On fine structure of strings: the universal correction to the Veneziano amplitude
HJE
On fine structure of strings: the universal correction
Amit Sever 0 1 2 3 5
Alexander Zhiboedov 0 1 2 4
0 17 Oxford Street, Cambridge, MA 20138 , U.S.A
1 Ramat Aviv 69978 , Israel
2 1211 Geneva 23 , Switzerland
3 School of Physics and Astronomy, Tel Aviv University
4 Department of Physics, Harvard University
5 CERN, Theoretical Physics Department
We consider theories of weakly interacting higher spin particles in flat spacetime. We focus on the fourpoint scattering amplitude at high energies and imaginary scattering angles. The leading asymptotic of the amplitude in this regime is universal and equal to the corresponding limit of the Veneziano amplitude. In this paper, we find that the first subleading correction to this asymptotic is universal as well. We compute the correction using a model of relativistic strings with massive endpoints. We argue that it is unique using holography, effective theory of long strings and bootstrap techniques.
1/N Expansion; Confinement

Meson spectrum
Meson scattering
PolchinskiStrassler mechanism
Massive ends approximation
Massive ends in the conformal gauge
Effective theory of long strings
Solving for the boundary metric Correction to the amplitude
Classical scattering of strings with massive ends
Discontinuity, Lorentzian segments and emergence of the s − u crossing 15
4.1
Derivation
Bootstrap
1 Introduction 2
Holographic setup 3 4 5
6
A
2.1
2.2
2.3
2.4
2.5
2.6
3.1
3.2
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
Discontinuities of the massive endpoints correction
Bootstrapping the correction
– 1 –
5.10 Fixing the solution
5.11 The role of crossing
Conclusions
A.1 Order m0
A.2 Order m1/2
A.3 Order m1
A.4 Order m3/2
A.5 Evaluation of A4
Mass perturbation for the fourpoint amplitude
B Bootstrap in Mellin space 40
Introduction
Weakly interacting higher spin particles (WIHS) appear in very different physical
contexts [1]. One is the YangMills theory at large N , where higher spin particles correspond
to approximately stable resonances (glueballs) with various masses and spins. Another is
HJEP06(218)54
treelevel scattering amplitudes of fundamental strings. In this case, higher spin particles
describe excitations of a fundamental string.
The Smatrix bootstrap aims to classify all such theories of weakly interacting higher
spin particles (WIHS) based on general principles. Twototwo scattering in such theories is
described by a meromorphic function A(s, t). It packs together in a neat way the spectrum
of particles mi and a set of threepoint couplings cijk.
Unitarity and causality, together with the presence of particles of spin higher than
two, results in a set of highly nontrivial constraints on the amplitude. At the present
moment only one solution to the problem is known explicitly — the celebrated Veneziano
formula [2] (and its cousins [3, 4]). This amplitude has several nongeneric properties such
as exact linearity of the Regge trajectories, an exact degeneracy of the spectrum and soft
high energy, fixed angle behavior [5].1
Without further assumptions, (i.e. the existence
of massless spin two particle in the spectrum), these properties are not expected to hold.
In other known theories of WIHS, such as confining gauge theories in the large N limit,
the high energy behavior for real scattering angles is powerlike [8] and no degeneracy at
all is expected in the spectrum [9–11]. Correspondingly, the details of A(s, t) may differ
significantly from one WIHS to another.
In [12] a systematic study of WIHS theories was initiated by studying the scattering
amplitudes at large energies and imaginary scattering angles. This kinematical regime is
controlled by the fastest spinning particles and was shown to be universal [12]. Namely, in
any WIHS the s, t ≫ 1 asymptotic of the amplitude A(s, t) coincides with the limit of the
Veneziano amplitude
s,t≫1
lim log A(s, t) = α′ [(s + t) log(s + t) − s log s − t log t] + . . . .
(1.1)
The analysis of [12] assumed that there is no accumulation point in the spectrum and
that there is a separation of scales in the large s, t limit. Correspondingly, the result (1.1)
shows that any WIHS theory satisfying these assumptions is a theory of strings. Specifically,
its spectrum contains an infinite set of asymptotically linear Regge trajectories.
To put this bootstrap program on a systematic computational path, we must
understand what the possible corrections to (1.1) are. Universality can only arise for the first
1In [6, 7] an ansatz for the scattering amplitude was analyzed for which the leading Regge trajectory
is piecewise linear and flat for negative t. It consists of an infinite sum of the Veneziano amplitudes with
different slopes approaching zero, all sharing the same mass spectrum.
– 2 –
few terms in the large s, t expansion. Beyond the universal terms, more assumptions will
be needed to fix the amplitude and the corresponding class of WIHS theories.
The aim of this paper is to study the universal corrections to (1.1). The result of our
analysis is that there is a unique, universal correction to (1.1) that grows with energy. This
correction scales as s1/4 in the limit s, t ≫ 1 with st held fixed. It takes the following form2
δ log A(s, t) = −m3/2 16√πα′
3
st
s + t
1/4
K
s
s + t
t
s + t
+ K
where K(x) is a complete elliptic integral of the first kind. A parameter m is the new scale
that enters the amplitude. On the string theory side, it could be thought of as adding a
point mass at the endpoint of the string that slows down its motion.3 The correction (1.2)
implies that the leading Regge trajectory takes the following form
8√π
3
which (1.1) and (1.2) are built.
On the bootstrap side the correction (1.2) is due to the fact that degeneracy of the
spectrum is lifted. We find that unitarity and crossing alone are not sufficient to uniquely
fix the correction (in contrast to [12]). Instead we consider a more restricted class of
corrections where the leading Regge trajectory in one of the two channels is not corrected.
We observe that the leading asymptotic of the amplitude found in [12] exhibit and emergent
symmetry between the s and uchannel. We assume that this symmetry is preserved by
the correction. With these assumptions we find that the correction is unique and coincides
with (1.2) upon symmetrisation over channels.
The paper is organized as follows. In section 2 we formulate the scattering problem in
a holographic setup. We review the PolchinskiStrassler mechanism [14] for the universality
of the high energy scattering at imaginary angles. When considering scattering of mesons,
this picture predicts a universal correction to the Veneziano asymptotic (1.1) due to massive
endpoints of the strings. We then use effective theory of rotating strings [15] to extend the
validity of this correction beyond the holographic setup. In section 3 we use the worldsheet
model of strings with massive ends to compute the correction to the npoint scattering
amplitude. The dynamics of relativistic strings with massive endpoints is highly
nonlinear, but the leading correction to the amplitude could be computed. For the fourpoint
amplitude with same masses at the ends, the result is (1.2). In section 4 we explain the
emergent s ↔ u symmetry of the asymptotic results (1.1) and (1.2). In section 5 we analyze
the result (1.2) using the bootstrap approach [12]. The new element when talking about
the correction to the leading solution is the sensitivity of the asymptotic amplitude to the
degeneracy of the spectrum. We further impose the emergent s ↔ u symmetry and show
that the massive ends correction is unique. We conclude in section 6.
2A formula for the npoint function could be found in the main text.
3For application of this model to the experimental data see [13]. There it has been observed that Regge
trajectories of mesons with s, c, and b quarks are best fitted to a rotating relativistic string model when
introducing nonzero endpoint masses.
– 3 –
Holography is a useful tool to study confining gauge theories, see [16, 17] for a recent review.
We start this section by reviewing the pioneering work of Polchinski and Strassler [14]
regarding the scattering in confining gauge theories with gravity duals. We explain how
the bootstrap results of [12] follow from the holographic consideration. We then use the
holographic setup to compute the first correction to the asymptotic form of the fourpoint
amplitude at large and positive s and t. In this discussion, we use only the gross features
of the holographic model. These features are the same in a wide range of models dual to
confining gauge theories [16, 17]. Correspondingly, in the regime of interest s, t ≫ 1 the
first deviation from the Veneziano amplitude is the same in all such theories. As advertised
in the introduction, this correction is a universal property of a generic theory of WIHS.
A generic holographic background dual to a confining gauge theory takes the form
ds2 = dr2 + f (r) dx12,d−1 ,
where xμ parametrizes R1,d−1 and r is a radial holographic direction. In the large r ≫ 1
UV region the background asymptotes to AdS, f (r) ≃ e2r. The background terminates in
the IR at some finite, positive warping factor f (rIR) > 0. Without loss of generality we
set rIR = 0. The value of the warping factor at r = 0 is dual to the confining scale or,
equivalently, the gap in the spectrum where the RG terminates. Hence, this is the part of
the background that is relevant for us. In this region
f (r) ≃ 1 + r2/R2 + . . .
where we have approximated the warping factor by its first quadratic expansion around
(2.1)
(2.2)
r = 0. For convenience, we set f (0) = 1.
2.1
Meson spectrum
To describe mesons in such a confining background, we add to it a probe flavor brane [16,
17]. It is a spacetime filling brane that extends from the boundary at r = ∞ down to some
radial position r0 ≥ 0. Strings ending on this brane are dual to mesons in the confining
gauge theory.
Mesons with low or zero spin are dual to short strings with energy of order the string
scale at r0, see figure 1.a. Mesons with very large spin are dual to large rotating open
strings. The main contribution to their energy is classical and can be computed by solving
for the classical rotating string in the background (2.1) [18].
At leading order in the large spin expansion, the problem reduces to a classical rotating
string in flat spacetime and the Regge trajectory is linear J = α′E2, where α′ is the effective
string scale in the IR. For large r0/ls, the first correction to this asymptotic trajectory is
also classical and takes the form [19]
J = α′E2
−
8√π
3
3 √
α′m 2 E + O(1) ,
E ≫ m ≫ 1/ls ,
(2.3)
where m = r0/α′.
– 4 –
When the spin of the string becomes large, it has the characteristic Ushape. The horizontal region
corresponds to a string rotating in flat space, and the vertical piece of the string is dual to a
constituent quark mass. b) The picture of a meson as a rotating string with massive ends that is
dual to the string depicted in a).
The corresponding classical string solution has two regions, see figure 1.a. One region is
localized at r = 0 where we have a rotating string solution in flat spacetime. In the second
region, the open string extends radially between r = 0 and the flavor brane at r = r0. For
large spins the string is large, and we may reliably describe it using the effective theory
of long strings. In the effective theory, the second vertical piece of the string reduces to a
pointlike mass m = r0/α′ at the ends of the open string, see figure 1.b [18].
In addition to the leading order classical correction to the asymptotic trajectory (2.3),
the trajectory receives a series of classical corrections as well as quantum corrections. The
quantum corrections are small compared to (2.3) when r0/ls = mls is large. Hence, this
is the limit we are considering here. We expect that at the next order the details of the
background will become important unless some extra symmetries are present.
2.2
Meson scattering
We are interested in the 2 → 2 scattering amplitude of scalar mesons in the limit s, t ≫
m2 ≫ 1/α′, where α′ is the IR string scale (at r = 0). This scattering amplitude is dual
to an amplitude of open stings in the background (2.1). The open strings end on the
flavor brane/branes. These strings satisfy Neumann boundary condition along the d flat
spacetime directions. In the radial direction they satisfy Neumann boundary condition for
r > r0 and Dirichlet boundary condition if the string stretches from r = r0 down to r < r0.
In the planar limit, the string worldsheet has a disk topology. At the boundary of the
disk, we have four vertex operators that inject four asymptotic meson states with momenta
{k1, k2, k3, k4}. Because these mesons have zero spin, their dual asymptotic string states
are not rotating. Masses of such asymptotic states are of order the string scale. Hence, in
the large s, t limit that we are interested in, the four asymptotic states are approximately
– 5 –
pointlike, and their momenta are null, ki2 = 0. Their vertex operators take the form
Vi(σ) = h (r(σ)) × eiki·x(σ)
(2.4)
where σ parametrize the boundary of the disk and h(r) is some radial profile.
large s and t the worldsheet path integral is wellapproximated by a saddle point. The
saddle corresponds to a large and smooth Euclidean classical string solution. There are two
kinematical regimes in which the saddle differs significantly. First, there is the regime of
real scattering angles. This regime was studied by Gross and Manes for open strings [20],
by Gross and Mende for closed strings [5] and by Alday and Maldacena in a holographic
setup [21]. In this kinematical regime, the corresponding amplitude is exponentially small
and represents a tunneling process. The other kinematical regime corresponds to purely
imaginary scattering angles. In this regime, the amplitude is exponentially large.
