diagram

# One-loop same helicity YM amplitudes from BG currents

Pratik Chattopadhyay and Kirill Krasnov

School of Mathematical Sciences, University of Nottingham, NG7 2RD, UK

(October 2021)

###### Abstract

We propose and prove a new formula for the one-loop all same helicity Yang-
Mills amplitudes. These amplitudes are seen to arise as a sum of products of
two tree-level Berends-Giele currents connected by an effective propagator. To
make sense of the propagators one needs to introduce the so-called region, or
dual momenta. The formula is proven by observing that it readily implies the
correct collinear limit properties. The only non-trivial part of the proof is
establishing that our formula for the amplitude is invariant under shifts of
the region momenta.

##  1 Introduction

The purpose of this paper is to propose a new formula for the same helicity
one-loop Yang-Mills amplitudes. A general formula for such amplitudes is known
from [1], [2], where it was conjectured and then proven by checking that it
satisfies the required collinear properties. An alternative derivation follows
from the results of [3].

The structure of the formula we propose is rather different from that in [2].
We show that the same helicity one-loop YM amplitudes can be built from the
same helicity Berends-Giele (tree-level) off-shell currents connected by
certain effective propagators. Importantly, the propagators can only be made
sense of if one introduces a parametrisation in terms of region momenta.

The new expression for the one-loop amplitudes is as follows. Consider a
colour ordered diagram with n external lines. Because the diagram is colour
ordered we can adopt the use of region momenta $p_{i}$ so that the null
momenta on the external legs are given by the difference of the region momenta

$\displaystyle k_{i}=p_{i}-p_{i-1},~{}~{}~{}p_{n}\equiv p_{0}.$ (1)

This ensures momentum conservation, but introduces indeterminacy in that all
region momenta can be shifted by an arbitrary amount without changing the
external momenta.

The amplitude can then be written as a sum over partitions of the set of
external states into two groups. Each group of states has a Berends-Giele (BG)
[4] current associated with it, and two such currents are glued by a certain
effective propagator. Berends-Giele currents are vectorial objects, but their
vector structure is simple and universal, see (8) below. Let $J(1,\ldots,n)$
be the most interesting scalar part of the current given in terms of the
momentum spinors as well as the reference spinor, see (9). The one-loop
amplitude can then be written pictorially as follows

$\displaystyle{\cal
A}=\sum_{part}\begin{gathered}~{}~{}~{}\fmfgraph*(120,70)\fmfleft{i,i1,i2,i3,i4}\fmfright{o1,o4,o5,o6,o7}\fmf{vanilla}{i,v}\fmf{vanilla}{i4,v}\fmfblob{.15w}{v}\fmf{vanilla,label=$p_j$,label.distance=1cm,label.side=left}{v,v1}\fmfdot{v1}\fmf{vanilla,label=$p_{i-1}$,label.distance=1cm,label.side=right}{v1,o}\fmfblob{.15w}{o}\fmf{vanilla}{o,o1}\fmf{vanilla}{o,o7}\fmflabel{$i$}{i}\fmflabel{$.$}{i1}\fmflabel{$.$}{i2}\fmflabel{$.$}{i3}\fmflabel{$j$}{i4}\fmflabel{$i-1$}{o1}\fmflabel{$.$}{o4}\fmflabel{$.$}{o5}\fmflabel{$.$}{o6}\fmflabel{$j+1$}{o7}\end{gathered}\quad\quad=\sum_{part}J(i,..,j)J(j+1,..,i-1)\langle
q|p_{j}\circ p_{i-1}|q\rangle^{2}.$ (3)

The large blobs in the pictorial representation of the amplitude stand for the
BG currents, with the off-shell legs pointing towards the internal line of
this diagram. The other quantities are as follows. The object $q$ is the
reference (auxiliary) spinor. The sum is taken over partitions of a cyclically
ordered set $1,\ldots,n$ with the convention $n+1=1$ into two subgroups
$i,\ldots,j$ and $j+1,\ldots,i-1$. The bra-ket notation is standard and
represents a particular 2-spinor contraction. The notation $p\circ q$
represents a contraction of two rank two spinors $(p\circ
q)_{M}{}^{N}:=p_{M}{}^{M^{\prime}}q_{M^{\prime}}{}^{N}$, so that the result
can be inserted between two copies of the unprimed reference spinor $q_{M}$.
The momenta $p_{j},p_{i-1}$ are the region momenta for the two regions that
are separated by the internal line that partitions the two groups of external
states. It is thus clear that the object $\langle q|p_{j}\circ
p_{i-1}|q\rangle^{2}$ plays the role of the effective propagator connecting
the BG currents, and that this propagator can only be made sense of if region
momenta are introduced. There is also a version of our formula that is written
in terms of the vector-valued BG currents that are connected by an effective
gluon propagator. This will be spelled out in the main text.

The formula (3) is proven by establishing that it has the correct collinear
properties of a one-loop same helicity amplitude. This follows quite easily
from the construction of the amplitude if one assumes that the right-hand side
of (3) is invariant under shifts of all region momenta by the same amount. So,
the difficulty of the proof of the formula (3) shifts to the proof that it is
independent of a choice of the region momenta. These arguments are given in
Section 4. We also explicitly check that the new formula gives the correct
known expressions in the case of four and five particles.

There are several previous works that are strongly related to the context of
this work. First, Zvi Bern (unpublished) observed that in the computation of
the same helicity one-loop amplitude in Yang-Mills theory, the sum of box,
triangle and bubble integrands, with a particular choice of the loop momenta,
vanishes. We reproduce this argument in the Appendix, in our notations, for
completeness. This has later been explored by [5] in the context of world-
sheet formulation of YM, and by [6] in the related light-cone formalism. There
is also a very closely related discussion, again in the light-cone, in the
context of chiral higher spin theories in [7]. The upshot of these works is
that the same helicity one-loop amplitudes in YM can be obtained by performing
tree-level calculations, but with insertions of certain helicity violating
”bubble” counterterms. Our new formula (3) is the same observation phrased in
the covariant formalism. In particular, we will see that the effective
propagator in (3) arises as the result of a ”bubble” computation.

The organisation of the rest of this paper is as follows. We start by
reviewing in Section 2 the covariant Feynman rules of the self-dual Yang-Mills
theory. This theory is a truncation of the full YM, and has simpler Feynman
rules. At the same time the amplitudes that are of interest to us coincide in
the full and SDYM, and so it is sufficient to work with an easier truncation.
We also review in this section the by now standard results on the YM BG
currents, and also review the previously known formulas for the same helicity
one-loop amplitude. Section 3 computes the bubble diagram and thus extracts
the form of the effective propagator to be used in the construction of the
one-loop amplitude. We do checks of the new formula in Section 4. We end with
a discussion. Appendix spells out the details of the computation that shows
that the sum of integrands of the box, triangle and bubble diagrams, with a
specific consistent choice of the loop momentum, vanishes. This shows that the
4-point amplitude can be re-written in terms of the bubble insertions. Our
formula (3) is the generalisation of this statement to an arbitrary number of
external gluons.

##  2 Review of SDYM theory

Self-dual Yang-Mills is a truncation of the full YM theory that keeps only a
subset of the Feynman diagrams of the latter. For some processes, for example
those involving mostly gluons of same helicity, only the Feynman diagrams
already present in SDYM can be seen to contribute in the full YM calculations.
This means that some of the processes can be computed in the simpler SDYM
theory. In particular, the same helicity one-loop amplitudes of the full YM
are correctly captured by the SD theory. Our presentation in this section
follows [8], where we also direct the reader for more details.

###  2.1 Self-dual Yang-Mills

The most useful covariant formulation of SDYM is due to Clamers and Siegel
[9]. We will state the action in 2-component spinor notations, which are used
throughout this paper. Our 2-component spinor conventions are spelled out in
the Appendix of [8]. The action reads

$\displaystyle
S[\phi,A]=\int\Phi^{a\,AB}(\partial_{A}{}^{A^{\prime}}A^{a}_{A^{\prime}B}+f^{abc}A^{b}_{A}{}^{A^{\prime}}A^{c}_{A^{\prime}B}).$
(4)

The field $\Phi^{a\,AB}$ is a symmetric rank two spinor
$\Phi^{a\,AB}=\Phi^{a\,(AB)}$ with values in the Lie algebra of some (compact)
group $G$, and $a,b,c$ are the Lie algebra indices. The capital Latin letters
(unprimed and primed) at the 2-component Lorentz spinor indices. The field
$\Phi^{a\,AB}$ is an auxiliary field the variation of the action with respect
to which imposes the equation that the self-dual part of the field strength of
the connection $A^{a}_{AA^{\prime}}$ vanishes.

The action can be expanded around any self-dual connection configuration. The
simplest case is to expand around the zero connection. The first term is then
the kinetic term, whose inverse, after an appropriate gauge-fixing, gives the
propagator of the theory. The second term is a simple cubic non-derivative
interaction. The full YM is obtained from this theory by appending to the
action a term quadratic in $\Phi^{a\,AB}$, see e.g. [8].

The gauge-fixing of the above action is most usefully done by prolonging the
auxiliary field $\Phi^{a\,AB}$ to contain also the $AB$ anti-symmetric part.
The variation with respect to this anti-symmetric part imposes the sharp
Landau-type gauge $\partial^{AA^{\prime}}A^{a}_{AA^{\prime}}=0$. The gauge-
fixed kinetic operator is a version of the (chiral) Dirac operator, is non-
degenerate, and can be inverted. The propagator of the theory thus connects
the $\Phi$ and $A$ legs of the vertices, and is, schematically, in momentum
space $1/k$. The operator in the denominator is the Dirac operator (in the
momentum space). Using the fact that Dirac operator squares to the Laplacian,
the propagator can be rewritten as $k/k^{2}$, where the Dirac operator is now
in the numerator, and the denominator contains the usual factor of $k^{2}$.
For details on the gauge-fixing procedure, see [8].

The variation of the (linearised) action with respect to the auxiliary field
gives the field equation $\partial_{A}{}^{A^{\prime}}A^{a}_{A^{\prime}B}=0$.
The solutions describe one of the two gluon helicity, in our conventions
negative. The corresponding polarisation vector is

$\displaystyle\epsilon_{-}^{MM^{\prime}}(k)=\frac{q^{M}k^{M^{\prime}}}{\langle
kq\rangle}.$ (5)

Here $\langle\lambda\nu\rangle=\lambda^{M}\nu_{M}$ is the spinor contraction
of two 2-component spinors $\lambda^{M},\nu^{M}$. The object $q^{M}$ is an
auxiliary spinor changing which amounts to gauge-transformations. The object
$k^{M^{\prime}}$ is the momentum spinor arising as the square root of a null
momentum $k^{MM^{\prime}}=k^{M}k^{M^{\prime}}$. The other gluon helicity is
described by the field $\Phi$, but we will never consider processes with this
other helicity on the external lines in this paper, and so we do not need to
define it.

###  2.2 BG current for all negative helicity states

The sum of all (colour-ordered) tree-level Feynman diagrams with the negative
helicity states inserted on all legs except one, the later being kept off-
shell, can be computed explicitly using the Berends-Giele recursion [4]. Using
the Feynman rules reviewed in the previous subsection, the computation is very
simple and is spelled out in e.g. [8]. The result is as follows.

Let us start by writing the single particle negative helicity state in the
form of a BG current. We have

$\displaystyle\epsilon_{-}^{MM^{\prime}}(k)=J(k)q^{N}k_{N}{}^{M^{\prime}}q^{M}\equiv
J(k)\langle q|\langle q|k|,$ (6)

where $k_{MM^{\prime}}=k_{M}k_{M^{\prime}}$ is the (null) momentum of the
particle and

$\displaystyle J(k)=\frac{1}{\langle qk\rangle\langle kq\rangle}.$ (7)

The BG current is then defined to be the sum of all (colour-ordered) Feynman
diagrams, with negative helicity states of momenta $k_{1},\ldots,k_{n}$
inserted (from the top), and the single remaining off-shell leg (at the
bottom). The convention is that the propagator is included in the off-shell
leg. It can then be shown by a recursion that all currents have the same form

$\displaystyle
J^{MM^{\prime}}(k_{1},\ldots,k_{n})=J(k_{1},\ldots,k_{n})q^{N}(\sum_{i=1}^{n}k_{i\,N}{}^{M^{\prime}})q^{M}\equiv
J(k)\langle q|\langle q|\sum_{i}k_{i}|,$ (8)

where

$\displaystyle J(k_{1},\ldots,k_{n})=\frac{1}{\langle q1\rangle\langle
12\rangle\langle 23\rangle\ldots\langle(n-1)n\rangle\langle nq\rangle},$ (9)

and we use the convention in which $k_{1}\equiv 1$ etc.

These BG currents could be converted to tree-level scattering amplitudes, but
all such amplitudes are easily seen to vanish. This is because they need to be
multiplied by the factor of $k^{2}$ for the off-shell leg, which then must be
taken on-shell. Thus, all BG currents get multiplied by the factor of zero in
the process of converting them to amplitudes. Thus, all tree-level amplitudes
in SDYM vanish.

