HU-EP-22/12

Tilings and Twist at $\mathbf{1/N^{4}}$

B. Eden†, T. Scherdin††

† Humboldt-Universität zu Berlin, Institut für Mathematik

†† Humboldt-Universität zu Berlin, Institut für Physik

E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>

We re-consider operator mixing in the so-called $SU(2)$ sector of ${\cal
N}\,=\,4$ super Yang-Mills theory with gauge group $SU(N)$. Where possible,
single-trace operators of moderate length are completed by higher-trace
admixtures so as to yield large $N$ tree level eigenstates.

We are particularly interested in parity pairs with three excitations. Since
parity is respected in the mixing, the odd single-trace operators at low
length cannot receive too many admixtures. We reproduce the tree-level norms
of a set of large $N$ eigenstates up to order $1/N^{4}$ by integrability
methods. This involves evaluating two-point functions on the sphere, the
torus, and the double-torus. A perfect match is found as long as descendents
are absent from the mixing.

Using twist to make the descendents appear in the integrability picture
immediately leads to the question how to modify the entangled states occurring
in the hexagon tessellations. We take a closer look at the double-trace
admixtures to the parity even three-excitation operator at length seven, which
are both products of a primary state and a descendent. Their two-point
functions are sensitive to the twist introduced into the Bethe equations. For
transverse scalar excitations we succeed in recovering the corresponding field
theory results. For longitudinal magnons our methods fail, pointing at a
potential weakness of the formalism.

##  1 Introduction

The AdS/CFT correspondence [1] entered a quantitative phase with the _BMN
construction_ [2] which directly linked a class of string theory states to a
dual set of composite operators in ${\cal N}\,=\,4$ super Yang-Mills theory.
All elementary fields of this superconformal field theory carry an adjoint
representation of a non-Abelian gauge group. The operators in question are
products of very many scalar fields of the same type with a few other fields
termed _impurities_ in between, made gauge invariant and cyclic by a single
trace over the gauge group. The one-loop anomalous dimensions of these long
operators can be matched with string theory predictions in the same expansion
parameter, although this would be small in gauge theory and large for the
strings [3].

In the simplest case, the impurities are all scalars of the same second type.
The spectrum of _planar_ one-loop anomalous dimensions in such an $SU(2)$
_sector_ is equivalent to the energy eigenvalues of the Heisenberg spin chain
[4, 5]. This observation has opened up an entire research agenda by the name
of _integrability_ in which one tries to include higher-loop corrections and
to extend the integrable system to all types of composite operators.

While the inclusion of higher genus corrections had been discussed in the
literature on the BMN operators from the very beginning [3], it remained an
open question for a long time how to address non-planar corrections in the
integrability framework [6]. The answer came in an unexpected fashion: an
efficient scheme for the computation of structure constants was designed in
[7] and extended to higher-point functions in [8, 9]. Feynman diagrams for
higher-point functions of single-trace operators can be drawn on closed
Riemann surfaces with punctures corresponding to the operator insertions. At
tree level in configuration space, propagators connecting the operators
naturally triangulate the surfaces. Each tile of such a triangulation is given
by the _hexagon operator_ of [7].

In a systematic large $N$ expansion, single-trace operators will acquire
admixtures of higher-trace operators. In [10] it was pointed that coincidence
limits of higher-point functions can be used to describe correlation functions
of operators with more than one colour trace. This is one step towards non-
planar integrability, and the one that we elaborate on in the present article
up to $1/N^{4}$ corrections. Eventually, it would be desirable to design a
formalism capable of calculating the mixing coefficients at non-planar order.
This second step will not be addressed here.

We will seek eigenoperators of the conformal group at one loop in the $SU(2)$
sector, albeit without sending the number of elementary fields to infinity.
The one-loop problem uniquely fixes the tree level eigenstates as long as the
operators are not protected from quantum corrections, i.e. when they have non-
vanishing anomalous dimension. We will then study how to combine these
operators in a large $N$ expansion. Ideally, we would like to construct finite
$N$ eigenstates; more realistically, we consider the first three terms of a
large $N$ expansion in a range of cases.

Building two-point functions of these exact (or $N$ expanded) eigenoperators
one has to compute a number of overlaps between their constituents. Here we
compare hexagon tessellations to tree level field theory with very good
success; problems arise where _descendents_ come in, thus operators that are
obtained from others by $SU(2)$ raising.

The article is organised as follows: in Section 2 we review the construction
of field theory eigenoperators using the one-loop _dilatation operator_ [11]
and discuss principle features of the spectrum, in particular the appearance
of _parity pairs_. In Section 3 we review the integrability formalism for the
one-loop spectrum and give a precise map form Bethe wave functions to
normalised field theory operators. The set of examples we will explicitly
discuss comprises an _exceptional operator_ and two _descendents_ whose
integrability description requires the notion of _twist_ , so the modification
of the Bethe equations of the Heisenberg chain by a small parameter, see [13]
and references therein. In Section 4 we introduce the hexagon operator for
three-point functions and comment on the effect of twist in this context.
Finally, in Section 5 we introduce tessellations for higher-point functions
and use them to re-calculate all two-point functions related to our set of
examples. Section 6 is dedicated to two-point functions of double-trace
operators one of whose parts is a descendent.

##  2 The spectrum in the $SU(2)$ sector from the field theory

The ${\cal N}\,=\,4$ model has the six complex scalar fields
$\phi^{[I,J]},\,I\in\\{1,2,3,4\\}$ obeying the reality constraint

$\left(\phi^{IJ}\right)^{*}\,=\,\phi_{IJ}\,=\,\frac{1}{2}\epsilon_{IJKL}\phi^{KL}$
(1)

So three of them, say, $\Phi^{i}\,=\,\phi^{1,i+1},\,i\in\\{1,2,3\\}$ are
independent with the reality constraint determining their complex conjugates.
To simplify the notation the scalar fields are often called $Z,\,X,\,Y$.

The scalar fields transform in the adjoint representation of a non-Abelian
gauge group: $\Phi^{i}\,=\,\Phi^{ia}\,T^{a}$ where $T$ is a generator of the
local symmetry. For definiteness we assume this _colour group_ to be of
$SU(N)$ type in this article. We will study large $N$ perturbation theory at
tree level w.r.t. to the gauge coupling $g_{YM}$. Gauge invariant composite
operators arise as products of the elementary fields under an $SU(N)$ trace.
Limiting the scope to $\Phi^{1},\,\Phi^{2}$ or $\Phi^{1},\,\Phi^{3}$ we obtain
what is called an $SU(2)$ sector, because the second type of field is obtained
from the first by the step operator $\partial_{2}^{3}$ or $\partial_{2}^{4}$,
respectively.

The ${\cal N}\,=\,4$ model is conjectured to be conformally invariant. Good
operators are then the eigenstates of the conformal group, which ought to be
orthonormal:

$\langle{\cal O}_{i}\bar{\cal O}_{j}\rangle\,=\,0\quad:\quad
i\,\neq\,j\,,\qquad\qquad\langle{\cal O}_{i}\bar{\cal
O}_{i}\rangle\,=\,\frac{1}{(x_{ij}^{2})^{L+g^{2}\,\gamma_{1,i}+\ldots}}\,,\quad
g^{2}\,=\,\frac{g^{2}_{\mathrm{YM}}N}{8\,\pi^{2}}$ (2)

The exponent in the last formula is the quantum corrected _dimension_ of the
operator, of which we have indicated the trivial part given by the length $L$,
so the number of elementary fields, and the _one-loop anomalous dimension_
$\gamma_{1}$.

Tree level two-point functions do not provide enough information to fix the
eigenstates even at tree level, whereas one-loop renormalisation does. In
terms of ${\cal N}\,=\,1$ supersymmetric Feynman rules this arises from the
contraction of the effective vertex

$g^{2}_{\mathrm{YM}}\int d^{4}x\,\mathrm{Tr}([\bar{Z},\bar{Y}][Z,Y])$ (3)

onto the two operators in the two-point functions.

The concept of the _dilatation operator_ 111For a point of view closer to the
original BMN literature see also [12]. was introduced in [11]: Wick
contraction of the barred fields in the vertex on one of the operators in the
two-point function removes a $Z$ and a $Y$, and puts another such pair back,
though not in all terms in the same position. This operation defines a linear
map, say, $D$ on the spaces of operators with identical length $L$ and number
$n$ of $Y$ fields. The one-loop anomalous dimensions $\gamma_{i}^{(1)}$ are
given by the eigenvalues of $D$ because the actual Feynman integral is
universal and can be factored out. Note that the $N$ dependence of the
eigenvalues can be quite involved whereby it is usually preferable to work
perturbatively in $1/N$.

The contraction on the second operator amounts to selecting matrix elements.
These do, of course, contain the same information so that one can construct
the orthogonal states also from the complete set of tree and one-loop two-
point functions, cf. [14]. Technically, this is harder because a system of
quadratic equations on the coefficients in the eigenstates has to be solved.

Now, $SU(N)$ generators satisfy $\mathrm{Tr}(T^{a})\,=\,0$ and the _cutting
and sewing rules_

$\mathrm{Tr}(T^{c}\,A\,T^{c}\,B)\,=\,\mathrm{Tr}(A)\,\mathrm{Tr}(B)-\frac{1}{N}\mathrm{Tr}(A\,B)\,,\qquad\mathrm{Tr}(T^{c}\,A)\,\mathrm{Tr}(T^{c}\,B)\,=\,\mathrm{Tr}(A\,B)-\frac{1}{N}\mathrm{Tr}(A)\,\mathrm{Tr}(B)\,,$
(4)

where $A,\,B$ are any products of generators. In particular, contracting the
effective vertex onto a _single-trace operator_ can cut it up into a _double-
trace operator_ , while the vertex can do both, sew a _double-trace operator_
to a single trace, or to further split it into three traces and so forth. Upon
computing with these rules, one finds that contractions of operators with
different numbers of traces are suppressed by powers of $1/N$.

We will try to complete any given leading $N$ single-trace eigenstate ${\cal
O}_{i}^{s}$ to a series of the form

${\cal O}_{i}^{s}+\left(\frac{a_{i}^{j}}{N^{2}}+\ldots\right){\cal
O}_{j}^{s}+\left(\frac{b_{i}^{k}}{N}+\frac{\hat{b}_{i}^{k}}{N^{3}}+\ldots\right){\cal
O}_{k}^{d}+\left(\frac{c_{i}^{l}}{N^{2}}+\ldots\right){\cal O}_{l}^{t}+\ldots$
(5)

with eigenvalues $\gamma_{1,i}^{s}+\hat{\gamma}_{1,i}^{s}/N^{2}+\ldots\,$. All
operators in this formula should be leading $N$ eigenstates; with the
subscripts $s,\,d,\,t$ we denote the number of traces222A study of non-planar
corrections to anomalous dimensions of operators related to giant gravitons
has been initiated in [15].. One expects the entire expansion to go in powers
of $1/N^{2}$ with an offset of $1/N$ per additional trace in the admixtures.
Last, to avoid trivial rescalings we impose $a_{i}^{i}\,=\,0$ (no sum).

The dilatation operator is a map of the type

$\displaystyle{\cal O}_{i}^{s}$ $\displaystyle\,\rightarrow\,$ $\displaystyle
N\gamma_{1,i}^{s}\,{\cal O}_{i}^{s}+d_{i}^{k}\,{\cal O}_{k}^{d}$
$\displaystyle{\cal O}_{k}^{d}$ $\displaystyle\,\rightarrow\,$ $\displaystyle
e_{k}^{j}\,{\cal O}_{j}^{s}+N\gamma_{1,k}^{d}\,{\cal O}_{k}^{d}+f_{k}^{l}{\cal
O}_{l}^{t}$ (6) $\displaystyle\ldots$

when all operators are leading N eigenstates. Hence in a hypothetical system
with one single- and one double-trace operator we have to solve the eigenvalue
equation

$D\left({\cal O}^{s}+\frac{b}{N}{\cal
O}^{d}+\ldots\right)\,=\,N\gamma_{1}^{s}\,{\cal O}^{s}+d\,{\cal
O}^{d}+\frac{b}{N}\left({\cal O}^{s}+N\gamma_{1}^{d}\,{\cal
O}^{d}\right)+\ldots\,=\,(N\,\gamma_{1}^{s}+\ldots)\left({\cal
O}^{s}+\frac{b}{N}{\cal O}^{d}+\ldots\right)$ (7)

which is trivial at the leading order while we learn from the $O(N^{0})$
coefficient of ${\cal O}^{d}$ that

$b\,=\,\frac{d}{\gamma_{1}^{s}-\gamma_{1}^{d}}\,.$ (8)

Consequently, if there is degeneracy $\gamma_{1}^{s}\,=\,\gamma_{1}^{d}$ in
the system while $d\,\neq\,0$, we will not succeed in constructing the desired
completion (5) of the single-trace eigenstate333In this situation further
problems arise, e.g. the expansion in terms of the coupling constant does not
commute over that in terms of the genus counting parameter [11]. Our point of
view is to expand in the coupling first. [3, 16, 11]. In fact, the whole
program seems doomed — wouldn’t a similar problem arise, if anywhere in the
expansion the one-loop anomalous dimension of any two multiple-trace operators
were degenerate? Despite the fact that our simple experiment with only one
operator of each type quickly becomes inconsistent going to higher orders, it
does indicate that only the one-loop anomalous dimension of any candidate
admixture and that of the single-trace operator to be completed have to be
distinct. This condition is not impossible to satisfy because the anomalous
dimensions of the higher states are roots of fairly complicated characteristic
polynomials and are therefore not too likely to reproduce previously
encountered values. In fact, scanning the spectrum of single-trace operators
up to length ten and four excitations to the order indicated in (5) we
observed a breakdown of the ansatz only for cases with
$\gamma_{1}^{s}\,\in\\{4,5,6\\}$.

Because there are as many eigenvectors in the single-trace sector of given
$L,\,n$ as there are states, we can always re-write the admixtures as products
of leading $N$ single-trace eigenstates. Incidentally, to leading $N$ the one-
loop anomalous dimension of such a product is the sum of that of its factors.

The one-loop mixing problem has the following features:

  * •

States of different one-loop anomalous dimension will be orthogonal at tree
level.

  * •

We can act on any state with length $L$ and $n$ excitations with the step
operator $\partial_{2}^{4}$ by global differentiation to produce a
_descendent_ of the same length but with one more excitation. If non-
vanishing, the descendent will be an eigenvector and has the same anomalous
dimension as the original operator. States that are not descendents are called
_primary_.

  * •

All the admixtures to renormalised single-trace operators (i.e. with
$\gamma_{1}^{s}\,\neq\,0$) also have non-vanishing anomalous dimensions. Since
the one-loop anomalous dimensions of all operators are non-negative, at least
one factor in each multi-trace operator is renormalised. This property should
be rooted in supersymmetry because non-renormalised operators are members of
shorter supermultiplets that could not consistently be added in.

  * •

At higher length and excitation number the leading $N$ single-trace spectrum
contains _degenerate pairs_. As a basis of such a $2\times 2$ cell we can
choose the even and odd part under _parity_ , i.e. reversal of the trace as in
$\mathrm{Tr}(ZZYZYY)\,\rightarrow\,\mathrm{Tr}(ZZYYZY)$. Parity is a strict
rule [11, 17]: the admixtures all have the same parity as the single-trace
state we complete.

  * •

When the two states of definite parity in a leading $N$ degenerate pair can be
completed by admixtures as advocated in (5) the degeneracy will be lifted by
the higher $N$ corrections444See also [18] for similar results in other
instances of the AdS/CFT correspondence. [11].