Holographically, there is a clear difference between these two kinematical regimes. For
real scattering angles, the classical solution extends towards the large r UV regime of the
holographic background. On the other hand, for imaginary scattering angles, the classical
solution extends towards the small r IR regime of the holographic background. Let us
elaborate on how these very different behaviors come about:
I. When the scattering angle is real, the four external momenta are real. Without loss of
generality, they can be placed in an R1,2 subspace. On the other hand, for imaginary
scattering angles, the external momenta are complex. They could also be thought of
as being real in a spacetime with two times and one space, R2,1.
II. In either case, in the coordinates in which the external momenta are real, the
classical solution is imaginary. This is because the source for these worldsheet fields is
imaginary. It comes from the vertex operators and is given by i P
example, for a free string in flat spacetime we have xμ(z, z¯) = iα′ P
where z = σ + iτ parametrize the upper half complex plane.
j kj · x(σj ). For
j kjμ log z − σj 2,
III. The induced metric on the worldsheet is Euclidean. To see this it is convenient to
switch to the Tdual coordinates ∂αyμ = iǫαβ∂βxμ. In these coordinates, the solution
is real and ends on a null polygon made of the external null momenta [21]. For
example, for a free string in flat spacetime we have yμ(z, z¯) = α′ P
real scattering angles the ordering of the edges is such that every two points on two
different edges of the polygon are spacelike separated. As a result, the induced metric
on the worldsheet is spacespace (∂yμ∂¯yμ > 0), and the Euclidean action SE > 0 is
positive. Hence, the amplitude is exponentially small (e−SE ≪ 1). For imaginary
scattering angles every two points on two different edges of the polygon are timelike
separated and the induced metric on the worldsheet is timetime (∂yμ∂¯yμ < 0).
Now, the Euclidean action is negative SE < 0 and the amplitude is exponentially
j kjμ log zz¯−−σσjj . For
large (e−SE ≫ 1).
– 6 –
IV. The action (in the conformal gauge) takes the form
radial equation of motion that follows from (2.5) is
∂∂¯r = − 2f (r)2
f ′(r)
∂yμ∂¯yμ
(2.6)
Let us consider the point in the bulk of the worldsheet where r is extremal (and
hence ∂∂¯r = 0). The holographic coordinate r changes as we go to the boundary of
the disk. For real scattering angles, ∂∂¯r is negative and hence r decreases while for
imaginary angles the picture is reversed.
We note that this is precisely the expected behavior that follows from unitarity [12].
Namely, for positive s > 0 and t > 0, which corresponds to the imaginary scattering
angle, the amplitude is dominated by its exponentially large imaginary part.
Due to
unitarity, the contribution of every onshell state is strictly positive, and there cannot be
any cancellations. Hence, the amplitude is large. On the other hand, for s > 0 and t < 0,
that correspond to real scattering angles, the cancelations are possible, and the amplitude
could become small.4
2.4
Massive ends approximation
As a result of the discussion in the previous section, in the conformal gauge the string
action reads
S =
1
2πα′
Z
dz2 ∂r ∂¯r + (1 + r2/R2) ∂x · ∂¯x + i X kj · x(σj ) .
Due to the PolchinskiStrassler mechanism, the classical string configuration for imaginary
scattering angles satisfies Dirichlet boundary condition in the radial direction and Neumann
in the transverse directions
∂τ xμτ=0 = i2πα′ X kjμδ(σ − σj ) ,
rτ=0 = r0 .
j
The equations of motion that follow from (2.7) take the form
R2 ∂∂¯r − r ∂xμ∂¯xμ = 0
(R2 + r2)∂∂¯xμ − r(∂r∂¯xμ + ∂¯r∂xμ) = i2πα′R2 X kjμ δ2(z − σj )
4The exponentially small result in the usual string theory requires finetuning and, indeed, does not hold
in a generic confining gauge theory [8].
(2.7)
(2.8)
(2.9)
j
j
– 7 –
These equations are subject to the Virasoro constraint
∂r∂r + (1 + r2/R2)∂xμ∂xμ = ∂¯r∂¯r + (1 + r2/R2)∂¯xμ∂¯xμ = 0
(2.10)
We are interested in solving these equations in the limit where α′s, α′t ≫ R/r0 ≫ 1.
In this limit there are three regimes, where we can solve the equations of motion:
to ∂∂¯xμ in (2.9) dominates and
a) Near the vertex operators insertion points {σj }. At these points the term proportional
xμ(z, z¯) ≃ iα˜′ kjμ log z − σj 2 ,
r = r0 ,
HJEP06(218)54
where α˜′ = α′ 1 + r02/R2 is the string scale at r0.
b) Deep inside the bulk of the string, where r = 0 and we have the flat space solution
xμ(z, z¯) = iα′√s (f μ(z) + c.c) .
c) In between region (a) and region (b) where r varies from r = r0 to r = 0.
In the limit s, t ≫ (r0/α′)2 we are working at one may approximate region (c) as a
pointlike mass. This point particle is a domain wall between two different tensions, one at
r = 0 (α′) and the other at r0 (α˜′). Its mass is given by the radial distance between the
flavor brane and the IR wall
m ≃ r0/α′ .
After we replace region (c) by a point mass, regions (a) and (b) could be glued together.
The gluing condition is equivalent to equating the difference in momentum flow between
the two regions to the acceleration of the mass. This approximation may break down close
to the vertex insertion. This fact, however, will not be relevant to us. The reason is that
in the limit s, t ≫ m2 ≫ 1/α′ region (c) is pushed towards the boundary and region (a)
shrinks correspondingly.
One can confirm the validity of the above approximation by working out the size and
gluing of these regions in more detail. In the end, we are left with an effective description
of a free string in flat spacetime with massless vertex operators insertions and a massive
point particle at the boundary
SENG =
1
2πα′
Z
Z
dσdτ pdet ∂αxμ∂βxμ + m
Here, the induced metric on the worldsheet is Euclidean, and the path integral weight
is e−SE . Note that the point particle action is reparametrization invariant. Using this
freedom, here we have already set a parametrization in which the boundary of the string
is at τ = 0.
Above, we have written the worldsheet action back in its gauge unfixed NambuGoto
form. The reason for that is that conformal gauge in the holographic background and the
one we will use in the effective theory are not the same.5
5This is because in the holographic background the worldsheet stress tensor also receives a contribution
from the radial direction.
– 8 –
(2.11)
(2.12)
(2.13)
We now fix the conformal gauge
(∂σx)2 − (∂τ x)2 = ∂σxμ∂τ xμ = 0 .
The action takes the form
SEconf =
1
2πα′
Z
d2z ∂zxμ∂z¯xμ + m
Z
S =
1
2πα′
Z
where
The dynamics of a string with massive ends (2.16) is highly nonlinear. Even though this
model was first introduced by Chodos and Thorn in 1974 [19] very little regarding its
properties is known, see [22]. One known exact solution of (2.16) which will be important
to us describes the rotating string and, thus, allows one to compute the correction to the
Regge trajectory. In analyzing this model we find it convenient to rewrite the worldline
action using the boundary metric6
1 Z
2
.
log A(s, t) ≃ −Sclassical
As we explain below introducing e(σ) makes perturbative expansion in m very natural.
Evaluation of the amplitude in terms of e(σ) will turn out very natural as well.
To summarize, in the limit s, t ≫ m2
≫ 1/α′ the amplitude is given by
where Sclassical is the extremum of the action (2.17). The first correction to this saddle point
due to the massive endpoints is universal because we have only used the gross features of
the holographic model which are shared by all models dual to a confining gauge theory.
Concretely, our derivation applies to any holographic background of the form (2.1) that
caps off in the IR (2.2).
It turns out that this effective description is also valid when the quantum corrections
are not small compared to the (bare) boundary mass. Namely, (2.19) also holds when
m2α′ is not large, but the numerical value of m2α′ may change. Even more than that, we
expect (2.19) to hold for closed strings too. To understand why it is so we will now discuss
this correction using the effective theory of long strings.
2.6
Effective theory of long strings
One can systematically study all possible corrections to the dynamics of very long
rotating strings. Understanding these corrections is the subject of the effective theory of long
strings [23]. The leading order result in this expansion is fixed by symmetry and is given by
6Some signs here may not look standard. This is because here we are considering a Euclidean solution.
(2.15)
(2.16)
(2.17)
(2.18)
(2.19)
experience the centrifugal potential and are pushed to the boundary of the string. The leading
effect is making the endpoints of the string massive. b) Imagine a rotating closed string. It has
folds, points where the extrinsic curvature diverges. Regularizing it by adding higher derivative
terms in the worldsheet action makes the endpoints move slower than the speed of light. The result
is again the string with the massive endpoints.
the NambuGoto action. The spectrum of long rotating open strings was studied using the
effective theory in [15].7 In this study, one starts from the free long rotating string and
considers all possible higher derivative operators that are consistent with the symmetries of the
problem. There are two types of corrections — due to the boundary and the bulk operators.
The first operator one can add on the boundary has the effect of adding mass at the
endpoint of the string. As discussed above, it leads to a
E correction to the asymptotic trajec
tory (2.3). There are no other boundary corrections to the trajectory that grow with energy.
√
The bulk operator that gives the leading correction to the asymptotic trajectory is the
socalled PolchinskiStrominger term [23]
LPS = β
(∂2xμ∂¯xμ)(∂xν ∂¯2xν ) .
(∂xμ∂¯xμ)2
(2.20)
In the holographic scenarios, this operator is generated by integrating out quantum
fluctuations to the quadratic order. It turns out that in the conformal gauge the
PolchinskiStrominger term (2.20) is singular at the boundary of the rotating string [15]. As a result,
it induces a boundary operator that has the effect of introducing a boundary mass.8 Hence,
we are back in the massive endpoints case (2.3). After canceling the boundary divergence,
the remaining bulk contribution is regular and only lead to power suppressed corrections
to the asymptotic trajectory.
There is an intuitive way to understand this effect. Suppose we start with no bare
boundary mass (r0 = 0) and consider a quantum fluctuation of the string in directions
transverse to the rotating plane. This fluctuation is then pushed by the centrifugal force
towards the boundary, see figure 5.a. Since it carries energy, it has the same leading order
effect as adding a boundary mass.9
To summarize — the leading correction to the asymptotic linear trajectory comes
from slowing down the boundary endpoints of the rotating string so that they are no
7For the long static string, the set of universal corrections was classified in [24].
8In [15] this mass was finetuned to zero.
9We thank Sergei Dubovsky for explaining this point to us.
longer moving at the speed of light. As a result, the asymptotic trajectory receives a E1/2
correction. The sign of this correction is negative because the end of the string slows down
so that it carries less angular momentum.
Let us briefly discuss the case of closed strings. The free closed rotating string has two
tips where the extrinsic curvature diverges. Similar to the open string case for the rotation
in one plane the PolchinskiStrominger term diverges. As a result, the effective theory
breaks down, and the tools used in [15] become inapplicable. That is why we leave the
consideration of closed string to the future and only present here our intuitive expectation.
Imagine we regularize the cusps divergences in some generic way. After regularization, (and
without finetuning), the curvature will be everywhere finite. As a result, the tips of the
rotating string will be smoothed out, see figure 5.b. Such tips carry some energy and can
no longer move at the speed of light. So again, the leading effect at large spin is to slow
down the tips, effectively adding there a mass. That is why we expect the correction to be
much more general than what we have shown here. In particular, the bootstrap analysis in
section 5 is applicable for both open and closed strings. There we find that the correction
is indeed unique under certain assumptions.
