Light stringy states - NBI Conference and Meetings (Indico) · 2017. 9. 26. · For the sake of...

Nordic String Meeting - 20/02/2012

Light stringy states Robert Richter

Based on: 1110.5359 [hep-th], 1110.5424 [hep-th] with P. Anastasopoulos & M. Bianchi

Monday, February 20, 12

Introduction and Motivation

c

R

b

a

dE

RD

L

LQRU


Motivation

✤ D-brane compactifications provide a promising framework for model building

✤ They allow for large extra dimensions which imply a significantly lower string scale, even of just a few TeV.

✤ Scenarios of these kinds may explain the hierarchy problem, but also allow for stringy signatures that can be observed at LHC.

✤ There exists a class of amplitudes containing arbitrary number of gauge bosons and maximal two chiral fermions that exhibit a universal behaviour independently of the specifics of the compactification

✤ Due to their universal behaviour they have predictive power.

✤ The observed poles correspond to the exchanges of Regge excitations of the standard model gauge bosons, whose masses scale with the string mass .

✤ Such poles might be observable at LHC if one has low string scale

Ms

Lust, Stieberger, Taylor, et. al.

Antoniadis, Arkani-Hamed, Dimopoulos, Dvali


Motivation

✤ On the other hand there exist a tower of stringy excitations localized at the intersections of two stacks of D-branes.

✤ Their masses depend on the string mass and the intersection angle ! and thus can be significantly lighter than the Regge excitations of the gauge bosons.

✤ As we will see those light stringy states show up as poles in the scattering amplitudes containing four fermions.

✤ Such amplitudes are very model dependent, thus do not have the predictive power of the universal amplitudes.

✤ However the poles corresponding to light stringy states should be observed primary to Regge excitations for the universal amplitudes and thus may provide a first step towards evidence for string theory.

Ms


For the sake of calculability we assume two intersectng D6-branes on , where the D6-branes wrap in each torus a one-cycle

✤ Supersymmetry translates into:

✤ We will take a closer look at the (massless and massive) states appearing at such an intersection.

Intersecting D6-branes

5X

6X

3X

4X

1X

3ș2ș1ș

4M

2X


Quantization at angles✤ Boundary conditions for strings between branes at angles:

✤ Mode expansion ( ):

for " = 0, 1/2 for R and NS respectively.

✤ The commutator/anticommutators:

@�Xp(⌧, 0) = Xp+1(⌧, 0) = 0@�Xp(⌧,⇡) + tan (⇡✓) @�Xp+1(⌧,⇡) = 0Xp+1(⌧,⇡)� tan (⇡✓) Xp(⌧,⇡) = 0

Zp = Xp + iXp+1

[↵In±✓, ↵

I0

m⌥✓] = (m± ✓) �n+m �II0

I(z) =X

r2Z+⌫

Ir�✓I

z�r� 12+✓I I(z) =

X

r2Z+⌫

Ir+✓I

z�r� 12�✓I

{ Im�✓I

, I0

n+✓I} = �m,n�

II0

pX

p+1X

ș


The vacuum✤ Intersecting branes in 10D:

✤ NS sector (4D bosons):

- Positive angle

- Negative angle

✤ Recall: GSO-projection requires odd number of fermionic excitations.

5X

6X

3X

4X

1X

3ș2ș1ș

4M

2X

↵m�✓I | ✓I iNS = 0 m � 1 r�✓I | ✓I iNS = 0 r � 1/2

↵m+✓I | ✓I iNS = 0 m � 0 r+✓I | ✓I iNS = 0 r � 1/2

↵m�✓I | ✓I iNS = 0 m � 0 r�✓I | ✓I iNS = 0 r � 1/2

↵m+✓I | ✓I iNS = 0 m � 1 r+✓I | ✓I iNS = 0 r � 1/2


Lightest string states✤ Recall the mass formula:

✤ Concrete setup: with (SUSY).

✤ The lowest fermionic excitations of this configuration:

✓ab1 < 0 , ✓ab2 < 0 , ✓ab3 < 0 ✓ab1 + ✓ab2 + ✓ab3 = �2

� 12�✓ab

I| ✓ab1,2,3 iNS M2 =

1

2(✓abI �

X

J 6=I

✓abJ )M2s =

�1 + ✓abI

�M2

s

Y

I

� 12�✓ab

I| ✓ab1,2,3 iNS M2 = (1 +

1

2(✓ab1 + ✓ab2 + ✓ab3 ))M2

s = 0

M2 = M2s

X

I

X

m✏Z

: ↵I�m+✓I↵

Im�✓I : +

X

m✏Z

(m� ✓I) : I�m+✓I

Im�✓I : +✏I0

!