###  2.3 Literature representation for one-loop amplitudes

The only non-vanishing amplitudes in SDYM are the one-loop ones, with all
external lines being states of negative helicity. The all same helicity one
loop amplitudes $A_{n}(1^{+},2^{+},...n^{+})$ in YM are finite rational
functions of the momenta involved and has cyclical symmetry in the arguments.
These amplitudes are singular in the region where two adjacent momenta become
collinear or a momentum become soft. These amplitudes were explicitly computed
at 4 and 5 points in [10], [1], and then an expression conjectured for them at
arbitrary multiplicity. This was then proved in [2] by showing that the
proposed formula has the correct collinear properties of a one-loop amplitude,
and also by [3] by an explicit computation in a simpler theory of massless
QED.

The amplitudes at 4 and 5 points are given by

$\displaystyle\mathcal{A}_{4}(1^{+},2^{+},3^{+},4^{+})=\frac{s_{12}s_{23}}{\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 41\rangle},$ (10)
$\displaystyle\mathcal{A}_{5}(1^{+},2^{+},3^{+},4^{+},5^{+})=\frac{s_{12}s_{23}+s_{23}s_{34}+s_{34}s_{45}+s_{45}s_{51}+s_{51}s_{12}+\epsilon(1,2,3,4)}{\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 45\rangle\langle
51\rangle},$ (11)

where $\epsilon(i,j,m,n)=[ij]\langle jm\rangle[mn]\langle ni\rangle-\langle
ij\rangle[jm]\langle mn\rangle[ni]$. In this paper we are not interested in
the overall multiplicative factors in the amplitudes, and these are omitted.

There are many other equivalent ways to write the 5-point amplitude. For
example, the paper [1] contains the following more compact expression

$\displaystyle\mathcal{A}_{5}(1^{+},2^{+},3^{+},4^{+},5^{+})$
$\displaystyle={s_{12}s_{23}+s_{45}s_{51}-\langle 23\rangle[34]\langle
45\rangle[52]}.$ (12)

The notation that is used everywhere is $s_{ij}=\langle ij\rangle[ij]$.

The formula that is valid at an arbitrary multiplicity is [3], [2]

$\displaystyle\mathcal{A}_{n}(1^{+},2^{+},....n^{+})=\sum_{1\leq
i_{1}<i_{2}<i_{3}<i_{4}\leq n}\frac{\langle
i_{1}i_{2}\rangle[i_{2}i_{3}]\langle i_{3}i_{4}\rangle[i_{4}i_{1}]}{\langle
12\rangle\langle 23\rangle....\langle n1\rangle}.$ (13)

##  3 Bubble computation and effective propagator

The purpose of this section is to spell out the computation of the one-loop
”bubble” correction to the self-energy diagram. We will see that while this
diagram can be argued to vanish, it is in fact undefined. If the loop momentum
is shifted, the difference between the unshifted and shifted value of this
diagram is non-zero an can be computed. Thus, this diagram must in general be
assigned a non-zero value. In our formula for the one-loop amplitudes it is
this diagram, with a specific prescription for the value of the shifted
momentum, that is inserted between the two BG currents.

###  3.1 Shift computation

(200,100) i o v u photon,tension=3,label=-ki,v1 photonv1,u vanillau,v2
photonv2,v vanillav,v1 photon,tension=3,label=kv2,o lu l+kv

Consider the self-energy diagram as pictured above, where the external lines
are projected to two negative helicity states and the convention being all
external momenta incoming. Using the Feynman rules reviewed in the previous
section, and the bra-ket spinor notation, the amplitude can be written as

$i\Pi^{--}=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l|k]\langle
q|l+k|k]}{l^{2}(l+k)^{2}\langle qk\rangle^{2}}.$ (14)

In the form it is written this integral can be argued to vanish. Indeed, there
is no linear in $l$ part of the integrand as it is proportional to $\langle
q|k|k]=\langle qk\rangle[kk]=0$. The only non-vanishing contribution thus
comes from

$\displaystyle\int\frac{d^{4}l}{(2\pi)^{4}}\frac{l_{\mu}l_{\nu}}{l^{2}(l+p)^{2}}$
(15)

Any Lorentz invariant regularisation of this will yield $x_{\mu}x_{\nu}$ to be
proportional to $k^{2}\eta_{\mu\nu}$ which is zero because $k$ is null or to
$k_{\mu}k_{\nu}$, which gives the numerator factor $\langle q|k|k]=0$ by using
$[kk]=0$. Thus, the self-energy diagram (14) can be argued to be zero.

However, (14) is a quadratically divergent integral, and so one must be
careful in reaching the conclusion that this object is zero. Let us consider
shifting the loop momentum as in $l=x+\tilde{s}$, where $\tilde{s}$ is some
momentum and $x$ is the new integration variable. The argument above depends
on the specific form of the integrand and is no longer applicable to the
shifted integrand. In fact, below we shall compute the effect of the shift by
$\tilde{s}$ and see that the shift is non-vanishing. What this means is that
the self-energy diagram projected onto two negative helicity states cannot in
general be assumed to vanish. Instead, it is given by a finite quantity,
depending on the shift parameter.

Let us compute the shift dependence of the self-energy diagram. We will use
the region momenta so that $k=s-\tilde{s}$, and

$\displaystyle\begin{split}l&=x+\tilde{s}\\\ l+k&=x+s.\end{split}$ (16)

We then have

$\displaystyle i\Pi^{--}=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|x+\tilde{s}|k]\langle q|x+s|k]}{(x+\tilde{s})^{2}(x+s)^{2}\langle
qk\rangle^{2}}$ (17)

We have already see that this integral vanishes after the shift $x\rightarrow
x-\tilde{s}$. We then compute the result of the shift. This is done using the
standard techniques, which are reviewed in the Appendix of [17]. The linear
part of the shift is given by

$\displaystyle-i\lim_{x\to\infty}\int\frac{d\Omega}{(2\pi)^{4}}\tilde{s}_{\mu}x^{\mu}\frac{\langle
q|x+\tilde{s}|k]\langle
q|x+s|k]}{x^{2}}\Bigg{(}1-\frac{2x.(\tilde{s}+s)}{x^{2}}\Bigg{)}$ (18)

The non-zero contribution can only come from the quadratic and quartic in $x$
terms. The quadratic term is

$\displaystyle\tilde{s}_{\mu}x^{\mu}\Big{(}\langle q|x|k]\langle
q|s|k]+\langle q|x|k]\langle q|\tilde{s}|k]\Big{)}$ (19)

Integrating over the directions of $x^{\mu}$ produces

$\displaystyle-\frac{i}{32\pi^{2}}\Big{(}\langle q|\tilde{s}|k]\langle
q|s|k]+\langle q|\tilde{s}|k]\langle q|\tilde{s}|k]\Big{)}$ (20)

The quartic in $x$ part is given by

$\displaystyle
2i\lim_{x\to\infty}\int\frac{d\Omega}{(2\pi)^{4}}\tilde{s}_{\mu}x^{\mu}(\tilde{s}+s)_{\nu}x^{\nu}\frac{\langle
q|x|k]\langle q|x|k]}{x^{4}}$ (21)

The integral is computed using

$\displaystyle\int\frac{d\Omega}{(2\pi)^{4}}\frac{x_{\mu}x_{\nu}x_{\rho}x_{\sigma}}{x^{4}}=\frac{1}{32.6\pi^{2}}(\eta_{\mu\nu}\eta_{\rho\sigma}+\eta_{\mu\rho}\eta_{\nu\sigma}+\eta_{\mu\sigma}\eta_{\rho\nu})$
(22)

This results in the following two contributions

$\displaystyle\begin{split}\frac{i}{32.3\pi^{2}}\Big{(}\langle
q|\tilde{s}|k]\langle q|\tilde{s}+s|k]+\langle q|\tilde{s}+s|k]\langle
q|\tilde{s}|k]\Big{)}=\frac{i}{16.3\pi^{2}}\Big{(}\langle
q|\tilde{s}|k]\langle q|\tilde{s}|k]+\langle q|\tilde{s}|k]\langle
q|s|k]\Big{)}\end{split}$ (23)

For the quadratic part of the shift, the integral is given by

$\displaystyle\frac{i}{2}\lim_{x\to\infty}\int\frac{d\Omega}{(2\pi)^{4}}\tilde{s}_{\mu}\tilde{s}_{\nu}x^{\mu}x^{2}\frac{\partial}{\partial
x_{\nu}}\frac{\langle q|x+\tilde{s}|k]\langle
q|x+s|k]}{(x+\tilde{s})^{2}(x+s)^{2}}$ (24)

When the derivative hits the denominator, it produces a factor proportional to

$\displaystyle\frac{i}{2}(-4)\lim_{x\to\infty}\int\frac{d\Omega}{(2\pi)^{4}}\tilde{s}_{\mu}\tilde{s}_{\nu}x^{\mu}x^{\nu}\frac{\langle
q|x+\tilde{s}|k]\langle q|x+s|k]}{x^{4}}$ (25)

The quartic in $x$ part of the numerator is the only which contributes. Using
(22) we find one of the contractions vanish and the other two contractions are
equal, giving

$\displaystyle-\frac{2i}{32.3\pi^{2}}\langle q|\tilde{s}|k]\langle
q|\tilde{s}|k]$ (26)

When the derivative in (24) hits the numerator, in the large $x$ limit, we get

$\displaystyle\frac{i}{2}\lim_{x\to\infty}\int\frac{d\Omega}{(2\pi)^{4}}\tilde{s}_{\mu}x^{\mu}\frac{\langle
q|x+\tilde{s}|k]\langle q|\tilde{s}|k]+\langle q|x+s|k]\langle
q|\tilde{s}|k]}{x^{2}}$ (27)

Using the relevant contraction, this gives

$\displaystyle\frac{i}{32\pi^{2}}\langle q|\tilde{s}|k]^{2}$ (28)

Adding all the contributions, we have for this amplitude

$\displaystyle\Pi^{--}=-\frac{i}{32.3\pi^{2}}\frac{\langle
q|\tilde{s}|k]\langle q|s|k]}{\langle qk\rangle^{2}}.$ (29)

Using the fact that $k=s-\tilde{s}$ we could write this result in terms of
only $s$ or $\tilde{s}$, but the form we chose will be most convenient below.

###  3.2 Bubble as an effective propagator

As we have just seen, a direct computation of the shift gives the result (29)
for the self-energy diagram. This result can be given the following
interpretation. We represent it as the effective propagator

$\displaystyle\begin{gathered}\fmfgraph*(140,70)\fmfleft{i}\fmfright{o}\fmf{vanilla}{i,v1}\fmfdot{v1}\fmf{vanilla}{o,v1}\fmflabel{$MM'$}{i}\fmflabel{$NN'$}{o}\end{gathered}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}:=~{}~{}~{}~{}~{}s^{~{}M}_{N^{\prime}}\tilde{s}_{N}^{~{}M^{\prime}},$
(31)

into which the polarisation state

$\displaystyle\epsilon^{-}_{MM^{\prime}}(k)=\frac{q_{M}k_{M^{\prime}}}{\langle
qk\rangle}$ (32)

is inserted on both sides. This results in (29).

##  4 New formula

It is now easy to see that the the sum of two copies of the Berends-Giele
currents (8) connected by the effective propagator (31) where $s,\tilde{s}$
stand for the region momenta separated by the propagator line, is given by the
right-hand side of the formula (3). We then conjecture that this gives the
correct formula for the one-loop same helicity YM amplitudes. A proof of this
conjecture, as well as explicit computations in the case of 4 and 5 points are
collected in this section.

###  4.1 Collinear limit

####  4.1.1 Review of the collinear limit properties

In [2], the form (13) of the one-loop amplitude has been fixed using the
requirement that it has the correct collinear limit behaviour. The general
form of the collinear limit is, see [2]

$\displaystyle\mathcal{A}^{\textrm{loop}}_{n;1}\xrightarrow{a\parallel
b}\sum_{\lambda=\pm}\textrm{Split}^{\textrm{tree}}_{-\lambda}(a^{\lambda_{a}},b^{\lambda_{b}})\mathcal{A}^{\textrm{loop}}_{n-1;1}(....(a+b)^{\lambda},...)$
$\displaystyle+\textrm{Split}^{\textrm{loop}}_{-\lambda}(a^{-},b^{-})\mathcal{A}^{\textrm{tree}}_{n-1}(....,(a+b)^{\lambda},...).$
(33)

In the same helicity sector the second term in (4.1.1) drops out because the
split factor in this case multiples the same helicity tree amplitude that
vanishes. Then, for the case of the all same helicity one loop amplitudes, the
collinear limit takes the simple form

$\displaystyle\mathcal{A}^{\textrm{one-
loop}}_{n}(1^{-},2^{-},...,n^{-})\xrightarrow{a\parallel
b}\textrm{Split}^{\textrm{tree}}_{-}(a^{-},b^{-})\mathcal{A}^{\textrm{one-
loop}}_{n-1}(....(a+b)^{-},...),$ (34)

where the split function in this case is given by

$\displaystyle\textrm{Split}_{-}^{\textrm{tree}}(a^{-},b^{-})=\frac{1}{\sqrt{z(1-z)}\langle
ab\rangle}.$ (35)

Here $z$ is a parameter that describes how the two momenta $a,b$ become
collinear, such that $a\rightarrow zP$ and $b\rightarrow(1-z)P$, where $P$ is
null momentum.