So far we have summarized the results of the effective theory for the spectrum of large
rotating strings. In this paper, however, we are interested in the scattering problem. When
studying a scattering process, one cannot assume a priori that we have a long string plus
small corrections. Indeed, as follows from the PolchinskiStrassler consideration, the long
string description is only valid for large and positive s, t. The effective theory does not
have any way of knowing that and this is partly the reason we did not start with this point
of view. On the other hand, the effective theory allows us to extend the validity of the
massive endpoints model beyond what one may deduct from the holographic setup.
3
Classical scattering of strings with massive ends
We shall now use the effective massive endpoints model (2.17) to compute the first
nontrivial correction to the asymptotic Veneziano amplitude. This computation amounts to
extremizing the action (2.17) with respect to xμ, e, {σj } and then evaluating the action
onshell, (2.19). Solving the corresponding nonlinear equations of motion is hard. However,
we will only need to solve them to the first nontrivial order in the mass. As discussed
above, higher orders are nonuniversal and will not be of interest to us.
Let us start by listing the equations of motion we would like to solve. First, we have
the boundary conditions for x. Due to the mass at the boundary of the string, the usual
Neumann boundary condition is modified to
1
2πα′ ∂τ xμ + m ∂σ √
∂σxμ
∂σxν ∂σxν
= i X kjμ δ(σ − σj ) .
j
(3.1)
This equation has a simple interpretation. The momentum flow from the bulk of the string
is equal to the acceleration of the mass at the boundary. In addition, the vertex operators
inject momenta kj at σ = σj .
In the bulk of the string, xμ is a free field satisfying
(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
√m.
(3.8)
(away from the insertion points), and is given by10
where x0μ is the solution with no boundary mass. It satisfies Neumann boundary conditions,
xμ = x0μ + yμ.
x0μ(z, z¯) = iα′ X kiμ log z − σi2
i
where at this point, the σj ’s are undetermined. Now, the boundary tangent is given by
∂σx · ∂σx = ∂τ x · ∂τ x = ∂τ y · ∂τ y
for
τ = 0 ,
σ 6= σj
where in the first step we have used the Virasoro constraint (3.3). By plugging the boundary
tangent (3.6) into the boundary conditions for xμ, (3.1), we learn that
y(z, z¯) ∝
√m
∂∂¯ xμ(z, z¯) = 0
for
τ = (z + z¯)/2 > 0 .
In addition to the worldsheet field xμ, we have to minimize the action with respect to
the vertex operators insertion points {σj }. Doing so is equivalent to imposing the Virasoro
constraints
∂xμ ∂xμ = ∂¯xμ ∂¯xμ = 0 .
Naively, we have a standard perturbative expansion in integer powers of the mass.
However, the main difficulty in the calculation comes from the fact that the naive small
m expansion is singular. In the absence of the mass, the ends of the string move at the
speed of light. Hence, even though the second term in the l.h.s. of (3.1) is proportional to
the mass, it is inversely proportional to the boundary velocity of the string √ 1
infinite when v = 1. As a result, the expansion reorganizes itself in powers o1f−√v2mw. hich is
To see how this fractional scaling emerges, we first split xμ as
plus higher order corrections.
In what follows, we will first derive a general expression for the correction to the
amplitude in terms of the leading order solution. In particular, our result holds for an
arbitrary number of particles. We will then focus on the case of the fourpoint amplitude.
It would be convenient to use the worldline metric (2.18) that also scales as e(σ) ∝
3.1
Solving for the boundary metric
We first note that the perturbation yμ is also a free field in the bulk of the string. Namely,
∂∂¯ yμ(z, z¯) = 0. At the boundary, it satisfies the boundary condition
10Recall that ∂z∂z¯ log zz¯ = 2πδ(
2
)(z, z¯).
∂τ yμ = −2πα′ ∂σ(e ∂σ[x0μ + yμ]).
We may think about the right hand side as a continuous momentum source for yμ. Hence,
similarly to (3.5), we can express the correction yμ as
Z
Z
yμ(z, z¯) = −α′
dσ ∂σ(e ∂σ[x0μ + yμ]) log z − σ2
= α′
dσ e ∂σ[x0μ + yμ]
1
σ − z
+
1
,
where in the second step we have preformed an integration by parts. Here, z (z¯) takes
values in the upper (lower) halfplane.
get the following equation for e(σ)
1
m2
1
(2πα′)2 e2 =
(2πα′)2 ∂σxμ∂σxμ
Next we can use (3.9) together with the expression for the boundary velocity (3.6), to
So far, we have not done any small mass approximation. That is, (3.9) and (3.10) are
exact equations, valid for any value of m. Next, we note that the leading order correction
corresponds to
e(σ) ∝
√m.
In this way, the first term in the r.h.s. of (3.10) is of order m, while other two terms are
of order m3/2 and m2 correspondingly. Therefore, we get that the first correction to the
boundary metric is
where x0μ here is simply the usual GrossMende solution (3.5) and the σj ’s being the solution
e∗(σ)2 =
m
2πα′p∂σ2x0 · ∂σ2x0
,
X
j
ki · kj
σi − σj
= 0
of scattering equations
for every i = 1, . . . , n [5].
we find that
en=4(σ) = c
4
√m Y q
j=1
The expression for the boundary metric (3.12) is general. In particular, it holds for an
arbitrary number of particles. Once we restrict to the case of the fourparticle amplitude,
σ − σj 
where
c−4 = (4π)2α′4(s + t) Y
σj+1 − σj  (3.14)
4
j=1
Importantly, we see that the induced metric vanishes at the vertex operators insertion
points. This gauge independent property comes hand in hand with the fact that at these
points we inject null momenta into the string.
Next, we turn to the evaluation of the onshell action.
3.2
To compute the correction to the amplitude we need to evaluate the onshell action (2.17)
on the classical solution for xμ, e and the insertion points {σj }. At the leading order,
m = 0, we have the GrossMende (GM) solution (3.5), (or more precisely, its analog for
imaginary scattering angles). The onshell action is given by one half times the value of
the source, evaluated onshell
SGM =
j
2
i X kj · x0(σj )
where the σj ’s solve the scattering equations (3.13). For the fourpoint amplitude, (3.15)
evaluates to (1.1).
For the correction, the situation is similar. We first subtract from (2.17) the leading
GM order (3.15). We remain with an action for the correction. It is a functional of yμ, e and
the corresponding shifts in the vertex operators insertion points {σj } → {σj + δj }. That is,
δS[m; e, yμ, {δj }] = S − SGM
Now the source in δS is the mass term in (2.17), R dσ me2 . The rest of the terms in the
action are functionals of yμ, e and the δj ’s, but not of m. Therefore, yμ and the δj ’s only
depend on m through their interaction with e. After solving for them in terms of e, we
remain with a functional of e only. The details of this are not important, all we need is that
yμ and the δj ’s are determined by e (and of course, the zeroth order solution). We have
(3.15)
HJEP06(218)54
(3.16)
(3.17)
δS[m; e] = δS [m; e, yμ[e], {δj [e]}]
In appendix A we have calculated δS[m; e] explicitly for the case of four external
particles. We know that the correction scales as δS[m; e] ∝ m3/2 and e ∝
√m. Therefore,
δS[m; e] =
for some function Λ. We can now apply an integrated version of the virial theorem. Namely,
we have
0 =
dσ e(σ)
Z
= − 2
We conclude that (see appendix A for a detailed derivation of this result)
Son−shell = SGM +
dσ
m2
e∗(σ)
= SGM +
2 √
3
2πα′ m3/2 Z
Note that even though this correction depends on the second derivative of x0, it is
reparametrization invariant. The reason for this is that before we turn on the mass, the
endpoints of the open string moves at the speed of light and hence ∂σx0 · ∂σ2x0 = 0.
Equation (3.20) is the main result of this section. It is the first correction to any
classical string configuration due to a small mass at the endpoints of the string. Once we
plug in (3.20) the GM solution for four external particles we get
log A(s, t) = −Son−shell = α′ [(s + t) log(s + t) − s log s − t log t]
−
Let us take the Regge limit of this result s ≫ t. The amplitude takes the standard
form log A(s, t) = j(t) log s with the Regge trajectory being
8√π
3
This correction to the Regge trajectory matches precisely with the one obtained before for
the explicit rotating string solution, see for example formula (87) in the review [17].
For us, the correction in (3.22) comes from the first elliptic function in (3.22). Namely,
the first elliptic function captures the correction to the spectrum in the tchannel and the
second elliptic function captures the correction to the spectrum in the schannel. It turns
out that this have a good physical reason. Going back to the holographic setup, we may
consider several flavor branes instead of one. Each flavor brane can end at its independent
radial position, r0,j . We can now scatter the open strings that stretch between different
types of flavor branes. Correspondingly, we will end in the effective theory with different
mass between different vertex operators insertions. We label the mass between vertices
with momentum kj , kj+1 by mj,j+1. The masses m2,3 and m4,1 correct the spectrum in
the schannel and the masses m1,2 and m3,4 correct the spectrum in the tchannel. The
correction to the amplitude now reads
δS[m; e∗] =
8√πα′
3
st
s + t
1
4
Similarly, for the npoint amplitude we have
m13,/22 + m33,/42
K
+
m43,/12 + m23,/32
K
δS[m; e∗] =
2 √
3
2πα′
X
i
mi3,/i+21
Discontinuity, Lorentzian segments and emergence of the s−u crossing
Consider an amplitude where external particles that transform in the bifundamental
representation of a flavor symmetry, so that their corresponding partial amplitude takes the
form Tr(T1T2T3T4)A(
1, 2, 3, 4
). The three Mandelstam invariants are
s = −(k1 + k2)2 ,
t = −(k1 + k4)2
and
u = −(k1 + k3)2 .
(4.1)
Such an amplitude is crossing symmetric with respect to the s and tchannels
(A(s, t) = A(t, s)), but not the s and u channels (A(s, t) 6= A(u, t)).11
11To make it concrete the reader can think of the Veneziano amplitude A(s, t) = Γ(Γ−(−s)sΓ−(−t)t) .
s + t
σi+1
Z
σi
(3.21)
+ . . . .
(3.22)
(3.24)
with ordered external states hTr V1(k1)V2(k2)V3(k3)V4(k4)i. b) The worldsheet after the analytic
continuation s → u. This new classical solution can be obtained from the amplitude with ordering
of points 2 and 3 is interchanged hTr V1(k1)V3⊺(k2)V2⊺(k3)V4(k4)i in two steps. First, we cut it in the
middle of the tchannel and then insert a piece of the Lorentzian evolution. This Lorentzian piece
contributes an imaginary piece to the Euclidean action. Its sign depends on the way we analytically
continue s → u.
Still, the asymptotic of the amplitude, both (1.1) and (3.23), is symmetric under the
exchange of s and s → u = −s − t.12 More precisely, if we analytically continue s → u, the
real part of the amplitude (1.1) stays intact. The same is true about the correction (3.23).
We can formulate this property as a vanishing of the double discontinuity
dDiscs log A(s, t) ≡ log A(−s − t + iǫ, t) + log A(−s − t − iǫ, t) − 2 log A(s, t) = 0 . (4.2)
This is an emergent symmetry and it calls for an explanation. The purpose of this
section is to provide such an explanation. As we will see it involves several ingredients:
t, t)] = iSLorentzian.13
I. The analytic continuation of the saddle point could be interpreted as an insertion of
the piece of the string that evolves in the imaginary worldsheet Euclidean time, see
vertical segment in 3 b. The onshell action on this Lorentzian solution is equal to
the imaginary part of the analytically continued asymptotic amplitude Im[log A(−s−
II. The effect of this Lorentzian evolution is twofold, see 3 b. First, the momenta k2
and k3 are mapped one into each other. Second, if we project to the Euclidean
pieces of the solution, it looks as their correspondent vertex operators are transposed
and their order is interchanged. Due to these effects, the Euclidean saddle maps
into itself, hTr V1(k1), V2(k2), V3(k3), V4(k4)i = hTr V1(k1)V3⊺(k2)V2⊺(k3)V4(k4)i and
the Re[log A(−s − t, t)] which is the onshell action on the analytically continued
Euclidean solution is the same as the original one.