�1

8± 1

2✓I


Lightest string states

✤ Some additional light states (for small ):

✤ These scalars are potentially very light, depending on the intersection angles.

✤ If the string scale is low, and the angles are small, such states have very low masses.

✤ Additional states, such as Higher Spin states, but even massive.

↵✓1

Y

I

� 12�✓I | ✓

ab1,2,3 iNS M2 = (1 +

1

2

X

I

✓I � ✓1)M2s = �✓1M2

s

(↵✓1)2Y

I

� 12�✓I | ✓

ab1,2,3 iNS M2 = (1 +

1

2

X

I

✓I � 2✓1)M2s = �2✓1M

2s


The vacuum✤ Intersecting branes in 10D:

✤ R sector (4D fermions):

- Positive angle

- Negative angle

5X

6X

3X

4X

1X

3ș2ș1ș

4M

2X

↵m�✓I | ✓I iR = 0 m � 1 r�✓I | ✓I iR = 0 r � 1

↵m+✓I | ✓I iR = 0 m � 0 r+✓I | ✓I iR = 0 r � 0

↵m�✓I | ✓I iR = 0 m � 0 r�✓I | ✓I iR = 0 r � 0

↵m+✓I | ✓I iR = 0 m � 1 r+✓I | ✓I iR = 0 r � 1


✤ Recall the mass formula:

✤ Concrete setup: with

✤ Massless state: the vaccum:

✤ Light states: ( is small):

✤ These states are the corresponding superpartners to the NS-scalars.

✤ Question: Can they be observed?

Lightest string states

M2 = M2s

X

I

X

m✏Z

: ↵I�m+✓I

↵Im�✓I

: +X

m✏Z

(m� ✓I) : I�m+✓I

Im�✓I

:

!+ ✏0

!0

| ✓ab1,2,3 iR

✓ab1 < 0 , ✓ab2 < 0 , ✓ab3 < 0 ✓ab1 + ✓ab2 + ✓ab3 = �2

↵✓1 | ✓ab1,2,3 iR M2 = �✓1M2s

(↵✓1)2 | ✓ab1,2,3 iR M2 = �2✓1M2

s


The Amplitude

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

Ȥ

Ȥ

ȥ

ȥ

�

�

�

�

disk


✤ Consider three stacks of D-branes within a semi-realistic brane configuration:

✤ For the sake of concreteness we choose the setup

✤ At the intersections live chiral fermions: , , , .

Amplitude

1acș

c

a

b

ȥȥ

Ȥ

1acș

c

a

bȤ

⇒✓1

ab + ✓2ab + ✓3

ab = 0✓1

bc + ✓2bc + ✓3

bc = 0✓1

ca + ✓2ca + ✓3

ca = �2

✓1ab > 0 , ✓2

ab > 0 , ✓3ab < 0

✓1bc > 0 , ✓2

bc > 0 , ✓3bc < 0

✓1ca < 0 , ✓2

ca < 0 , ✓3ca < 0

� �


✤ We want to compute the scattering amplitudes of four chiral fermions:

✤ The corresponding diagram contains different channels:

✤ Two difficulties: 1. Vertex operators

2. Bosonic Twist field correlator

Amplitude

�

�

�

�

gauge boson + ... scalar + ...

D � �

E=

�

�

�

�

�

�

�

�

⇒

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

s-channel t-channel


V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

Vertex Operators

disk


✤ To each state there is a corresponding vertex operator, whose form crucially depends on the intersection angle

✤ challenging part is the internal part bosonic and fermionic twist fields

✤ Method: We determine the OPE’s of and with vacuum and excitations of it

✤ Example:

✤ analogous for fermionic degrees of freedom and higher excitations

✤ OPE’s give us detailed knowledge of the conformal twist fields.

