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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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
✤ 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
Monday, February 20, 12
The Amplitude
V� 1
2
V� 1
2
V� 1
2�
V� 1
2�
Ȥ
Ȥ
ȥ
ȥ
�
�
�
�
disk
Monday, February 20, 12
✤ 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
� �
Monday, February 20, 12
✤ 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
Monday, February 20, 12
V� 1
2
V� 1
2
V� 1
2�
V� 1
2�
Vertex Operators
disk
Monday, February 20, 12
✤ 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)
Monday, February 20, 12
✤ 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 ��
�✓
Monday, February 20, 12
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
Monday, February 20, 12
✤ 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
Monday, February 20, 12
✤ 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
Monday, February 20, 12
✤ 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
[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
Monday, February 20, 12
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
Monday, February 20, 12
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
Monday, February 20, 12
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
�.
Monday, February 20, 12
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)
Monday, February 20, 12
✤ 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
Monday, February 20, 12
✤ 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
Monday, February 20, 12
✤ Computing:
which takes the form:
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
Monday, February 20, 12
✤ Computing:
which takes the form:
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
✓
Monday, February 20, 12
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]
�
Monday, February 20, 12
✤ 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
Monday, February 20, 12
✤ 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
Monday, February 20, 12
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...
Monday, February 20, 12
✤ 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
Monday, February 20, 12
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.
Monday, February 20, 12
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
Monday, February 20, 12