All states with $L\,=\,2,3$ are _non-renormalised_ or _protected_ , i.e. have
$\gamma_{1}^{s}\,=\,0$. At every $(L,n)$ there will be one protected single-
trace state; at the multi-trace level there are in general several protected
operators according to how $L$ is partitioned into shorter parts and how the
excitations are distributed over these. Especially at short length most states
are parity even. Therefore the mixing in the odd sector is significantly
simpler.

In fact, all states with $n\,=\,0,1$ $Y$ fields are protected. Also, all
states with $n\,=\,2$ are parity even. We are therefore particularly
interested in the next simplest case $n\,=\,3$. Scanning states of length up
to and including $L\,=\,9$ we are able to point out all properties that are
new in the integrability picture w.r.t. to the previously studied operators
with two $Y$ fields [10].

As an illustration, let us comment on the states of length 5: in the protected
sector we have the single-trace operators

$\frac{1}{\sqrt{5}}\mathrm{Tr}(Z^{5}),\,\mathrm{Tr}(Z^{4}Y),\,\frac{1}{\sqrt{2}}\mathrm{Tr}(ZZ\\{Z,Y\\}Y),\,\frac{1}{\sqrt{2}}\mathrm{Tr}(Z\\{Z,Y\\}YY),\,\mathrm{Tr}(ZY^{4}),\,\frac{1}{5}\mathrm{Tr}(Y^{5})\,.$
(9)

All of these can be generated by derivatives on the first operator in the
list. Hence they are _vacuum descendents_. Then there are two states of one-
loop anomalous dimension $\gamma_{1}^{s}\,=\,4$:

$\frac{1}{\sqrt{2}}Tr(ZZ[Z,Y]Y),\,\frac{1}{\sqrt{2}}\mathrm{Tr}(Z[Z,Y]YY)\,,$
(10)

which we will refer to as (5,2) and (5,3), respectively. The (5,3) operator is
a descendent of the (5,2) case. Double-trace states must be built from factors
of length 2 and 3. They are all protected at leading order in $N$ since their
constituents are. We can conclude that both renormalised states, (5,2) and
(5,3), respectively, receive no admixtures. As a consequence, their one-loop
anomalous dimension will receive no corrections in $1/N$. Notice that all
length 5 states are parity even.

Beyond (5,2) and (5,3) we will encounter the following operators in our study:

  1. 1.

The first parity odd operator, which comes at $(6,3)$:

$(6,3)^{e}\,=\,\frac{i}{\sqrt{2}}\mathrm{Tr}(ZZY[Z,Y]Y)$ (11)

For want of other parity odd states at this length this is a true eigenstate
of the mixing problem also at finite $N$, its anomalous dimension
$\gamma_{1}^{s}\,=\,6$ does not receive corrections in $1/N$. We have given
the operator the label $e$ for _exceptional_ for reasons that will become
apparent in the next section.

  2. 2.

At length 7, a first degenerate pair of primaries with $\gamma_{1}^{s}\,=\,5$
appears with three $Y$ fields:

$\displaystyle(7,3)^{-}$ $\displaystyle=$
$\displaystyle\frac{i}{\sqrt{2}}\,(0,-1,1,0,0)\,,$ (12)
$\displaystyle(7,3)^{+}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{2}}{\sqrt{15}}\left(-1,\frac{3}{2},\frac{3}{2},-1,-1\right)$
(13)

The coefficient vectors in these formulae refer to the basis

$\mathrm{Tr}(Z^{4}Y^{3}),\,\mathrm{Tr}(ZZZYZYY),\,\mathrm{Tr}(ZZZYYZY),\,\mathrm{Tr}(ZZYZZYY),\,\mathrm{Tr}(ZZYZYZY)\,.$
(14)

A remark on the norms is in order: we take out a number coefficient so that
the leading $N$ part of the two-point functions will be $N^{L}$. This will be
useful in comparing to integrability.

The parity odd case above experiences no mixing, while the parity even one
could have the double-trace admixtures $(5,3)(2,0),\,(5,2)(2,1),\,(4,2)(3,1)$.
The third of these is a descendent. It does in fact not enter the mixing. The
other two come in the special combination

${\cal
O}^{\perp}\,=\,\frac{1}{\sqrt{3}}\left(\sqrt{2}\,(5,3)(2,0)-(5,2)(2,1)\right)\,.$
(15)

We remark that ${\cal O}^{\perp}$ is a primary state: it is exactly tree
orthogonal to the descendent $\partial_{2}^{4}\,(5,2)(2,0)$. Thus the mixing
takes place exclusively between the primaries at $(7,3)$. For now another
property is more important: $(7,3)^{+}$ and ${\cal O}^{\perp}$ map into each
other under the dilatation operation $D$:

$\frac{D}{N}\left(\begin{array}[]{c}(7,3)^{+}\\\\[2.84526pt] {\cal
O}^{\perp}\end{array}\right)\,=\,\left(\begin{array}[]{cc}5&-\frac{2\sqrt{5}}{N}\\\\[2.84526pt]
-\frac{4\sqrt{5}}{N}&4\end{array}\right)\left(\begin{array}[]{c}(7,3)^{+}\\\\[2.84526pt]
{\cal O}^{\perp}\end{array}\right)$ (16)

Notice that this matrix is not symmetric, whereby its left- and right-
eigenvectors are distinct. Worse, the members of either set are not tree
orthogonal in contradiction to the definition (2). An approriate change of
basis is discussed in Appendix A.

Here we content ourselves with constructing an eigenoperator
$(73)^{+}+\,b\,{\cal O}^{\perp}$ determining the left-eigenvectors $(1,b)$ of
the matrix in the last equation:

$\gamma_{1}\,=\,\frac{1}{2}\left(9\pm\sqrt{1+\frac{160}{N^{2}}}\right)\,,\qquad
b\,=\,\frac{N}{8\,\sqrt{5}}\left(1\mp\sqrt{1+\frac{160}{N^{2}}}\right)$ (17)

The upper sign in front of the root corresponds to the mixing problem in (5)
with

$\gamma_{1}^{s}\,=\,5,\,\quad\hat{\gamma}_{1}^{s}\,=\,40,\qquad
b_{1}\,=\,-2\,\sqrt{5},\quad\hat{b}_{1}\,=\,80\,\sqrt{5}$ (18)

while the other sign yields an eigenoperator in which the double-trace part is
dominant. Notice that the factor $\sqrt{5}$ arises from the unit normalisation
of the operators. To convey the central point of this discussion we juxtapose
the anomalous dimensions of the ($N$ corrected) single-trace operators:

$(7,3)^{-}\,:\,\gamma_{1}\,=\,5\,,\qquad(7,3)^{+}\,:\,\gamma_{1}\,=\,5+\frac{40}{N^{2}}+\ldots$
(19)

Hence the $N$ dependence lifts the degeneracy of the eigenvalues, and the
parity eigenstates are the good operators at finite $N$ or in a large $N$
expansion. These results were originally derived in [19, 11] and recently
recovered in [20].

  3. 3.

At length 8, there are three primaries with three $Y$’s:

$\displaystyle\gamma_{1}^{s}\,=\,6\quad:\quad(8,3)^{e}\,$ $\displaystyle=$
$\displaystyle\frac{i}{2}\left(\mathrm{Tr}(Z^{4}Y[Z,Y],Y)-\mathrm{Tr}(ZZ[Z,Y]ZZYY)\right)\,,$
$\displaystyle\gamma_{1}^{s}\,=\,4\quad:\quad(8,3)^{+}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(\mathrm{Tr}(Z^{5}Y^{3})-\mathrm{Tr}(Z^{4}Y\\{Z,Y\\}Y)+\mathrm{Tr}(ZZYZZYZY)\right)\,$
(20) $\displaystyle\gamma_{1}^{s}\,=\,4\quad:\quad(8,3)^{-}$ $\displaystyle=$
$\displaystyle\frac{i}{2}\left(\mathrm{Tr}(Z^{4}Y[Z,Y],Y)+\mathrm{Tr}(ZZ[Z,Y]ZZYY)\right)\,.$

The leading $N$ degeneracy of $\gamma_{1}^{s}\,=\,6,4$ with that of
$(6,3)^{e}(2,0)$ and $(5,3)(3,1),\,(5,2)(2,1)$, respectively, prevents us from
discussing the first two cases in this list. On the other hand, $(8,3)^{-}$ is
an exact eigenoperator of $D$; it could mix with $(6,3)^{e}(2,0)$ but does
not. In Appendix A we further comment on this example, too.

  4. 4.

At $L\,=\,9,\,n\,=\,3$ the primary single-trace operators come in three
degenerate pairs. The three parity even states have a host of admixtures
including all four possible triple-trace operators carrying leading $N$
anomalous dimension. Interestingly, there is re-mixing also amongst the parity
even primaries at the level of the $a_{2}/N^{2}$ coefficients whereas single-
trace descendent states do not come in.

The parity odd single-trace states $(9,3)_{i}^{-}$ are

$\displaystyle\gamma_{1,1}^{s}\,=\,8.25342\,,\qquad(9,3)_{1}^{-}$
$\displaystyle=$ $\displaystyle i\,(0.0532393,-0.17321,0.683494)\,,$
$\displaystyle\gamma_{1,2}^{s}\,=\,5.51997\,,\qquad(9,3)_{2}^{-}$
$\displaystyle=$ $\displaystyle i\,(0.616784,-0.320709,-0.129317)\,,$ (21)
$\displaystyle\gamma_{1,3}^{s}\,=\,3.22661\,,\qquad(9,3)_{3}^{-}$
$\displaystyle=$ $\displaystyle i\,(0.341676,0.605924,0.126938)\,,$

where the coefficient vectors refer to the three parity odd combinations

$\mathrm{Tr}(Z^{5}Y[Z,Y]Y),\,\mathrm{Tr}(Z^{4}Y[Z^{2},Y]Y),\,\mathrm{Tr}(Z^{3}YZ[Z,Y]ZY)\,.$
(22)

Curiously the one-loop anomalous dimensions of the descendent states
$6.82843,4,1.17157,0$ lie in between.

The $(9,3)_{i}^{-}$ operators can and do mix with
$(6,3)^{e}(3,0),\,(7,3)^{-}(2,0)$ whereas there are no parity-odd triple-trace
operators at length 9. Like their even cousins they display re-mixing among
themselves at $O(N^{-2})$. Up to $1/N^{3}$ we obtain

$\displaystyle(9,3)_{1}^{-}+\frac{0.303442}{N^{2}}(9,3)_{2}^{-}-\frac{0.0353558}{N^{2}}(9,3)_{3}^{-}$
$\displaystyle\quad+\left(\frac{0.738458}{N}-\frac{1.72311}{N^{3}}\right)(6,3)^{e}(3,0)+\left(\frac{0.461192}{N}-\frac{1.32898}{N^{3}}\right)(7,3)^{-}(2,0)+\ldots\,,$
$\displaystyle(9,3)_{2}^{-}+\frac{49.9031}{N^{2}}(9,3)_{1}^{-}+\frac{5.98334}{N^{2}}(9,3)_{3}^{-}$
(23)
$\displaystyle\quad-\left(\frac{14.3515}{N}-\frac{1041.83}{N^{3}}\right)(6,3)^{e}(3,0)-\left(\frac{4.00861}{N}+\frac{189.762}{N^{3}}\right)(7,3)^{-}(2,0)+\ldots\,,$
$\displaystyle(9,3)_{3}^{-}-\frac{2.33609}{N^{2}}(9,3)_{1}^{-}+\frac{1.47962}{N^{2}}(9,3)_{2}^{-}$
$\displaystyle\quad+\left(\frac{0.700161}{N}-\frac{1.85918}{N^{3}}\right)(6,3)^{e}(3,0)+\left(\frac{1.31087}{N}+\frac{4.92937}{N^{3}}\right)(7,3)^{-}(2,0)+\ldots\,,$

with the anomalous dimensions

$\displaystyle\gamma_{1,1}^{s}\,=\,8.25342+\frac{8.18190}{N^{2}}+\ldots\,,$
$\displaystyle\gamma_{1,2}^{s}\,=\,5.51997-\frac{39.8239}{N^{2}}+\ldots\,,$
(24)
$\displaystyle\gamma_{1,3}^{s}\,=\,3.22661+\frac{1.64201}{N^{2}}+\ldots\,.$

For comparison, the parity even partners555To the same accuracy, all six
single-trace primaries are given in Appendix A. We do not display the
admixtures. acquire the one-loop anomalous dimensions

$\displaystyle\gamma_{1,1}^{s}\,=\,8.25342+\frac{53.8338}{N^{2}}+\ldots\,,$
$\displaystyle\gamma_{1,2}^{s}\,=\,5.51997+\frac{72.4183}{N^{2}}+\ldots\,,$
(25)
$\displaystyle\gamma_{1,3}^{s}\,=\,3.22661+\frac{50.2636}{N^{2}}+\ldots\,.$

In the main part of the article we will look at two-point functions of these
operators up to $1/N^{4}$ by integrability methods, that is via hexagon
tessellations [7, 8, 9]. It was advocated in [10] that each $m$-trace operator
can be realised as a coincidence limit of $m$ vertices of a triangulation of a
Riemann surface whose genus matches the order in $N$, i.e. a sphere at leading
order, a torus for $1/N^{2}$ corrections etc. One purpose of the present work
is to check this conjecture on a wider and more complicated range of examples.
Given the list of cases above we will have to analyse

  * •

single-trace/single-trace: two-point functions on the sphere, the torus, and
the double torus

  * •

single-trace/double-trace: three-point functions in a point identification
limit on the sphere and on the torus

  * •

double-trace/double-trace: four-point functions in a double coincidence limit
on the sphere; the relevant torus part can be read out of previous work [10].

The focus is not so much on the actual coefficients of the first few terms in
the large $N$ expansion of the two-point functions; like the mixing
coefficients for the $(9,3)_{i}^{-}$ operators these are some rather
uninspiring numbers, cf. Appendix A. Instead we embark on checking the hexagon
amplitudes against the field theory results for every colour factor, so
whether the integrability construction can reproduce field theory term by
term.

##  3 The spectrum problem from integrability

The leading $N$ part of $D$ arises by nearest neighbour interactions, so when
the vertex (3) is contracted onto two adjacent fields within the same colour
trace. In this case the action is proportional to $\mathbb{I}-\mathbb{P}$
(i.e. identity minus permutation), which is the Heisenberg $XXX$ Hamiltonian
acting on a chain of down spins $Z$ and up spins $Y$ [4, 5]. The leading $N$
eigenstates of our map $D$ in the single-trace sector are therefore the energy
eigenstates of the Heisenberg spin chain. The famous Bethe ansatz can be used
to construct these: the up spins or $Y$ fields are viewed as excitations
moving along the sites of a chain of length $L$. Every such _magnon_ is given
a momentum $p$ with which it moves along the chain. The step operator

$e^{i\,p}\,=\,\frac{u+\frac{i}{2}}{u-\frac{i}{2}}\,,\qquad
u\,=\,\frac{1}{2}\cot\left(\frac{p}{2}\right)$ (26)

takes it from one site of the chain to the next. It is customary to use the
_rapidity_ (or _Bethe root_) $u$ instead of the momentum $p$ in order to
obtain algebraic equations. Pulling a magnon over another one induces a phase

$S_{jk}\,=\,\frac{u_{j}-u_{k}+i}{u_{j}-u_{k}-i}$ (27)

which goes by the name of _scattering matrix_. The Bethe-Yang equations

$e^{i\,p_{j}\,L}\prod_{k\neq j}S_{jk}\,=\,1\,,\qquad j\in\\{1\ldots n\\}$ (28)

express that transporting a magnon once around the chain should not have any
effect because the original configuration is recovered. Translation invariance
around the chain implies the usual momentum conservation constraint
$\sum_{j}p_{j}\,=\,0$ or in terms of the Bethe rapidities

$\prod_{j=1}^{n}\frac{u_{j}+\frac{i}{2}}{u_{j}-\frac{i}{2}}\,=\,1\,.$ (29)

Finally, the energy is given by

$\gamma_{1}^{s}\,=\,\sum_{j=1}^{n}\frac{1}{u_{j}^{2}+\frac{1}{4}}\,.$ (30)

An all $Z$ state — so one without any magnon — has energy 0. For one magnon
the situation is not much different: the zero momentum constraint implies
$p_{1}\,=\,0$ and so $u_{1}$ is infinite such that again
$\gamma_{1}^{s}\,=\,0$. For $n\,>\,\lfloor L/2\rfloor$ excitations we should
discover the same spectrum as for $L-n$ excitations because we might then
regard $Y$ as a vacuum site and $Z$ as an excitation. These arguments rule out
states of non-vanishing energy for $L\,=\,2,3$.