Vertex Operators

@Z, @Z,

|✓IiNS ⇠ s✓I�✓I

@ZI(z)| ✓I iNS =1X

n=�1↵In�✓I z�n+✓I�1| ✓I iNS =

0X

n=�1↵In�✓I z�n+✓I�1| ✓I iNS

! z✓I�1 ↵I�✓I | ✓I iNS = z✓I�1 ⌧+✓I (0)

@ZI(z)| ✓I iNS =

1X

n=�1↵In+✓I z�n�✓I�1| ✓I iNS =

�1X

n=�1↵In+✓I z�n�✓I�1| 0 iNS

! z�✓I ↵I�1+✓I | ✓I iNS = z�✓I e⌧+✓I (0)


✤ For the NS-sector apply the following dictionary

positive angle negative angle

✤ For the R-sector apply the following dictionary

positive angle negative angle

Vertex Operators

✓ ✓

✓✓

| ✓ iNS : ei✓H �+✓ | ✓ iNS : ei✓H ��

�✓

↵�✓| ✓ iNS : ei✓H ⌧+✓ ↵✓| ✓ iNS : ei✓H ⌧��✓

(↵�✓)2 | ✓ iNS : ei✓H !+

✓ (↵✓)2 | ✓ iNS : ei✓H !�

�✓

� 12+✓| ✓ iNS : ei(✓�1)H �+

✓ � 12�✓| ✓ iNS : ei(✓+1)H ��

�✓

↵�✓ � 12+✓| ✓ iNS : ei(✓�1)H ⌧+✓ ↵✓ � 1

2�✓| ✓ iNS : ei(✓+1)H ⌧��✓

(↵�✓)2 � 1

2+✓| ✓ iNS : ei(✓�1)H !+✓ (↵✓)

2 � 12�✓| ✓ iNS : ei(✓+1)H !�

�✓

| ✓ iR : ei(✓�1/2)H �+✓ | ✓ iR : ei(✓+1/2)H ��

�✓


An example✤ Consider with .

✤ A massless state in the NS-sector and the corresponding VO:

Conformal dimension World-sheet charge

✤ A massless state in the R-sector and the corresponding VO:

Conformal dimension World-sheet charge

h = 2� 1

2

X

I

✓abI +k2

2= 1 +

k2

2

✓ab1 < 0 , ✓ab2 < 0 , ✓ab3 < 0 ✓ab1 + ✓ab2 + ✓ab3 = �2

U(1)WS =3X

I=1

�✓abI + 1

�= 1

U(1)WS =3X

I=1

✓✓abI +

1

2

◆= �1

2h =

3

8+

1

4+

3

8+

k2

2= 1 +

k2

2

3Y

I=1

�1/2�✓abI|✓ab1,2,3iNS : V (�1) = ⇤ab � e�'

3Y

I=1

ei(✓abI +1)HI��

�✓abIeikX

|✓ab1,2,3iR : V � 12 = ⇤ab ↵S

↵e�'/23Y

I=1

ei(✓abI + 1

2 )HI��✓ab

IeikX


✤ For this configuration, the amplitude is

with the corresponding vertex operators:

Amplitude

ab : V� 1

2 = ⇤ab

↵ e�'/2S↵

2Y

I=1

�+✓I

abei(✓I

ab� 12 )HI ��✓3ab

ei(✓3ab+12 )H3 eikX

ba : V� 1

2

= ⇤ba ↵ e�'/2S↵2Y

I=1

��✓I

abei(�✓I

ab+12 )HI �+

�✓3abei(�✓3ab� 1

2 )H3 eikX

bc : V� 1

2� = ⇤bc�

↵ e�'/2S↵

2Y

I=1

�+✓I

bcei(✓I

bc� 12 )HI ��✓3

bcei(✓3

bc+12 )H3 eikX

cb : V� 1

2� = ⇤cb�↵ e�'/2S↵

2Y

I=1

��✓I

bcei(�✓I

bc+12 )HI �+

�✓3bc

ei(�✓3bc� 1

2 )H3 eikX

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

✓1ab > 0 , ✓2

ab > 0 , ✓3ab < 0

✓1bc > 0 , ✓2

bc > 0 , ✓3bc < 0

✓1ca < 0 , ✓2

ca < 0 , ✓3ca < 0


✤ Computing:

which takes the form:

Amplitude

A = Tr (⇤ba ⇤ab ⇤bc ⇤cb) ↵ ↵�

��

Z 1

0dx

De

�'/2(0)e

�'/2(x)e

�'/2(1)e

�'/2(1)E

⇥DS

↵(0) S↵(x) S�(1)S�(1)ED

e

ik1X(0)e

ik2X(x)e

ik3X(1)e

ik4X(1)E

⇥D�

+�✓3

ab(0)��✓3

ab(x)��✓3

bc(1)�+

�✓3bc

(1)E

⇥2Y

I=1

D�

�✓I

ab(0)�+

✓Iab

(x)�+✓I

bc(0)��

✓Ibc

(1)E

⇥De

i(�✓3ab� 1

2 )H3(0)e

i(✓3ab+

12 )H3(x)

e

i(✓3bc+

12 )H3(1)

e

i(�✓3bc� 1

2 )H3(1)E

⇥2Y

I=1

De

i(�✓Iab+

12 )HI(0)

e

i(✓Iab� 1

2 )HI(x)e

i(✓Ibc� 1

2 )HI(1)e

i(�✓Ibc+

12 )HI(1)

E

A =D (0) (x) �(1) �(1)

E

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk


✤ Computing:


Amplitude

A = Tr (⇤ba ⇤ab ⇤bc ⇤cb) ↵ ↵�

��

Z 1

0dx

De

�'/2(0)e

�'/2(x)e

�'/2(1)e

�'/2(1)E

⇥DS

↵(0) S↵(x) S�(1)S�(1)ED

e

ik1X(0)e

ik2X(x)e

ik3X(1)e

ik4X(1)E

⇥D�

+�✓3

ab(0)��✓3

ab(x)��✓3

bc(1)�+

�✓3bc

(1)E

⇥2Y

I=1

D�

�✓I

ab(0)�+

✓Iab

(x)�+✓I

bc(0)��

✓Ibc

(1)E

⇥De

i(�✓3ab� 1

2 )H3(0)e

i(✓3ab+

12 )H3(x)

e

i(✓3bc+

12 )H3(1)

e

i(�✓3bc� 1

2 )H3(1)E

⇥2Y

I=1

De

i(�✓Iab+

12 )HI(0)

e

i(✓Iab� 1

2 )HI(x)e

i(✓Ibc� 1

2 )HI(1)e

i(�✓Ibc+

12 )HI(1)

E

[x(1� x)]�14

x

� 341

✏↵� ✏↵� (1� x)�12

x

� 121

x

k1·k2 (1� x)k2·k3x

k4(k1+k2+k3)1

x

(�✓3ab� 1

2 ) (✓3ab+

12 )(1� x)(✓

3ab+

12 ) (✓3

bc+12 )

x

(�✓3bc� 1

2 )((�✓3ab� 1

2 )+(✓3ab+

12 )+(✓3

bc+12 ))

1

?

A =D (0) (x) �(1) �(1)

E

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk


The correlator: h�+1�✓(0) �

+✓ (x) �

+1�⌫(1) �

+⌫ (1)i

b

1-ș

c

a

a

ı��(0)

șı��(x)

Ȟı��(�)

1-Ȟı��(1)

Ȟ ș


Recipe A✤ Extend via the “doubling trick” the upper half plane to the whole complex plane.

✤ Quantum part is then computed by employing conformal field theory techniques (energy momentum tensor method) analogous to the closed string derivation:

✤ The classical part is given by the sum over all quadrangles connecting the four chiral fields .

✤ The final result is then given in the so-called Lagrangian Form.Cvetic, Papadimitriou, Abel, Owen

limz!z2

✓hT (z)�↵(z1)��(z2)��(z3)��(z4)i

h�↵(z1)��(z2)��(z3)��(z4)i�

h��

(z � z2)2

◆

= @z2 lnh�↵(z1)��(z2)��(z3)��(z4)i


Recipe A✤ One independent angle:

✤ Two independent angles:

with:

x

⌫(1�⌫)1 h�1�✓(0)�✓(x)�1�⌫(1)�⌫(1)i = x

�✓(1�✓)(1� x)

�✓(1�⌫)

2F1[✓, 1� ⌫, 1, x]

ssin(⇡✓)

t(x)

⇥X

ep,q

exp

�⇡

sin(⇡✓)

t(x)

L

2a

↵

0 ep2 � ⇡

t(x)

sin(⇡✓)

R

21 R

22

↵

0L

2a

q

2

�.


Recipe B✤ A generic closed twisted-field correlator takes the form:

where is holomorphic, and

✤ The open twisted-field correlator will look like (Hamiltonian Form):

where for the simple case of a D1-brane we have:

✤ Our task is to bring the closed correlator to the above form and get the open one.