####  4.1.2 Verification of the correct collinear properties of the new
formula

Our expression (3) has the required collinear properties by construction.
Indeed, we first note that in a colour-ordered amplitude only two adjacent
momenta can contribute to the collinear singularity. Consider a pair $a,b$
that is colour-adjacent and goes collinear. In the sum in (3) this pair
belongs to two types of terms. In the first type, both $a,b$ belong to the
same Berends-Giele current. As we shall see in a moment, Berends-Giele
currents have the correct collinear properties. So, these terms satisfy the
collinear properties automatically.

The second type of terms is when $a,b$ belong to two groups of momenta that in
(3) are separated by the bubble insertion. Let $a=j,b=j+1$. Let $p_{j}$ be the
region momentum between $a,b$. Now, by region momentum independence of the
amplitude (that should be established separately), we can set to zero any one
of the region momenta. Let us choose to set to zero $p_{j}=0$. It is then easy
to see that all the terms where $a,b$ do not belong to the same BG current get
multiplied by $\langle q|p_{j}\circ p_{i}|q\rangle^{2}$, where $p_{i}$ is some
other region momentum. Thus, with our choice $p_{j}=0$ all these terms do not
contribute. All other terms have $a,b$ belong to the same BG current, i.e. all
other terms are those of the first type. This shows that one can always write
the one-loop amplitude in a form that exhibits the $a,b$ collinear property
manifestly.

To complete the argument we can see that

$\displaystyle
J(1,\ldots,j,j+1,\ldots,n)\rightarrow\frac{1}{\sqrt{z(1-z)}\langle
ab\rangle}J(1,\ldots,P,\ldots,n),$ (36)

so that the BG currents have the collinear property manifest. This follows
from the formula (9) and the fact that the momentum spinors for $a,b$ are
given by

$\displaystyle\lambda_{a}\rightarrow\sqrt{z}\lambda_{P},~{}~{}~{}~{}\lambda_{b}\rightarrow\sqrt{1-z}\lambda_{P}.$
(37)

The only property that remains to be shown is the momentum region independence
of the formula (3). We do this in the following subsection.

###  4.2 Alternative ways of writing the formula

Using the fact that

$\displaystyle p_{j}=p_{i-1}+\sum_{l=i}^{j}k_{l},$ (38)

we can rewrite the formula (3) lowering the power of the region momenta in it

$\displaystyle{\cal A}=\sum_{part}J(i,\ldots,j)J(j+1,\ldots,i-1)\langle
q|p_{j}\circ(\sum_{l=i}^{j}k_{l})|q\rangle^{2}.$ (39)

We can also rewrite the sum over partitions as a sum over cyclic permutations
of the set $1,\ldots,n$. Indeed, it is easy to check that

$\displaystyle 2{\cal A}=\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)\langle
q|p_{i}\circ(k_{i+1}+\ldots+k_{n})|q\rangle^{2}+{\rm cyclic},$ (40)

where we need to add all cyclic permutations of the set $(1,\ldots,n)$. This
form of the formula is particularly convenient for estabslishing the region
momentum independence. Written in this way the amplitude formula is very
similar to the one that appears in [7] in the light-cone gauge.

###  4.3 Region momentum independence

The purpose of this subsection is to argue that the amplitude is invariant
under shifts of all region momenta by the same amount. Combined with our
collinear limit argument, this gives a proof of the formula (3).

####  4.3.1 Quadratic part of the dependence

When we shift all region momenta by some value $x$ there are both quadratic
and linear in $x$ terms that appear. Using (40), the part quadratic in the
shift can be written as

$\displaystyle{\cal
A}^{x^{2}}=\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)\langle
q|x\circ(k_{i+1}+\ldots+k_{n})|q\rangle^{2}+{\rm cyclic}.$ (41)

Because $x$ here is an arbitrary vector, so is the primed spinor $\langle
q|x|:=\mu$. This means that we must consider

$\displaystyle\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)[\mu|(k_{1}+\ldots+k_{i})|q\rangle[\mu|k_{i+1}+\ldots+k_{n}|q\rangle+{\rm
cyclic},$ (42)

where we wrote the expression more symmetrically. This can be computed using
the identity

$\displaystyle\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)[\mu|k_{1}+\ldots+k_{i}|q\rangle[\mu|k_{i+1}+\ldots+k_{n}|q\rangle=\frac{[\mu|\sum_{i<j}i\circ
j|\mu]}{\langle q1\rangle\langle 12\rangle\ldots\langle(n-1)n\rangle\langle
nq\rangle},$ (43)

which holds for arbitrary momenta $1,\ldots n$. The momenta in this formula
are not assumed to add up to zero. This formula is proven analogously to how
the recursive formula for the Berends-Giele currents is established.

We will also need the identity

$\displaystyle\sum_{i=1}^{n-1}\frac{\langle i(i+1)\rangle}{\langle
iq\rangle\langle(i+1)q\rangle}=\frac{\langle 1n\rangle}{\langle
1q\rangle\langle nq\rangle},$ (44)

which is a simple consequence of Schouten identity. It can also be written as

$\displaystyle\sum_{i=1}^{n}\frac{\langle i(i+1)\rangle}{\langle
iq\rangle\langle(i+1)q\rangle}=0,$ (45)

with the convention that $n+1=1$. Using this identity we have

$\displaystyle\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)[\mu|(k_{1}+\ldots+k_{i})|q\rangle[\mu|k_{i+1}+\ldots+k_{n}|q\rangle+{\rm
cyclic}=$ (46) $\displaystyle\frac{2}{\langle
12\rangle\ldots\langle(n-1)n\rangle\langle
n1\rangle}\left(\sum_{i<j}\frac{[\mu|i\circ j|\mu](ij)}{\langle
iq\rangle\langle jq\rangle}\right).$

No momentum conservation has yet been used. It is then easily checked that
when the momentum conservation e.g. in the form $-n=1+\ldots+(n-1)$ is used,
the coefficients in front of independent $[\mu|i\circ j|\mu]$ factors with
$i,j=1,\ldots,n-1$ vanish. Thus, the quadratic in $x$ part of dependence of
the amplitude on the region momentum vanishes.

####  4.3.2 Linear part of the dependence

Taking the first variation of the amplitude as all region momenta vary, and
denoting $\langle q|x|=[\mu|$ as before, we get a multiple of

$\displaystyle\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)[\mu|(k_{1}+\ldots+k_{i})|q\rangle\langle
q|p_{i}\circ(k_{i+1}+\ldots+k_{n})|q\rangle+{\rm cyclic}.$ (47)

The idea is again to compute the sum here explicitly, similar to what one does
in the check of the Berends-Giele formula for the all same helicity currents.
This is an exercise in applying Schouten identity. The result is

$\displaystyle\sum_{i=1}^{n-1}J(1,\ldots,i)J(i+1,\ldots,n)[\mu|(k_{1}+\ldots+k_{i})|q\rangle\langle
q|p_{i}\circ(k_{i+1}+\ldots+k_{n})|q\rangle=$ (48)
$\displaystyle\frac{1}{\langle 1q\rangle\langle
12\rangle\ldots\langle(n-1)n\rangle\langle
nq\rangle}\left(\sum_{i=1}^{n-1}\sum_{j>i}[\mu|i\circ j\circ
p_{i}|q\rangle-\sum_{i=1}^{n-2}[\mu|i|q\rangle\sum_{j>i}^{n}\sum_{l>j}^{n}s_{jl}\right),$

where the last terms contain Mandelstam variables $s_{ij}:=\langle
ij\rangle[ij]$ and arise from relating the region momenta to each other via
relations of the type $p_{i+1}=p_{i}+k_{i+1}$.

It remains to add the cyclic permutations, and then apply the momentum
conservation. Taking the cyclic permutation of the first set of terms in
brackets in (48), and using (44) gives

$\displaystyle-\sum_{i=1}^{n-1}\sum_{j>i}\frac{\langle ij\rangle}{\langle
iq\rangle\langle jq\rangle}([\mu|i\circ j\circ p_{i}|q\rangle-[\mu|j\circ
i\circ p_{j}|q\rangle),$ (49)

where from now on we omit the common prefactor

$\displaystyle\frac{1}{\langle 12\rangle\ldots\langle(n-1)n\rangle\langle
n1\rangle}.$ (50)

We now use the momentum conservation to express the last momentum $k_{n}$ in
terms of all the rest. After this, we collect the terms in front of similar
$[\mu|i\circ j\circ p|q\rangle$ expressions. Using (44) one more time we get
for these terms

$\displaystyle-\sum_{i=1}^{n-1}\sum_{j=1}^{n-1}\frac{\langle
in\rangle}{\langle iq\rangle\langle nq\rangle}[\mu|j\circ
i\circ(p_{n}-p_{j})|q\rangle.$ (51)

The sum here is taken over $i\not=j$. This is now written in terms of
differences of region momenta, and so depends just on the external momenta
$p_{n}-p_{j}=-(k_{1}+\ldots+k_{j})$. We can thus rewrite the above as

$\displaystyle\sum_{i=1}^{n-1}\sum_{j=1,j\not=i}^{n-1}\sum_{l=1}^{j}\frac{\langle
in\rangle}{\langle iq\rangle\langle nq\rangle}[\mu|j\circ i\circ l|q\rangle.$
(52)

Let us now consider the quantity $[\mu|j\circ i\circ l|q\rangle$ where $l$ is
one of the momenta. We can exchange $i\circ l=-l\circ i-s_{il}{\mathbb{I}}$,
where $s_{il}$ is the Mandelstam variable. On the other hand, we have

$\displaystyle\frac{\langle in\rangle}{\langle iq\rangle\langle
nq\rangle}[\mu|j\circ l\circ i|q\rangle=\frac{1}{\langle nq\rangle}[\mu
j]\langle jl\rangle[li]\langle in\rangle=\frac{\langle jn\rangle}{\langle
jq\rangle\langle nq\rangle}[\mu|j|q\rangle\frac{\langle j|l\circ
i|n\rangle}{\langle jn\rangle}.$ (53)

This is linear in $i$, and so the sum over $i$ can be easily taken. Using the
momentum conservation gives

$\displaystyle\sum_{i=1,i\not=j}^{n-1}\frac{\langle jn\rangle}{\langle
jq\rangle\langle nq\rangle}[\mu|j|q\rangle\frac{\langle j|l\circ
i|n\rangle}{\langle jn\rangle}=-\frac{\langle jn\rangle}{\langle
jq\rangle\langle nq\rangle}[\mu|j|q\rangle\frac{\langle
j|l\circ(j+n)|n\rangle}{\langle jn\rangle}=\frac{\langle jn\rangle}{\langle
jq\rangle\langle nq\rangle}[\mu|j|q\rangle s_{lj}.$ (54)

Using these identities (52) becomes

$\displaystyle-\sum_{i=1}^{n-1}[\mu|i|q\rangle\sum_{j=1}^{n-1}\frac{\langle
jn\rangle}{\langle jq\rangle\langle nq\rangle}\sum_{l=1}^{i}s_{jl}$ (55)

Let us now do the same manipulations with the second set of terms in (48). We
add cyclic permutations, then use the momentum conservation to express
$[\mu|n|q\rangle$ in terms of the other quantities of this type, use (44) as
well as momentum conservation for the Mandelstam variables. We get precisely
the terms in (55) with opposite signs, so these terms cancel each other and
the linear part is zero. This finishes the proof.

###  4.4 4-point amplitude

The first non-trivial case is that of four gluons. For 3 gluons our formula
produces an expression that can be rewritten in terms of Mandelstan variables,
and these vanish at 3 points.

In the case of 4 gluons, substituting the expressions for the currents we get

$\displaystyle\mathcal{A}_{4}=\frac{1}{\langle q1\rangle\langle
1q\rangle}\frac{1}{\langle q2\rangle\langle 23\rangle\langle 34\rangle\langle
4q\rangle}\langle q|p_{1}\circ p_{4}|q\rangle^{2}+\frac{1}{\langle
q2\rangle\langle 2q\rangle}\frac{1}{\langle q3\rangle\langle 34\rangle\langle
41\rangle\langle 1q\rangle}\langle q|p_{2}\circ p_{1}|q\rangle^{2}+$ (56)
$\displaystyle\frac{1}{\langle q3\rangle\langle 3q\rangle}\frac{1}{\langle
q4\rangle\langle 41\rangle\langle 12\rangle\langle 2q\rangle}\langle
q|p_{3}\circ p_{2}|q\rangle^{2}+\frac{1}{\langle q4\rangle\langle
4q\rangle}\frac{1}{\langle q1\rangle\langle 12\rangle\langle 23\rangle\langle
3q\rangle}\langle q|p_{4}\circ p_{3}|q\rangle^{2}+$
$\displaystyle\frac{1}{\langle q1\rangle\langle 12\rangle\langle
2q\rangle}\frac{1}{\langle q3\rangle\langle 34\rangle\langle 4q\rangle}\langle
q|p_{4}\circ p_{2}|q\rangle^{2}+\frac{1}{\langle q2\rangle\langle
23\rangle\langle 3q\rangle}\frac{1}{\langle q4\rangle\langle 41\rangle\langle
1q\rangle}\langle q|p_{3}\circ p_{1}|q\rangle^{2}.$

We know from general grounds that this expression must be region momentum and
auxiliary spinor $q$ independent, and should match the known 4-point
amplitude. But it is instructive to see how this happens.