Normally, states with non-vanishing energy are given by a set of distinct,
finite rapidities. The first non-trivial case we encounter is (4,2) with Bethe
rapidities $\pm 1/\sqrt{12}$ and energy 6. According to the above all other
length 4 states are protected.

Let us have a look at the length 5 problem: in the field theory the two states
with anomalous dimension 4 are (5,2) and (5,3) as defined in the last section.
For the (5,2) state we have the regular non-degenerate solution
$u_{1}\,=\,1/2\,=\,-u_{2}$ with energy 4. For the (5,3) case there is no
regular solution. Yet, one can put, say, $u_{3}\,=\,\infty$ whereby it scales
out of the equations reducing them to the (5,2) case. One thus has the three
rapidities $\pm 1/2,\,\infty$ similar to what we said about the one-magnon
case at any length. Indeed, the one-magnon case at length $L$ is always a
descendent of the spin chain vacuum $\mathrm{Tr}(Z^{L})$. In both descendent
cases, $(L,1)$ and (5,3) we found an infinite Bethe root.

Descendents states can be systematically obtained introducing _twist_ into the
Bethe equations, see [13] and the references given there. In the present work
we do so by an extra factor $e^{i\,\epsilon}$ in every Bethe equation:

$e^{i\,\epsilon}e^{i\,p_{j}\,L}\prod_{k\neq j}S_{jk}\,=\,1$ (31)

The coupled equations are then solved order by order in a series expansion in
$\epsilon$. The two states we need for the set of correlators chosen in the
last section are $L\,=\,2,\,n\,=\,1$ with a single rapidity

$u_{1}\,=\,-\frac{2}{\epsilon}+\frac{1}{24}\epsilon+\frac{1}{5760}\epsilon^{3}+\ldots$
(32)

and $L\,=\,5,\,n\,=\,3$:

$\displaystyle u_{1}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}-\frac{1}{10}\epsilon-\frac{11}{80}\epsilon^{2}-\frac{761}{6000}\epsilon^{3}-\frac{12829}{96000}\epsilon^{4}-\frac{453889}{3000000}\epsilon^{5}+\ldots\,,$
$\displaystyle u_{2}$ $\displaystyle=$
$\displaystyle\phantom{-}\frac{1}{2}-\frac{1}{10}\epsilon+\frac{11}{80}\epsilon^{2}-\frac{761}{6000}\epsilon^{3}+\frac{12829}{96000}\epsilon^{4}-\frac{453889}{3000000}\epsilon^{5}+\ldots\,,$
(33) $\displaystyle u_{3}$ $\displaystyle=$
$\displaystyle-\frac{1}{\epsilon}+\frac{29}{60}\epsilon+\frac{4657}{18000}\epsilon^{3}+\ldots\,.$

In the $\epsilon\,\rightarrow\,0$ limit we do indeed obtain the set of
rapidities of the primary (e.g. (20) and (52), respectively) and an infinite
rapidity.

Beyond missing the descendents, the Bethe equations (28) do not have regular
solutions corresponding to certain exceptional operators [5] like
$(6,3)^{e},\,(8,3)^{e}$ etc. Introducing twist we can construct such
solutions, too. For instance, for the $(6,3)^{e}$ operator we have

$\displaystyle u_{1}$ $\displaystyle=$
$\displaystyle-\frac{i}{2}+\frac{1}{3}\epsilon-\frac{19}{432}\epsilon^{3}+\frac{811}{34560}\epsilon^{5}-\frac{i}{243}\epsilon^{6}-\frac{262357}{17418240}\epsilon^{7}+\frac{11\,i}{1944}\epsilon^{8}+\ldots\,,$
$\displaystyle u_{2}$ $\displaystyle=$
$\displaystyle\phantom{-}\frac{i}{2}+\frac{1}{3}\epsilon-\frac{19}{432}\epsilon^{3}+\frac{811}{34560}\epsilon^{5}+\frac{i}{243}\epsilon^{6}-\frac{262357}{17418240}\epsilon^{7}-\frac{11\,i}{1944}\epsilon^{8}+\ldots\,,$
(34) $\displaystyle u_{3}$ $\displaystyle=$
$\displaystyle-\frac{1}{24}\epsilon+\frac{97}{3456}\epsilon^{3}-\frac{2843}{138240}\epsilon^{5}+\frac{4889627}{278691840}\epsilon^{7}+\ldots$

Looking at the leading terms $\pm i/2,\,0$ we see that the first two are
simple poles/zeroes of the momentum factor which is raised to the sixth power
in the Bethe equations, whereas the $u_{1}-u_{2}+i\,=\,O(\epsilon^{6})$ so
that $S_{12}$ can offset this singularity. The energy of the solution is
$6+O(\epsilon^{2})$ as expected for $(6,3)^{e}$. There are similar solutions
at any even length.

For $n\,=\,2$ the zero momentum constraint always implies $u_{2}\,=\,-u_{1}$.
Degenerate pairs arise for sets of roots that do not go into themselves under
$\\{u_{i}\\}\,\rightarrow\,\\{-u_{i}\\}$, which can occur only if
$n\,\geq\,3$. The (undeformed) Bethe equations, the momentum constraint, and
the energy are invariant under the simultaneous sign flip of all rapidities so
that asymmetric root sets do come in pairs. Yet, the two sets of rapidities do
yield two distinct states.

Let $0\,<\,m_{j}\,\leq\,L$ denote the position of the magnon $j$ with rapidity
$u_{j}$ along the chain. The _Bethe wave function_ is

$\psi(L,n)\,=\,\sum_{m_{1},\ldots,m_{n}\,=\,1}^{L}\
\prod_{j\,=\,1}^{n}e^{i\,p_{j}\,m_{j}}\ \prod_{j<k}T_{jk}\ \,|m_{1}\ldots
m_{j}\rangle\,,\qquad\qquad
T_{jk}\,=\,\Biggl{\\{}\begin{array}[]{ccl}m_{j}\,<\,m_{k}&:&1\\\
m_{j}\,=\,m_{k}&:&0\\\ m_{j}\,>\,m_{k}&:&S_{jk}\end{array}$ (35)

where the ket state has $Y$ magnons at the sites $m_{j}$ and otherwise $Z$’s.
In this definition the chain is a priori open with a designated beginning and
end. The magnons do not yet have definite rapidities. The Bethe equations
assure that we can close the chain and so make it cyclic. To that end the
rapidities must solve (28).

A normalised, cyclic state is given by [21]

${\cal O}\,=\,\frac{\psi(L,n)}{\sqrt{{\cal
G}\,L\,\prod_{j\,<\,k}S_{jk}\,\prod_{j}(u_{j}^{2}+\frac{1}{4})}}\,.$ (36)

Obviously, any permutation of the rapidities of a Bethe solution is also a
solution. The normalised state does not depend on the ordering of the
rapidities as long as the same one is chosen in the wave function as well as
the phase factor (the product of $S$ matrices) in the denominator. The last
missing piece of information is the definition of the _Gaudin norm_ :

${\cal G}\,=\,\mathrm{Det}\
\phi_{jk}\,,\qquad\phi_{jk}\,=\,\frac{\partial\log\left(e^{i\,p_{j}\,L}\prod_{l\,\neq\,j}S_{jl}\right)}{\partial\,u_{k}}$
(37)

All of this perfectly works for the states involving twist, to leading order
in $\epsilon$. By way of example, in our simplest case the state is

$(2,1)\,=\,\frac{\mathrm{Tr}(YZ)+\mathrm{Tr}(ZY)}{\sqrt{\frac{\epsilon^{2}}{2}*2*1*\frac{4}{\epsilon^{2}}}}\,=\,\mathrm{Tr}(ZY)$
(38)

as expected for the unit norm descendent in field theory. For the length 5
descendent we find

$(5,3)\,=\,\frac{5(1+i)(\mathrm{Tr}(ZZYYY)-\mathrm{Tr}(ZYZYY))}{\sqrt{(80\,\epsilon^{2})*5*i*\frac{1}{4\,\epsilon^{2}}}}\,=\,\frac{1}{\sqrt{2}}\mathrm{Tr}(Z[Z,Y]YY)$
(39)

and finally

$(6,3)^{e}\,=\,\frac{\frac{486\,i}{\epsilon^{5}}(\mathrm{Tr}(ZZYZYY)-\mathrm{Tr}(ZZYYZY))}{\sqrt{\frac{11664}{\epsilon^{6}}*6*\frac{243}{\epsilon^{6}}*\frac{\epsilon^{2}}{36}}}\,=\,\frac{i}{\sqrt{2}}\mathrm{Tr}(ZZY[Z,Y]Y)\,.$
(40)

For the degenerate pairs, the situation is particularly interesting: let us
consider the (7,3) states normalised as in (36). With $j\,=\,1,2$ we find

$(7,3)^{j}=\frac{1}{\sqrt{15}}\left(-1,\frac{3}{2},\frac{3}{2},-1,-1\right)-(-1)^{j}\frac{i}{2}\left(0,-1,1,0,0\right)$
(41)

in the basis (14) so that comparing to (12), (13)

$(7,3)^{\pm}\,=\,\frac{1}{\sqrt{2}}\bigl{(}(7,3)^{1}\pm(7,3)^{2}\bigr{)}\,.$
(42)

Similarly, $(8,3)^{\pm}\,=\,\bigl{(}(8,3)^{1}\pm(8,3)^{2}\bigr{)}/\sqrt{2}$.
In fact, a relation

${\cal O}^{\pm}\,\propto\,\frac{1}{\sqrt{2}}({\cal O}^{1}\pm{\cal O}^{2})$
(43)

seems to hold in general, where ${\cal O}^{\pm}$ have purely real/imaginary
coefficients and are othonormal at leading $N$.

Last, let us comment on the degeneracy of the spectrum in the $SU(2)$ sector.
The two-magnon situation is particularly prone to falling victim to this
problem: as the zero momentum constraint implies $u_{2}\,=\,-u_{1}$ the
scattering matrix $S_{12}$ becomes equal to $e^{-i\,p}$. The one remaining
Bethe equation is

$e^{i\,p\,(L-1)}\,=\,1\qquad\Rightarrow\qquad
p\,=\,2\,\pi\frac{m}{L-1},\,m\,\in\\{0\ldots\lfloor(L-1)/2\rfloor\\}\,.$ (44)

As a consequence, there are degenerate momenta between the $L$ and the
$L^{\prime}$ problem with two magnons whenever $L-1,\,L^{\prime}-1$ are not
co-prime. For instance, the renormalised (42) state has $L-1\,=3,\,m\,=\,1$
and there is a length 7 two magnon state with $L^{\prime}-1\,=\,6,\,m\,=2$
with the same energy $\gamma_{1}^{s}\,=\,6$. At length 8 the operators
$(4,2)(4,0)$ and $(4,2)(2,0)(2,0)$ have degenerate leading $N$ first anomalous
dimension, at length 9 even $(4,2)(5,0),\,(4,2)(3,0)(2,0),\,(7,2)(2,0)$, and
all of these operators are parity even so that they can mix. These are not
even the only degeneracies in the $SU(2)$ sector spectrum — e.g. $(6,3)^{e}$
has $\gamma_{1}^{s}\,=\,6$ exactly as the primary (4,2) state. This made us
worry about the success of our program of completing single-trace eigenstates
by higher-trace admixtures. Fortunately, the spectrum of higher states is much
more irregular than the two-magnon example, and — as we could convince
ourselves in Section 2 — the degeneracy of the multi-trace operators amongst
themselves is apparently not an obstacle.

##  4 Hexagons and twist

In the spectrum problem, the full set of excitations is treated as follows:

$\phi^{aa^{\prime}}\,\rightarrow\,\phi^{a}\bar{\phi}^{a^{\prime}},\quad\psi^{\alpha
a^{\prime}}\,\rightarrow\,\psi^{\alpha}\bar{\phi}^{a^{\prime}},\quad\bar{\psi}^{a\dot{\alpha}}\,\rightarrow\,\phi^{a}\bar{\psi}^{\dot{\alpha}},\quad
D^{\alpha\dot{\alpha}}\,\rightarrow\,\psi^{\alpha}\bar{\psi}^{\dot{\alpha}}$
(45)

where $a\,=\,1,2,\,a^{\prime}\,=\,3,4$ and $\alpha,\,\dot{\alpha}$ are two-
component spinor indices. The _left excitations_
$\\{\phi^{a},\,\psi^{\alpha}\\}$ and the _right excitations_
$\\{\bar{\phi}^{a^{\prime}},\bar{\psi}^{\dot{\alpha}}\\}$ transform under
separate $PSU(2,2)_{L,R}$ groups that share a central extension. The left and
the right excitation from any two-index scalar, fermion, or derivative put on
the chain are given the same rapidity. The excitations now scatter separately
on both chains with a 10 component $S$ matrix derived in [22] times a phase
factor [23]. Every such tensor product of $S$ matrices is divided by a single
copy of (27).

One recovers the simpler prescription we had given for the $SU(2)$ sector when
only scalars of the same type are considered: on the left as well as on the
right chain solely the component $A$ of the complete matrix [22] is solicited,
and this is exactly equal to (27) at tree level. Normalising by one more $A$
element we recover what we had before because the phase factor goes to unity
at tree level.

The idea of the hexagon approach [7] for three-point functions owes much to
the earlier paper [24], in which a string theory three-point interaction was
regarded as $\langle\mathrm{in}|{\cal O}|\mathrm{out}\rangle$. The authors
were trying to impose form factor axioms as is done in two-dimensional physics
[25]. In [7] the three-vertex is sliced through the middle resulting into two
hexagonal patches. Three of these edges are string pieces, or in the dual
field theory picture one half of a single-trace state (i.e. of a spin chain as
we learned in the last section). These are called _physical edges_. Between
them lie _virtual edges_ that can in the field theory be understood as the
bunches of configuration space propagators stretching between the three
operators at tree level. In [7], axioms for these hexagonal patches were not
only formulated but also solved.

Figure 1: splitting a three-point function into two hexagons

The resulting prescription for the computation of a structure constant is as
follows:

  * •

Find the Bethe solutions desribing the three selected primary operators.

  * •

Cut each Bethe state into two halves corresponding to its portions on the two
hexagons.

  * •

Move all excitations to the same edge of the hexagon by _crossing
transformations_.

  * •

Split each two-index field or derivative from an operator into a left and a
right magnon.

  * •

Scatter only on the left or only on the right chain with the 10 component
$PSU(2,2)$ invariant $S$ matrix [22] supplemented by an adapted phase factor
[7].

  * •

Use a certain contraction rule to put back together magnons with the same
rapidity.