III. The fact that as we analytically continue the vertex operators get interchanged and
transposed is a kind of Stokes phenomenon. Indeed, if we are to do the analytic
continuation of the full amplitude A(s, t) before taking the limit, we would get
hTr V1(k1)V2(k3)V3(k2)V4(k4)i 6= hTr V1(k1)V2(k2)V3(k3)V4(k4)i.
12Since we are in the regime s, t ≫ 1 we can neglect the external masses.
13We thank Shota Komatsu for useful discussions on related topics.
HJEP06(218)54
The punchline is that the emergent symmetry (4.2) is a manifestation of some kind of
stringy Stokes phenomenon which turns an analytic continuation to an a priori different
physical regime into a symmetry transformation.
The fact that after the continuation to the Lorentzian time we end up with a periodic
solution is nontrivial. In the full microscopic theory, we have a combination of all states
propagating in the tchannel and there is no reason for this superposition of eigenstates to
have a periodicity in the Lorentzian “time.” This periodicity emerges when we look at the
logarithm of the amplitude in the classical limit s, t ≫ 1. Similarly, the fact that the real
part of the result is s ↔ u invariant does not hold at the full microscopic level and is an
emergent symmetry of the classical limit s, t ≫ 1.
Let us now see how this mechanism comes about in detail. We start with the explicit free
string classical solution (3.5) and follow the analytic continuation. We then argue that
any boundary correction cannot affect the result. Finally, we verify that the Lorentzian
piece of the solution indeed matches with the imaginary part of the analytically continued
corrected amplitude (3.22).
For convenience, we work in the conformal gauge, with z = eτ+iσ parameterizing the
upper halfplane. We fix the frame both on the worldsheet and in spacetime. In spacetime
we go to the center of mass frame, where ~k1 + ~k2 = 0. On the boundary of the worldsheet,
we have four vertex operators located at {z1, z2, z3, z4}. We do an SL(
2
) transformation to
place these vertex operators at
{z1, z2, z3, z4} = {−ea, −e−a, e−a, ea}
where
z1,4z2,3
z1,2z3,4 = sinh2(a) = s/t .
(4.3)
(4.4)
With this choice, the energy flow across any point on the slice at z = 1 is zero. We
choose to analytically continue s → s
this analytic continuation, we arrive at
= −s − t in the upper complex s plane.14 After
sinh2(a ) = u/t = −1 − sinh2(a) ,
a
= a + i
π
2
Correspondingly, we have
{z1 , z2 , z3 , z4 } = {−ea+i π2 , −e−a−i π2 , e−a−i π2 , ea+i π2 } ∝ {−ea, −e−a−iπ, e−a−iπ, ea} (4.5)
where in the last step we did an overall resealing all the insertion points. After the analytic
continuation we have that z1 < z
3 < z
2 < z4 , so it seems as if the vertex operators
at z2 and z3 have interchanged their order. However, as we explain next, the contour of
integration in the complex τ plane respects the original ordering.
At this point we only have specified the location of four vertex operators in the τ plane.
We should now specify the rest of the string boundary. To fix it we impose that the string
satisfies Neumann boundary conditions. These are satisfied at σ = 0 and σ = π, with
14If we instead analytically continue s → s = −s − t in the lower complex s plane, then we get that
a = a − i π2 .
arbitrary τ , (recall that z = eτ+iσ and z¯ = eτ−iσ weather τ is real or not). To have a clear
separation into the Euclidean and Lorentzian regions we choose the boundary as follows
σ ∈ {0, π} ,
τ ∈ {[−∞ − iπ, τ∗ − iπ], [τ∗ − iπ, τ∗], [τ∗, ∞]}
(4.6)
where −a < τ∗ < a. In addition, we choose τ∗ = 0. (As discussed above, at any point
along this slice the energy flow between the two sides of the string vanishes.)
By letting σ to vary continuously between σ = 0 and σ = π, this new boundary curve
continues nicely into the bulk of the string. Correspondingly, depending on whether we
analytically continue s → u through the upper or lower complex plane, we end with the
new classical string solutions
x
x
0 = x0(eτ+iσ, eτ−iσ)
0 = x0(eτ+iσ, eτ−iσ)
with
with
σ ∈ [0, π] , τ ∈ {[−∞ − iπ, −iπ], [−iπ, 0], [0, ∞]}
σ ∈ [0, π] , τ ∈ {[−∞ + iπ, +iπ], [+iπ, 0], [0, ∞]}
(4.7)
The string solution x
0 has three regions discussed above and is plotted in figure 3.b.
The Lorentzian region corresponds to the piece where τ ∈ [0, iπ]. The two Euclidean
regions can be glued into the original solution, (where {V2(k2), V3(k3)} is interchanged
with {V3⊺(k2), V2⊺(k3)}). This is evident once we plug (4.5) into (3.5) and noticing that
x0μ(eτ+iσ, eτ−iσ) = x0μ(e(τ+iπ)+i(π−σ), e(τ+iπ)−i(π−σ)) + const .
Hence, the real part of the logarithm of the amplitude is equal to the amplitude with s and
u interchanged. It remains to evaluate the action on the Lorentzian region. We find that
dσ
h
∂τ x0(eτ+iσ, eτ−iσ) 2
+ ∂σx0(eτ+iσ, eτ−iσ) 2i
= iπα′ t
1
2πα′
iπ
Z
0
dτ
π
Z
0
iπ
Z
0
(4.8)
(4.9)
(4.10)
in agreement with the discontinuity of (1.1).
Next, we turn on the massive endpoints correction. Our result for the correction to
the action (3.20) was general, for any classical solution and any parametrization of the
boundary. Hence, we just need to evaluate it on the two new boundary segments at σ = 0
and at σ = π with τ ∈ [0, iπ]. We find that
dτ ∂τ2x0(eτ , eτ ) · ∂τ2x0(eτ , eτ ) 1/4 = 2√2[s t(s + t)] 41 K√( s +tt )
This is precisely the imaginary part of the analytic continuation of the corresponding term
s
s+t in (3.23). Note that only the m21,2 and m23,4 boundaries contribute to the
imaginary part.15 Indeed, the other piece
This explains why the double discontinuity of the mass correction had to vanish.
maps to itself under s → −s − t.
1
s+t
s t 4 K
t
s+t
15In the discussion above we assumed that m12,2 = m32,4 and m22,3 = m42,1. This restriction was used to
simplify the discussion but is not necessary.
The property of the classical solution that its double discontinuity vanishes is not
bounded to the first correction. It is a general property valid for any boundary interaction.
The reason is that no matter what the boundary interaction is, the bulk of the string is
free. In the frame where the four vertex operators are at
the solution is a linear combinations of terms of the form log z −z∗2, sourced at the
boundary points z∗ and −z∗. As before, under the analytic continuation τ → τ ± iπ, this function
goes back to itself plus a constant shift. Moreover, after the analytic continuation (4.5)
the solution goes to the one where vertices V2 and V3 are interchanged and the details are
exactly the same as above. To find the analytic continuation of the vertex insertion points,
parametrized by a, we have used their relation to the Mandelstam invariants (4.3). Even
though this relation is now modified, the analytic continuation of a stays the same. To see
that we note that after the continuation a
= a + i π2 , the solution still extremize the real
part of the action and looks as the original saddle point problem with k2 and k3.
This picture breaks down when we start including interactions in the bulk of the
string. However, since the classical string is of the size √s, in the large s, t limit all such
bulk corrections are power suppressed.
To summarize the vanishing of the double discontinuity (4.2) is a stringy property that
emerges in the classical large s, t limit. We will use it in our discussion of the bootstrap
below.
5
In the previous sections, we started from a theory of strings and derived the first correction
to the asymptotic form of the amplitude. Hence, both the leading contribution and the
correction are fingerprints of the string. In this section, we take an opposite point of view
and do not assume a theory of strings. Instead, we analyze the computation of previous
sections using the asymptotic bootstrap initiated in [12]. One of the goals of this bootstrap
program is to classify all possible theories of weakly interacting higher spin particles (WIHS)
using unitarity and causality. Comparing the asymptotic form of the amplitude in a general
theory of WIHS to the one obtained above is an indication to what extent all such theories
are theories of strings.
We start by reviewing our definition of theories of WIHS. We summarize the main ideas
of the bootstrap program of [12] and the argument that fixes the leading asymptotic of
the amplitude to be the one of Veneziano. Next, we turn to the discussion of corrections.
We check that the correction we found using the string with massive ends model fits
perfectly into bootstrap expectations. The main novelty in the discussion of corrections is
sensitivity to the microscopic degeneracy of the spectrum. In particular, the massive ends
correction corresponds to a nondegenerate microscopic spectrum of particles (this should
be juxtaposed with the Veneziano amplitude whose spectrum is maximally degenerate).
As explained in the previous section, both the leading solution and the massive ends
correction have an extra property of the s−u crossing or zero double discontinuity (4.2). For
leading Regge trajectories, some of the zeros escape to infinity and come back to the unit disk. This
effect due to nondegeneracy of the spectrum is responsible for the nonzero support of the excess
zeros distribution outside of the unit circle.
As in [12], if we assume separation of scales we get
Z
δ log A(s, t) = m2−2ktk
d2z δρ(z, z¯) log 1 +
s
tz
,
where δρ(z, z¯) is real and has positive support outside of the unit circle. After putting back
the dimensionfull parameters, the corrected Regge trajectory is
j(t) = α′ t + c m2−2ktk + . . . ,
where
d2z δρ(z, z¯) = c .
Z
(5.9)
(5.10)
Here, m has mass dimensions and c is dimensionless constant.
Recall that the leading distribution (5.4) has an emergent symmetry between the s and
the uchannels, (5.6). As discussed above, this is also a symmetry of zeros of any individual
partial wave. Once we consider the correction due to the nondegenerate spectrum, that
symmetry could be broken. However, such breaking requires finetuning. To break the
symmetry, the nondegenerate spectrum should have two scales. At the first scale, only
part of the degeneracy is removed by grouping together even and odd partial waves so that
the zeros of their sums do not respect the z ↔ 1 − z symmetry at the macroscopic level.
As a result, in between, the excess zeros escape to infinity in a nonsymmetric way. At the
second scale, all the degeneracy is removed.
In a generic situation, we expect to have one scale where all the degeneracy is lifted.
The corresponding pattern of excess zeros outside of the unit circle then should respect the
symmetry to the u channel. Namely, in a generic situation, we expect that
δρ(1 − z, 1 − z¯) = δρ(z, z¯) = δρ(z¯, z) ,
(5.11)
where the last equality follows from the reality of the amplitude.
5.6
Before proceeding to the massive endpoints correction, let us discuss shortly a nongeneric
scenario in which the spectrum is degenerate, and the symmetry (5.11) is broken. In such
a scenario there are of course no excess zeros outside of the unit circle. The only effect of
the correction is to redistribute zeros inside the unit circle. Microscopically, the amplitude
is still controlled by the sum of Legendre polynomials, as in (5.3), but with a slightly
corrected coefficients and spectrum.
Given k in (5.10), analyticity and crossing fix the solution uniquely, [12]. The
corresponding correction to the density of excess zeros has support in the interval 0 < x < 1
δρk(x) = t2−k × δρ(t, tx, tx) =
π
k sin πk xk−1
− log x +
∞
m=1
X (k)m m m!
1 − xm !
Importantly, for any k 6= 1 we have δρk(x) 6= δρk(1 − x), so the symmetry to the
uchannel is broken.
On the other hand, for scattering of identical particles, the correction (5.12) is ruled out
even if the spectrum is degenerate. The reason is that in the asymptotic limit, (5.3) contains
only even partial waves. As a result, the symmetry to the uchannel is explicit. Hence, a
Regge trajectory that consists of degenerate spectrum of even spins only takes the form
(5.12)
(5.13)
(5.14)
(5.15)
j(t) = α′t + α0 + . . . ,
where the dots stand for terms that are suppressed at large t.
Massive ends and excess zeros
δjs(s) = 0 ,
δjt(t) = −t1/4 ,
In this section, we analyze the massive endpoints correction (3.23) using bootstrap. We
find that it indeed satisfies all constraints that were discussed above and, hence, is captured
by the dynamics of excess zeros. We then discuss additional properties of the result and
the uniqueness of the amplitude of this type.