We parametrise all region momenta in terms of one of them, e.g. $p_{1}=x$, and
the external momenta. We have

$\displaystyle p_{1}=x,\quad p_{2}=2+x,\quad p_{3}=3+2+x,\quad p_{4}=x-1.$
(57)

We can then drop all terms containing $x$ because these terms vanish to render
the result region momentum independent. This collapses the result to

$\displaystyle\mathcal{A}_{4}=\frac{1}{\langle q3\rangle\langle
3q\rangle}\frac{1}{\langle q4\rangle\langle 41\rangle\langle 12\rangle\langle
2q\rangle}\langle q|3\circ 2|q\rangle^{2}+\frac{1}{\langle q4\rangle\langle
4q\rangle}\frac{1}{\langle q1\rangle\langle 12\rangle\langle 23\rangle\langle
3q\rangle}\langle q|1\circ 4|q\rangle^{2}+$ (58)
$\displaystyle\frac{1}{\langle q1\rangle\langle 12\rangle\langle
2q\rangle}\frac{1}{\langle q3\rangle\langle 34\rangle\langle 4q\rangle}\langle
q|1\circ 2|q\rangle^{2}=\frac{\langle 2q\rangle[32]^{2}}{\langle
4q\rangle\langle 12\rangle\langle 41\rangle}+\frac{\langle
1q\rangle[14]^{2}}{\langle 12\rangle\langle 23\rangle\langle
3q\rangle}+\frac{\langle 1q\rangle\langle 2q\rangle[12]^{2}}{\langle
3q\rangle\langle 12\rangle\langle 34\rangle\langle 4q\rangle}.$

We now eliminate $q$ dependence using the momentum conservation

$\displaystyle\frac{[32]}{\langle 41\rangle}=\frac{[12]}{\langle
34\rangle},\qquad\frac{[14]}{\langle 23\rangle}=\frac{[12]}{\langle
34\rangle}.$ (59)

This gives

$\displaystyle\mathcal{A}_{4}=\frac{[12]}{\langle 12\rangle\langle
34\rangle\langle 3q\rangle\langle 4q\rangle}\left(\langle 3q\rangle\langle
2q\rangle[32]+\langle 1q\rangle\langle 4q\rangle[14]+\langle 1q\rangle\langle
2q\rangle[12]\right)=$ (60) $\displaystyle\frac{[12][43]}{\langle
12\rangle\langle 34\rangle}=\frac{[12][23]}{\langle 34\rangle\langle
41\rangle}=\frac{s_{12}s_{23}}{\langle 12\rangle\langle 23\rangle\langle
34\rangle\langle 41\rangle},$

which is the correct answer (10) for this amplitude.

###  4.5 Five point amplitude

The expression for the amplitude in terms of currents is

$\displaystyle{\mathcal{A}}_{5}=J(1)J(2,3,4,5)\langle q|p_{1}\circ
p_{5}|q\rangle^{2}+J(2)J(3,4,5,1)\langle q|p_{2}\circ
p_{1}|q\rangle^{2}+J(3)J(4,5,1,2)\langle q|p_{3}\circ p_{2}|q\rangle^{2}$ (61)
$\displaystyle+J(4)J(5,1,2,3)\langle q|p_{4}\circ
p_{3}|q\rangle^{2}+J(5)J(1,2,3,4)\langle q|p_{5}\circ p_{4}|q\rangle^{2}$
$\displaystyle+J(1,2)J(3,4,5)\langle q|p_{2}\circ
p_{5}|q\rangle^{2}+J(2,3)J(4,5,1)\langle q|p_{3}\circ
p_{1}|q\rangle^{2}+J(3,4)J(5,1,2)\langle q|p_{4}\circ p_{2}|q\rangle^{2}$
$\displaystyle+J(4,5)J(1,2,3)\langle q|p_{5}\circ
p_{3}|q\rangle^{2}+J(5,1)J(2,3,4)\langle q|p_{1}\circ p_{4}|q\rangle^{2}.$

Again, we know that it must reproduce the known answer, but would like to see
explicitly how this happens. This requires much more work as compared to the
4-point case.

####  4.5.1 Extracting the region momentum independent result

We again parametrise the region momenta in terms of one of them, and the
external momenta

$\displaystyle p_{1}=x,\quad p_{2}=2+x,\quad p_{3}=3+2+x,\quad
p_{4}=4+3+2+x,\quad p_{5}=x-1.$ (62)

All terms containing $x$ must drop out by region momentum independence. This
gives the following expression

$\displaystyle{\mathcal{A}}_{5}=J(3)J(4,5,1,2)\langle q|3\circ
2|q\rangle^{2}+J(4)J(5,1,2,3)\langle
q|4\circ(3+2)|q\rangle^{2}+J(5)J(1,2,3,4)\langle q|1\circ 5|q\rangle^{2}$ (63)
$\displaystyle+J(1,2)J(3,4,5)\langle q|2\circ
1|q\rangle^{2}+J(3,4)J(5,1,2)\langle q|(4+3)\circ
2|q\rangle^{2}+J(4,5)J(1,2,3)\langle q|(4+5)\circ 1|q\rangle^{2}.$

Substituting the expressions for the currents we get

$\displaystyle{\mathcal{A}}_{5}=\frac{[23]^{2}\langle 2q\rangle}{\langle
4q\rangle\langle 45\rangle\langle 51\rangle\langle
12\rangle}+\frac{([41]\langle 1q\rangle+[45]\langle 5q\rangle)^{2}}{\langle
5q\rangle\langle 51\rangle\langle 12\rangle\langle 23\rangle\langle
3q\rangle}+\frac{[15]^{2}\langle 1q\rangle}{\langle 12\rangle\langle
23\rangle\langle 34\rangle\langle 4q\rangle}$ (64)
$\displaystyle+\frac{[12]^{2}\langle 1q\rangle\langle 2q\rangle}{\langle
3q\rangle\langle 12\rangle\langle 34\rangle\langle 45\rangle\langle
5q\rangle}+\frac{([21]\langle 1q\rangle+[25]\langle 5q\rangle)^{2}\langle
2q\rangle}{\langle 3q\rangle\langle 4q\rangle\langle 5q\rangle\langle
34\rangle\langle 51\rangle\langle 12\rangle}+\frac{([14]\langle
4q\rangle+[15]\langle 5q\rangle)^{2}\langle 1q\rangle}{\langle
3q\rangle\langle 4q\rangle\langle 5q\rangle\langle 45\rangle\langle
12\rangle\langle 23\rangle}.$

Let us start by bringing it all to the common denominator

$\displaystyle{\mathcal{A}}_{5}=\frac{1}{\langle 3q\rangle\langle
4q\rangle\langle 5q\rangle\langle 12\rangle\langle 23\rangle\langle
34\rangle\langle 45\rangle\langle 51\rangle}\times$
$\displaystyle\Big{(}[23]^{2}\langle 2q\rangle\langle 3q\rangle\langle
5q\rangle\langle 23\rangle\langle 34\rangle+[15]^{2}\langle 1q\rangle\langle
3q\rangle\langle 5q\rangle\langle 45\rangle\langle 51\rangle+[12]^{2}\langle
1q\rangle\langle 2q\rangle\langle 4q\rangle\langle 23\rangle\langle 51\rangle$
$\displaystyle+([41]\langle 1q\rangle+[45]\langle 5q\rangle)^{2}\langle
4q\rangle\langle 34\rangle\langle 45\rangle+([23]\langle 3q\rangle+[24]\langle
4q\rangle)^{2}\langle 2q\rangle\langle 23\rangle\langle 45\rangle+([14]\langle
4q\rangle+[15]\langle 5q\rangle)^{2}\langle 1q\rangle\langle 34\rangle\langle
51\rangle\Big{)}.$

We then expand the squares and collect the terms next to common square bracket
factors. One then notices that such terms can be rewritten more compactly
using Schouten identity

$\displaystyle[15]^{2}\langle 1q\rangle\langle 3q\rangle\langle
5q\rangle\langle 45\rangle\langle 51\rangle+[15]^{2}\langle
5q\rangle^{2}\langle 1q\rangle\langle 34\rangle\langle
51\rangle=[15]^{2}\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle\langle
51\rangle\langle 35\rangle$ (65) $\displaystyle[23]^{2}\langle
2q\rangle\langle 3q\rangle\langle 5q\rangle\langle 23\rangle\langle
34\rangle+[23]^{2}\langle 3q\rangle^{2}\langle 2q\rangle\langle
23\rangle\langle 45\rangle=[23]^{2}\langle 2q\rangle\langle 3q\rangle\langle
4q\rangle\langle 23\rangle\langle 35\rangle$ $\displaystyle[14]^{2}\langle
1q\rangle^{2}\langle 4q\rangle\langle 34\rangle\langle
45\rangle+[14]^{2}\langle 4q\rangle^{2}\langle 1q\rangle\langle
34\rangle\langle 51\rangle=[14]^{2}\langle 1q\rangle\langle 4q\rangle\langle
5q\rangle\langle 34\rangle\langle 41\rangle.$

This gives for the amplitude

$\displaystyle{\mathcal{A}}_{5}=\frac{1}{\langle 3q\rangle\langle
4q\rangle\langle 5q\rangle\langle 12\rangle\langle 23\rangle\langle
34\rangle\langle 45\rangle\langle 51\rangle}\times$ (66)
$\displaystyle\Big{(}s_{23}[23]\langle 2q\rangle\langle 3q\rangle\langle
4q\rangle\langle 35\rangle+s_{15}[15]\langle 1q\rangle\langle 4q\rangle\langle
5q\rangle\langle 53\rangle+[12]^{2}\langle 1q\rangle\langle 2q\rangle\langle
4q\rangle\langle 23\rangle\langle 51\rangle$ $\displaystyle+s_{45}[45]\langle
5q\rangle^{2}\langle 4q\rangle\langle 34\rangle+[24]^{2}\langle
4q\rangle^{2}\langle 2q\rangle\langle 23\rangle\langle
45\rangle+s_{14}[14]\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle\langle
43\rangle$ $\displaystyle 2s_{45}[41]\langle 1q\rangle\langle 5q\rangle\langle
4q\rangle\langle 34\rangle+2s_{23}\langle 3q\rangle[24]\langle
4q\rangle\langle 2q\rangle\langle 45\rangle+2s_{15}[14]\langle
4q\rangle\langle 5q\rangle\langle 1q\rangle\langle 43\rangle\Big{)},$

where we used the notation $s_{ij}:=\langle ij\rangle[ij]$.

To understand the steps that follow we start by writing the amplitude that we
want to reproduce. Our starting point is the form (12) of the amplitude. It is
clear that in this expression there are terms that can be written in terms of
Mandelstam variables, but there is always a remainder that cannot be written
in this way. In the formula (12) this is the last term. This term, however,
can be written in many different ways. Let us first massage it into the form
that will be useful later.

We use the momentum conservation in the form $-\langle 23\rangle[34]=\langle
21\rangle[14]+\langle 25\rangle[54]$ to rewrite

$\displaystyle\langle 12\rangle\langle 23\rangle\langle 34\rangle\langle
45\rangle\langle
51\rangle{\mathcal{A}}_{5}=s_{12}s_{23}+s_{45}s_{51}+s_{25}s_{45}+\langle
21\rangle[14]\langle 45\rangle[52].$ (67)

Finally, we use Schouten identity in the last term to rewrite the amplitude as

$\displaystyle\langle 12\rangle\langle 23\rangle\langle 34\rangle\langle
45\rangle\langle
51\rangle{\mathcal{A}}_{5}=s_{12}s_{23}+s_{45}s_{51}+s_{25}s_{45}+s_{25}s_{14}+\langle
24\rangle[14]\langle 15\rangle[52].$ (68)

The idea now is to see which of the terms in the amplitude (66) can reproduce
the last term in (68). Most of the terms in (66) already contain factors of
Mandelstam variables, and so cannot be responsible for this term. The only
terms that can be responsible are the ones containing $[12]^{2}$ and
$[24]^{2}$. To massage these terms into the desired form we use

$\displaystyle-[21]\langle 1q\rangle=[23]\langle 3q\rangle+[24]\langle
4q\rangle+[25]\langle 5q\rangle.$ (69)

This gives, using the momentum conservation in terms proportional to $[24]$

$\displaystyle[12]^{2}\langle 1q\rangle\langle 2q\rangle\langle
4q\rangle\langle 23\rangle\langle 51\rangle+[24]^{2}\langle
4q\rangle^{2}\langle 2q\rangle\langle 23\rangle\langle 45\rangle=$ (70)
$\displaystyle s_{23}[21]\langle 2q\rangle\langle 3q\rangle\langle
4q\rangle\langle 15\rangle+[21][25]\langle 2q\rangle\langle 4q\rangle\langle
5q\rangle\langle 23\rangle\langle 15\rangle-s_{23}[24]\langle
4q\rangle^{2}\langle 2q\rangle\langle 35\rangle.$

We then use for the middle term

$\displaystyle[21]\langle 23\rangle=[14]\langle 43\rangle+[15]\langle
53\rangle,$

to get

$\displaystyle[12]^{2}\langle 1q\rangle\langle 2q\rangle\langle
4q\rangle\langle 23\rangle\langle 51\rangle+[24]^{2}\langle
4q\rangle^{2}\langle 2q\rangle\langle 23\rangle\langle 45\rangle=$
$\displaystyle s_{23}[21]\langle 2q\rangle\langle 3q\rangle\langle
4q\rangle\langle 15\rangle-s_{23}[24]\langle 4q\rangle^{2}\langle
2q\rangle\langle 35\rangle+[25][14]\langle 43\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle\langle 15\rangle+s_{15}[25]\langle 53\rangle\langle
2q\rangle\langle 4q\rangle\langle 5q\rangle.$

As the last step, we extract the $q$-independent part of the third term using
Schouten identity.