The various hexagon scattering amplitudes needed for our purposes are appended
in an ancillary file. We will not delve into their derivation; the techniques
are well-explained in [7]. Further comments relevant in the present context
can also be found in [8, 10].

Nonetheless, we need to explain two features of the construction, the first of
which is the _twisted translation_ of the vacuum $Z(0)$ to the rotated field

$\hat{Z}(a)\,=\,Z+a\,(Y-\bar{Y})+a^{2}\bar{Z}$ (46)

at some position $a$ along a real line. Likewise, for the _longitudinal_
magnons $Y,\,\bar{Y}$ one has

$\hat{Y}\,=\,Y+a\,\bar{Z}\,,\qquad\hat{\bar{Y}}\,=\,\bar{Y}-a\,\bar{Z}\,.$
(47)

In constructing structure constants from elements of the Bethe ansatz it is
assumed that the three operators are located at $a\,=\,0,\,1,\,\infty$. To be
able to compare we must project the propagators of free field theory according
to (46) and (47). For the _transverse_ $X$ nothing happens because it is not
involved in the twisted translation:

$\langle X(a_{1})\bar{X}(a_{2})\rangle\,=\,\frac{1}{(a_{1}-a_{2})^{2}}$ (48)

Here we have omitted the colour indices and the normalisation factor
$-1/(2\,\pi)^{2}$ that decorates every massless configuration space
propagator. There is no off-diagonal propagator tying $X$ to any of the other
scalars so that $X$ amplitudes will only exist if there are equal numbers of
$X$ and $\bar{X}$ fields. In fact, due to the contraction prescription and the
crossing rules this is also valid on a single hexagon. Norm-like hexagon
amplitudes involve an $SU(2)$ sector operator built from $X$’s at one point,
the same operator at a second point though realised with $\bar{X}$ scalars,
and a unit operator at the third end of the three-point function, cf. [7].

On the hexagon the conjugate particle to $Y$ is $Y$ itself, again owing to the
crossing rules in combination with the contraction prescription. Strangely
enough, one can apparently insert the same operator at two ends of the three-
point function and unity at the third to compute norms with the _longitudinal_
$Y$ magnons [10].

The _effective scalar propagators_ relevant to this exercise are (fields at
$a_{1},a_{2}$):

$\langle ZZ\rangle\,=\,1\,,\qquad\langle
ZY\rangle\,=\,\frac{1}{a_{2}-a_{1}}\,,\qquad\langle
YZ\rangle\,=\,\frac{1}{a_{1}-a_{2}}\,,\qquad\langle YY\rangle\,=\,0\,.$ (49)

In computing norms with these rules there are apparently no problems with
primary single-trace states; the projected field theory results and the
hexagon amplitudes agree. Yet, the last rule in (49) has an effect that will
be problematic for our purposes: we saw in Section 2 that the mixing of the
operator $(7,3)^{+}$ involves the descendent (5,3). Forming two-point
functions using two identical operators realised from $\hat{Z},\,\hat{Y}$
fields we calculate

$\langle(5,2)(5,2)\rangle\,=\,N(N^{2}-1)(N^{2}-4)\,,\qquad\langle(5,3)(5,3)\rangle\,=\,0\,.$
(50)

The way the operators are defined in Section 2 the two norms must be equal in
field theory; a zero result would be non-unitary. In the hexagon picture, the
twisted translation is the reason for this quirky accident because in the
second case a $\langle YY\rangle$ contraction cannot be avoided. The issue
will not only affect the (5,3) state but indeed any $SU(2)$ sector operator
beyond _half filling_. These are always descendents, but as the example shows
they will occur in higher-trace admixtures to single-trace primaries. In this
situation we should unfortunately expect a breakdown of the integrability
picture we are trying to develop, or at the very least we have to accept that
all $Y$ correlators cease to be norm-like despite of the interpretation of $Y$
as its own conjugate. The problem is a new type of _finite size effect_.

Second, the notion of an _entangled state_ allows us to split Bethe states
[26, 7]. This is best explained on the simplest non-trivial example: let us
assume there are two magnons and bringing these from one hexagon to the other
we would cross a bunch of $l$ propagators. The splitting is accomplished by:

$\displaystyle\psi(\\{u_{1},u_{2}\\})\,\rightarrow\,$ (51) $\displaystyle\ \
\psi_{1}(\\{u_{1},u_{2}\\})\psi_{2}(\\{\\})-e^{i\,p_{2}\,l}\psi_{1}(\\{u_{1}\\})\psi_{2}(\\{u_{2}\\})-e^{i\,p_{1}\,l}S_{12}\psi_{1}(\\{u_{2}\\})\psi_{2}(\\{u_{1}\\})+e^{i\,(p_{1}+p_{2})}\psi_{1}(\\{\\})\psi_{2}(\\{u_{1},u_{2}\\})$

Higher cases work analogously. Like in the normalisation condition (36) it
crucial to adhere to the ordering in which the magnons are originally put on
the first few sites of the Bethe state we want to split. An $S$ matrix must be
inserted for every magnon overtaking another one upon jumping to the second
portion of the state. Last, the process may be iterated splitting the second
portion again to arrive at a triple _partition_ etc.

When calculating a structure constant with more than one renormalised operator
the questions arises whether the result depends on the relative starting
points of the partitions. For instance, for two non-trivial operators in a
three-point function we could start both on the same hexagon or on different
ones, also we can flip back and front of the figure. It is the Bethe equations
and the momentum constraint that guarantee consistency: if we started the
partitioning in (51) on the other hexagon, the magnons would transverse the
other edge of width $L-l$. Keeping track of
$\psi_{1}(\alpha)\,\psi_{2}(\bar{\alpha})$ we find the conditions

$e^{i\,p_{1}\,l}\,S_{12}\,=\,e^{i\,p_{2}\,(L-l)}\,,\qquad
e^{i\,p_{2}\,l}\,=\,e^{i\,p_{1}\,(L-l)}\,S_{12}$ (52)

which reduce to the Bethe equation for the particle with $p_{1}$ upon putting
$p_{2}\,=\,-p_{1}$.

Seeking regular solutions of the Bethe equations at given $L,\,n$ one has as
many equations as unknown rapidities. Due to the high order of the equations
(written as polynomials) there will be several solutions, but the solution set
is always discrete. The momentum constraint selects the cyclic cases, so those
that are able to describe single-trace operators. In deriving the Bethe
solutions for our $(2,1),\,(5,3),\,(6,3)^{e}$ solutions with twist we have
used the momentum constraint in that vein: it is violated at
$\epsilon\,\neq\,0$ but its leading order singles out the desired solution.
Indeed, multiplying all Bethe equations we find

$e^{i\,n\,\epsilon}\,e^{i\,L\sum_{j=1}^{n}p_{j}}\,=\,1\qquad\Rightarrow\qquad
e^{i\sum_{j=1}^{n}p_{j}}\,=\,e^{-i\,\frac{n}{L}\,\epsilon}\,.$ (53)

The hexagon itself is an off-shell object like the Bethe wave function prior
to imposing cyclicity. On the other hand, we might guess that the twist of the
Bethe equations should appear in building the entangled state as

$e^{i\,p\,l}\,\rightarrow\,e^{i\,n\,\epsilon}\,e^{i\,p\,l}\,.$ (54)

Let us modify the argument about the partition independence of the two-magnon
entangled state. We define that any magnon crossing the edge of width $l$
picks up $n_{A}$ units of twists and any magnon crossing the edge of width
$L-l$ should acquire $n_{B}$ units of twist. Equating the coefficients of
$\psi_{1}(\\{u_{1},u_{2}\\})\psi_{2}(\\{\\})$ in the two situations we find
can fix $n_{A}$, looking at $\psi_{2}(\\{\\})\psi_{1}(\\{u_{1},u_{2}\\})$ we
determine $n_{B}$. The result is

$n_{A}\,=\,\frac{l}{L}\,,\quad n_{B}\,=\,1-\frac{l}{L}\,.$ (55)

The coefficients of
$\psi_{1}(\\{u_{1}\\})\psi_{2}(\\{u_{2}\\}),\,\psi_{1}(\\{u_{2}\\})\psi_{2}(\\{u_{1}\\})$
are then automatically partition independent. Also in a range of cases with
more edges (so multiple partitions) and more magnons we could confirm that
$n_{i}\,=\,l_{i}/L$ (where $\sum l_{i}\,=\,L$) assures partition independence.
This solution realises $\sum n_{i}\,=\,1$ as expected from the Bethe
equations. On the other hand, $n_{i}$ is given different values at the
opposite ends of an edge connecting operators of unequal length. Last, the
outcome $n\,=\,l/L$ is perhaps not a surprise — not catering e.g. for some
jump in the twist at a certain point of the chain we have implicitly assumed
an even distribution of the twist.

Unfortunately, this elegant manner of regularising the hexagon amplitudes does
not reproduce the results of tree level field theory, cf. Section 6 on the
descendent correlators relevant to the $(7,3)^{+}$ mixing problem. In trying
to design a viable scheme one can develop some fantasy: should the $n$’s
around a chain really sum up to 1? Should $n$ depend on the edge width? On the
same edge, do we have the same $n$ at either side? Should we put in a twist
factor also for a magnon from an operator with a straight, so untwisted
solution? Should the twist factors be more general functions?

For the time being we dropped the last idea because experiments with twist
factors like $e^{i\,p\,\epsilon}$ do not seem to yield a suitable series
expansion. For transverse magnons $X,\bar{X}$ the following principles proved
helpful:

  * •

Whenever a magnon comes from a solution with twist $e^{i\epsilon}$ in the
Bethe equation we associate extra factors $e^{i\,n_{j}\,\epsilon}$ to the
edges crossed. So for untwisted Bethe solutions these extra factors are
trivially 1. Below we will occasionally scale
$\epsilon\,\rightarrow\,c_{0}\,\epsilon$ in both, the Bethe
equations/solutions and the extra factors.

  * •

We will be pragmatic and put in such factors everywhere, not necessarily
assuming that they are equal at both ends of the same edge. The condition that
all $n$’s of any one operator must add up to the twist in its Bethe equation
is then usually empty, because the last edge is not crossed in forming the
partitions.

  * •

We impose consistency conditions: the independence of scaling the twist of any
operator by $c_{0}$, and (where that becomes relevant) the existence of the
two-point limit in calculating two-point functions of double-trace operators.

  * •

We allow $\\{n_{j}\\}$ to be specific to the partitioning chosen on a given
diagram. For fixed partitioning we rather look for solutions independent of
the edge width.

For the transverse magnons this program is quite successful: the $n$’s are
usually 0 and sometimes 1; we emphasise that the non-trivial case does
occasionally appear. The coefficients at the ends of the same edge should
apparently be equal. Therefore, the condition $\sum n_{i}\,=\,1$ around a
chain does yield a helpful and consistent constraint from time to time.
Unfortunately, as the aforementioned problems with the interpretation of all
$Y$ correlators might have suggested, nothing works to plan for longitudinal
magnons, cf. Appendix C.

Let us illustrate the procedure and its problems on the example of the leading
$N$ norm of the $(5,3)$ operator: we use the original hexagon construction for
a three-point functions on the sphere with two three-magnon operators and an
identity. Let the non-trivial operators both be (5,3), once realised with $X$
magnons and once with $\bar{X}$ excitations. The fact that the two Bethe
solutions are equal makes the computation run into the particle creation poles
$1/(u-v)$ with $u$ a rapidity of the first operator and $v$ one of the second.
These poles have to cancel between the various partitions. In practice,
factoring them out of the complete amplitude is quite hard already for two
sets of three off-shell rapidities. Introducing twist factors on the edges
further complicates the issue. On the other hand, we can use the twist
$\epsilon$ as a regulator if we rescale
$\epsilon\,\rightarrow\,c_{0}\,\epsilon$ for the second operator. The particle
creation poles will then appear as $1/(1-c_{0})$. Organising the entangled
states such that the magnons from both operators are shifted over the common
width 5 edge we calculate the hexagon amplitude

${\cal
A}\,=\,\frac{20\,i\,c_{0}\,\epsilon^{2}(1-c_{0}-n_{1}+c_{0}\,n_{2})(1-c_{0}+n_{1}-c_{0}\,n_{2})(4-4\,c_{0}+5\,n_{1}-5\,c_{0}\,n_{2})}{(1-c_{0})^{3}(a_{1}-a_{2})^{6}}+O(\epsilon^{3})\,.$
(56)

For every operator, this is to be divided by the root of the same phase factor
and Gaudin determinant that we have encountered in the normalisation of the
Bethe state given in (36). The factor $\sqrt{\prod_{i}u_{i}^{2}+1/4}$ is not
needed in the hexagon picture. To leading order in $\epsilon$ we compute $i$
for the phase and for the Gaudin norm $80\,\epsilon^{2}$ or
$80\,c_{0}^{2}\,\epsilon^{2}$ at the first and second point, respectively. We
should thus normalise by $80\,i\,c_{0}\,\epsilon^{2}$.

Therefore, we demand

$\frac{\partial\,{\cal A}/c_{0}}{\partial\,c_{0}}\,=\,0\,.$ (57)

to guarantee a well-defined coincidence limit $c_{0}\,\rightarrow\,1$. This
imposes $n_{1}\,=\,n_{2}$ for the coefficients at opposite ends of the edge.
With that

${\cal
A}\,=\,\frac{20\,i\,c_{0}\,\epsilon^{2}(1-n_{1})(1+n_{1})(4+5\,n_{1})}{(a_{1}-a_{2})^{6}}\,,$
(58)

whereby the desired result is obtained putting $n_{A,1}\,=\,0$. This looks as
if the regulator is not needed at all. It is instructive to repeat the
exercise building the two entangled states starting on the same hexagon. To
this end, let us move the magnons of the second operator over the
corresponding width 0 edge. Now, condition (57) reads $n_{2}\,=\,1-n_{1}$.
With that (58) is reproduced. Next, let us move the magnons of both states
over the respective width zero edges. Instead of formula (56) we find the same
with $n_{j}\,\rightarrow\,1-n_{j}$. The consistency reads again
$n_{1}\,=\,n_{2}$ and to reproduce the desired outcome we must set
$n_{1}\,=\,n_{2}\,=1$.

In conclusion, even if the twist is invisible organising the partitions as we
first did, there is one unit of twist bringing each magnon once around the
chain. This is lumped on the edges of width 0, so away from the edge
connecting the two operators with equal rapidities.

Repeating the exercise for two equal operators with $Y$ magnons we obtain

${\cal
A}\,=\,\frac{20\,i\,c_{0}\,\epsilon^{2}(1-c_{0}+n_{1}-c_{0}\,n_{2})(4-4\,c_{0}+5\,n_{1}-5\,c_{0}\,n_{2})(c_{0}\,n_{1}-n_{2}+n_{1}n_{2}-c_{0}\,n_{1}n_{2})}{(1-c_{0})^{3}(a_{1}-a_{2})^{6}}\,.$
(59)

The independence of $c_{0}$ requires $n_{1}\,=\,n_{2}$ as before, with which

${\cal
A}\,=\,\frac{20\,i\,c_{0}\,\epsilon^{2}(1-n_{1})(1+n_{1})(4+5\,n_{1})\,n_{1}}{(a_{1}-a_{2})^{6}}\,.$
(60)

As for the transverse magnons, the other patterns of partitioning yield the
same upon successively mapping $n_{j}\,\rightarrow\,1-n_{j}$. Our observations
about the location of the twist carry over.

Interestingly, the amplitude with longitudinal magnons is the one for
transverse magnons times another factor $n_{1}$. Clearly, $n_{1}\,=\,0$
consistently reproduces the unwanted field theory result. Unfortunately, there
is no real value for $n_{1}$ yielding $80\,i\,c_{0}\,\epsilon^{2}$.