Because the terms in (3.23) are independent, we will focus on one of them only. Namely,
we set m2,3 = m4,1 = 0 and m1,2 = m3,4 = m in (3.23)
δ log A(s, t) = −
s t
s + t
1/4
K
s
s + t
This amplitude is of course no longer crossing symmetric. When talking about the
correction, crossing is not an important symmetry. It can always be restored by summing over
the two independent corrections due to the mass in the s and the t channels, (3.22).
Now, (5.14) represents an amplitude where the string that is been exchanged in the
tchannel (that is giving sj(t)) has massive endpoints while the ends of the string that is been
exchanged in the schannel are still massless. As a result, the correction to the trajectories
in the two channels takes the form
distribution, ρt,h. It is bounded inside the unit circle and is of indefinite sign.
The other is
a vertical distribution, ρt,v. It is a strictly positive distributed that is diffused along the line
β = s/t = −1/2 + iy and goes all the way to infinity.
where for convenience, we work in units where 8√πα′ m3/2 = 1. It is easy to verify that
Because the trajectory in the schannel is not corrected, the corresponding correction
to the distribution of excess zeros should integrate to zero. Importantly, the degeneracy in
the schannel is not lifted. Therefore, this correction to the distribution is expected to be
bounded to the unit disk. Indeed, we find that
t→∞
lim δ log A(s, t) = −π s 4 + . . .
1
and
lim δ log A(s, t) = − t 4 log s + . . .
(5.16)
− 2
s t
s + t
1
4
K
s
s + t
= s 4
dx ρs(x) log 1 +
where
and
ρs(x) = r(x) + r(1 − x) ,
r(x) = − √
1
2 2π [x(1 − x)]3/4 ,
K(x) − 2E(x)
1
0
K(x) = R √
dt
1−t2√1−xt2
,
1
E(x) = R dt √11−−xtt22 ,
0
√
are complete elliptic integrals of the first and second kind correspondingly. One can check
following form
− 2
st
s + t
1
4
K
s
s + t
1
1Z
0
The correction to the distribution in the tchannel is more interesting. It takes the
= t 4 dx ρt,h(x) log 1 +
+ t 4
dy ρt,v(y) log 1 +
3
s→∞
1
0
1
Z
0
dx ρs(x) = 0 .
s
tx
−∞
1
t
sx
(5.17)
(5.18)
(5.19)
(5.20)
s
(5.21)
where the horizontal (first) term is given by a symmetric ρt,h(x) = ρt,h(1 − x) distribution
of zeros between (−1, 0)
ρt,h(x) = − r(x)
− r(1 − x) for
< x < 1 .
for 0 < x <
1
2
,
1
2
The vertical (second) piece, running along x = 21 + iy, is given by
ρt,v(y) =
π
4 z1/2(1 − z)1/2 (E(z) − (1 − z)K(z)) ,
1
2 −
s
y2
1 + 4y2
where
z =
. (5.23)
Again to correctly reproduce the Regge limit s ≫ 1, namely −t1/4 log s, we have
1
Z
0
Z
∞
−∞
dx ρt,h(x) +
dy ρt,v(y) = −1.
Moreover, as one can easily check ρt,v(y) > 0 for any y and the symmetry x → 1 − x is
manifest. Finally, we note that this distribution saturate the condition on its first moment.
Namely, δM1 = 0 in (5.8).
Recall that from the holographic and EFT points of view the sign of the correction to
the Regge trajectory δj(t) = ct1/4 is fixed because it corresponds to the slowdown of the
string, namely c < 0. From the bootstrap point of view, this condition follows neatly from
positivity of ρt,v(y).
To summarize, we see that the massive endpoints correction is welldescribed by the
effective classical distribution of excess zeros. It satisfies all the known bootstrap restrictions.
Since it has support outside of the unit disk, it results from a nondegenerate spectrum.
5.8
Discontinuities of the massive endpoints correction
The correction δ log A(s, t) in (5.14) has some remarkable properties that we now describe.
As discussed above, it corresponds to a situation where in the tchannel the Regge trajectory
is corrected and in the schannel it is not. Correspondingly, in the tchannel we have excess
zeros outside of the unit disk and in the schannel we do not.
It is instructive to introduce the effective electric field in both channels and express
them in terms of the distribution in the schannel [12]
(5.22)
(5.24)
(5.25)
fs(β) ≡ s1−k∂tδ log A(s, t) =
β =
s
t
,
ft(β) ≡ t1−k∂sδ log A(s, t) = βk−1 k
dx ρs(x) log 1 +
where for massive endpoints, we have k = 41 and ρs(x) is given in (5.18).
1
Z
0
dx ρs(x)
x + β1 ,
1
Z
0
1
βx
−
1
Z
0
dx
ρs(x)
1 + βx ,
It is easy to check that for these specific k and ρs(x), the electric field (5.26) have an
intriguing property
1
2i
[ft(−β + i0) − ft(−β − i0)] = fs − β
1
,
β > 1 .
(5.26)
1
2
One way to interpret it is the following. The l.h.s. of (5.26) effectively computes the vertical
density of zeros ρt,v(y), whereas the r.h.s. relates it to the electric field created by the zeros
inside the unit circle ρs(x) in the dual channel.
Another property is a direct consequence of (4.2) which after differentiation takes the
form20
HJEP06(218)54
dDisc[ft(β)] = ft(β) +
[ft(−1 − β + i0) + ft(−1 − β − i0)] = 0 ,
β > 1 .
(5.27)
This property of the electric field also holds for the logarithm of the amplitude. As we
explained this is a condition expected for any worldsheet theory. In terms of the bootstrap it
is related to the s−u channel crossing and the statement that there is no Stokes phenomenon
when we connect two regions of scattering at purely imaginary angles in each of which the
amplitude is exponentially large.
Bootstrapping the correction
Above we analyzed properties of the massive endpoint correction. Next we would like to see
how this correction emerges from the bootstrap analysis. We restrict ourselves to the case
when the Regge trajectory is not corrected in one of the channel. Within this class we will
argue that further imposing (5.27) fixes the solution uniquely to be given by (5.18) and k =
1/4. To prepare the ground, we will first analyze the large β structure of ft(β) for general k.
The double discontinuity constraint is simpler in terms of the symmetric variable
– 28 –
b = β − 2
.
1
In terms of this variable, the analytic continuations in (5.27) simply rotates b → −b. The
general large b power expansion of the electric field that is consisting with zero double
discontinuity is of the form
b→∞
lim ft(b) =
∞
X
1
n b1+2n (cn + dn log b) .
Lets now see what this implies for the small x expansion of ρs(x). Using the symmetry
ρs(x) = ρs(1 − x) we rewrite ft(β) in (5.26) as
1
2
Z
0
ft(β) = βk−1
dx [F (β, x) + F (β, 1 − x)] ρs(x) ,
F (β, x) = k log 1 +
(5.28)
(5.29)
1
βx
1
− 1 + βx
of ρs(x) or its mirror point x → 1. Since, when writing (5.30) we are already imposing the
ρs(x) = ρs(1 − x) reflection symmetry, its enough to focus on the small x expanssion.
We start by recalling that from the Regge limit we already know that
xli→m0 ρs(x) ∼ xk−1 log x + [less singular terms] .
(5.31)
Plugging xk−1 log x in (5.30) and expanding at large b we get both, terms of the type bk−1−n
and of the type b−n, with n integer.
We conclude that the subleading terms in the small x expansion (5.31) should be such
a) They should produce terms dnb−2n−1 log b in (5.29) with some dn 6= 0. These are
the terms that correspond to a distribution of excess zeros outside the unit disk and
represents the removal of the degeneracy.21
b) It has to be such that all terms of the type b−2n log b must cancel.
c) It has to be such that all terms of the type b−2n must cancel.
d) The terms of the type bk−1−n that are generically produced from (5.30) should all be
canceled.
These conditions are satisfied as follows
of the type bnxk−1+n log x.
fixes b2k+1 in terms of b2k.
satisfy c) and d).
A) To generate the terms dnb−2n−1 log(g) with dn 6= 0 we add to ρs(x) subleading terms
B) To cancel dn with even n we tune the coefficients bn. One can easily check that it
C) Next we can ask: which terms can we add to the distribution so that it generates the
same terms that we already have? The only other terms that have this property are
terms anxk−1+n. These generate terms bk−1−n and b−n and, thus, could be used to
Finally, one can try and add terms of the type xn to the small x expansion of ρs(x).
However, it is easy to see that these terms generate new undesired terms of the type
bk−1−n log b, which should be all set to zero. This condition sets the coefficients of such xn
terms to zero.
Thus, we arrive at the following ansatz for the distribution
xli→m0 ρs(x) = xk−1 X∞ xn(an log x + bn) .
n=0
(5.32)
Next, we explore this expansion in greater detail.
are considering here.
21If the degeneracy is not removed then as was shown in [12] there is no solution to the problem that we
1
Z
0
δρk(y) =
dx [K(y, x) + K(1 − y, 1 − x)] δρk(x) ,
where, after some algebra, the kernel can be written as22
K(y, x) =
cot πk
π
+
(1 − y)k−1
π sin πk
1
y P
x − y − k log
y
x + y − x y
x
x − y
+ k log
x (1 − y)
x + y − x y
.
It is straightforward to check that the distribution (5.18) indeed solves this integral
equation. To proceed we rewrite the integral equation (5.33) as follows
We shall now utilize the s−u crossing (4.2) in the form of (5.27) to fix k and the distribution.
That is, we consider a flavor amplitude with a generic k and try to find ρk(x) = ρk(1 − x)
such that (5.27) holds. This constraint leads us to the massive ends amplitude as the
unique solution.
To start, we use (5.26) to turn the zero double discontinuity condition (5.27) into an
integral equation. More precisely, we can compute ft(−1 − β ± i0) by doing an analytic
continuation under the integral (5.26). As we do that 1+xβ and log(1 + β1x ) develop an
1
imaginary part. This imaginary part, however, cancels in the combination (5.27). As
the result we get an integral equation of the type R01 dx K˜ (β, x)ρk(x) = 0 for β > 1.
Introducing y = β1 and taking the combination of the equation together with its image
under y → 1 − y and x → 1 − x, we turn it into a simpler integral equation. Using the
symmetry ρk(x) = ρk(1 − x), we can write the result as follows
(5.33)
(5.34)
(5.36)
(5.37)
Z
0
1
2
ρk(y) =
dx [K(y, x) + K(1 − y, x) + K(x, 1 − y) + K(1 − y, 1 − x)] ρk(x),
(5.35)
where we again used the symmetry ρk(x) = ρk(1 − x). At this stage we can plug the
ansatz (5.32) to both sides of the equation and match the coefficients.
We focus on the terms an yk+n−1 log y in (5.33). For the first few n’s one finds the
following result
The second equation has two solutions
a1 = 0 ,
1 − cot2 πk = 0 .
1
4
k =
and
k =
3
4
Higher order orders terms in n lead to an infinite number of relations between different
an’s, again fixing a2n+1 in terms of a2n.
22Here, P x −1y = limǫ→0 21 x−y+iǫ + x−y−iǫ stands for the principle value.
1 1
After this point we have not found an analytic way to solve the equation (5.35). Its
structure, however, suggests a straightforward numerical strategy, as we now describe. We
truncate the sum (5.33) at some nmax
We then consider the positivedefinite functional by squaring the difference between the
correspondent coefficients in the l.h.s. and r.h.s. of (5.35)
nmax
n=0
ρnum(x) = xk−1 X xn(an log x + bn) .
k
χ2 =
nmax
X (aln.h.s.
n=0
− arn.h.s.)2 + (bln.h.s.
− brn.h.s.)2
(5.38)
(5.39)
and minimize it for arbitrary values of an>1 and bn. If a solution exists then we expect
that by increasing nmax we will get χ2 approaching zero. This is, indeed, exactly what we
got for k = 1/4. It is however not the case for k = 3/4. Hence, we conclude that only
solutions with k = 1/4 exists.