$\displaystyle[12]^{2}\langle 1q\rangle\langle 2q\rangle\langle
4q\rangle\langle 23\rangle\langle 51\rangle+[24]^{2}\langle
4q\rangle^{2}\langle 2q\rangle\langle 23\rangle\langle
45\rangle=[52][14]\langle 24\rangle\langle 15\rangle\langle 3q\rangle\langle
4q\rangle\langle 5q\rangle$ (71) $\displaystyle+s_{23}[21]\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle\langle 15\rangle-s_{23}[24]\langle
4q\rangle^{2}\langle 2q\rangle\langle 35\rangle+s_{15}[25]\langle
53\rangle\langle 2q\rangle\langle 4q\rangle\langle 5q\rangle$
$\displaystyle+[25][14]\langle 23\rangle\langle 4q\rangle^{2}\langle
5q\rangle\langle 15\rangle.$

The term on the right-hand side of the first line (after dividing by the
$q$-dependent terms in the denominator) is precisely the last term in (68)
that can not be written in terms of Mandelstam variables. The term on the last
line can also be written in terms of Mandelstam variables. Indeed, we first
use Schouten identity $\langle 15\rangle\langle 4q\rangle=\langle
14\rangle\langle 5q\rangle-\langle 1q\rangle\langle 54\rangle$ to write

$\displaystyle[25][14]\langle 23\rangle\langle 4q\rangle^{2}\langle
5q\rangle\langle 15\rangle=s_{14}[25]\langle 23\rangle\langle 4q\rangle\langle
5q\rangle^{2}-[25][14]\langle 23\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle\langle 54\rangle.$ (72)

We then use $-[25]\langle 54\rangle=[21]\langle 14\rangle+[23]\langle
34\rangle$ to finally get

$\displaystyle[25][14]\langle 23\rangle\langle 4q\rangle^{2}\langle
5q\rangle\langle 15\rangle=s_{14}[25]\langle 23\rangle\langle 4q\rangle\langle
5q\rangle^{2}+s_{14}[21]\langle 23\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle+s_{23}[14]\langle 34\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle.$ (73)

It thus remains to reproduce the other $s$-containing terms in the formula
(68) for the amplitude. We substitute the terms in the second and third line
of (71) into (66) instead of the $[12]^{2},[24]^{2}$ terms. This gives a part
of the amplitude that is supposed to contain all terms with Mandelstam
variables

$\displaystyle\langle 3q\rangle\langle 4q\rangle\langle 5q\rangle\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 45\rangle\langle
51\rangle{\mathcal{A}}^{\prime}_{5}=s_{23}[23]\langle 35\rangle\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle+s_{15}[15]\langle 53\rangle\langle
1q\rangle\langle 4q\rangle\langle 5q\rangle$ (74)
$\displaystyle+s_{45}[45]\langle 34\rangle\langle 4q\rangle\langle
5q\rangle^{2}+s_{14}[14]\langle 43\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle$ $\displaystyle+2s_{45}[14]\langle 43\rangle\langle
1q\rangle\langle 4q\rangle\langle 5q\rangle+2s_{23}[24]\langle
45\rangle\langle 2q\rangle\langle 3q\rangle\langle
4q\rangle+2s_{15}[14]\langle 43\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle$ $\displaystyle+s_{23}[21]\langle 15\rangle\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle-s_{23}[24]\langle 35\rangle\langle
2q\rangle\langle 4q\rangle^{2}+s_{15}[25]\langle 53\rangle\langle
2q\rangle\langle 4q\rangle\langle 5q\rangle$ $\displaystyle+s_{14}[25]\langle
23\rangle\langle 4q\rangle\langle 5q\rangle^{2}+s_{14}[21]\langle
23\rangle\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle-s_{23}[14]\langle
43\rangle\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle.$

There are some immediate simplifications. The terms containing $s_{23}\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle$ simplify using $[23]\langle
35\rangle+[21]\langle 15\rangle=-[24]\langle 45\rangle$. The terms containing
$[14]\langle 43\rangle\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle$
simplify using $s_{14}+2s_{45}+2s_{15}-s_{23}=s_{23}-s_{14}$, and so

$\displaystyle\langle 3q\rangle\langle 4q\rangle\langle 5q\rangle\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 45\rangle\langle
51\rangle{\mathcal{A}}^{\prime}_{5}=s_{23}[24]\langle 45\rangle\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle$ (75)
$\displaystyle+s_{15}[15]\langle 53\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle+s_{45}[45]\langle 34\rangle\langle 4q\rangle\langle
5q\rangle^{2}$ $\displaystyle-s_{23}[24]\langle 35\rangle\langle
2q\rangle\langle 4q\rangle^{2}+s_{15}[25]\langle 53\rangle\langle
2q\rangle\langle 4q\rangle\langle 5q\rangle$ $\displaystyle+s_{14}[25]\langle
23\rangle\langle 4q\rangle\langle 5q\rangle^{2}+s_{14}[21]\langle
23\rangle\langle 1q\rangle\langle 4q\rangle\langle
5q\rangle+(s_{23}-s_{14})[14]\langle 43\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle.$

The two terms in the last line containing $s_{14}\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle$ simplify using $-[12]\langle 23\rangle-[14]\langle
43\rangle=[15]\langle 53\rangle$, and so

$\displaystyle\langle 3q\rangle\langle 4q\rangle\langle 5q\rangle\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 45\rangle\langle
51\rangle{\mathcal{A}}^{\prime}_{5}=s_{23}[24]\langle 45\rangle\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle$ (76)
$\displaystyle+(s_{15}+s_{14})[15]\langle 53\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle+s_{45}[45]\langle 34\rangle\langle 4q\rangle\langle
5q\rangle^{2}$ $\displaystyle-s_{23}[24]\langle 35\rangle\langle
2q\rangle\langle 4q\rangle^{2}+s_{15}[25]\langle 53\rangle\langle
2q\rangle\langle 4q\rangle\langle 5q\rangle$ $\displaystyle+s_{14}[25]\langle
23\rangle\langle 4q\rangle\langle 5q\rangle^{2}+s_{23}[14]\langle
43\rangle\langle 1q\rangle\langle 4q\rangle\langle 5q\rangle.$

We then again use the same momentum conservation formula on the very last term
to get

$\displaystyle\langle 3q\rangle\langle 4q\rangle\langle 5q\rangle\langle
12\rangle\langle 23\rangle\langle 34\rangle\langle 45\rangle\langle
51\rangle{\mathcal{A}}^{\prime}_{5}=s_{23}[24]\langle 45\rangle\langle
2q\rangle\langle 3q\rangle\langle 4q\rangle$ (77) $\displaystyle-
s_{45}[15]\langle 53\rangle\langle 1q\rangle\langle 4q\rangle\langle
5q\rangle+s_{45}[45]\langle 34\rangle\langle 4q\rangle\langle 5q\rangle^{2}$
$\displaystyle-s_{23}[24]\langle 35\rangle\langle 2q\rangle\langle
4q\rangle^{2}+s_{15}[25]\langle 53\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle$ $\displaystyle+s_{14}[25]\langle 23\rangle\langle
4q\rangle\langle 5q\rangle^{2}-s_{23}[12]\langle 23\rangle\langle
1q\rangle\langle 4q\rangle\langle 5q\rangle.$

We can now use Schouten identity to extract the $q$-invariant pieces, and
match these to those in (68). We have

$\displaystyle-s_{23}[12]\langle 23\rangle\langle 1q\rangle\langle
4q\rangle\langle 5q\rangle=s_{23}s_{12}\langle 3q\rangle\langle
4q\rangle\langle 5q\rangle+s_{23}[12]\langle 31\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle,$ (78) $\displaystyle-s_{45}[15]\langle
53\rangle\langle 1q\rangle\langle 4q\rangle\langle
5q\rangle=s_{45}s_{51}\langle 3q\rangle\langle 4q\rangle\langle
5q\rangle+s_{45}[15]\langle 31\rangle\langle 4q\rangle\langle 5q\rangle^{2},$
$\displaystyle s_{14}[25]\langle 23\rangle\langle 4q\rangle\langle
5q\rangle^{2}=s_{14}s_{25}\langle 3q\rangle\langle 4q\rangle\langle
5q\rangle+s_{14}[25]\langle 53\rangle\langle 2q\rangle\langle 4q\rangle\langle
5q\rangle.$

These gives three of the four Mandelstam variable containing terms in (68).
The remainder, which is supposed to give the last $s_{25}s_{45}$ term is

$\displaystyle s_{23}[24]\langle 45\rangle\langle 2q\rangle\langle
3q\rangle\langle 4q\rangle+s_{45}[15]\langle 31\rangle\langle 4q\rangle\langle
5q\rangle^{2}+s_{45}[45]\langle 34\rangle\langle 4q\rangle\langle
5q\rangle^{2}$ (79) $\displaystyle-s_{23}[24]\langle 35\rangle\langle
2q\rangle\langle 4q\rangle^{2}+s_{15}[25]\langle 53\rangle\langle
2q\rangle\langle 4q\rangle\langle 5q\rangle$ $\displaystyle+s_{14}[25]\langle
53\rangle\langle 2q\rangle\langle 4q\rangle\langle 5q\rangle+s_{23}[12]\langle
31\rangle\langle 2q\rangle\langle 4q\rangle\langle 5q\rangle.$

The second and third terms here, using the momentum conservation, give

$\displaystyle s_{45}[25]\langle 23\rangle\langle 4q\rangle\langle
5q\rangle^{2}=s_{45}s_{25}\langle 3q\rangle\langle 4q\rangle\langle
5q\rangle+s_{45}[25]\langle 53\rangle\langle 2q\rangle\langle 4q\rangle\langle
5q\rangle.$ (80)

The first term gives the last term in (68). Thus, we have the remainder which
is

$\displaystyle s_{23}[24]\langle 45\rangle\langle 2q\rangle\langle
3q\rangle\langle 4q\rangle+s_{45}[25]\langle 53\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle$ (81) $\displaystyle-s_{23}[24]\langle
35\rangle\langle 2q\rangle\langle 4q\rangle^{2}+s_{15}[25]\langle
53\rangle\langle 2q\rangle\langle 4q\rangle\langle 5q\rangle$
$\displaystyle+s_{14}[25]\langle 53\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle+s_{23}[12]\langle 31\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle.$

Applying Schouten identity once more on the first term in the first and second
lines

$\displaystyle s_{23}[24]\langle 45\rangle\langle 2q\rangle\langle
3q\rangle\langle 4q\rangle-s_{23}[24]\langle 35\rangle\langle 2q\rangle\langle
4q\rangle^{2}=s_{23}[24]\langle 43\rangle\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle,$ (82)

we get a set of terms all proportional to $\langle 2q\rangle\langle
4q\rangle\langle 5q\rangle$

$\displaystyle s_{23}[24]\langle 43\rangle+s_{23}[21]\langle
13\rangle+s_{45}[25]\langle 53\rangle+s_{15}[25]\langle
53\rangle+s_{14}[25]\langle 53\rangle=[25]\langle
53\rangle(s_{45}+s_{15}+s_{14}-s_{23})=0,$

where we applied momentum conservation to the first two terms. Thus, the
correct expression (68) for the 5-point amplitude is reproduced.

##  5 Discussion

This paper establishes a new way of understanding the same helicity one-loop
Yang-Mills amplitudes as arising from tree-level Berends-Giele currents
connected by an effective propagator. This new interpretation is important for
several reasons.

The so-called CSW [11] tree-level formalism gives a prescription for how to
compute tree-level YM amplitudes with effective Feynman rules that use MHV
amplitudes as vertices. An attempt to extend this to loop amplitudes was made
in [12], where it was in particular suggested that one-loop same helicity
amplitudes that are the subject of this paper can be added as new interaction
vertices, in addition to the tree level MHV vertices. The interpretation of
these one-loop amplitudes that emerges from our work is that they are build
from more elementary tree-level blocks. It thus seems unnatural to try to
build more involved amplitudes from the amplitudes that are already composed.
Our results show that these amplitudes are made from simpler tree-level
amplitudes connected by an effective propagator. Thus, it appears that it is
the region momenta-dependent effective propagator that should be added into
the set of CSW rules if one is to reproduce loop amplitudes.

Our new interpretation is also relevant for the problem of UV divergence of
quantum gravity. In papers [13], [14] this divergence was linked to the non-
vanishing of the same helicity one-loop amplitude in GR. Given that the story
with GR is likely mirroring that in YM, see more on this below, the
interpretation of this paper shows that the non-vanishing of the same helicity
one-loop amplitudes can in turn be linked to the non-vanishing of the one-loop
bubble. It would thus be very interesting to better understand the
significance and interpretation of the non-vanishing of the bubbles.