Aclosed

= |K(z)|2X

~

k,~v

c~

k~v

w(z)↵0p2

L4 w(z)

↵0p2R

4

p2L =

✓~k +

~v

↵0

◆2

p2R =

✓~k � ~v

↵0

◆2

Aopen

= K(x)X

p,q

c

p,q

w(x)↵

0p

2open

p2open

=1L2

p2 +1

↵02R2

1R22

L2q2

~k, ~v 2 ⇤⇤w(z)


✤ The closed string result has the form:

where and .

✤ The open string result has the form:

where and .

after Poisson resummation the same result as via recipe A.

Correlator: One angle

w(z) = exp

i⇡⌧(z)

sin(⇡✓)

�⌧(z) = ⌧1 + i⌧2 = i 2F1[✓, 1� ✓; 1� z]

2F1[✓, 1� ✓; z]

w(x) = exp

� ⇡t(x)

sin(⇡✓)

�t(x) =

12i

(⌧(x)� ⌧(x)) = 2F1[✓, 1� ✓; 1� x]2F1[✓, 1� ✓;x]

|z1|2✓(1�✓)D�1�✓(0) �✓(z, z) �1�✓(1) �✓(1)

E

=

CV⇤

|z(1� z)|�2✓(1�✓)

|2F1[✓, 1� ✓; z]|2 ⇥X

k2⇤⇤,v12⇤c

exp [�2⇡if23 · k] w(z)

12 (k+ v

2 )2 w(z)

12 (k� v

2 )2

b

ș1- ș

1-ș

b

a

a

ı��(0)

șı��(x)

șı��(�)

1-șı��(1)

Dixon, Friedan, Martinec, Shenker


✤ Following the same recipe, we get (original closed string result is far too complicated to display here):

with the and

✤ Again after Poisson resummation the same result as obtained via recipe A.

Correlator: Two angles

x

⌫(1�⌫)1

D�1�✓(0) �✓(x) �1�⌫(1) �⌫(1)

E=

p↵

0

La

x

�✓(1�✓)(1� x)�✓(1�⌫)

2F 1[✓, 1� ⌫, 1;x]

X

p,q

w(x)

„↵0L2

ap2+ 1

↵0R2

1 R22

L2a

q2«

b

1-ș

c

a

a

ı��(0)

șı��(x)

Ȟı��(�)

1-Ȟı��(1)

Ȟ ș

w(x) = exp

� ⇡t(x)

sin(⇡✓)

�

t(x) =sin(⇡✓)

2⇡

✓�(✓) �(1� ⌫)�(1 + ✓ � ⌫)

2F 1[✓, 1� ⌫, 1 + ✓ � ⌫; 1� x]2F 1[✓, 1� ⌫, 1;x]

+�(⌫) �(1� ✓)�(1 + ⌫ � ✓)

2F 1[1� ✓, ⌫, 1� ✓ + ⌫; 1� x]2F 1[1� ✓, ⌫, 1;x]

◆

Burwick, Kaiser, Muller


✤ Computing:


Amplitude (Back)

A = Tr (⇤ba ⇤ab ⇤bc ⇤cb) ↵ ↵�

��

Z 1

0dx

De

�'/2(0)e

�'/2(x)e

�'/2(1)e

�'/2(1)E

⇥DS

↵(0) S↵(x) S�(1)S�(1)ED

e

ik1X(0)e

ik2X(x)e

ik3X(1)e

ik4X(1)E

⇥D�

+�✓3

ab(0)��✓3

ab(x)��✓3

bc(1)�+

�✓3bc

(1)E

⇥2Y

I=1

D�

�✓I

ab(0)�+

✓Iab

(x)�+✓I

bc(0)��

✓Ibc

(1)E

⇥De

i(�✓3ab� 1

2 )H3(0)e

i(✓3ab+

12 )H3(x)

e

i(✓3bc+

12 )H3(1)

e

i(�✓3bc� 1

2 )H3(1)E

⇥2Y

I=1

De

i(�✓Iab+

12 )HI(0)

e

i(✓Iab� 1

2 )HI(x)e

i(✓Ibc� 1

2 )HI(1)e

i(�✓Ibc+

12 )HI(1)

E

[x(1� x)]�14

x

� 341

✏↵� ✏↵� (1� x)�12

x

� 121

x

k1·k2 (1� x)k2·k3x

k4(k1+k2+k3)1

x

(�✓3ab� 1

2 ) (✓3ab+

12 )(1� x)(✓

3ab+

12 ) (✓3

bc+12 )

x

(�✓3bc� 1

2 )((�✓3ab� 1

2 )+(✓3ab+

12 )+(✓3

bc+12 ))

1

?