To probe the uniqueness of the k = 1/4 solution we subtract the massive end solution
conditions. We checked for few low nmax’s that χ2
from the other potential solution δρˆk, so that a0 = 0. Since the integral equation (5.35)
is linear, the subtracted solution δρˆ − δρmassive ends is again a solution. We now assume
that there is such aˆn − anmassive ends 6= 0 and repeat the minimization procedure for this
≃ 0 solution does not exist. We conclude
that aˆn = an for all n. Repeating the same procedure for bn’s we conclude that the massive
ends solution is unique.
Let us emphasize that given a δρk(x) that satisfies (5.33) we can easily use it to
construct a fully crossing symmetric amplitude by simply taking a sum of
1
Z
0
δ log A(s, t) =
dx ρk(x) tk log 1 +
+ sk log 1 +
.
(5.40)
t x
t
s x
This amplitude could be still rewritten in terms of zeros in one channel only and moreover
by construction it has zero double discontinuity (4.2).
5.11
The role of crossing
picture of the string we assumed:
Note that in the discussion above crossing did not play any role. Indeed, motivated by the
1. The correction to the amplitude is a sum of the t and schannel corrections, (5.40).
2. The double discontinuity (4.2) of the correction vanishes.
Given these two assumptions, we showed that the massive ends correction is unique.
Both of these assumptions are tied to the fact that the massive endpoint correction is a
boundary interaction on the string worldsheet. The first one comes about because in
perturbation theory on the worldsheet we are instructed to sum over the boundary interaction
in the s and t channels. The second one comes about because the bulk of the string is free,
(and hence, holds at any order in the boundary interaction).
In general, we would like to relax both assumptions and still show that the solution is
unique. Since both are associated with a boundary interaction on the worldsheet, they may
be not independent of each other. From our discussion of the structure of the degeneracy
removal, it seems that assumption 1 is not enough to fix the solution. Instead, we expect
assumption 1 to follow from assumption 2.
Dropping assumption 1, we can proceed as follows. Instead of trivializing crossing
and analyzing the double discontinuity constraint as we did above, we could trivialize
the latter and let the nontrivial constraint to come from crossing. More precisely, we
imagine the following problem. Consider a crossing symmetric correction to the amplitude
δ log A(s, t) = δ log A(t, s) with zero double discontinuity (4.2). Based on our discussion
we expect the correction to take the form
δ log A(s, t) = tk dx ρh(x) log 1 +
+ tk
dy ρv(y) log 1 +
. (5.41)
1
Z
0
1
Z
0
1
Z
0
tx
Z
∞
−∞
π π
ρv(yei 2 ) + ρv(−ye−i 2 ) = 0 ,
y >
1
2
.
ρv(y) =
cn
X
n y2n+1 .
We can now first impose the vanishing of the double discontinuity. The double discontinuity
outside of the unit disk takes the form
This condition fixes the form of the large y expansion of ρv(y) to be
(5.42)
(5.43)
Now the crossing symmetry constraint becomes very nontrivial. Indeed, we have
k dx ρh(x) log 1 +
+ tk
dy ρv(y) log 1 +
tx
Z
∞
−∞
sx
Z
∞
−∞
t( 21 + iy)
s( 12 + iy)
= sk dx ρh(x) log 1 +
+ sk
dy ρv(y) log 1 +
, (5.44)
where ρv(y) is subject to (5.43).
Showing that the massive ends solution is the unique solution to this problem is
equivalent to relaxing assumption 1 above. We expect this to be the case, but we leave the
analysis of this problem to the future.
6
In this paper, we considered the 2 → 2 scattering amplitude in theories of weakly interacting
higher spin particles. We have computed the first correction to the asymptotic form of the
amplitude (1.1). The result is (1.2). We have argued that this is the unique universal
correction in the large energy, imaginary scattering angle regime. In the string theory
description, this correction due to massive endpoints that slow down string’s motion. On
the bootstrap side, (1.2) captures the nondegeneracy of the microscopic spectrum, which
the leading solution (1.1) does not reflect. Similarly, the correction to the npoint amplitude
takes the form (3.20). In principle, one can compute further corrections which are
subleading in m, but we do not expect these to be universal. It is also trivial to generalize the
result to the case of scattering of mesons with different flavors (3.24).
Both the leading solution (1.1) and the correction (1.2) exhibit an emergent s ↔ u
channel symmetry. We explained how it comes about from the analytic continuation of
a free string classical solution with boundary interactions. The continuation develops an
imaginary part which computes the onshell action on the Lorentzian continuation of the
Euclidean solution. The real part of the onshell action, however, stays the same as a result
of a certain stringy Stokeslike phenomenon. The consequence of this is the fact that the
double discontinuity vanishes (4.2).
We have analyzed the result using the bootstrap techniques of [12]. The new feature
of the correction is nonboundedness of the support of the excess zeros. This property
reflects the sensitivity of the asymptotic amplitude to the nondegeneracy of the microscopic
spectrum (the leading order answer is insensitive to it). From the bootstrap point of view,
the property (4.2) could be interpreted as absence of the Stokes phenomenon when we
connect different channels that describe scattering of identical particles at imaginary angles.
It would be interesting to understand this symmetry better and see if it could be violated.
When (4.2) is imposed, we have shown that the massive endpoints solution is unique.
We have argued that (1.2) is the only universal, modelindependent correction to (1.1)
which grows with energy. To bootstrap the amplitude beyond this correction we believe
one should start adding modeldependant input. It would be interesting to understand
what is the minimal input needed to fix the next correction and what is the most natural
way to specify it. One might hope that akin to the 3d Ising model [29], there exist special
corners in the space of parameters of the spectrum that describe physical theories. One
natural assumption is to have graviton in the spectrum. More generally, it would be most
rewarding if one could formulate the problem of fixing the amplitude using a numerical
Smatrix bootstrap approach, (see [30, 31] for a different, but related recent progress).
There is one important characteristic feature of strings that so far our analysis did not
touch. That is, the high energy density of states in the string models exhibits characteristic
Hagedorn growth. For the Veneziano amplitude, due to the degeneracy of the spectrum,
this property could not be seen at the level of the fourpoint amplitude.23 There, the
Hagedorn growth was derived by considering higherpoint functions [32, 33]. We expect this to be
a necessary step to establish Hagedorn from the bootstrap as well. We note, however, that
the massive endpoints correction results from the removal of degeneracy in the spectrum.
Hence, it may be tied to the Hagedorn growth. To understand if this is indeed the case one
should study what is the minimal density of states that could lead to the correction (1.2).
We leave a more thorough investigation of the degeneracy removal pattern to the future.
23One can argue for it based on the fourpoint function if one assumes that all the threepoint coefficients
are of the same order. This extra assumption, however, needs further justification.
Another natural direction is to consider full quantum amplitudes [30, 31]. In this case,
akin to the PolchinskiStrassler mechanism, the large energy limit of the gravitational
amplitudes seems to have universal features due to the production of large black holes. It
would be fascinating to explore this regime using bootstrap.
Finally, let us mention an interesting extension of the present analysis. That is the
analysis of WIHS theories in AdS. Such theories are dual to large N CFTs. A natural
space for their bootstrap is Mellin space [34]. It is quite easy to recover the usual results
about the gauge theories, see appendix B. In AdS, however, we expect to have more
solutions, corresponding to vector models, see e.g. [35], and SYKlike models, see e.g. [36]
and references therein. These models have an accumulation point in the twist spectrum,
which is an analog of the accumulation in the spectrum discussed in the flat space context.
It would be very interesting to understand what are all possible WIHS in AdS and bootstrap
them. We leave this problem for the future.
Acknowledgments
We are grateful to O. Aharony, S. CaronHuot, T. Cohen, S. Dubovsky, R. Gopakumar, Z.
Komargodski, S. Komatsu, C. Sonnenschein, M. Strassler and S. Yankielowicz for useful
discussions. The work of AZ is supported by a Simons Investigator Award of X. Yin from the
Simons Foundation. AS has been supported by the ICORE Program of the Planning and
Budgeting Committee, The Israel Science Foundation (grant No. 1937/12) and by the Israel
Science Foundation (grant number 968/15) and the EUFP7 Marie Curie, CIG fellowship.
A
Mass perturbation for the fourpoint amplitude
In this appendix, we develop the small mass perturbation theory, order by order. We
confirm (3.18) and (3.20) through a detailed stepbystep computation. This exercise also
demonstrates the power of the shortcut that we took in the main text.
Recall that the small mass perturbation theory is organized in powers of √m (this fact
was derived in section 3). Hence, we expand the action
S =
1
2πα′
Z
2
d2z ∂zxμ∂z¯xμ +
m2
e
+ i X kj xμ(σ˜j ) ,
μ
(A.1)
as well as all variables in powers of √m. We then extremize the action order by order. We
have
S = S0 + √mS1 + . . . ,
xμ = x0μ + √mx1μ + . . . ,
e = e0 + √me1 + . . .
(A.2)
We also have to expand the vertex operators insertion points which enter the onshell action
through an SL(
2
) invariant cross ratio σσ˜˜11,,24σσ˜˜32,,43 .
Indeed, the action (A.1) has an SL(
2
) gauge symmetry that acts on the vertex operator
insertion points σ˜j , leaving their conformal cross ratio σσ˜˜11,,42σσ˜˜23,,34 invariant. It is convenient
to pick an SL(
2
) frame. We choose
{σˇ1, σˇ2, σˇ3, σˇ4} =
0, ,
1 s + t
2 s + 2t
, 1
− {√mλ1 + mλ2 + . . . , 0, 0, 0}
≡ {σ1, σ2, σ3, σ4} − {√mλ1 + mλ2 + . . . , 0, 0, 0} ,
where we have used the fact that σσ˜˜11,,42σσ˜˜23,,34 m=0 ≡ λ0 = s/t, [5, 20]. In terms of these λ’s the
cross ratio takes the form
σ˜1,4σ˜2,3
σ˜1,2σ˜3,4 = λ0 + √mλ0λ1 + . . . .
The action is Poincare invariant, and hence only depends on the Mandelstam invariants
s = −(k1 + k2)2 and t = −(k1 + k4)2. We can always do a Poincare transformation that
places the four external momenta in an R1,2 subspace of R1,d−1. A convenient choice is
k1 =
k2 =
,
2
, − 2
, 0 ,
, 0 ,
k3 =
k4 =
,
s + 2t
2 s
√ , i
s + 2t
, −i
pt(s + t) !
pt(s + t) !
(A.3)
(A.4)
(A.5)
(A.6)
(A.7)
(A.8)
(A.9)
where the signature is (−, +, +). Note that the third component is purely imaginary. This
is because we are considering imaginary scattering angles, where s, t > 0. As mentioned in
the main text, one may think of the momenta (A.5) as being real, with signature (−, +, −).
A.1
Order m0
At leading order m = 0 and the action becomes free. The corresponding classical solution
in this case is the famous one of [5, 20], (or its analog for s, t > 0). Here, we present it as
a preparation for higher orders.
At zero mass the equations of motion simplify to
∂∂¯x0μ = 0 ,
1
2πα′ ∂τ x0 τ=0 = i X kjμδ(σ − σj) ,
μ
j
e0 = 0 .
S0 = α′ [s log s + t log t − (s + t) log(s + t)]
Correspondingly, the free classical solution is given by
x0μ = iα′ X kjμ log z − σj2 .
j
By plugging this back into the free action we arrive at
S0 =
4πα′
dσdτ ∂αx0 · ∂αx0 + i X kj · x0(σj) = − 21 α′ [t log(1 + λ0) + s log(1 + 1/λ0)]
Finally, extremizing S0 with respect to λ0 we find that λ0 = s/t. By plugging this back
into the action we get
This order simply vanishes by the e.o.m. for x0 and λ0. Explicitly, we have
dσdτ ∂αx0 · ∂αx1 + i X kj · x1(σj) + iλ1∂λ X kj · x0(σˇj)λ=0
(A.11)
where σˇ was defined in (A.3).
