes like h
! CD , there are
Rule for duplex-double pole
rule is conjectured to be
RIuIlIe[pA; pB; pE; pC ; pD] =
(2pA pD)(2pB pC )
(2pA pC )(2pB pD)
(p2E)(2pA pD + 2pB pC
CHY con guration
E =
poles are simple poles except two double poles s 1 and s 2 ( 1 T
2 = ;). Furthermore, to
2 is just a single node e. With above speci cations, we can derive the following
are two points a; b having a line connecting to subset
1. Similarly, for the subset
2, there are two points c; d having a line connecting to subset 2
as can be seen in gure 4a.
the same argument.
b 2 B, and similarly
2 will split into two simple corners C and D with c 2 C, d 2 D,
(4) Using (3.5), there is no single pole
with true subsets
(5) Using (3.5), there is no single pole
S E with the true subset
reason is that if [ ] = 0, then L[ ; E]
1 since b; a can not belong to same single
pole. In other words, we will have L[ ; E] + [ ]
1. Similarly there is no single pole =
S E with the true subset
(6) For the case
S E with true subsets
with L[ ; ] = 0, L[ ; E] + [ ]
1 and L[ ; E] + [ ]
1, we get [ ] < 0 always,
i.e., there is no such a simple pole.
1 and 2. The reason is that there are at most (j 1j 2) compatible combinations
2) compatible combinations from
2, thus at most we can get (n
number (n
3) for all poles. In other words, all allowed Feynman diagrams will contain
picture, the integrated result should be13
s2 1 s2 2 C[A]C[B]C[C]C[D]C[E] ;
1; b 2 B =
1=A; c 2 2=D; d 2 D
Now we need to determine the expression X to get the Feynman rule.
identity is given by
1 =
i2 1nfag j2 2nfdg
where we have split the sum j 2
simple poles. Firstly, it can not create new simple poles of the form
S E with
(a) Original integrand
(b) Cross-ratio identity: the
(c) Cross-ratio identity: the
red lines represent terms zzijjdzzaiad and zzieedzzaiad provided by the cross-ratio identity.
true subsets of either
1 or 2. Secondly, it can not create new simple poles of the form
the true subsets of 1 and
2 respectively. Using (3.5), one see that after
not create new simple poles of the form
the true subsets of 1 and
straightforwardly. Putting all considerations together, the rst part of (5.4) gives
C[Ae]C[Be] = @ s 1 hABi
C[A]C[B]A C[E] :
C[A]C[B]C[C]C[D]C[E] :
Now we consider the second part of (5.4). Because of the factor zzieedzzaiad , especially the
denominator zie, it is easy to see from (3.5) that now the subset
points in e1 and d; c are the special points in e2. Using result (3.21) we get
S E =
2 and e2 =
(2pB pE )
C[Ae]C[Be]C[C]C[D] ;
where we have used the fact e2 =
2. Now coming to the key point: the allowed splitting
1. Furthermore,
the Ae split into two corners A; B and we have
Putting it back, we get
T2;1 =
(2pB pE) (2(pA + pB) pC ) 1
C[A]C[B]C[C]C[D]C[E] :
C[Ae] =
PhABi C[A]C[B] . Here node e can be attached to some propagators in the
Putting (5.5), (5.8) and (5.9) together with proper sign, we get the nal result14
T2;2 =
(2pA pC )s 1 C[A]C[B]C[C]C[D]C[E] :
2
which implies that the X in (5.3) is given by
XhABCDi = (2pB pC )s 2
(2pB pE)(2(pA + pB) pC )
these propagators Bt in the Feynman diagram
, as can be seen in gure 5b. Now we can
write down the expression when node e is attached to propagator Bt as
T ;Bt = <
>> sBEsB1E:::sBtEsBt sBt+1 :::sBm 1
t = m :
sBtE, sBt 1E, sBt 2E until sBE . Thirdly,
= Q
t Bt is the other part of propagators,
which are not a ect by node e, as shown in gure 5. It will be the same for all T ;Bt .
14The relative minus sign is because for the second part we have inserted another cross ratio identity.
sBm−1
(a) The Feynmann diagram
(b) e attaches to the propagator sB1t
of the sequence Bm
B and the node
e attaches to this diagram.
Having obtained the expression, we can carry out the sum. It is easy to see that
T ;Bm 1 +T ;Bm =
sBE sB1E : : : sBm 2E sBm 1E
sBE sB1E : : : sBm 2E sBm 1E
sBE sB1E : : : sBm 2E sBm 1
Adding T ;Bm 2 , we get
T ;Bm 2 + T ;Bm 1 + T ;Bm
sBE sB1E : : : sBm 2E
sBE sB1E : : : sBm 2E
Now we can see the recursive pattern, which leads to
sBE sB1E : : : sBm 3E sBm 2 sBm 1
X T ;Bt =
sBsB1 : : : sBm 3 sBm 2 sBm 1
out when summing over all terms.
second splitting of e1 as
T2;2 =
e1=A S Be
2pA pC C[A]C[C]C[D] X
X T ;Bt
at the wanted result (5.9).
Conclusion
combinations are gauge invariant.
new poles. Thus we have provided a treatment of new poles for this rule.
analytic calculation.
draft has been edited.
Open Access.
This article is distributed under the terms of the Creative Commons
any medium, provided the original author(s) and source are credited.
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