A =D (0) (x) �(1) �(1)

E

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk


✤ Computing:


Amplitude (Back)

A = Tr (⇤ba ⇤ab ⇤bc ⇤cb) ↵ ↵�

��

Z 1

0dx

De

�'/2(0)e

�'/2(x)e

�'/2(1)e

�'/2(1)E

⇥DS

↵(0) S↵(x) S�(1)S�(1)ED

e

ik1X(0)e

ik2X(x)e

ik3X(1)e

ik4X(1)E

⇥D�

+�✓3

ab(0)��✓3

ab(x)��✓3

bc(1)�+

�✓3bc

(1)E

⇥2Y

I=1

D�

�✓I

ab(0)�+

✓Iab

(x)�+✓I

bc(0)��

✓Ibc

(1)E

⇥De

i(�✓3ab� 1

2 )H3(0)e

i(✓3ab+

12 )H3(x)

e

i(✓3bc+

12 )H3(1)

e

i(�✓3bc� 1

2 )H3(1)E

⇥2Y

I=1

De

i(�✓Iab+

12 )HI(0)

e

i(✓Iab� 1

2 )HI(x)e

i(✓Ibc� 1

2 )HI(1)e

i(�✓Ibc+

12 )HI(1)

E

[x(1� x)]�14

x

� 341

✏↵� ✏↵� (1� x)�12

x

� 121

x

k1·k2 (1� x)k2·k3x

k4(k1+k2+k3)1

x

(�✓3ab� 1

2 ) (✓3ab+

12 )(1� x)(✓

3ab+

12 ) (✓3

bc+12 )

x

(�✓3bc� 1

2 )((�✓3ab� 1

2 )+(✓3ab+

12 )+(✓3

bc+12 ))

1

A =D (0) (x) �(1) �(1)

E

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

✓


Amplitude✤ Combining all together we get:

where

✤ Finally, we need to normalize the amplitude.

e

�Scl(✓,⌫)=

X

epi,qi

exp

"�⇡

sin(⇡✓)

t(✓, ⌫, x)

L

2b

i

↵

0 ep2i

� ⇡

t(✓, ⌫, x)

sin(⇡✓)

R

2xi

R

2yi

↵

0L

2b

i

q

2i

#

A = igs C Tr (⇤ba ⇤ab ⇤bc ⇤cb) · � · �(2⇡)4�(4)

4X

i

ki

!

⇥Z 1

0dx

x

�1+k1·k2 (1� x)�32+k2·k3

e

�Scl(✓1ab,1�✓1

bc)e

�Scl(✓2ab,1�✓2

bc)e

�Scl(1+✓3ab,�✓3

bc)

[I(✓1ab, 1� ✓1bc, x) I(✓2ab, 1� ✓2bc, x) I(1 + ✓

3ab,�✓3bc, x)]

12

I(✓, ⌫, x) =12⇡

⇢�(✓) �(1� ⌫)�(1 + ✓ � ⌫) 2F 1[1� ✓, ⌫, 1;x]2F 1[✓, 1� ⌫, 1 + ✓ � ⌫; 1� x]

+�(⌫) �(1� ✓)�(1 + ⌫ � ✓) 2F 1[✓, 1� ⌫, 1;x]2F 1[1� ✓, ⌫, 1� ✓ + ⌫; 1� x]

�


✤ At the limit the amplitude factorizes on the exchange of a gauge boson:

✤ that allows to normalize the amplitude ( )

with

✤ Comparing the two results we normalize the amplitude to: .

gD6b

Amplitude at the s-channel ( )x! 0

pi = qi = 0

A4(k1, k2, k3, k4) = iZ

d4k d4k0

(2⇡)4

Pg Ag

µ(k1, k2, k)Ag,µ(k3, k4, k0)�(4)(k � k

0)

k2 � i✏

C = 2⇡

Agµ(k1, k2, k3) = i

s

(2⇡)4↵03/2gsQ3i=1 2⇡Lbi

(2⇡)4�(4)

3X

i=1

ki

! �µ Tr(⇤ba ⇤ab⇤bb)

x! 0

⇒V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

Ȥ

Ȥ

ȥ

ȥ


✤ At the limit there are other (higher order) poles corresponding to other massive exchanges:

✤ Other poles that arise from , they correspond to KK and winding states exchanges.