This leads to the question of interpretation of such a simple construction as
our formula (3). This formula cries for an interpretation other than just an
observation that the formula works. The fact that region momentum variables
play such a prominent role points to both the worldsheet interpretation as in
[5], as well as the (momentum) twistor space interpretation as in [15]. In any
case, it would be very interesting to understand a deeper origin of the
formula (3).

Another natural question is how much of what was described here can be
generalised to the case of gravity. As in the case of YM, there exists a
truncation of the full theory of General Relativity that keeps only its self-
dual sector. In its version that most parallels the story of SDYM, this has
been described in [8]. The corresponding ”flat” version relevant for GR with
zero scalar curvature has been recently described in [16]. The ”flat” version
that is most relevant for computing flat space amplitudes has properties very
similar to that of SDYM. There is a single kinetic term, and then a single
cubic interaction, which however in the gravity case contains two derivatives.
The Feynman rules for this SDGR theory are immediate, and can be used for
computing amplitudes. In particular, the all-negative BG currents can be
computed by a recursion, see e.g. [8] and references therein for the
description of the result and the procedure. The integrand of the one-loop 4
point amplitude can also be written down without any difficulty.

What can also be studied without difficulty is the ”bubble” self-energy
diagram. Analogously to the SDYM case, this amplitude can be argued to be
zero, while at the same time shift dependent. The shift can be computed, with
the result being the most natural generalisation of the effective propagator
(31) from spin one two spin two case. What is far from clear, however, is how
to interpret the shift parameters in the gravity case. In the case of YM we
could associate these with the region momenta. There is no cyclic ordering in
the case of gravity, and so the use of region momenta does not seem justified
any longer. Thus, while there is strong evidence that there must exist a
gravitational analog of the formula (3), it is not clear how to to give
meaning to the momentum variables appearing in the gravitational effective
propagator, and so it is not clear what form the GR analog of (3) should take.
We hope to return to all these questions in a future publication.

## Acknowledgements

We are grateful to Eugene Skvortsov for pointing out the ”sum over bubbles”
interpretation of the one-loop amplitudes, and in particular for the reference
[5].

##  Appendix A Sum of box, triangle and bubble diagrams

The purpose of this Appendix is to spell out, in our notations, the argument
that the sum of the box, triangle and bubble one-loop integrands, with a
specific choice of the loop momentum, equivalent to the parametrisation in
terms of the region momenta, vanishes. This means that the 4-point one-loop
amplitude can be reduced to ”bubble” insertions. Our formula (3) is then the
generalisation of this statement to an arbitrary number of external states.

###  A.1 Box Diagram

The box diagram

$\displaystyle\fmfgraph*(110,80)\fmfbottom{i1,i2}\fmftop{o1,o2}\fmf{photon,tension=3}{i1,v1}\fmf{photon,tension=3}{i2,v2}\fmf{photon,tension=3}{o1,v3}\fmf{photon,tension=3}{o2,v4}\fmf{photon}{v3,v3^{\prime}}\fmf{vanilla}{v3^{\prime},v1}\fmf{photon}{v1,v1^{\prime}}\fmf{vanilla}{v1^{\prime},v2}\fmf{photon}{v2,v2^{\prime}}\fmf{vanilla}{v2^{\prime},v4}\fmf{photon}{v4,v4^{\prime}}\fmf{vanilla}{v4^{\prime},v3}\fmflabel{$1^{+}$}{o1}\fmflabel{$4^{+}$}{o2}\fmflabel{$2^{+}$}{i1}\fmflabel{$3^{+}$}{i2}\fmflabel{$l+1+2$}{v1^{\prime}}\fmflabel{$l+1$}{v3^{\prime}}\fmflabel{$l-4$}{v2^{\prime}}\fmflabel{$l$}{v4^{\prime}}$

has already been studied by us in [17]. It is given by

$\displaystyle i\mathcal{M}=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l|4]\langle q|l+1|1]\langle q|l+1+2|2]\langle
q|l-4|3]}{l^{2}(l+1)^{2}(l+1+2)^{2}(l-4)^{2}\prod_{j=1}^{4}\langle
qj\rangle}.$ (83)

We use manipulations described in [17] to reduce it to a sum of triangle-like
and then bubble-like diagrams, using various identities to cancel the scalar
propagator factors of the type $l^{2}$ from the denominator. The first stage
of reduction is described by

$\displaystyle
2i\mathcal{M}=2i\mathcal{M}_{1}+2i\mathcal{M}_{2}+2i\mathcal{M},_{3}$ (84)

with

$\displaystyle\begin{split}2i\mathcal{M}_{1}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|3\circ(l-4)|q\rangle\langle q|l+1|1]\langle
q|l+1+2|2]}{(l+1)^{2}(l+1+2)^{2}(l-4)^{2}\prod_{j=1}^{4}\langle
qj\rangle\langle 43\rangle}\\\
2i\mathcal{M}_{2}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(l-4)|q\rangle\langle q|l+1|1]\langle
q|l+1+2|2]}{l^{2}(l+1)^{2}(l-4)^{2}\prod_{j=1}^{4}\langle qj\rangle\langle
43\rangle}\\\ 2i\mathcal{M}_{3}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(l-3-4)|q\rangle\langle q|l+1|1]\langle
q|l+1+2|2]}{l^{2}(l+1)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle qj\rangle\langle
43\rangle}.\end{split}$

In the second stage of reduction one gets, see Section 3 of [17] for
derivation

$\displaystyle 4i\mathcal{M}_{1}=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l+1|1]\langle q|(l-4)\circ(1+4)|q\rangle}{(l+1)^{2}(l-4)^{2}\langle
23\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q4\rangle}$ (85) $\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l+1|1]\langle q|(l+1+2)\circ 3|q\rangle}{(l+1+2)^{2}(l-4)^{2}\langle
23\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q4\rangle}$ $\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l+1|1]\langle q|(l+1+2)\circ 2|q\rangle}{(l+1)^{2}(l+1+2)^{2}\langle
23\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q4\rangle}$ $\displaystyle
4i\mathcal{M}_{2}=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l+1|2]\langle
q|l\circ 1|q\rangle}{l^{2}(l+1)^{2}\langle 41\rangle\langle 43\rangle\langle
q1\rangle\langle q2\rangle\langle q3\rangle}$ (86)
$\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l+1|2]\langle
q|(l-4)\circ(2+3)|q\rangle}{(l+1)^{2}(l-4)^{2}\langle 41\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q3\rangle}$
$\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l+1|2]\langle
q|l\circ 4|q\rangle}{l^{2}(l-4)^{2}\langle 41\rangle\langle 43\rangle\langle
q1\rangle\langle q2\rangle\langle q3\rangle}$ $\displaystyle
4i\mathcal{M}_{3}=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(3+4)|q\rangle\langle q|(1+2)\circ l|q\rangle}{l^{2}(l+1+2)^{2}\langle
12\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}$ (87)
$\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(3+4)|q\rangle\langle q|(l+1)\circ 1|q\rangle}{l^{2}(l+1)^{2}\langle
12\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}$
$\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(3+4)|q\rangle\langle q|(l+1)\circ
2|q\rangle}{(l+1)^{2}(l+1+2)^{2}\langle 12\rangle\langle 43\rangle\langle
q1\rangle\langle q2\rangle\langle q3\rangle\langle q4\rangle}$

We now implement the use of region momenta $p_{i}$, which we assign as follows

$\displaystyle k_{1}=p_{1}-p_{4},\quad k_{2}=p_{2}-p_{1},\quad
k_{3}=p_{3}-p_{2},\quad k_{4}=p_{4}-p_{3}.$ (88)

We also assign the region momentum for the the loop interior and re-express
the loop momentum in terms of region momenta. Since the loop momentum has to
be integrated, there is an arbitrariness in its assignment. Indeed, we can add
any momenta $q$ to the loop. We thus make a specific choice $l=x-p_{4}$, where
$x$ is the region momentum inside the loop and $p_{4}$ is one of the external
region momenta. The momentum $x$ is now to be integrated over. We then re-
write the quadratically divergent integrals in terms of the dual momentum
variables and usual momenta, given this particular choice of the loop
momentum. This gives

$\displaystyle
4i\mathcal{M}_{1}=4i\mathcal{M}_{11}+4i\mathcal{M}_{12}+4i\mathcal{M}_{13},$
(89)

where

$\displaystyle\begin{split}4i\mathcal{M}_{11}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|1]\langle q|(p_{2}-x)\circ
2|q\rangle}{(p_{2}-x)^{2}(p_{1}-x)^{2}\langle 23\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q4\rangle}\\\
4i\mathcal{M}_{12}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|1]\langle
q|(p_{3}-x)\circ(1+4)|q\rangle}{(p_{1}-x)^{2}(p_{3}-x)^{2}\langle
23\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q4\rangle}\\\ 4i\mathcal{M}_{13}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|1]\langle q|(p_{2}-x)\circ
3|q\rangle}{(p_{3}-x)^{2}(p_{2}-x)^{2}\langle 23\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q4\rangle},\end{split}$
$\displaystyle
4i\mathcal{M}_{2}=4i\mathcal{M}_{21}+4i\mathcal{M}_{22}+4i\mathcal{M}_{23},$
(90)

where

$\displaystyle\begin{split}4i\mathcal{M}_{21}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|2]\langle q|(p_{4}-x)\circ
1|q\rangle}{(p_{4}-x)^{2}(p_{1}-x)^{2}\langle 41\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q3\rangle}\\\
4i\mathcal{M}_{22}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|2]\langle
q|(p_{3}-x)\circ(3+2)|q\rangle}{(p_{1}-x)^{2}(p_{3}-x)^{2}\langle
41\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle}\\\ 4i\mathcal{M}_{23}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|p_{1}-x|2]\langle q|(p_{4}-x)\circ
4|q\rangle}{(p_{4}-x)^{2}(p_{3}-x)^{2}\langle 41\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q3\rangle},\end{split}$

and finally

$\displaystyle
4i\mathcal{M}_{3}=4i\mathcal{M}_{31}+4i\mathcal{M}_{32}+4i\mathcal{M}_{33},$
(91)

where

$\displaystyle\begin{split}4i\mathcal{M}_{31}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ(3+4)|q\rangle\langle
q|(p_{2}-x)\circ(3+4)|q\rangle}{(p_{4}-x)^{2}(p_{2}-x)^{2}\langle
12\rangle\langle 43\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}\\\
4i\mathcal{M}_{32}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ(3+4)|q\rangle\langle q|(p_{1}-x)\circ
1|q\rangle}{(p_{4}-x)^{2}(p_{1}-x)^{2}\langle 12\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q3\rangle\langle
q4\rangle}\\\ 4i\mathcal{M}_{33}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ(3+4)|q\rangle\langle q|(p_{2}-x)\circ
2|q\rangle}{(p_{1}-x)^{2}(p_{2}-x)^{2}\langle 12\rangle\langle
43\rangle\langle q1\rangle\langle q2\rangle\langle q3\rangle\langle
q4\rangle}\end{split}$ (92)

###  A.2 Triangle Diagrams

There are four in-equivalent triangle diagrams to consider. We start by
considering the one where the legs $3$ and $4$ are attached to two of the
vertices.