Finally, we recall that the hexagon amplitudes do not contain colour factors
[10].

##  5 Hexagon tessellations for higher-point functions

In [8, 9, 10] it was argued that the hexagon operator can be used to compute
higher-point functions of the ${\cal N}\,=\,4$ theory from integrability, too.
For our purposes it will be good enough to stick to the restricted kinematics
on the real line introduced above. In [8, 10] an orbital factor

$v_{i;jk}\,=\,\frac{1}{a_{i}-a_{j}}-\frac{1}{a_{i}-a_{k}}$ (61)

per scalar magnon on edge $i$ was included into every hexagon. These labels
refer to the physical edges, and in the turning sense of the hexagon $j(k)$
are to the left(right) of $i$. Since $v_{i;jk}$ is antisymmetric under
$j\,\leftrightarrow\,k$ it flips sign switching from the first hexagon in
Figure 1 to the second which has the opposite orientation. This explains the
minus sign in (51) for transferring a magnon from the first portion of the
entangled state to the second. Including the position vectors into the
hexagons we have to write a plus sign instead.

The claim is then that any triangulation of a punctured Riemann surface can be
used to compute the contribution of a given tree level graph, if the latter
can be drawn along the edges of the tiles.

Figure 2: a four-point diagram and the associated colour graph for edge widths
$\\{2,1,2,1,2,1\\}$

Consider the four-point graph in the left panel of Figure 2 with two single-
trace operators of length 6 at the punctures 1 and 3 and two operators of
length 3 at the two other points 2,4. In the right panel we have opened up the
graph on the plane to make it simpler to draw the seven propagators composing
a planar graph on the sphere. We can translate the sample _ribbon graph_ on
the right panel into the colour factor

$\mathrm{Tr}(T^{1}T^{2}T^{3}T^{4}T^{5}T^{6})\,\mathrm{Tr}(T^{2}T^{1}T^{7}T^{4}T^{8}T^{9})\,\mathrm{Tr}(T^{6}T^{5}T^{7})\,\mathrm{Tr}(T^{3}T^{9}T^{8})$
(62)

where $T^{i}$ is a short for $T^{a_{i}}$. Note that the indices from lines
connecting two operators (above 1,2) occur anticyclically within the second
trace when they are cyclically positioned on the first. We can stratify the
contributions to free field theory correlators by such colour factors. Their
coefficients in the tree QFT results with effective propagators are faithfully
reproduced by hexagon amplitudes with the right set of edge widths [8, 10].
The latter should neither depend on the tiling nor the way the entangled
states are organised.

We will try to reproduce double-trace/double-trace contributions from the
diagram in Figure 2, Panel 1. As a first example, consider two $(63)^{e}(30)$
operators. As above, the single-trace parts of length 6 are placed at points
1,2 and the length 3 parts at points 3,4 with the intention to take a
coincidence limit $a_{3}\,\rightarrow\,a_{1},\,a_{4}\,\rightarrow\,a_{2}$. As
usual, for normal-ordered operators there should be no self-contractions.
Therefore we are looking for tree graphs of the form of an _empty square_ with
$l_{E,b}\,=\,0\,=\,l_{E,f}$. There are the four possibilities
$\\{l_{A},\,l_{B},\,l_{C},\,l_{D}\\}\,=\,\\{6,0,3,0\\},\,\\{5,1,2,1\\},\,\\{4,2,1,2\\},\,\\{3,3,0,3\\}$.
The ribbon graph for the third of these is illustrated on the right panel of
Figure 2.

To handle the particle creation poles we scale the twist of the second
operator by $c_{0}$ as explained above. The $(6,3)^{e}$ operators behaves like
any other primary state: potential twist factors in the entangled state do not
even appear in the $O(\epsilon^{0})$ term of the normalised hexagon
amplitudes. For both, transverse and longitudinal magnons we obtain the
results $1,\,1/2,\,1/6,\,0$, respectively. For comparison, omitting the space-
time factor, free field theory yields
$N^{9}-10\,N^{7},\,9\,N^{7},\,3\,N^{7},\,0$, respectively, at the relevant
orders in $N$. These results are also valid for both, transverse and
longitudinal magnons. As observed in [10], the connected part of the hexagon
amplitudes needs to be scaled up by a factor
$\sqrt{L_{1}L_{2}L_{3}L_{4}}\,=\,18$ to obtain a match. Note that the product
of the sphere colour factors in the disconnected part is
$N^{9}-9\,N^{7}+\ldots$ The discrepancy is accounted for by the torus part of
the $(6,3)^{e}$ two-point functions times the sphere part of the $(3,0)$ two
point-function and vice versa, cf. Appendix B.

Within the hexagon framework we can draw another class of planar tree graphs
on the sphere: placing the two length 6 operators at points 1 and 3 of the
diagram, and the two length 2 operators at points 2 and 4, we can put
$l_{A}\,=\,0\,=\,l_{C},\,l_{B}\,=\,3\,=l_{D}$. The coincidence limit is now
$a_{2}\,\rightarrow\,a_{1},\,a_{4}\,\rightarrow\,a_{3}$. Below, we will refer
to this topology as _belt around the belly_. Now, the three propagators
connecting the length 6 operators can be distributed around the equator of the
sphere as
$\\{l_{E,b},l_{E,f}\\}\,=\,\\{3,0\\},\,\\{2,1\\},\,\\{1,2\\},\,\\{0,3\\}$.
Here, the $\\{3,0\\}$ and $\\{0,3\\}$ cases should both be equal to the
$\\{3,3,0,3\\}$ case in the last paragraph. We need to take into account only
one of the three realisations of this graph. Yet, generically the $\\{2,1\\}$
and $\\{1,2\\}$ cases would both be needed, cf. [8]. This is in fact a feature
of the tree field theory computation, too. In the case at hand these graphs
all vanish, though.666The twist regulator could bring them back by the
$n_{E,b},n_{E,f}$ coefficients. We learn by comparison with field theory and
the empty square case that these $n$ coefficients must vanish.

The complete two-point functions of the $(9,3)_{i}^{-}+\ldots$ operators
contain a cross term $\langle(6,3)^{e}(3,0)*(7,3)^{-}(2,0)\rangle$. We compute
separately for the two $(7,3)^{j}$ Bethe states and then form a difference. In
field theory as well as in the hexagon picture the amplitudes neither depend
on the twist regulator nor on which type of excitation is chosen. The possible
tree diagrams are $\\{6,0,2,1\\},\,\\{5,1,1,2\\},\\{4,2,0,3\\}$. In the first
case, the colour factor identically vanishes. Field theory yields
$-2\sqrt{3},\,0$ for the first Bethe state in the two other cases,
$+2\sqrt{3},\,0$ for the second.

Upon multiplication by $\sqrt{2367}$ the hexagon amplitudes yield the same, on
the condition that we reverse the sign of the amplitude for the first (7,3)
Bethe state. We have no explanation for this rule. The _belt_ diagrams non-
trivially add up to zero in a surprisingly non-trivial manner:

$(l_{E,b},l_{E,f})$ | hexagon tiling
---|---
(4,0), (0,4) | $\frac{1}{2\sqrt{21}}$
(3,1) | $-\frac{1}{4\sqrt{21}}-i\frac{\sqrt{5}}{12\sqrt{7}}$
(2,2) | 0
(1,3) | $-\frac{1}{4\sqrt{21}}+i\frac{\sqrt{5}}{12\sqrt{7}}$

The four colour factors can be distinguished by the ribbon structure but are
identically equal upon evaluation. Again, a sign needs to be amended on the
integrability side for $\langle(6,3)^{e}(3,0)*(7,3)^{+}(2,0)\rangle$ to be
absent and $\langle(6,3)^{e}(3,0)*(7,3)^{-}(2,0)\rangle$ to exist.

Last, at the leading order in $N$, the two-point functions of $(7,3)^{j}$ in
free field theory as well as integrability (times 14 in the connected part)
for the $\\{7,0,2,0\\},\,\\{6,1,1,1\\},\,\\{5,2,0,2\\}$ tree graphs have the
matrix form

$N^{9}\,\left(\begin{array}[]{cc}1&0\\\ 0&1\end{array}\right),\qquad\
8\,N^{7}\,\left(\begin{array}[]{cc}1&0\\\
0&1\end{array}\right),\qquad\frac{20}{3}N^{7}\,\left(\begin{array}[]{cc}1&1\\\
1&1\end{array}\right)\,.$ (63)

Again, in the integrability picture an extra minus sign on one of the states
is required to match field theory. Considering the _belt_ cases we see a
cancellation of imaginary parts within the pairs $\\{4,1\\},\,\\{1,4\\}$ and
$\\{3,2\\},\,\\{2,3\\}$ and the sum
$\\{5,0\\}+\\{4,1\\}+\\{3,2\\}+\\{2,3\\}+\\{1,4\\}$ is as stated. Since there
is no twist regulator at our command to control the particle creation poles we
have calculated numerically to high precision shifting the second set of
rapidities by a small increment. The error comes at the expected order so that
it is easy to truncate appropriately and match by the exact results stated
above.

Our $(7,3)^{+}$ mixing example includes double-trace/double-trace two-point
functions of the (5,3)(2,0) and (5,2)(2,1) operators involving descendents.
The situation here is more complicated in two ways: with transverse magnons we
occasionally need the twist regulator with non-trivial $n$’s as shown in
Section 6. The regularised $Y$ correlators in the integrability picture could
(almost) fit the $X$ combinatorics but hardly the results for longitudinal
magnons themselves, which deviate from the $X$ correlators. This is so even
for $\langle(5,2)(2,1)*(5,2)(2,1)\rangle$, which is not beyond half filling in
any obvious way. Tables are given in Appendix C.

Figure 3: diagrams for single-trace/double-trace correlators

Next, we also wish to check the agreement between field theory and
integrability for the single-trace/double-trace parts of correlators at
leading order and on the torus. Here the first panel of Figure 3 may be
employed, with a length 2 or 3 operator at point 3 and an identity insertion
at point 4. For the primary operators we have prevalently calculated with
longitudinal magnons; checks indicate that transverse magnons yield identical
results, both in field theory and integrability. The space time dependence of
the correlators is the expected factor $1/(a_{1}-a_{2})^{6}$ whereby no
problem with the coincidence limit can occur. We emphasise the following two
points: first, when the two operators carrying excitations are linked by only
one edge of the maximum width (or equivalents by cyclicity) the colour factor
is of order $N^{L-1}$ with $L$ the greatest length. Yet, as these diagrams are
connected (albeit _extremal_) the same combinatorics applies as in the
connected part of the double-trace/double-trace analysis above: the sphere
part of the integrability computation needs to be scaled up by
$\sqrt{L_{1}L_{2}L_{3}}$ like the true torus cases777The hexagon amplitude for
the graph in Panel 2 of Figure 3 with edge widths $\\{2,1,2,1,2,1\\}$
overcounts the field theory result by a factor of 3 in this normalisation due
to its threefold cyclic symmetry. The effect was first noticed in [10]..
Second, for the parity pairs we have to introduce an extra sign on
$(9,3)^{1},\,(7,3)^{1}$ again.

However, introducing the third non-trivial puncture into the game, two more
types of colour factors can arise: first, the part of the double trace
operator with the shorter length can lie inside one of the ribbons between the
two longer operators, see Panel 2 of Figure 3. Second, in the $(9,3)^{-}$
problem we have mixing with $(6,3)^{e}(3,0)$. Obviously, we need to able to
connect the length 9 and the length 3 operator with three single lines, which
cannot be realised on the first two tilings. Here we have used the diagram in
Panel 3.

The parity even $\langle(7,3)^{+}*{\cal O}_{\perp}\rangle$ correlator also
requires the sign flip on the $(7,3)^{1}\ldots$ hexagon amplitudes. Then
everything works out with transverse as well as longitudinal magnons without
any twist-like modification of the entangled states. We obtain

$\braket{(7,3)^{+}*(5,3)(2,0)}\,=\,2\,\frac{\sqrt{10}}{\sqrt{3}}\,(N^{6}-5\,N^{4}+\ldots)\,=\,-\sqrt{2}\,\braket{(7,3)^{+}*(5,2)(2,1)}\,.$
(64)

Finally, we need single-trace/single-trace two-point functions on the sphere,
the torus, and the double torus. In principle, tessellating the latter works
in the obvious way and could be handled as described above. Yet, the workload
quickly grows out of hand because there are relatively many distinct sets of
edge widths. Consequently, we used a Mathematica script to generate all Wick
contractions, to classify these by the colour factors as in the right panel of
Figure 2 or equation (62), and to automatically embed the corresponding graph
on the surface of the smallest possible genus. We have chiefly studied
amplitudes for longitudinal magnons.

Where twist is not around to help with the particle creation poles we have
regularised by a small shift of one set of rapidities as was already done to
obtain (63). For the double torus computations we have used minimally 100
digits of precision and a shift of order $10^{-25}$ to arrive at 20 digits of
precision in the output; we kept the precision fairly low to save on computer
time.

Figure 4: a tessellation of the double torus with a colour ribbon graph for
the $\braket{(5,2)*(5,2)}$ correlator marked on it. The length five operators
are at points 1 and 2. Four identity operators were added to achieve a
triangulation. Actual propagators are depicted as solid lines, auxiliary edges
of width zero as dotted lines.

The simplest example in which a genus 2 diagram contributes is the norm of the
(5,2) operator. Due to its low length the operator does not receive admixtures
so that all diagrams can be embedded on the double torus. Integrability
supplemented by colour factors exactly reproduces the norm stated in (50).

##  6 Twisting $\langle XXX\,\bar{X}\bar{X}\bar{X}\rangle$

###  6.1 $\mathbf{\langle(5,3)(2,0)*(5,3)(2,0)\rangle}$

Like in the $(6,3)^{e}(3,0)$ two-point function we put the single-trace
operators carrying excitations at points 1 and 2 and their complements (2,0)
in the double trace products at points 2 and 4. The coincidence limit
$3\,\rightarrow\,1,\,4\,\rightarrow\,2$ is even trivial for the _empty square_
diagram with $l_{E,b}\,=\,0\,=\,l_{E,f}$ in Figure 2, Panel 1. To regularise
the particle creation poles we rescale once again the regulator at the second
point, i.e. we employ $c_{0}\,\epsilon$ there. In the
$\epsilon\,\rightarrow\,0$ limit there are infinite rapidities so that the
situation resembles the norm example at the end of Section 4. The $c_{0}$
independence condition (57) forces

$n_{A,1}\,=\,n_{A,2}$ (65)

which is common to all four cases in the table below, even prior to the two-
point limit. Omitting the obvious space time factor $1/(a_{1}-a_{2})^{6}$:

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | 1 | $\phantom{10*}\ \ \frac{1}{4}\,(1-n_{A,1})(1+n_{A,1})(4+5\,n_{A,1})$
4111 | 4 | $10*\frac{1}{20}(1+n_{A,1})(2-5\,n_{A,1})(4+5\,n_{A,1})$
3202 | 1 | $10*\frac{1}{20}(2+5\,n_{A,1})(1-8\,n_{A,1}-5\,n_{A,1}^{2})$

In the field theory results we have indicated the leading $N$ coefficient
only. Clearly, for a match we must put $n_{A,1}\,=\,0$. For a test we have
calculated the $\langle(5,3)(2,0)*(5,3)(2,0)\rangle$ hexagon amplitudes also
sending the magnons of the operator at point 2 over the edge $B$. Then $c_{0}$
independence implies $n_{A,1}\,=\,1-n_{B,2}$ as in the norm example in Section
4. Turning the entangled state at point 1 nothing changes as long as the edge
$A$ is crossed. On the other hand, if the magnons go over the edges
$E_{f},\,D,\,E_{B}$ we find the same with
$n_{A,1}\,\rightarrow\,1-n_{E,f}-n_{D,1}-n_{E,b}$ and to match QFT one has to
assign $n_{E,f}+n_{D,1}+n_{E,b}\,=\,1$. Hence we cannot resolve exactly where
the twist has to go.