The first two terms are zero because x0μ is onshell. The last term is zero because S0
was minimized with respect to variations of λ0. Hence, S1 = 0 and at this order we get no
in agreement with (1.1).
Note that we did not have to impose the Virasoro constraints
∂zx0 · ∂zx0 = ∂z¯x0 · ∂z¯x0 = 0 .
(A.10)
It is automatically satisfied after we minimize the action (A.8) with respect to the vertices
insertion points. This will be true for higher orders in m as well.
4πα′
4πα′
2 1 λ
+ λ ∂
i 2 2 X kj · x0(σˇj)λ=0 .
The first line vanishes for the same reason as (A.11). Indeed, exchanging xμ
λ1 → λ2 the two are identical. In the second line we have a free bulk field, x1μ with two
1 → x
2μ and
boundary sources. We split it accordingly
Z
y1μ(z, z¯) = −α′
dσ ∂σ (e ∂σx0μ) log z − σ2 = α′
dσ e1 ∂σx0
μ
1
σ − z
1
σ − z¯
We see that x˜1μ ∝ k1μ and therefore P kj · x˜1(σˇj) is independent of λ. As a result
x˜1μ(z, z¯) = i α2′ k1μ λ1
z − σ1
1
z¯ − σ1
.
1 Z
2πα′
d2z∂x˜1 · ∂¯x˜1 + iλ1∂λ X kj · x˜1(σˇi)λ=0 = 0 .
A.2
Order m1/2
S1 =
4πα′
constraint on x1μ, λ1.
A.3
Order m1
At order m1 we have
where
and
S2 =
dσdτ ∂αx0 · ∂αx2 + i X kj · x2(σj) + iλ2∂λ X kj · x0(σˇj)λ=0
dσ e1 ∂σx0 · ∂σx1 + iλ1∂λ X kj · x1(σˇj)λ=0
j
j
Z
x1μ = x˜1μ + y1μ ,
j
j
(A.12)
(A.13)
, (A.14)
(A.15)
(A.16)
They evaluate to
where
1 Z
2
Z
2 1 λ
j
j
2i λ21∂λ2 X kj · x0(σˇi)λ=0 = − 2 s + t 1
α′ s t
λ2,
dσ e1 ∂σx0 · ∂σy1 = −2α′(s + t)E[e1]2,
dσ µ (σ) ,
µ (σ) ≡ α′
Q (σ − σj)
pQ σi+1 − σi e1(σ).
S2 = −α′
s t
λ1 +
√s t
2
E[e1] .
Combining all pieces together we get that S2 is a complete square
Extremizing (A.21) with respect to λ1, we arrive at
2πα′
d2z∂x˜1 · ∂¯y˜1 +
Z
dσ e1 ∂σx0 · ∂σx˜1 = 0 ,
(A.17)
where we have viewed x˜1μ as a perturbation of y1μ. We remain with three terms
S2 = iλ1∂λ X kj · y1(σj) + λ ∂
i 2 2 X kj · x0(σˇj)λ=0 +
dσ e1 ∂σx0 · ∂σy1 .
(A.18)
2
λ1 = −2 √s t
s + t
S2 = 0 ,
We conclude that
and, hence, the correction starts at order m3/2.
Note that (A.22) is precisely the Virasoro constraint at order √m. Namely,
∂zx1 · ∂zx0 = − 2
Q (z − σj)
α′ pQ σi+1 − σi √s t λ1 + 2(s + t)E[e1]
is automatically satisfied once we extremize with respect to all variables.
A.4
Order m3/2
At order m3/2 we organize all the contributions as follows
S3 =
dσ
m2
e1
+ A1 + A2 + A3 + A4
(A.19)
(A.20)
(A.21)
(A.22)
(A.23)
(A.24)
(A.25)
HJEP06(218)54
where
A1 =
A2 =
A3 =
A4 =
2πα′
2πα′
Z
2
We find that: around this extremum.
Z
j
dσdτ ∂αx0 · ∂αx3 + i X kj · x3(σj) + iλ3∂λ x0(σˇj)λ=0 ,
(A.26)
dσdτ ∂αx1 · ∂αx2 +
dσ e1 ∂σx0 · ∂σx2 + iλ1∂λ X kj · x2(σˇj)λ=0 ,
dσ e2 ∂σx1 · ∂σx0 + iλ2∂λ X kj · x1(σˇj)λ=0 + iλ2λ1∂λ2 X kj · x0(σˇj)λ=0 ,
dσ e1 ∂σx1 · ∂σx1 +
i λ2∂2 X kj · x1(σˇj)λ=0 +
i λ3∂3 X kj · x0(σˇj)λ=0 .
j
1) A1 = 0 because x0 and λ0 are the extremum of S0. A1 is a linear variation of S0
2) A2 = 0 for any x2. This is because x1 is the extremum of S2, (A.12), and in A2 we
have a linear variation of x1 in S2 around this point.
3) The first term in A3 vanishes because ∂σx1 · ∂σx0 = 0. The last two terms sum up
to zero because
iλ2∂λ X kj · x1(σˇj)λ=0 = −iλ2λ1∂λ2 X kj · x0(σˇj)λ=0 =
j
α′
2 λ2√s t E[e1] (A.27)
The reason for this cancellation is that at order m1 we concluded that S2 is given by
S2 =
4πα′
= −α′
s t
2(s + t)
λ1 +
√s t
E[e1]
Z
2 1 λ
2
.
dσdτ ∂αx1 · ∂αx1 +
dσ e1 ∂σx0 · ∂σx1
+iλ1∂λ X kj · x1(σˇi)λ=0 + λ ∂
1 2 2 X kj · x0(σˇi)λ=0
(A.28)
If we vary this as λ1 → λ1 + λ2 or e1 → e1 + e2 to linear order and adjust the
classical solution accordingly x
μ
1 → x1μ + δxμ(e1, λ2) then the result vanishes because
we started at the extremum. As we seen above, any linear variation x1 → x1μ + δxμ of
μ
S2 vanishes by itself, just the same as A2 = 0. So we see that A3 = 0 too. Note that
S3 is therefore independent of e2 and λ2. Hence, we can evaluate it without solving
for these new variables.
3i! λ31∂λ3 X kj · x0(σˇi)λ=0 = − 6 √s t
α′ s + t
(5s + 4t)E[e1]3 ,
i λ2∂2 X kj · x1(σˇi)λ=0 =
√s t
2
dσ e1 ∂σx1 · ∂σx1 = − 4
µ (σ)
σ − σ1
,
Z
2
Z
Z
(s + t)
dσ µ (σ)
dχ µ (χ)
(A.29)
dξ e1(ξ)
(σ − χ)2
(σ − ξ)2(ξ − χ)2 − (σ − ξ)2(ξ − σ1)2
HJEP06(218)54
integral
I(z) =
Note that the onshell action is precisely of the form (3.18) that we used in the text.
In this way, we can run a simple argument and conclude that the result is given by (3.20).
To run the argument, however, we should make sure that (A.30) is not zero.
Surprisingly, if we symmetrize the sum of three terms above over integration variables the sum
vanishes for any e1.24 The integrals, however, contain divergences on the integration
contour and we need to treat them carefully with iǫ prescription coming from the worldsheet.
After this is done the result is indeed what we expected. Showing this explicitly is the
subject of the next section.
A.5
Evaluation of A4
We now evaluate A4 on the solution for e1(σ) (3.14) explicitly and confirm the result (3.22).
All we need for the derivation in the main text is that A4 6= 0. Still, it does not harm to
confirm the result by an independent explicit calculation.
To evaluate the integrals in (A.30) explicitly it is convenient to introduce the following
Z
σ − z
Qi σ − σi1/2
Qi(σ − σi)
4
i=1
Ii(z) =
σi+1
Z
σi
σ − z Qj σ − σj 1/2 ,
1
where we assume z to be in the upper halfplane. It is also convenient to introduce
E ≡ − zl→im∞ z I(z) =
Z
Qi σ − σi1/2
Qi(σ − σi)
= 4
r 2(2s + t)
s + t
K
t
s + t
− K
s
s + t
To perform the integral in Ii(z), (A.30), we make the following change of variables
σ = σi + (σi+1 − σi) sin2 φ
dσ
σ − σi√σi+1 − σ
= 2dφ
24We thank T. Cohen for pointing this to us.
(A.30)
(A.31)
(A.32)
where
α′2
α′2
− 4
Z
Z
2
2
π
2
Z
Z
Z
. After this change Ii(z) becomes
Ii(z) = − (σi+1 − σi)2
dφ q σi+1−σi−1
1
σi+1 − σi−1σi+2 − σi(σi+2 − z)
1
1
σi+1−σi − cos2 φ q σi+2−σi+1 + cos2 φ σzi−+σ1−i+σ1i + cos2 φ
σi+1−σi
K(u) + σi+2 − σi
σi − z
1 − u
Π
−w,
u − 1
1
u =
(σi+2 − σi+1)(σi − σi−1) ,
(σi+2 − σi)(σi+1 − σi−1)
w =
(z − σi+2)(σi+1 − σi) .
(z − σi)(σi+2 − σi+1)
We can now plug the solution for e1(σ) (3.14) into (A.30) to get
(A.33)
(A.34)
Z
Z
(s + t)
dσ µ (σ)
dχ µ (χ)
dξ e1(ξ)
= α′2
pst(s + t) Z
π(2s + t)
dξ e1(ξ)
Re[I(ξ)]2 − E Re
(s + t)
dσ µ (σ)
dχ µ (χ)
dξ e1(ξ)
= α′2
pst(s + t) Z
π(2s + t)
dξ
µ (ξ)
ξ − σ1
dξ e1(ξ) E
Re
= −α′2 √st(s + t)3/2 Z
2π(2s + t)2
(σ − χ)2
(σ − ξ)2(ξ − χ)2
− (σ − ξ)2(ξ − σ1)2
dI(ξ)
dξ
σ1 − ξ
dξ Q(ξ − σi) σ1 − ξ
,
E
(σ1 − ξ)2
1
,
(A.35)
where to get these results one should be careful with the iǫ prescription which we kept
implicit in the formulas for the onshell action. We perform the last integral over ξ
numerically. To do this we need to understand the behavior of the function close to the vertex
operators. The result is that it has a (σ − σi)− 21 integrable singularity close to σ2,3,4 and
has a simple pole at σ1. These statements are not manifest when looking at (A.35). For
1
example the last line has a singularity (ξ−σ1)3/2 close to ξ = σ1, it however is canceled
against the other terms. The iǫ prescription instructs us to treat the pole at ξ = σ1 as a
principal value. When combined with the first line in (A.30) we get that
A4 =
m2
S3 =
+ A4 =
2
m2
m2
e1
as expected.
B
Bootstrap in Mellin space
It is natural to ask what is the analog of the flat space Smatrix asymptotic bootstrap [12]
in AdS. A natural idea is to consider the Mellin amplitude M (s, t) and focus on the regime
where s and t are larger than any relevant scale in the problem, say √λ. Let us recall what
is the Mellin representation of the planar connected fourpoint function F (u, v) [34, 37].
The standard version of Mellin transform takes the form
Mc(γ12, γ14) = R∞ dudv uγ12 vγ14 F (u, v) .
uv
(B.1)
inverse transformation
F (u, v) =
cZ+i∞dγ12dγ14 u−γ12 v−γ14 Mc(γ12, γ14) ,
a < c < b .
The relation between γij and Mandelstamlike variables we mentioned above is the
followγ12 = Δ − 2
t
γ13 = − 2 ,
s
γ14 = Δ − γ12 − γ13 =
s + t
2
= − 2
where Δ is the conformal dimension of the external identical scalar operators.
In the discussion of planar theories it is common to define the Mellin amplitude by
writing down explicitly the gammafunctions prefactor
Mc(γ12, γ14) = [Γ(γ12)Γ(γ13)Γ(γ14)]2 M (s, t) .
The prefactor has singularities at γ12, γ13, γ14 = −integer, which corresponds to the
exchange of the double trace operators.