(150,100) i1,i2 o1,o2 photon,tension=3i1,v1,o1 photon,tension=3i2,v3
photon,tension=3o2,v4 vanillav1,v1’ photonv1’,v2 photon,label=$l$v2,v2’
vanillav2’,v4 photon,label=$l-4$v4’,v4 vanillav4’,v3
photon,label=$l+1+2$v3’,v3 vanillav3’,v2 $1^{+}$o1 $4^{+}$o2 $2^{+}$i1
$3^{+}$i2 v1’ v2’ v3’ v4’

$\displaystyle 2i\mathcal{T}^{12}=\frac{1}{\langle
12\rangle}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l|4]\langle
q|l-4|3]\langle
q|(l+1+2)\circ(1+2)|q\rangle}{l^{2}(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle}$ (93)

Let us start by multiplying the numerator and denominator by $\langle
34\rangle$. This allows to write the integral as

$\displaystyle 2i\mathcal{T}^{12}=-\frac{1}{\langle 12\rangle\langle
34\rangle}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l\circ 4\circ
3\circ(l-4)|q\rangle\langle
q|l\circ(1+2)|q\rangle}{l^{2}(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle},$ (94)

where we used $\langle q|(1+2)\circ(1+2)|q\rangle=0$ to get rid of $(1+2)$
from the last factor in the numerator. We now replace $4=l-(l-4)$ and use the
identity $l\circ l=\frac{1}{2}l^{2}\mathds{1}$ to write the integral as

$\displaystyle 2i\mathcal{T}^{12}=-\frac{1}{\langle 12\rangle\langle
34\rangle}\Bigg{[}-\frac{1}{2}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(l-4)\circ 3|q\rangle\langle
q|l\circ(1+2)|q\rangle}{(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle qj\rangle}$
$\displaystyle-\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l\circ(l-4)\circ
3\circ(l-4)|q\rangle\langle
q|l\circ(1+2)|q\rangle}{l^{2}(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle}\Bigg{]}$ (95)

Next we use the spinor identity

$\displaystyle A\circ B=-B\circ A+(A.B)\mathds{1}$ (96)

where $(A.B)$ is the metric pairing. This identity holds for any two arbitrary
rank two mixed spinors $A$ and $B$. Using this, we have

$\displaystyle(l-4)\circ 3\circ(l-4)=-(l-4)\circ(l-4)\circ 3+(l-4)((l-4).3)$
$\displaystyle=-\frac{1}{2}(l-4)^{2}3+(l-4)((l-4).3)$ (97)

The second term in the last line of (A.2) can be written as

$\displaystyle(l-4).3=\frac{1}{2}((l-4)^{2}-(l-4-3)^{2})$ (98)

Therefore, this gives us

$\displaystyle(l-4)\circ
3\circ(l-4)=\frac{1}{2}(l-4)^{2}(l-4-3)-\frac{1}{2}(l-4-3)^{2}(l-4)$ (99)

We now replace $l-3-4=l+1+2$ and cancel denominators to get

$\displaystyle 2i\mathcal{T}^{12}=\frac{1}{\langle 12\rangle\langle
34\rangle}\Bigg{[}\frac{1}{2}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(l-4)\circ 3|q\rangle\langle
q|l\circ(1+2)|q\rangle}{(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle qj\rangle}$
$\displaystyle+\frac{1}{2}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(l+1+2)|q\rangle\langle
q|l\circ(1+2)|q\rangle}{l^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle qj\rangle}$
$\displaystyle-\frac{1}{2}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(l-4)|q\rangle\langle
q|l\circ(1+2)|q\rangle}{l^{2}(l-4)^{2}\prod_{j=1}^{4}\langle
qj\rangle}\Bigg{]}$ (100)

One of the factors of $l$ in the second and third integrals can be immediately
eliminated owing to the identity $\langle q|l\circ l|q\rangle=0$. We then
choose to rewrite it as

$\displaystyle i\mathcal{T}^{12}=\frac{1}{4\langle 12\rangle\langle
34\rangle}\Bigg{[}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|(l-4)\circ
3|q\rangle\langle
q|(l+1+2)\circ(1+2)|q\rangle}{(l-4)^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle}$ $\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(1+2)|q\rangle\langle
q|(l+1+2)\circ(1+2)|q\rangle}{l^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle}$ $\displaystyle+\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle q|l\circ
4|q\rangle\langle
q|(l+1+2)\circ(1+2)|q\rangle}{l^{2}(l-4)^{2}\prod_{j=1}^{4}\langle
qj\rangle}\Bigg{]}$ (101)

Let us then rewrite this in the dual momentum variables. As before, we choose
the loop momentum variable $l=p_{4}-x$. We have

$\displaystyle
4i\mathcal{T}^{12}=4i\mathcal{T}^{12}_{1}+4i\mathcal{T}^{12}_{2}+4i\mathcal{T}^{12}_{3}$
(102)

where

$\displaystyle\begin{split}4i\mathcal{T}^{12}_{1}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(p_{3}-x)\circ 3|q\rangle\langle
q|(p_{2}-x)\circ(1+2)|q\rangle}{(p_{3}-x)^{2}(p_{2}-x)^{2}\langle
12\rangle\langle 34\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{12}_{2}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ(1+2)|q\rangle\langle
q|(p_{2}-x)\circ(1+2)|q\rangle}{(p_{4}-x)^{2}(p_{2}-x)^{2}\langle
12\rangle\langle 34\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{12}_{3}&=\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ 4|q\rangle\langle
q|(p_{2}-x)\circ(1+2)|q\rangle}{(p_{4}-x)^{2}(p_{3}-x)^{2}\langle
12\rangle\langle 34\rangle\prod_{j=1}^{4}\langle qj\rangle}\end{split}$ (103)

The diagram with legs 1 and 2 inserted to two of the vertices is

(150,100) i1,i2 o1,o2 photon,tension=3i1,v1,o1 photon,tension=3i2,v3
photon,tension=3o2,v4 vanillav1,v1’ photonv1’,v2 photon,label=$l$v2,v2’
vanillav2’,v4 photon,label=$l-2$v4’,v4 vanillav4’,v3
photon,label=$l+3+4$v3’,v3 vanillav3’,v2 $3^{+}$o1 $2^{+}$o2 $4^{+}$i1
$1^{+}$i2 v1’ v2’ v3’ v4’

It can be obtained from (103) by substitutions

$\displaystyle 1\to 3,\quad 2\to 4,\quad 3\to 1,\quad 4\to 2.$ (104)

The loop momentum in terms of region momenta is $l=p_{2}-x$. We have

$\displaystyle
4i\mathcal{T}^{34}=4i\mathcal{T}^{34}_{1}+4i\mathcal{T}^{34}_{2}+4i\mathcal{T}^{34}_{3},$
(105)

where

$\displaystyle\begin{split}4i\mathcal{T}^{34}_{1}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{1}-x)\circ 1|q\rangle\langle
q|(p_{4}-x)\circ(3+4)|q\rangle}{(p_{1}-x)^{2}(p_{4}-x)^{2}\langle
12\rangle\langle 34\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}\\\
4i\mathcal{T}^{34}_{2}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{2}-x)\circ(3+4)|q\rangle\langle
q|(p_{4}-x)\circ(3+4)|q\rangle}{(p_{2}-x)^{2}(p_{4}-x)^{2}\langle
12\rangle\langle 34\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}\\\
4i\mathcal{T}^{34}_{3}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{2}-x)\circ 2|q\rangle\langle
q|(p_{4}-x)\circ(3+4)|q\rangle}{(p_{2}-x)^{2}(p_{1}-x)^{2}\langle
12\rangle\langle 34\rangle\langle q1\rangle\langle q2\rangle\langle
q3\rangle\langle q4\rangle}\end{split}$ (106)

We note that the terms arising here cancel precisely the terms in
${\mathcal{M}}_{3}$.

The diagram with legs 2 and 3 inserted to two of the vertices is

(150,100) i1,i2 o1,o2 photon,tension=3i1,v1,o1 photon,tension=3i2,v3
photon,tension=3o2,v4 vanillav1,v1’ photonv1’,v2 photon,label=$l$v2,v2’
vanillav2’,v4 photon,label=$l-3$v4’,v4 vanillav4’,v3
photon,label=$l+1+4$v3’,v3 vanillav3’,v2 $4^{+}$o1 $3^{+}$o2 $1^{+}$i1
$2^{+}$i2 v1’ v2’ v3’ v4’

It can be obtained from (103) by substitutions

$\displaystyle 1\to 4,\quad 4\to 3,\quad 3\to 2,\quad 2\to 1.$ (107)

The loop momentum in terms of region momenta is $l=p_{3}-x$. We get

$\displaystyle
4i\mathcal{T}^{41}=4i\mathcal{T}^{41}_{1}+4i\mathcal{T}^{41}_{2}+4i\mathcal{T}^{41}_{3},$
(108)

where

$\displaystyle\begin{split}4i\mathcal{T}^{41}_{1}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{2}-x)\circ 2|q\rangle\langle
q|(p_{1}-x)\circ(1+4)|q\rangle}{(p_{2}-x)^{2}(p_{1}-x)^{2}\langle
41\rangle\langle 23\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{41}_{2}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{3}-x)\circ(1+4)|q\rangle\langle
q|(p_{1}-x)\circ(1+4)|q\rangle}{(p_{3}-x)^{2}(p_{1}-x)^{2}\langle
41\rangle\langle 23\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{41}_{3}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{3}-x)\circ 3|q\rangle\langle
q|(p_{1}-x)\circ(1+4)|q\rangle}{(p_{2}-x)^{2}(p_{3}-x)^{2}\langle
41\rangle\langle 23\rangle\prod_{j=1}^{4}\langle qj\rangle}\end{split}$ (109)

The diagram with legs 1 and 4 inserted to two of the vertices is

(150,100) i1,i2 o1,o2 photon,tension=3i1,v1,o1 photon,tension=3i2,v3
photon,tension=3o2,v4 vanillav1,v1’ photonv1’,v2 photon,label=$l$v2,v2’
vanillav2’,v4 photon,label=$l-1$v4’,v4 vanillav4’,v3
photon,label=$l+2+3$v3’,v3 vanillav3’,v2 $2^{+}$o1 $1^{+}$o2 $3^{+}$i1
$4^{+}$i2 v1’ v2’ v3’ v4’

It can be obtained from (103) by substitutions

$\displaystyle 1\to 2,\quad 2\to 3,\quad 3\to 4,\quad 4\to 1.$ (110)

The loop momentum in terms of region momenta is $l=p_{1}-x$. This gives

$\displaystyle
4i\mathcal{T}^{23}=4i\mathcal{T}^{23}_{1}+4i\mathcal{T}^{23}_{2}+4i\mathcal{T}^{23}_{3},$
(111)

where

$\displaystyle\begin{split}4i\mathcal{T}^{23}_{1}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ 4|q\rangle\langle
q|(p_{3}-x)\circ(2+3)|q\rangle}{(p_{4}-x)^{2}(p_{3}-x)^{2}\langle
23\rangle\langle 41\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{23}_{2}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{1}-x)\circ(2+3)|q\rangle\langle
q|(p_{3}-x)\circ(2+3)|q\rangle}{(p_{1}-x)^{2}(p_{3}-x)^{2}\langle
23\rangle\langle 41\rangle\prod_{j=1}^{4}\langle qj\rangle}\\\
4i\mathcal{T}^{23}_{3}&=\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{1}-x)\circ 1|q\rangle\langle
q|(p_{3}-x)\circ(2+3)|q\rangle}{(p_{1}-x)^{2}(p_{4}-x)^{2}\langle
23\rangle\langle 41\rangle\prod_{j=1}^{4}\langle qj\rangle}.\end{split}$ (112)

###  A.3 Bubble Diagrams

####  A.3.1 Internal Bubbles

There are two inequivalent permutations of the bubbles on internal lines. In
one diagram, the legs 1 and 2 join on one side of the internal line while 3
and 4 on the other side. This is given by

(180,50) i1,i3,i2 o1,o3,o2 photoni1,v2,o1 vanillav2,v1’ photonv1’,v1
photonv1,i3 vanillai3,v4 photonv4,o3 vanillao3,v1 photonv4,v3 vanillav3,v3’
photoni2,v3’,o2 li3 l+1+2o3 $1^{+}$i1 $2^{+}$o1 $3^{+}$o2 $4^{+}$i2

$\displaystyle 4i\mathcal{B}^{12}=\frac{1}{\langle 12\rangle\langle
34\rangle}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|l\circ(1+2)|q\rangle\langle
q|(l+1+2)\circ(3+4)|q\rangle}{l^{2}(l+1+2)^{2}\prod_{j=1}^{4}\langle
qj\rangle}$ (113)

Writing it in the dual momentum variables with $l=p_{4}-x$ we get

$\displaystyle 4i\mathcal{B}^{12}=\frac{1}{\langle 12\rangle\langle
34\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ(1+2)|q\rangle\langle
q|(p_{2}-x)\circ(3+4)|q\rangle}{(p_{4}-x)^{2}(p_{2}-x)^{2}\prod_{j=1}^{4}\langle
qj\rangle}.$ (114)

The other internal bubble diagram is

(180,50) i1,i3,i2 o1,o3,o2 photoni1,v2,o1 vanillav2,v1’ photonv1’,v1
photonv1,i3 vanillai3,v4 photonv4,o3 vanillao3,v1 photonv4,v3 vanillav3,v3’
photoni2,v3’,o2 li3 l+1+4o3 $4^{+}$i1 $1^{+}$o1 $2^{+}$o2 $3^{+}$i2

It can be obtained from the above by permutations, with $l=p_{3}-x$ and the
result being

$\displaystyle 4i\mathcal{B}^{23}=\frac{1}{\langle 41\rangle\langle
23\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{3}-x)\circ(1+4)|q\rangle\langle
q|(p_{1}-x)\circ(3+2)|q\rangle}{(p_{3}-x)^{2}(p_{1}-x)^{2}\prod_{j=1}^{4}\langle
qj\rangle}.$ (115)

####  A.3.2 Bubbles on External Lines

There are four distinct diagrams here, corresponding to the different ways of
insertion of bubbles to four external lines. One can write each contribution
as an insertion of a current to one of the legs of the bubble. For example, we
have for the insertion of the bubble between the particle 1 and the $J(2,3,4)$
current

$\displaystyle\begin{gathered}~{}~{}~{}\fmfgraph*(130,30)\fmfleft{i}\fmfright{o}\fmfbottom{v}\fmftop{u}\fmf{photon}{i,v1}\fmf{photon}{v1,u}\fmf{vanilla}{u,v2}\fmf{photon}{v2,v}\fmf{vanilla}{v,v1}\fmf{photon}{v2,v3}\fmf{vanilla}{v3,o}\fmfblob{.15w}{o}\fmflabel{l}{u}\fmflabel{l+1}{v}\fmflabel{$1^+$}{i}\end{gathered}$
(117)
$\displaystyle 4i\mathcal{B}^{1}=\frac{\langle q3\rangle}{\langle
23\rangle\langle 34\rangle}\int\frac{d^{4}l}{(2\pi)^{4}}\frac{\langle
q|(l+1)\circ 1|q\rangle\langle q|l|1]}{l^{2}(l+1)^{2}\prod_{j=1}^{4}\langle
qj\rangle}$ (118)