For the _belt around the belly_ version we put the two non-BPS operators at
points 1 and 3 and the vacua at 2 and 4. On the equator lines of the diagram
we have in principle four different indices — one $n_{E,b}$ and one $n_{E,f}$
at each point. For brevity we have put all of these equal. The $c_{0}$
independence constraint is then empty. We organise the partitions such that
the magnons are clockwise brought over the backwards equator edge first. The
excitations thus never cross the two edges $B,\,D$ of non-vanishing length, so
that their $n$ coefficients cannot appear. We actually find results without
the $n_{A},\,n_{C}$ either:

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | 1 | $10*\frac{1}{10}(1+9\,n_{E,b}+10\,n_{E,b}^{2})$
{2,1}0202 | 2 | $10*\frac{1}{10}(1-3\,n_{E,b}+10\,n_{E,b}^{2})$

Once again, we must drop the regulator. Note that the second line in the table
shows the necessity of doubling the $\\{2,1\\}$ graph. Upon tracking the
colour factors the field theory result actually contains both, a $\\{2,1\\}$
and a $\\{1,2\\}$ part while $\\{3,0\\}$ can only be realised in one way.

In the light of our previous experience an interesting experiment to make is
to move the magnons of the left operator over the edges $D,\,E_{b},\,A$ and
those of the right operator over $C,\,E_{f},\,B$. From (57) we find the two
constraints

$n_{A,1}+n_{D,1}\,=\,1-n_{E,b}-n_{E,f}\,,\qquad
n_{B,3}+n_{C,3}\,=\,1-n_{E,b}-n_{E,f}\,.$ (66)

If we now assume $n_{E,f}\,=\,n_{E,b}$ we fall upon the results in the table
whereby $n_{E,b}\,=\,0$. Nonetheless, we have learned $n_{A,1}+n_{D,1}\,=\,1$
and $n_{B,3}+n_{C,3}\,=\,1$ and so the twist exists, obeys the familiar sum
rule from the Bethe equations. Once again, we cannot precisely localise it.
Here it is distributed around the square frame, i.e. not on the lines
connecting operators with identical rapidities.

###  6.2 $\mathbf{\langle(5,3)(2,0)*(5,2)(2,1)\rangle}$

We will now put the (5,3) operator at point 1, its partner (2,0) at the
opposite point 3, and the (5,2) primary at point 2 its partner, the vacuum
descendent (2,1) at point 4 of the _empty square_ in Figure 2, panel 1. We
will associate $n$ coefficients to the states at points 1 and 4; the (5,2)
operator at point 2 is undeformed, so it seems logical to not associate it
with $\epsilon$ at all.888In a more general ansatz one might artificially
introduce twist into the (5,2) Bethe equations and drag such coefficients
along. At point 1 we organise the partition as before, while at point 4 we
move the magnons clockwise over the edge $C$. Coefficients to be expected in
the result are now thus $n_{A,1},\,n_{D,1}$ and $n_{E,f}$ at point 1 as well
as $n_{C,4}$.

Again, we will distinguish the twist at point 4 from that at point 1 by a
scale $c_{0}$ as one must be able to move both parameters independently. The
amplitudes at $O(\epsilon^{2})$ are polynomials of $c_{0}$ and the
$n^{\prime}s$ over a denominator $2-c_{0}$. Clearly we should impose (57)
which yields

$n_{D,1}\,=\,2\,(1-n_{C,4})$ (67)

as a simple solution common to all three cases in the table below. Remarkably,
had we assumed equal coefficients at both ends of the $D$ egde, so
$n_{D,1}\,=\,n_{D,4}$, the twisted Bethe equation (31) would have made us
expect $n_{D,1}\,=\,1-n_{C,4}$. Reconciling this with the $c_{0}$ independence
of the formula singles out $n_{C,4}\,=\,1$, which will indeed be our result,
see below.

Employing (67) we obtain

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | 0 | $\phantom{10*}\ \,\frac{\sqrt{2}}{4}(1+n_{A,1})(4+5\,n_{A,1})(-1+n_{C,4})$
4111 | $\sqrt{2}$ | $10*\frac{\sqrt{2}}{40}(4+5\,n_{A,1})(-5-5\,n_{A,1}+6\,n_{C,4}+10\,n_{A,1}n_{C,4})$
3202 | $\sqrt{2}$ | $10*\frac{\sqrt{2}}{20}(2+5\,n_{A,1})(-2+3\,n_{C,4}+5\,n_{A,1}n_{C,4})$

for the leading $N$ parts. Demanding equality of field theory and normalised
hexagon amplitudes simultaneously in all three cases is possible and uniquely
determines

$n_{A,1}\,=\,0\,,\qquad n_{C,4}\,=\,1\,.$ (68)

It is also possible to keep the rapidities of the (5,2) operator undetermined
during the $\epsilon$ expansion. The particle creation pole factors out upon
which we can substitute the Bethe solution $u_{2}\,=\frac{1}{2}\,=\,-u_{1}$.
The $\epsilon$ expansion does not bring out $n_{A,1}$ in this case whereas
$c_{0}$ independence still creates the same constraint. We directly land on
the three results in the table evaluated at $n_{A,1}\,=\,0$. With hindsight,
this and the Bethe argument uniquely pin down the three amplitudes. In
conclusion, this example requires introducing a single factor
$e^{i\,c_{0}\,\epsilon}$, namely when the magnon of (2,1) transverses the edge
$C$ independently of the edge width. We could carry away a second message:
there should be no unit of twist on lines connecting deformed operators.

Reorganising the partitions we can generate the transformations
$n_{D,1}\,\rightarrow\,1-n_{E,b}-n_{A,1}-n_{E,f}$ and/or
$n_{C,4}\,\rightarrow\,1-n_{D,4}$ on the set of parameters. As expected by
now, the content of the equations does not change. In particular, the second
move allows us to drop all $n$ coefficients although the twist is inherently
present. The example illustrates very well how the descendent amplitudes may
collapse or come out wrong for some choice of partitions when no $n$’s are
used.

In the _belt_ embedding we put the (5,3) operator at point 1, (2,0) at point
2, (5,2) at point 3 and finally (2,1) at point 4. The magnons at point 1 go
over $l_{E,b},\,l_{A},\,l_{E,f}$ as before and we also move that of (2,1) over
edge $C$. Again, for simplicity we assume $n_{E,b}\,=\,n_{E,f}$. The $c_{0}$
independence constraint valid in both cases is

$n_{E,b}\,=\,-\frac{1}{2}-\frac{n_{A,1}}{2}+n_{C,4}\,,$ (69)

with which:

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | $\sqrt{2}$ | $10*\frac{\sqrt{2}}{20}(1+5\,n_{A,1}+n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})$
{2,1}0202 | $-2\sqrt{2}$ | $10*\frac{\sqrt{2}}{20}(-5-5\,n_{A,1}+13\,n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})$

Taking into account the doubling of the second diagram the two equations have
the simultaneous solutions

$n_{A,1}\,=\,1,\ n_{C,4}\,=\,1\qquad\lor\qquad n_{A,1}\,=\,\frac{2}{5},\
n_{C,4}\,=\,\frac{1}{2}\,.$ (70)

and so $n_{E,b}\,=\,0$ or $n_{E,b}\,=\,-\frac{1}{5}$. In the second solution,
the value of $n_{A,1}$ suggests to try the homogeneous distribution of twist
$n\,=\,l/L$ initially discussed. However, once again that fails to yield the
right results.

Aesthetically, the first solution seems more appealing — also, it is more in
line with what we have seen so far: the two sum rules we expect are
$1-n_{C,4}-n_{D,4}\,=\,0,\,=\,1-n_{A,1}-n_{E,f}-n_{D,1}-n_{E,b}$. Assuming
$n_{E,b}\,=\,0\,=n_{E,f}$ (equal rapidities at $\epsilon\,\rightarrow\,0$) and
$n_{D,j}\,=\,0$ (the edge connects two operators carrying twist) we could have
conjectured $n_{A,1}\,=\,1\,=\,n_{C,4}$. We can hide the twist choosing the
partitions such that the edges $A,\,C$ are avoided and putting all $n$’s to 0.

###  6.3 $\mathbf{\langle(5,2)(2,1)*(5,2)(2,1)\rangle}$

For the _empty square_ we put the two (5,2) operators at points 1,2 and the
two vacuum descendents at 3,4. To the latter we assign twist
$\epsilon,\,c_{0}\,\epsilon$ to regularise their particle creation pole in the
amplitude. We move both magnons over the $C$ edge. The (5,2) Bethe solution is
undeformed so that we cannot rescale any epsilon here; we will substitute
$u\,=\,\mp\frac{1}{2}$ for the first operator, expand in $\epsilon$, factor
the amplitude, and finally put the rapidities of the second (5,2) operator on
shell. As expected, the amplitudes have a pole $1/(1-c_{0})$ and regulator
independence implies $n_{C,3}\,=\,n_{C,4}$. We find the table

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | 1 | $\phantom{10*\frac{1}{10}}(1-n_{C,4})$
4111 | 5 | $10*\frac{1}{10}(5-6\,n_{C,4})$
3202 | 2 | $10*\frac{1}{10}(2-3\,n_{C,4})$

For a leading $N$ match it follows $n_{C,4}\,=\,0$, so the twist is hidden on
the edges $B,\,D$.

The _belt around the belly_ version has the two (5,2) operators at points 1
and 3 and the vacuum descendents at 2 and 4. The results are $n$ independent:

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | 2 | $10*\frac{1}{5}$
{2,1}0202 | 0 | 0

Fortunately, this comes out right!

##  7 Conclusions

In this article we have studied the operator mixing in the so-called $SU(2)$
sector in ${\cal N}\,=\,4$ super Yang-Mills theory. Specifically, we have
focused on multi-trace admixtures to single-trace operators in a number of
examples with the aim of constructing a large $N$ expansion of the form ${\cal
O}^{s}+1/N\,{\cal O}^{d}+1/N^{2}\,{\cal O}^{t}+\ldots$, where the subscripts
refer to single-, double-, and triple-trace operators.

As is well-known [3, 16, 11], the degeneracy of the spectrum poses a problem:
if a single-trace and a double-trace leading $N$ eigenstate have the same
anomalous dimension, we naively find the sum and difference of the two
operators as eigenstates, without an offset of $1/N$ in the relative
coefficients. Hence the large $N$ expansion is not of the desired form.
Degenerate perturbation theory can improve on this. It has also been suggested
to look at the $\beta$ deformed theory [27] instead in which the degeneracy is
lifted. The idea would be to define the states of the undeformed theory by the
limit $\beta\,\rightarrow\,0$ [20]. It would be interesting to address the
operator mixing also in such cases; clearly hexagon tilings will still
correctly reproduce the various overlaps.

Degeneracies are particularly frequent in the two-excitation spectrum. For
operators with more excitations the eigenvalue problems have characteristic
polynomials of higher degree whose roots rarely coincide with with those of
lower cases. This is so in particular for so-called primary states, i.e. those
that cannot be derived from others by $SU(2)$ raising. Adding in higher-trace
admixtures we found that their leading $N$ anomalous dimensions should also
differ from that of the single-trace state in question. Importantly,
degeneracies within the relevant set of multi-trace operators are apparently
irrelevant. Scanning the space of single-trace operators from length 2 to 10
with up to four excitations we found that the systematic $N$ expansion can be
constructed in most cases, at least to the order indicated above.

An interesting feature of the higher-excitation spectrum is the existence of
parity pairs, i.e. pairs of single-trace operators with degenerate leading $N$
anomalous dimension. In the exact planar limit we are free to choose any basis
for such a $2\,\times\,2$ cell. One can always assume this basis to consist of
the odd and even part under parity, here defined as the reversal of all colour
traces. Parity is in fact strictly respected in the operator mixing, and the
even and odd operators acquire distinct $1/N^{2}$ corrections to the common
leading $N$ anomalous dimensions.

To spare some work we studied three-excitation cases up to length 9 in which
no triple-trace operators occur, and in which the mixing does not involve too
many distinct operators. In order to compute two-point functions of the $N$
completed single-trace operators we obviously need to evaluate those of all
their constituents, which we were able to reproduce by hexagon tilings of the
sphere, the torus, and the double-torus. In a coincidence limit, three-point
functions on the sphere and the torus can be used to compute single-
trace/double-trace mixing to the required order, and four-point functions on
the sphere should reproduce the double-trace/double-trace contributions.

In a similar vein, the single-trace/double-trace (and vice versa) transitions
caused by the one-loop dilatation operator can be captured by an overlap
formula [20] in the spin chain picture. For simplicity, we have rather
imported the one-loop mixing matrices from field theory and then derived their
eigenvectors by traditional means. A third ingredient is a similarity
transformation by the (root of) the matrix of tree level two-point functions.
This is given by the very pieces that we did reproduce from integrability
supplemented by colour factors [10]. It would be worthwhile looking for a more
systematic approach to the entire diagonalisation process.

Admittedly, the current technology constitutes at most a proof of principle
because the integrability calculations are more cumbersome than tree level
field theory. Nonetheless, for primary operators — including an exceptional
case, which the Bethe ansatz misses unless twist is introduced — the
tessellation method works perfectly. Interestingly, we need to introduce an
extra sign on one of the Bethe states describing each degenerate pair.

In our picture, multi-trace operators are products of single-trace
eigenstates. As our $(7,3)^{+}$ example shows, higher-trace admixtures to
primary states can have multi-trace admixtures in which one or more of the
factors are descendents. The Bethe ansatz projects out descendent states, but
like the aforementioned exceptional operator they may be brought back
introducing twist into the Bethe equations.

The original hexagon construction [7] as well as its application to higher-
point functions [8, 9] cuts up the Riemann surface on which a graph can be
drawn. In this process also the single trace operator viz Bethe state at a
punture is cut into various pieces. As a consequence, we have to address the
question as to how this can be made compatible with twist. For the single-
trace/single trace and single-trace/double-trace two-point function we studied
there is no need to alter the cutting rules to reproduce field theory from
integrability.

The situation changes w.r.t. the double-trace/double-trace two-point functions
in our $(7,3)^{+}$ example: for transverse magnons in the terminology of [7]
an integrability/field theory match can be achieved using a number of
empirical rules:

  * •

We introduce a factor $e^{i\,\epsilon}$ into each Bethe equation for a
descendent state. Bringing a magnon once around the entangled state at the
corresponding puncture we should include a factor $e^{i\,n_{j}\,\epsilon}$
when the edge $j$ is crossed.

  * •

Homogeneously distributing the twist putting $n_{j}\,=\,l_{j}/L$ (with $l_{j}$
the width of the edge and $L$ the length of the operator) generically fails.
Instead, the twist is inhomogeneously distributed depending on the
tessellation and the position of the operators on it. Note that we did not
have to include twist into our map from Bethe states to field theory operators
which implies that the twist is lumped at one site of the chain.

  * •

In the examples studied, we cannot uniquely assign $n_{j}$ coefficients. Yet,
around an entangled state they should add up to 1 as expected from the Bethe
equations. If an edge connects two operators with twist, or two operators with
Bethe rapidities degenerate at zero twist, the corresponding $n$ coefficient
ought to vanish.