✤ Additional poles from higher order poles of the “quantum part” corresponding to Regge excitations.

✤ Similar pole structure than the behavior of amplitudes containing at most two chiral fermions. Thus “universal behavior” dressed with poles arising from KK and winding states.

Amplitude at the s-channel ( )x! 0

x! 0

⇒V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

Ȥ

Ȥ

ȥ

ȥ


Amplitude at the t-channel ( )x! 1✤ In the limit the amplitude factorizes on the exchange of scalar particles

✤ In this limit the amplitude takes the form (SUSY preserved, ignore WS-instantons):

�

�

�

�

⇒

x! 1

A = · � · �Z 1

1�✏dx (1� x)�1+k2·k3 ��

12

1�✓1ab,1�✓1

bc,✓1ab+✓1

bc��

12

1�✓2ab,1�✓2

bc,✓2ab+✓2

bc��

12

�✓3ab,�✓3

bc,2+✓3ab+✓3

bc

⇥✓

1 +�1�✓1

ab,1�✓1bc,✓1

ab+✓1bc

�✓1ab,✓1

bc,2�✓1ab�✓1

bc

(1� x)2(1�✓1ab�✓1

bc)

◆

⇥✓

1 +�1�✓2

ab,1�✓2bc,✓2

ab+✓2bc

�✓2ab,✓2

bc,2�✓2ab�✓2

bc

(1� x)2(1�✓2ab�✓2

bc)

◆

⇥✓

1 +��✓3

ab,�✓3bc,2+✓3

ab+✓3bc

�1+✓3ab,1+✓3

bc,�✓3ab�✓3

bc

(1� x)2(�✓3ab�✓3

bc�1)

◆�� 12

massless scalar exchange

V� 1

2

V� 1

2

V� 1

2�

V� 1

2�

disk

�↵,�,� =�(↵)�(�)�(�)

�(1� ↵)�(1� �)�(1� �)etc...


✤ Assuming that is small, the amplitude becomes:

✤ Thus we have the exchange of:

- a massless scalar :

- a massive scalar : with .

✤ Note that there is no coupling to the lightest massive field with mass .

✤ This is can be traced back to the fact that the two bosonic twist fields and do not couple to the excited twist field , but only to an even excited twist field.

Subdominant poles

A = · � · �Z 1

1�✏dx (1� x)�1+k2·k3

Y

2 ��

⇣1 + c1(1� x)2(1�✓

1ab�✓

1bc) + ...

⌘

1� ✓1ab � ✓1

bc = �✓1ca

M2 = �2✓1caM2

s

M2 = �✓1caM2s

Y

I

�1/2�✓Ica| ✓ca1,2,3iNS

�↵✓1

ca

�2 Y

I

�1/2�✓Ica| ✓ca1,2,3iNS


Further poles✤ The exchange particle is a scalar field, thus the signatures induced by the

tower resembles signatures of KK states in extra-dimensional theories.

✤ Above just the first sub-dominant poles, but there are many more poles.

✤ In case the fermions are too much separated in the internal manifold WS-instantons cannot be ignored; poles arising from them correspond to exchanges of KK and winding excitations.

✤ There are also integer poles, that correspond to exchange of Higher Spin states.

✤ Other poles that correspond to exchanges of massive scalar fields whose mass is non-vanishing even for vanishing intersection angles.

✤ Rich spectrum of signatures, but the first once to be observed correspond to lightest string states.


Conclusions✤ We have studied the spectrum of open strings localized at the intersections of D6-

branes.

✤ The masses of such states scale as and can thus be parametrically smaller than the string scale if the relevant angle is small.

✤ We have considered scattering amplitudes that expose such light stringy states.

Along the computation

Give a description to formulate the vertex operators for states localized at intersections.

Rederived the four bosonic twist field correlator with one and two independent angles.

✤ Investigated s- and t-channel and found poles corresponding to light stringy states.

✤ Assuming a scenario with a low string scale, these states may be observable at LHC.

✤ However further poles corresponding to KK and winding states, as well as Higher Spin states

Rich spectrum of signatures

M2 ⇡ ✓M2s


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Light stringy states - NBI Conference and Meetings (Indico) · 2017. 9. 26. · For the sake of...

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