As is the case with the usual scattering amplitudes, the Mellin amplitude M (s, t) has
poles dictated by the twists of the exchanged operators τ = Δ − J . The residues are fixed
in terms of Δ and J and the correspondent threepoint couplings. It takes the form
M (s, t) ≃ t − (τ + 2m)
2
cΔΔτ QJ,m
+ . . . ,
QJ,m
Δ,τ,d(s) = K(Δ, J, m) QJΔ,m,τ,d(s) ,
where m = 0, 1, 2, . . . stands for the level of the descendent, QJΔ,m,τ,d(s) are polynomials
in s of degree J , c2ΔΔτ labels the correspondent threepoint coupling and K(Δ, J, m) is a
kinematical prefactor that can be found in [38]. As in flat space, unitarity implies that
for s, t ≫ 1 the Mellin amplitude develops large imaginary part.
Let us assume that the states that dominate the Mellin amplitude have large spin
J ≫ s, t. In this case the crossing equation to leading order takes the form
M (s, t) ≃
J(t)
X cj js =
j
J(s)
j
X cj jt ≃ M (t, s) ,
where we have used the large J asymptotic of the Mack Polynomials
lim QJ,m
2J−m+τ
π Γ(m + 1)Γ( 2Δ−22m−τ )2
J s+ 12 (sm + O(sm−1))
Γ( s+τ+2m )2
2
1
J
+ O
(B.7)
The crossing equation (B.6) is basically telling us that J (t)s ≃ J (s)t. It has the trivial
solution
A posteriori we see that, indeed, J (t) ≫ t as we assumed.
J (t) = e c .
t
(B.3)
HJEP06(218)54
(B.4)
(B.6)
(B.8)
Notice that close to the pole at t = τ + 2m, the Mellin amplitude is controlled by
J (t) = J (τ + 2m). At large twist, the solution (B.8) corresponds to
τ (J ) = c log J + . . . .
(B.9)
Of course, this formula is very familiar from the studies of the gauge theories [39]. Here
we see how this result comes out of the simple bootstrap consideration. It implies that the
asymptotic behavior of the Mellin amplitude takes the form
s,t→∞
lim log M (s, t) =
s t + . . . ,
1
c
(B.10)
Identical formulas can be written for the other channels: s, u ≫ 1 or t, u ≫ 1.
This result is related to the double lightcone limit of the connected fourpoint function
discussed in [40]. Indeed, the limit γ12, γ14 → ∞ of (B.1) is controlled by the u, v → 0 of
F (u, v). In [40] it was argued that in planar gauge theory, this limit is controlled by the
factor
in Mellin space.
u,v→0
lim F (u, v) ≃ e− 4
Γcusp log u log v ,
which is related to the AldayMaldacena solution [21]. Let us see what this behavior implies
We start from the kinematical region 0 ≤ u, v ≤ 1 consider the small u, v contribution
to Mc(γ12, γ14) in (B.1). Using (B.11) we arrive at
Mc(γ12, γ14) ≃
0
1
u v
Z du dv uγ12 vγ14 e− 4
Γcusp log u log v =
Z
γ12γ14 + Γc4usp V
This integral is welldefined in the γ12, γ14 > 0 region. To connect with (B.10) we
analytically continue to the region γ12, γ14 < 0 (and hence positive s and t). As we analytically
continue γ12 and γ14, the pole in (B.12) moves from the negative V axis towards the
positive axis and then back to the negative axis dragging the contour of integration together
with it. The leading asymptotic is given by the residue at the pole. It gives
s,t→∞
lim log M (s, t) =
s t + . . . .
1
Γcusp
Comparing this with (B.10) we learn that c = Γcusp. This is precisely what we expect
from (B.9). Notice that regions other than 0 ≤ u, v ≤ 1 produce smaller contributions at
large γ12, γ14 < 0, which justifies our approximation above.
Let us conclude this appendix with two future directions. First, it would be interesting
to understand how the vector models and SYKlike models fit into this picture. In these
cases, it is not true that J (t) ≫ t and there is an accumulation point in the twist spectrum.
Finally, it would also be interesting to understand the relation of our considerations in this
appendix to the recent work on bootstrap in Mellin space [41, 42].
Open Access.
This article is distributed under the terms of the Creative Commons
Attribution License (CCBY 4.0), which permits any use, distribution and reproduction in
any medium, provided the original author(s) and source are credited.
(B.11)
(B.12)
(B.13)
References
[2] G. Veneziano, Construction of a crossingsymmetric, Regge behaved amplitude for linearly
rising trajectories, Nuovo Cim. A 57 (1968) 190 [INSPIRE].
[3] M.A. Virasoro, Alternative constructions of crossingsymmetric amplitudes with Regge
behavior, Phys. Rev. 177 (1969) 2309 [INSPIRE].
Phys. Lett. B 197 (1987) 129 [INSPIRE].
[5] D.J. Gross and P.F. Mende, The highenergy behavior of string scattering amplitudes,
[6] O. Andreev and W. Siegel, Quantized tension: stringy amplitudes with Regge poles and
parton behavior, Phys. Rev. D 71 (2005) 086001 [hepth/0410131] [INSPIRE].
[7] G. Veneziano, S. Yankielowicz and E. Onofri, A model for pionpion scattering in largeN
QCD, JHEP 04 (2017) 151 [arXiv:1701.06315] [INSPIRE].
[8] J. Polchinski and M.J. Strassler, Hard scattering and gauge/string duality,
Phys. Rev. Lett. 88 (2002) 031601 [hepth/0109174] [INSPIRE].
[9] H.B. Meyer and M.J. Teper, Glueball Regge trajectories and the pomeron: a lattice study,
Phys. Lett. B 605 (2005) 344 [hepph/0409183] [INSPIRE].
[10] S. Dubovsky, R. Flauger and V. Gorbenko, Effective string theory revisited,
JHEP 09 (2012) 044 [arXiv:1203.1054] [INSPIRE].
[11] O. Aharony and N. Klinghoffer, Corrections to NambuGoto energy levels from the effective
string action, JHEP 12 (2010) 058 [arXiv:1008.2648] [INSPIRE].
[12] S. CaronHuot, Z. Komargodski, A. Sever and A. Zhiboedov, Strings from massive higher
spins: the asymptotic uniqueness of the Veneziano amplitude, JHEP 10 (2017) 026
[arXiv:1607.04253] [INSPIRE].
[13] J. Sonnenschein and D. Weissman, Rotating strings confronting PDG mesons,
JHEP 08 (2014) 013 [arXiv:1402.5603] [INSPIRE].
[14] J. Polchinski and M.J. Strassler, The string dual of a confining fourdimensional gauge
theory, hepth/0003136 [INSPIRE].
[15] S. Hellerman and I. Swanson, String theory of the Regge intercept,
Phys. Rev. Lett. 114 (2015) 111601 [arXiv:1312.0999] [INSPIRE].
[16] J. Erdmenger, N. Evans, I. Kirsch and E. Threlfall, Mesons in gauge/gravity duals — a
review, Eur. Phys. J. A 35 (2008) 81 [arXiv:0711.4467] [INSPIRE].
[17] J. Sonnenschein, Holography inspired stringy hadrons, Prog. Part. Nucl. Phys. 92 (2017) 1
[arXiv:1602.00704] [INSPIRE].
[18] M. Kruczenski, L.A. Pando Zayas, J. Sonnenschein and D. Vaman, Regge trajectories for
mesons in the holographic dual of largeNc QCD, JHEP 06 (2005) 046 [hepth/0410035]
Nucl. Phys. B 72 (1974) 509 [INSPIRE].
[19] A. Chodos and C.B. Thorn, Making the massless string massive,
[20] D.J. Gross and J.L. Manes, The highenergy behavior of open string scattering,
JHEP 06 (2007) 064 [arXiv:0705.0303] [INSPIRE].
[21] L.F. Alday and J.M. Maldacena, Gluon scattering amplitudes at strong coupling,
[22] B.M. Barbashov and V.V. Nesterenko, Relativistic string with massive ends,
[arXiv:1302.6257] [INSPIRE].
[24] O. Aharony and Z. Komargodski, The effective theory of long strings, JHEP 05 (2013) 118
[25] M.A. Vasiliev, Consistent equation for interacting gauge fields of all spins in
(3 + 1)dimensions, Phys. Lett. B 243 (1990) 378 [INSPIRE].
[26] I.R. Klebanov and A.M. Polyakov, AdS dual of the critical O(N ) vector model,
Phys. Lett. B 550 (2002) 213 [hepth/0210114] [INSPIRE].
hepth/9404025] [INSPIRE].
[28] G.W. Moore, Symmetries of the bosonic string S matrix, hepth/9310026 [Addendum
[29] S. ElShowk, M.F. Paulos, D. Poland, S. Rychkov, D. SimmonsDuffin and A. Vichi, Solving
the 3D Ising model with the conformal bootstrap, Phys. Rev. D 86 (2012) 025022
[arXiv:1203.6064] [INSPIRE].
[30] M.F. Paulos, J. Penedones, J. Toledo, B.C. van Rees and P. Vieira, The Smatrix bootstrap.
Part I: QFT in AdS, JHEP 11 (2017) 133 [arXiv:1607.06109] [INSPIRE].
[31] M.F. Paulos, J. Penedones, J. Toledo, B.C. van Rees and P. Vieira, The Smatrix bootstrap.
Part II: two dimensional amplitudes, JHEP 11 (2017) 143 [arXiv:1607.06110] [INSPIRE].
[32] S. Fubini and G. Veneziano, Level structure of dualresonance models,
Nuovo Cim. A 64 (1969) 811 [INSPIRE].
[33] S. Fubini, D. Gordon and G. Veneziano, A general treatment of factorization in dual
resonance models, Phys. Lett. B 29 (1969) 679 [INSPIRE].
JHEP 03 (2011) 025 [arXiv:1011.1485] [INSPIRE].
[34] J. Penedones, Writing CFT correlation functions as AdS scattering amplitudes,
[35] S. Giombi and X. Yin, The higher spin/vector model duality, J. Phys. A 46 (2013) 214003
[arXiv:1208.4036] [INSPIRE].
[36] J. Murugan, D. Stanford and E. Witten, More on supersymmetric and 2d analogs of the
SYK model, JHEP 08 (2017) 146 [arXiv:1706.05362] [INSPIRE].
[37] G. Mack, Dindependent representation of conformal field theories in D dimensions via
transformation to auxiliary dual resonance models. Scalar amplitudes, arXiv:0907.2407
[38] M.S. Costa, V. Goncalves and J. Penedones, Conformal Regge theory, JHEP 12 (2012) 091
[INSPIRE].
[arXiv:1209.4355] [INSPIRE].
JHEP 11 (2007) 019 [arXiv:0708.0672] [INSPIRE].
functions to Wilson loops, JHEP 09 (2011) 123 [arXiv:1007.3243] [INSPIRE].
bootstrap, JHEP 05 (2017) 027 [arXiv:1611.08407] [INSPIRE].
2 (s + t) [1] A.M. Polyakov , Fine structure of strings, Nucl . Phys. B 268 ( 1986 ) 406 [INSPIRE].
[4] J.A. Shapiro , Electrostatic analog for the Virasoro model , Phys. Lett. B 33 ( 1970 ) 361 Theor . Math. Phys. 31 ( 1977 ) 465 [Teor . Mat. Fiz . 31 ( 1977 ) 291] [INSPIRE].
[23] J. Polchinski and A. Strominger , Effective string theory , Phys. Rev. Lett . 67 ( 1991 ) 1681 [27] D.J. Gross , Highenergy symmetries of string theory , Phys. Rev. Lett . 60 ( 1988 ) 1229 [39] L.F. Alday and J.M. Maldacena , Comments on operators with large spin , [40] L.F. Alday , B. Eden , G.P. Korchemsky , J. Maldacena and E. Sokatchev , From correlation [41] R. Gopakumar , A. Kaviraj , K. Sen and A. Sinha , Conformal bootstrap in Mellin space , Phys. Rev. Lett . 118 ( 2017 ) 081601 [arXiv: 1609 .00572] [INSPIRE]. [42] R. Gopakumar , A. Kaviraj , K. Sen and A. Sinha , A Mellin space approach to the conformal