  * •

Partition invariance — the independence of the outcome of the way the Bethe
states are cut — is guaranteed in that different choices only lead to
reparametrisations but do not alter the results.

Imposing these rules of thumb the method does have predictive power. Yet, a
more comprehensive study is clearly needed: for once we have not considered
double or even higher descendents. Second, for higher-point functions with
descendents at many punctures it is not clear that the rules above can be
consistently imposed.

Furthermore, realising the $SU(2)$ sector with longitudinal magnons [7] we
have not found a way to reconcile integrability with field theory results for
our double-trace/double-trace two-point functions involving descendents at
both ends. The tree field theory results deviate form those for transverse
magnons in these cases, which is an effect of the twisted translation built
into the hexagon approach. Curiously, the hexagon amplitudes listed in
Appendix C fall upon the expressions for transverse magnons. If not resolved,
the issue causes a new type of finite size problem for integrability: a
systematic large $N$ expansion cannot be dealt with when single-trace
operators mix with descendents of shorter operators as factors of some multi-
trace admixture.

The symmetry underlying the twisted translation is at odds with the existence
of twist in this sector. It is therefore of vital interest to study for what
other excitations of the complete ${\cal N}\,=\,4$ super spin chain similar
difficulties arise. We might hope that a nested Bethe ansatz for the Dynkin
diagram employed in [5, 22] opens a way of removing the difficulties. In a
putative hexagon approach to the $\beta$ deformed theory this point will be of
central importance.

## Acknowledgements

B. Eden is supported by Heisenberg funding of the Deutsche
Forschungsgemeinschaft, grant Ed 78/7-1 or 441791296. We are grateful to D. le
Plat and T. Mc Loughlin for discussions about related material, and to C.
Kristjansen, A. Spiering and T. Mc Loughlin for comments on the manuscript.

## Appendix A: two-point functions up to $1/N^{4}$

#### Exceptional operator at length 6

The $(6,3)^{e}$ operator of equations (11), (40) has the two-point function

$\braket{(6,3)^{e}*(6,3)^{e}}\,=\,N^{6}\left(1-5\,N^{-2}+4\,N^{-4}\right).$
(71)

#### Degenerate pair at (7,3)

The Bethe states $(7,3)^{j}$ in (41) have overlaps

$\displaystyle\braket{(7,3)^{1}*(7,3)^{1}}\,=\,\braket{(7,3)^{2}*(7,3)^{2}}$
$\displaystyle=\,N^{7}\left(1-8\,N^{-2}+19\,N^{-4}+{\cal O}(N^{-6})\right),$
(72) $\displaystyle\braket{(7,3)^{1}*(7,3)^{2}}$
$\displaystyle=\,N^{7}\left(0+6\,N^{-2}-30\,N^{-4}+{\cal O}(N^{-6})\right)\,.$
(73)

From here it follows

$\displaystyle\braket{(7,3)^{-}*(7,3)^{-}}$
$\displaystyle=\,N^{7}\left(1-14\,N^{-2}+49\,N^{-4}+{\cal
O}(N^{-6})\right)\,,$ (74) $\displaystyle\braket{(7,3)^{+}*(7,3)^{+}}$
$\displaystyle=\,N^{7}\left(1-\phantom{1}2\,N^{-2}-11\,N^{-4}+{\cal
O}(N^{-6})\right)\,.$

While we are done with the parity odd two-point function, we still need to
discuss the effect of the admixtures to the parity even one. From (64) we find

$\braket{(7,3)^{+}*{\cal
O}_{\perp}}\,=\,-2\,\sqrt{5}\,(N^{6}-5\,N^{4}+\ldots)$ (75)

and adding the various results for transverse magnons (we drop the
longitudinal case for now due to the difficulties displayed in Appendix C):

$\braket{{\cal O}_{\perp}*{\cal O}_{\perp}}\,=\,N^{7}+N^{5}-26\,N^{3}+\ldots$
(76)

From here we could straightforwardly compute the two-point functions of the
two eigenstates (17) in Section 2. However, the mostly single-trace and the
mostly double-trace state are not orthogonal. As has been pointed out in the
literature [28], this can be mended by a similarity transform

$M\,\rightarrow\,S^{-\frac{1}{2}}\,M\,S^{\frac{1}{2}}\,,\qquad
S_{ij}\,=\,\braket{{\cal O}_{i}\,{\cal O}_{j}}_{\mathrm{tree}}\,.$ (77)

The root of the tree matrix $S$ has fairly complicated $N$ dependence. We
resort to expanding up to NNLO in $N^{-2}$:

$\sqrt{\frac{S}{N^{7}}}\,=\,\left(\begin{array}[]{ll}1-\frac{7}{2\,N^{2}}-\frac{3}{8\,N^{4}}&-\frac{\sqrt{5}}{N}+\frac{9\,\sqrt{5}}{4\,N^{3}}\\\\[2.84526pt]
-\frac{\sqrt{5}}{N}+\frac{9\,\sqrt{5}}{4\,N^{3}}&1-\frac{2}{N^{2}}-\frac{15}{4\,N^{2}}\end{array}\right)+\ldots$
(78)

The similarity transform maps $M$ as stated in (16) to

$M\,\rightarrow\,\left(\begin{array}[]{ll}5-\frac{5}{N^{2}}-\frac{5}{2\,N^{2}}&-\frac{3\,\sqrt{5}}{N}+\frac{3\,\sqrt{5}}{4\,N^{3}}\\\\[2.84526pt]
-\frac{3\,\sqrt{5}}{N}+\frac{3\,\sqrt{5}}{4\,N^{3}}&4+\frac{5}{N^{2}}+\frac{5}{2\,N^{4}}\end{array}\right)+\ldots\,.$
(79)

The root of the tree matrix is uniquely determined if we choose it real and
symmetric. Since it is not orthogonal the transformation can change the scalar
product of the eigenvectors; they become orthogonal.

The structure of the $N$ dependence is not blurred by the transformation,
although the one-loop matrix now has an infinite $N$ expansion. The closest to
what we have done in Section (2) is to look for an eigenvector of this new
mixing matrix of the form $\\{1,\,b\\}$, i.e. we complete the single-trace
state with admixtures without rescaling it. Our old mixing coefficients in
(17) are obviously related by the transformation (78) _up to an $N$ dependent
rescaling_ putting the first component of the eigenvector to 1. With this
definition

$b_{1}\,\rightarrow\,-3\,\sqrt{5}\,,\qquad\hat{b}_{1}\,\rightarrow\,\frac{423\,\sqrt{5}}{4}$
(80)

and finally:

$\braket{(7,3)^{+}_{c}*(7,3)^{+}_{c}}\,=\,N^{7}(1+45\,N^{-2}-\frac{6345}{2}\,N^{-4}+\ldots)$
(81)

#### Degenerate pair at (8,3)

The one-loop mixing matrix for the parity odd $L\,=\,8,\,n\,=\,3$ operators is

$M\,=\,\left(\begin{array}[]{ccc}4&0&0\\\ 0&6&\frac{6}{N}\\\
\frac{-2}{N}&\frac{6}{N}&6\end{array}\right)$ (82)

in the basis $\\{(8,3)^{-},\,(8,3)^{e},\,(6,3)^{e}(2,0)\\}$. This does have a
left-eigenvector $(1,0,0)$ as stated in Section 2. The similarity
transformation by the root of the tree level matrix results to:

$M\,\rightarrow\,\left(\begin{array}[]{lll}4+\frac{1}{2\,N^{2}}+\frac{209}{32\,N^{4}}&\frac{5}{4\,N^{2}}+\frac{4}{N^{4}}&\frac{-1}{N}+\frac{21}{8\,N^{3}}\\\\[2.84526pt]
\frac{5}{4\,N^{2}}+\frac{4}{N^{4}}&6-\frac{25}{32\,N^{4}}&\frac{6}{N}+\frac{11}{8\,N^{3}}\\\\[2.84526pt]
\frac{-1}{N}+\frac{21}{8\,N^{3}}&\frac{6}{N}+\frac{11}{8\,N^{3}}&6-\frac{1}{2\,N^{2}}-\frac{23}{4\,N^{4}}\end{array}\right)+\ldots$
(83)

Inspecting the latter matrix one might worry whether it is still possible to
complete the $(8,3)^{-}$ operator to a large $N$ eigenstate. In fact, there is
no problem:

$(8,3)^{-}_{c}\,=\,(8,3)^{-}-\frac{17}{8\,N^{2}}\,(8,3)^{e}+\left(\frac{1}{2\,N}+\frac{83}{16\,N^{3}}\right)\,(6,3)^{e}(2,0)+\ldots\,.$
(84)

It is clear that the state must survive because it can also be derived it by
acting on the vector of coefficients $(1,0,0)$ in the other basis by
multiplying with $\sqrt{S}$ and rescaling the first component. The two-point
function becomes

$\braket{(8,3)^{-}_{c}*(8,3)^{-}_{c}}\,=\,N^{8}\left(1+\frac{1}{4\,N^{2}}+\frac{621}{64\,N^{4}}+\ldots\right)\,.$
(85)

#### Degenerate pairs at (9,3)

Let us order the complete basis as

$\\{{\cal O}^{1,2,3},\,{\cal O}^{1,2,4},\,{\cal O}^{1,2,5},\,{\cal
O}^{1,2,6},\,{\cal O}^{1,2,7},\,{\cal O}^{1,2,8},\,{\cal O}^{1,3,5},\,{\cal
O}^{1,3,6},\,{\cal O}^{1,3,7},\,{\cal O}^{1,4,7}\\}\,.$ (86)

We have the paired states

$\displaystyle(9,3)_{1}^{j}$ $\displaystyle=$
$\displaystyle(0.019698,-0.061589\mp 0.037646\,i,-0.032810\pm
0.12248\,i,0.14940,-0.032810\mp 0.12248\,i,$ $\displaystyle-0.061589\pm
0.037646\,i,0.33666,-0.28492-0.48330\,i,-0.28492\pm 0.48330\,i,0.25288)\,,$
$\displaystyle(9,3)_{2}^{j}$ $\displaystyle=$
$\displaystyle(-0.13995,0.24631\pm 0.43613\,i,-0.29736\mp
0.22678\,i,0.38201,-0.29736\pm 0.22678\,i,$ $\displaystyle 0.24631\mp
0.43613\,i,-0.18337,0.0070425\mp 0.091441\,i,0.0070425\pm
0.091441\,i,0.029342)\,,$ $\displaystyle(9,3)_{3}^{j}$ $\displaystyle=$
$\displaystyle(0.38917,-0.23868\mp 0.24160\,i,-0.17869\mp
0.42845\,i,0.056392,-0.17869\pm 0.42845\,i,$ $\displaystyle-0.23868\pm
0.24160\,i,-0.15640,0.20049\mp 0.08976\,i,0.20049\pm 0.08976\,i,0.14458)\,.$

The parity odd states (21) have the two-point functions

$\displaystyle\braket{(9,3)_{1}^{-}*(9,3)_{1}^{-}}$
$\displaystyle=\,N^{9}\left(1-7.1317\,N^{-2}+14.706\,N^{-4}+{\cal
O}(N^{-6})\right),$ $\displaystyle\braket{(9,3)_{2}^{-}*(9,3)_{2}^{-}}$
$\displaystyle=\,N^{9}\left(1+0.3985\,N^{-2}-87.730\,N^{-4}+{\cal
O}(N^{-6})\right),$ $\displaystyle\braket{(9,3)_{3}^{-}*(9,3)_{3}^{-}}$
$\displaystyle=\,N^{9}\left(1-33.267\,N^{-2}+414.02\,N^{-4}+{\cal
O}(N^{-6})\right)\,,$ (88) $\displaystyle\braket{(9,3)_{1}^{-}*(9,3)_{2}^{-}}$
$\displaystyle=\,N^{9}\left(0+5.9542\,N^{-2}-29.236\,N^{-4}+O(N^{-6})\right)\,,$
$\displaystyle\braket{(9,3)_{1}^{-}*(9,3)_{3}^{-}}$
$\displaystyle=\,N^{9}\left(0-0.7722\,N^{-2}-0.4182\,N^{-4}+O(N^{-6})\right)\,,$
$\displaystyle\braket{(9,3)_{2}^{-}*(9,3)_{3}^{-}}$
$\displaystyle=\,N^{9}\left(0-1.4489\,N^{-2}+50.853\,N^{-4}+O(N^{-6})\right)\,.$

Further,

$\displaystyle\braket{(9,3)_{1}^{-}*(7,3)^{-}(2,0)}$ $\displaystyle=$
$\displaystyle\phantom{-}0.9411\,N^{8}-13.515\,N^{6}+O(N^{4})\,,$
$\displaystyle\braket{(9,3)_{2}^{-}*(7,3)^{-}(2,0)}$ $\displaystyle=$
$\displaystyle-5.1929\,N^{8}+79.646\,N^{6}+O(N^{4})\,,$ (89)
$\displaystyle\braket{(9,3)_{3}^{-}*(7,3)^{-}(2,0)}$ $\displaystyle=$
$\displaystyle-2.4795\,N^{8}+51.496\,N^{6}+O(N^{4})$

and

$\displaystyle\braket{(9,3)_{1}^{-}*(6,3)^{e}(3,0)}$ $\displaystyle=$
$\displaystyle\phantom{-}2.9140\,N^{8}-12.874\,N^{6}+O(N^{4})\,,$
$\displaystyle\braket{(9,3)_{2}^{-}*(6,3)^{e}(3,0)}$ $\displaystyle=$
$\displaystyle\phantom{-}5.7868\,N^{8}-60.161\,N^{6}+O(N^{4})\,,$ (90)
$\displaystyle\braket{(9,3)_{3}^{-}*(6,3)^{e}(3,0)}$ $\displaystyle=$
$\displaystyle-0.1467\,N^{8}-11.267\,N^{6}+O(N^{4})\,.$

In combination with formulae (95), (100), (101) of Appendix B these yield the
tree matrix $S_{ij}$ whose root defines the similarity transform to orthogonal
states. The two-point functions for the $(9,3)^{-}_{j,c}$ states are

$\displaystyle\braket{(9,3)_{1,c}^{-}*(9,3)_{1,c}^{-}}$
$\displaystyle=N^{9}\left(1+5.6881\,N^{-2}+7.5170\,N^{-4}+O(N^{-6})\right)\,,$
$\displaystyle\braket{(9,3)_{2,c}^{-}*(9,3)_{2,c}^{-}}$
$\displaystyle=N^{9}\left(1+174.91\,N^{-2}-\phantom{.}10337\,N^{-4}+O(N^{-6})\right)\,,$
(91) $\displaystyle\braket{(9,3)_{3,c}^{-}*(9,3)_{3,c}^{-}}$
$\displaystyle=N^{9}\left(1+0.3980\,N^{-2}+13.221\,N^{-4}+O(N^{-6})\right)\,.$

## Appendix B: disconnected double-trace two-point functions

In the disconnected part of the two-point functions of $(6,3)^{e}(3,0)$ we
encounter the norm

$\braket{(6,3)^{e}*(6,3)^{e}}\,=\,1*(N^{6}-6\,N^{4}+\ldots)+1*(N^{4}+\ldots)\,=\,N^{6}-5\,N^{4}+\ldots$
(92)

where we have separated by sphere and torus diagrams. The torus part involves
graphs that can be marked on Figure 3, Panel 1 with two length 6 and two
identity operators as in [10]. The relevant sets of edge widths are

$\\{0,1,1,4\\},\,\\{0,2,2,2\\},\,\\{1,1,1,3\\},\,\\{1,1,2,2\\},\,\\{1,2,1,2\\}\,.$
(93)

In field theory as well as in integrability, the amplitudes for the first and
third of these vanish. The hexagon computation has to be scaled up by a factor
$\sqrt{L_{1}\,L_{2}}\,=\,6$ upon which the second and the fifth case have to
receive extra factors $1/3,\,1/2$, respectively, due to their three- and
twofold cyclic symmetry, cf. [10].