Rewriting this in terms of the dual momentum variables with $l=p_{4}-x$ we get

$\displaystyle 4i\mathcal{B}^{1}=\frac{\langle q3\rangle}{\langle
23\rangle\langle 34\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{1}-x)\circ 1|q\rangle\langle
q|p_{4}-x|1]}{(p_{4}-x)^{2}(p_{1}-x)^{2}\prod_{j=1}^{4}\langle qj\rangle}.$
(119)

The other similar contributions are obtained by cyclic permutations. We have

$\displaystyle 4i\mathcal{B}^{2}=\frac{\langle q4\rangle}{\langle
34\rangle\langle 41\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{2}-x)\circ 2|q\rangle\langle
q|p_{1}-x|2]}{(p_{2}-x)^{2}(p_{1}-x)^{2}\prod_{j=1}^{4}\langle qj\rangle},$
(120) $\displaystyle 4i\mathcal{B}^{3}=\frac{\langle q1\rangle}{\langle
41\rangle\langle 12\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{3}-x)\circ 3|q\rangle\langle
q|p_{2}-x|3]}{(p_{2}-x)^{2}(p_{3}-x)^{2}\prod_{j=1}^{4}\langle qj\rangle},$
(121) $\displaystyle 4i\mathcal{B}^{4}=\frac{\langle q2\rangle}{\langle
12\rangle\langle 23\rangle}\int\frac{d^{4}x}{(2\pi)^{4}}\frac{\langle
q|(p_{4}-x)\circ 4|q\rangle\langle
q|p_{3}-x|4]}{(p_{4}-x)^{2}(p_{3}-x)^{2}\prod_{j=1}^{4}\langle qj\rangle}$
(122)

###  A.4 Sum of Integrands

As we previously noticed, the terms in ${\mathcal{M}}_{3}$ get cancelled by
one of the triangle diagrams ${\mathcal{T}}^{34}$. Our aim is now to show that
all other terms get cancelled agains each other as well. To this end, we will
group the terms according to their denominators of the form
$(p_{i}-x)^{2}(p_{j}-x)^{2}$ for some $i,j$.

We start with the denominator factor $(p_{1}-x)^{2}(p_{2}-x)^{2}$. The
contributions to this come from ${\mathcal{M}}_{11},{\mathcal{T}}^{41}_{1}$ as
well as one of the bubbles ${\mathcal{B}}^{2}$. The sum of the corresponding
numerators is given by

$\displaystyle\begin{split}&\frac{\langle q|(p_{2}-x)\circ 2|q\rangle}{\langle
23\rangle\langle 43\rangle\langle 14\rangle}\Big{(}\langle
q|(p_{1}-x)\circ(1+4)|4\rangle\langle q3\rangle-\langle
q|(p_{1}-x)\circ(1+4)|q\rangle\langle 43\rangle\\\
&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\langle
q|(p_{1}-x)\circ 2|3\rangle\langle q4\rangle\Big{)},\end{split}$ (123)

where we rewrote the first term in a suggestive way. We now use the momentum
conservation $2=-(1+3+4)$ in the last term, and then Schouten identity
$|4\rangle\langle q3\rangle=|q\rangle\langle 43\rangle+|3\rangle\langle
q4\rangle$ to see that the sum in brackets is zero.

Next consider the denominator factor $(p_{1}-x)^{2}(p_{3}-x)^{2}$. The
contributions come from
$\mathcal{M}_{12},\mathcal{M}_{22},{\mathcal{T}}^{41}_{2},{\mathcal{T}}^{23}_{2}$
and the bubble $\mathcal{B}^{23}$. The triangle contributions double each
other, and the sum of these numerators is

$\displaystyle\begin{split}&\frac{\langle
q|(p_{3}-x)\circ(1+4)|q\rangle}{\langle 23\rangle\langle 43\rangle\langle
14\rangle}\Big{(}\langle q|(p_{1}-x)\circ(1+4)|4\rangle\langle
q3\rangle+\langle q|(p_{1}-x)\circ(2+3)|3\rangle\langle q4\rangle\\\
&-2\langle q|(p_{1}-x)\circ(1+4)|q\rangle\langle 43\rangle+\langle
q|(p_{1}-x)\circ(1+4)|q\rangle\langle 43\rangle\Big{)},\end{split}$ (124)

where again we rewrote the first terms in a suggestive way. Relacing
$2+3=-(1+4)$ in the second term and using the same Schouten identity as above
we see the cancellation.

Next consider the denominator factor $(p_{1}-x)^{2}(p_{4}-x)^{2}$. The
contributions come from $\mathcal{M}_{21},\mathcal{T}^{23}_{3}$ and
${\mathcal{B}}^{1}$. The sum of these numerators is given by

$\displaystyle\begin{split}&\frac{\langle q|(p_{4}-x)\circ 1|q\rangle}{\langle
23\rangle\langle 43\rangle\langle 41\rangle}\Big{(}\langle q|(p_{1}-x)\circ
2|3\rangle\langle q4\rangle+\langle q|(p_{3}-1)\circ(2+3)|q\rangle\langle
43\rangle\\\
&~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}+\langle
q|(p_{4}-x)\circ 1|4\rangle\langle q3\rangle\Big{)}.\end{split}$ (125)

We now use the fact that $p_{3}=p_{1}+2+3$, and so $p_{3}$ can be replaced by
$p_{1}$ in the second term. Similarly, $p_{4}=p_{1}-1$, and so we can replace
$p_{4}$ with $p_{1}$ in the last term. We then similarly replace $2$ by
$(2+3)$ in the first term, and $1$ by $(1+4)$ in the last. Then again the same
Schouten identity implies the cancellation.

Let us now consider the denominator factor $(p_{2}-x)^{2}(p_{3}-x)^{2}$. The
situation is somewhat more interesting here. There are 4 integrands that
contribute, namely
${\mathcal{M}}_{13},{\mathcal{T}}^{12}_{1},{\mathcal{T}}^{41}_{3}$ and the
bubble ${\mathcal{B}}^{3}$. Let us start by considering the sum of
${\mathcal{T}}^{41}_{3}$ and ${\mathcal{B}}^{3}$. This can be written as

$\displaystyle\frac{\langle q|(p_{3}-x)\circ 3|q\rangle}{\langle
41\rangle\langle 12\rangle\langle 23\rangle}\left(\langle
q|(p_{1}-x)\circ(1+4)|q\rangle\langle 12\rangle-\langle q|(p_{2}-x)\circ
3|2\rangle\langle q1\rangle\right).$ (126)

Using $p_{3}-p_{1}=3+2$ in the first term, as well as $1+4=-(2+3)$, we can
replace $p_{1}$ there by $p_{3}$. Similarly, in the second term we can use
$p_{3}-p_{2}=3$ to replace $p_{2}$ with $p_{3}$. The expression in brackets is
then

$\displaystyle-\langle q|(p_{3}-x)\circ(2+3)|q\rangle\langle 12\rangle-\langle
q|(p_{3}-x)\circ(2+3)|2\rangle\langle q1\rangle=-\langle
q|(p_{3}-x)\circ(2+3)|1\rangle\langle q2\rangle$ (127)

by Schouten identity. This can be written as

$\displaystyle-\langle q|(p_{3}-x)\circ(2+3)|1\rangle\langle
q2\rangle=-\langle q|(p_{3}-x)\circ(1+2+3)|1\rangle\langle q2\rangle=\langle
q|(p_{3}-x)|4]\langle 41\rangle\langle q2\rangle.$ (128)

On the other hand, the sum of ${\mathcal{M}}_{13}$ and
${\mathcal{T}}^{12}_{1}$ is given by

$\displaystyle\frac{\langle q|(p_{3}-x)\circ 3|q\rangle}{\langle
43\rangle\langle 12\rangle\langle 23\rangle}\left(\langle q|(p_{1}-x)\circ
1|2\rangle\langle q3\rangle-\langle q|(p_{2}-x)\circ(1+2)|q\rangle\langle
23\rangle\right).$ (129)

We can replace $p_{1}$ by $p_{4}$ and $1$ by $(1+2)$ in the first term, and
$p_{2}$ by $p_{4}$ in the second term. This gives for the expression in the
brackets

$\displaystyle\langle q|(p_{4}-x)\circ(1+2)|2\rangle\langle q3\rangle-\langle
q|(p_{4}-x)\circ(1+2)|q\rangle\langle 23\rangle=\langle
q|(p_{4}-x)\circ(1+2)|3\rangle\langle q2\rangle.$ (130)

This can be written as

$\displaystyle\langle q|(p_{4}-x)\circ(1+2)|3\rangle\langle q2\rangle=\langle
q|(p_{4}-x)\circ(1+2+3)|3\rangle\langle q2\rangle=-\langle
q|(p_{4}-x)|4]\langle 43\rangle\langle q2\rangle.$ (131)

Given that we can replace here $p_{4}$ by $p_{3}$, it is clear that the terms
${\mathcal{M}}_{13},{\mathcal{T}}^{12}_{1},{\mathcal{T}}^{41}_{3},{\mathcal{B}}^{3}$
cancel each other.

For the denominator $(p_{2}-x)^{2}(p_{4}-x)^{2}$ there are only two
contributing terms ${\mathcal{T}}^{12}_{2}$ and ${\mathcal{B}}^{12}$, which
directly cancel each other. For the denominator $(p_{3}-x)^{2}(p_{4}-x)^{2}$
we have the terms
${\mathcal{M}}_{23},{\mathcal{T}}^{12}_{3},{\mathcal{T}}^{23}_{1},{\mathcal{B}}^{4}$
contributing, and the cancelation here is similar to the one encountered in
the case $(p_{2}-x)^{2}(p_{3}-x)^{2}$.

Therefore the total integrand, as a sum of box, four triangles, two internal
bubbles and eight external bubbles vanishes. Pictorially, this can be
represented as

$\displaystyle\begin{gathered}\fmfgraph*(60,45)\fmfbottom{i1,i2}\fmftop{o1,o2}\fmf{photon,tension=3}{i1,v1}\fmf{photon,tension=3}{i2,v2}\fmf{photon,tension=3}{o1,v3}\fmf{photon,tension=3}{o2,v4}\fmf{photon}{v3,v3^{\prime}}\fmf{vanilla}{v3^{\prime},v1}\fmf{photon}{v1,v1^{\prime}}\fmf{vanilla}{v1^{\prime},v2}\fmf{photon}{v2,v2^{\prime}}\fmf{vanilla}{v2^{\prime},v4}\fmf{photon}{v4,v4^{\prime}}\fmf{vanilla}{v4^{\prime},v3}\fmflabel{}{o1}\fmflabel{}{o2}\fmflabel{}{i1}\fmflabel{}{i2}\fmflabel{}{v1^{\prime}}\fmflabel{}{v3^{\prime}}\fmflabel{}{v2^{\prime}}\fmflabel{}{v4^{\prime}}\end{gathered}~{}~{}+~{}~{}4\times\begin{gathered}\fmfgraph*(70,40)\fmfbottom{i1,i2}\fmftop{o1,o2}\fmf{photon,tension=3}{i1,v1,o1}\fmf{photon,tension=3}{i2,v3}\fmf{photon,tension=3}{o2,v4}\fmf{vanilla}{v1,v1^{\prime}}\fmf{photon}{v1^{\prime},v2}\fmf{photon}{v2,v2^{\prime}}\fmf{vanilla}{v2^{\prime},v4}\fmf{photon}{v4^{\prime},v4}\fmf{vanilla}{v4^{\prime},v3}\fmf{photon}{v3^{\prime},v3}\fmf{vanilla}{v3^{\prime},v2}\fmflabel{}{o1}\fmflabel{}{o2}\fmflabel{}{i1}\fmflabel{}{i2}\fmflabel{}{v1^{\prime}}\fmflabel{}{v2^{\prime}}\fmflabel{}{v3^{\prime}}\fmflabel{}{v4^{\prime}}\end{gathered}~{}~{}+~{}~{}2\times\begin{gathered}\fmfgraph*(70,30)\fmftop{i1,i3,i2}\fmfbottom{o1,o3,o2}\fmf{photon}{i1,v2,o1}\fmf{vanilla}{v2,v1^{\prime}}\fmf{photon}{v1^{\prime},v1}\fmf{photon}{v1,i3}\fmf{vanilla}{i3,v4}\fmf{photon}{v4,o3}\fmf{vanilla}{o3,v1}\fmf{photon}{v4,v3}\fmf{vanilla}{v3,v3^{\prime}}\fmf{photon}{i2,v3^{\prime},o2}\end{gathered}~{}~{}+~{}~{}4\times\begin{gathered}\fmfgraph*(70,30)\fmfleft{i}\fmfright{o}\fmfbottom{v}\fmftop{u}\fmf{photon}{i,v1}\fmf{photon}{v1,u}\fmf{vanilla}{u,v2}\fmf{photon}{v2,v}\fmf{vanilla}{v,v1}\fmf{photon}{v2,v3}\fmf{vanilla}{v3,o}\fmfblob{.15w}{o}\end{gathered}~{}~{}=0$
(135)

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