Trivially,

$\braket{(3,0)*(3,0)}\,=\,1*\frac{(N^{2}-1)(N^{2}-2)}{N}+1*\frac{2\,(1-N^{2})}{N}\,=\,N^{3}-5\,N+\ldots$
(94)

so that the product yields $N^{9}-10\,N^{7}+\ldots$ as stated in the
paragraphs after (62) for the $\\{6,0,3,0\\}$ double-trace graph. From the
other formulae in Section 5 it follows

$\braket{(6,3)^{e}(3,0)*(6,3)^{e}(3,0)}\,=\,N^{9}+2\,N^{7}+\ldots$ (95)

As before, we will compute the $\\{7,0,2,0\\}$ sphere and torus parts of the
$(7,3)^{-}(2,0)$ two-point functions from the overlaps of the two Bethe states
$(7,3)^{j}$. The sphere part is obviously diagonal and comes with the colour
factor $(N^{7}-7\,N^{5}+\ldots)(N^{2}-1)\,=\,N^{9}-8\,N^{7}+\ldots$. In the
torus part of $\braket{(7,3)^{j}*(7,3)^{k}}$ we find the colour ribbon graphs

$\\{0,1,1,5\\},\,\\{0,2,2,3\\},\,\\{1,1,1,4\\},\,\\{1,1,2,3\\},\,\\{1,1,3,2\\},\,\\{1,2,1,3\\},\,\\{1,2,2,2\\}$
(96)

which evaluate to:

$N^{5}\,\\{-1,1,1,1,1,1,1\\}$ (97)

Now, w.r.t. this basis:

$\displaystyle\braket{(7,3)^{1}*(7,3)^{1}}=$
$\displaystyle\\{2,\,0,\,2,\,-1-\sqrt{15}\,i,\,-1+\sqrt{15}\,i,\,2,\,-1\\}$
$\displaystyle\rightarrow-N^{5}+\ldots$
$\displaystyle\braket{(7,3)^{2}*(7,3)^{2}}=$
$\displaystyle\\{2,\,0,\,2,\,-1+\sqrt{15}\,i,\,-1-\sqrt{15}\,\,i,\,2,\,-1\\}$
$\displaystyle\rightarrow-N^{5}+\ldots$ (98)
$\displaystyle\braket{(7,3)^{1}*(7,3)^{2}}\,=$
$\displaystyle\braket{(7,3)^{2}*(7,3)^{1}}\,=\
\\{0,\,2,\,0,\,3,\,3,\,0,\,-2\\}$ $\displaystyle\rightarrow\,6\,N^{5}+\ldots$

Again, the integrability results for the torus need to be scaled up by
$\sqrt{L_{1}L_{2}}\,=\,7$ to find a match, though there are no extra factors
due to cyclic invariance here. For the first Bethe state we must insert an
extra sign. In conclusion, the diagonal leading $N$ disconnected part of
equation (63) receives a correction

$-3\,N^{7}\left(\begin{array}[]{rr}3&-2\\\ -2&3\end{array}\right)\,.$ (99)

Collecting terms:

$\braket{(7,3)^{-}(2,0)*(7,3)^{-}(2,0)}\,=\,N^{9}-7N^{7}+\ldots$ (100)

For completeness, we recall:

$\braket{(6,3)^{e}(3,0)*(7,3)^{-}(2,0)}\,=\,-2\,\sqrt{6}\,N^{7}+\ldots$ (101)

Third, for the $(7,3)^{+},\,{\cal O}_{\perp}$ mixing example we need the norm
of the (5,3) state to torus order. We will use the torus diagram Figure 3,
Panel 1 with identity insertions at points 3,4 with transverse magnons. There
are only three contributing tree diagrams,

$\\{0,0,0,5\\},\,\\{0,1,1,3\\},\,\\{1,1,1,2\\}\,,$ (102)

where we include the sphere contribution as an additional test although it is
already covered by (58).

This offers an opportunity to apply the rules for twist developed above. We
should not put twist on the edges connecting the two length five operators by
what was said before. On the other hand, hiding the twist on the edges of
width zero the $n$ coefficients tend to drop. Hence, starting both partitions
in the centre of the figure so that edge A is crossed first it should be
possible to compute without any regulator. And indeed, putting
$a_{1}\,=\,0,\,a_{2}\,=\,1$ and normalising as in the derivation of (58) we
obtain

$-\frac{(a_{3}-a_{4})}{a_{3}\,(1-a_{4})}\ \\{1,1/5,\,1/5\\}\ .$ (103)

The complete space time factor would likely be
$-a_{34}/(a_{12}^{5}\,a_{13}\,a_{24})$. It comes as a surprise because of the
wrong sign, but also because the positions of the identity operators do not
drop, and because it has a preferred association between points 1,3 and 2,4,
respectively, which might arise from the turning sense of the partitioning999A
non-trivial space time factor arises also from the torus diagram in Figure 3,
Panel 3. However, in that instance the dependence on the fictious operators’
positions drops in the two-point limit.. As the computation impeccably yields
the correct amplitudes (as usual, the true torus part needs an extra factor
$\sqrt{L_{1}L_{2}}\,=\,5$) we clearly need to eliminate the space time factor
and the additional sign. The point deserves future attention.

The three colour factors associated with (102) are
$N^{5}-5\,N^{3},\,-N^{3},\,N^{3}$ up to the relevant order, so that the true
torus part does not even come in. Together with the $\braket{(2,0)*(2,0)}$
part $N^{2}-1$ the first entry in the and the table after (65) should thus
pick up a colour factor $N^{7}-6\,N^{5}+\ldots$

## Appendix C: the correlators of Section 6 with longitudinal magnons

We repeat the computations identically, though with $Y$ magnons. For ease of
comparison we stick to the partitions yielding the tables in Section 6. As we
know by now, different choices would only result in reparametrisations,

###  $\mathbf{\langle(5,3)(3,0)*(5,3)(3,0)\rangle}$ with longitudinal magnons

For the _empty square_ the $c_{0}$ independence condition (57) yields

$n_{A,1}\,=\,n_{A,2}.$ (104)

as before. With that

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | 0 | $\phantom{10*}\ \ \frac{1}{4}\,(1-n_{A,1})(1+n_{A,1})(4+5\,n_{A,1})(n_{A,1}+n_{D,1})$
4111 | -2 | $10*\frac{1}{20}(1+n_{A,1})(2-5\,n_{A,1})(4+5\,n_{A,1})(n_{A,1}+n_{D,1})$
3202 | -2 | $10*\frac{1}{20}(2+5\,n_{A,1})(1-8\,n_{A,1}-5\,n_{A,1}^{2})(n_{A,1}+n_{D,1})$

A ratio of the second and third entry shows that there is no universal, real
solution for $n_{A,1},\,n_{D,1}$. The most likely interpretation will then be
to put $n_{A,1}\,=\,0$ and to have non-universal
$n_{D,1}\,=\,\\{0,\,1/2,\,2\\}$. The twist trick unfortunately loses all
predictive power.

With identical $n$ coefficients at both ends of the equator lines, the _belt_
diagram does not have any constraint from $c_{0}$ independence. Yet, we do see
the $n_{A,1},\,n_{C,3}$ parameters now. The existence of the point
identification limit imposes a new constraint:

$n_{A,1}\,=\,0\qquad\lor\qquad n_{C,3}\,=\,0$ (105)

These yield exchangeable results, so let us assume $n_{C,3}\,=\,0$. We find

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | -2 | $10*\frac{1}{10}(1+9\,n_{E,b}+10\,n_{E,b}^{2})(1-n_{A,1})$
{2,1}0202 | 4 | $10*\frac{1}{10}(1-3\,n_{E,b}+10\,n_{E,b}^{2})(1-n_{A,1})$

There is no universal solution here either. One would presumably put
$n_{E,b}\,=\,0$ and tune $n_{A,1}$.

###  $\mathbf{\langle(5,3)(2,0)*(5,2)(2,1)\rangle}$ with longitudinal magnons

The _empty square_ has the $c_{0}$ independence constraint
$n_{D,1}\,=\,2\,(1-n_{C,4})$ as before. We compute:

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | $-\sqrt{2}$ | $\phantom{10*}\ \,\frac{\sqrt{2}}{4}(1+n_{A,1})(4+5\,n_{A,1})(-1+n_{C,4})(2+n_{A,1}-2\,n_{C,4})$
4111 | $-5\sqrt{2}$ | $10*\frac{\sqrt{2}}{40}(4+5\,n_{A,1})(-5-5\,n_{A,1}+6\,n_{C,4}+10\,n_{A,1}n_{C,4})(2+n_{A,1}-2\,n_{C,4})$
3202 | $-2\sqrt{2}$ | $10*\frac{\sqrt{2}}{20}(2+5\,n_{A,1})(-2+3\,n_{C,4}+5\,n_{A,1}n_{C,4})(2+n_{A,1}-2\,n_{C,4})$

In the _belt_ case we fall upon the known $c_{0}$ independence constraint
$n_{E,b}\,=\,-\frac{1}{2}-\frac{n_{A,1}}{2}+n_{C,4}$. The point identification
limit $2\,\rightarrow\,1,\,4\,\rightarrow\,3$ requires
$n_{A,1}\,n_{C,4}\,(11+5\,n_{A,1}-10\,n_{C,4})\,=\,0$ for the $\\{3,0\\}0202$
graph and $n_{A,1}\,n_{C,4}\,(3+5\,n_{A,1}-10\,n_{C,4})$ for $\\{2,1\\}0202$.
So for both graphs we have three possibilities:

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | $-2\sqrt{2}$ | $10*\frac{\sqrt{2}}{20}(1+5\,n_{A,1}+n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})(1-2\,n_{C,4})|_{n_{A,1}\,=\,0}$
|  | $10*\frac{\sqrt{2}}{20}(1+5\,n_{A,1}+n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})(1-n_{A,1})|_{n_{C,4}\,=\,0}$
|  | $10*\frac{\sqrt{2}}{10}(6+5\,n_{A,1})(1-n_{A,1})$
{2,1}0202 | 0 | $10*\frac{\sqrt{2}}{20}(-5-5\,n_{A,1}+13\,n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})(1-2\,n_{C,4})|_{n_{A,1}\,=\,0}$
|  | $10*\frac{\sqrt{2}}{20}(-5-5\,n_{A,1}+13\,n_{C,4}+5\,n_{A,1}n_{C,4}-10\,n_{C,4}^{2})(1-n_{A,1})|_{n_{C,4}\,=\,0}$
|  | $10*\frac{\sqrt{2}}{50}(-2+5\,n_{A,1})(1-n_{A,1})$

With the exception of the $\\{2,1\\}$ case in the second table none of this
can be satisfied with likely values for $n_{A,1}$ or $n_{C,4}$.

###  $\mathbf{\langle(5,2)(2,1)*(5,2)(2,1)\rangle}$ with longitudinal magnons

We have identified $n_{C,3}\,=\,n_{C,4}$ whereby no constraint arises from
(57).

$l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
5020 | -1 | $\phantom{10*}\ \,-(1-n_{C,4})(1-2\,n_{B,3}-2\,n_{C,4})$
4111 | -7 | $10*\frac{-1}{10}(5-6\,n_{C,4})(1-2\,n_{B,3}-2\,n_{C,4})$
3202 | -4 | $10*\frac{-1}{10}(2-3\,n_{C,4})(1-2\,n_{B,3}-2\,n_{C,4})$

Forming ratios we find $n_{C,4}\,=\,2$. From the absolute normalisation it
then follows that $n_{B,3}\,=\,-1$. For once there is a fairly appealing
solution!

The _belt around the belly_ comes with the constraint $n_{A,2}\,n_{C,4}\,=\,0$
if we let the magnons of the two vacuum descendents cross the edges $A,\,C$,
respectively, on their path through the entangled state. Assuming
$n_{C,4}\,=\,0$ the amplitudes are:

$\\{l_{E,b},l_{E,f}\\}l_{A}l_{B}l_{C}l_{D}$ | QFT | hexagon tiling
---|---|---
{3,0}0202 | -4 | $10*\frac{1}{5}(1-2\,n_{A,2})$
{2,1}0202 | +4 | 0

The $\\{3,0\\}$ case is trivial to satisfy — and the coefficient $3/2$ perhaps
still credible — but the second row is a blatant contradiction.

## Appendix D: a pretty hexagon amplitude for longitudinal magnons

All $Y$ hexagon amplitudes with magnons on only two of the physical edges
directly evaluate to products of $A$ and $(A-B)/2$ elements of the $PSU(2|2)$
scattering matrix [22]. Putting one of the excitations onto the third egde
radically changes the picture: the scattering involves all 10 elements of the
$S$ matrix and yields a large sum of terms. The most complicated amplitude of
the type we encounter in the present context is

${\cal
A}_{3|1|2}\,=\,\langle{\mathfrak{h}}|Y_{1}Y_{2}Y_{3}|Y_{4}|Y_{5}Y_{6}\rangle\,=\,-\langle{\mathfrak{h}}|Y_{1}^{4\gamma}Y_{2}^{4\gamma}Y_{3}^{4\gamma}\,\bar{Y}_{4}^{2\gamma}\,Y_{5}\,Y_{6}\rangle$
(106)

At tree level, it turns out to have a concise decomposition over the particle
creation poles:

$\displaystyle{\cal A}_{3|1|2}$ $\displaystyle=$ $\displaystyle
i\frac{S_{21}S_{31}}{u_{14}}\,\frac{h_{32}h_{65}}{h_{25}h_{26}h_{35}h_{36}}+i\frac{S_{32}}{u_{24}}\,\frac{h_{31}h_{65}}{h_{15}h_{16}h_{35}h_{36}}+i\frac{1}{u_{34}}\,\frac{h_{21}h_{65}}{h_{15}h_{16}h_{25}h_{26}}+$
$\displaystyle
i\frac{1}{u_{54}}\,\frac{h_{21}h_{31}h_{32}}{h_{16}h_{26}h_{36}}-i\frac{S_{65}}{u_{64}}\,\frac{h_{21}h_{31}h_{32}}{h_{15}h_{25}h_{35}}+\frac{h_{21}h_{31}h_{32}h_{65}}{h_{15}h_{16}h_{25}h_{26}h_{35}h_{36}}$

where $u_{ij}$ is a rapidity difference and the tree level $SU(2)$ S matrix is
defined in (27). Finally, the tree level dressing factor is [7]

$h_{ij}\,=\,\frac{u_{i}-u_{j}}{u_{i}-u_{j}-i}\,.$ (108)

Note that formula (Appendix D: a pretty hexagon amplitude for longitudinal
magnons) singles out the poles in $u_{4}$ and the finite part behind; it is
somewhat reminiscent of [29] in the amplitude literature.

For a proof one might start refining the triangulation by inserting an
identity operator in the middle of the hexagon. We obtain a partition into 64
terms with splitting factors consisting of $S$ matrices only because the new
edges are of width zero. Every new hexagon amplitude has excitations at only
two physical edges and thereby factors into products of $h$ factors as in
(Appendix D: a pretty hexagon amplitude for longitudinal magnons). The
argument is not sufficient to explain the still simpler final form of the
amplitude, though.

For longitudinal cases that do not have one preferred magnon (here the one
that is alone on its edge) we have so far not been able to spot a similar
pattern. Such techniques would be particularly useful in manipulating non-
vanishing amplitudes for transverse magnons, which are typically bulky sums.

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