Effects of D-instantons particle physics...Orientifold projection Ωaction is σ →−σ. There is an action on worldsheet fermions (−1)F.Z 2 involution in target space M identifies

Effects of D-instantons → particle physics

Instantons associated with Euclidean D2 configurations. The framework is very geometric.

CFT techique to write vertex operators.

1. Quantization of physical states in intersecting D-branes

2. Quantization of states with D-instantons: zero modes in this background. We will

refer to rigid O(1) instanton (stringy; direct corrections to the holomorphic part of

the D-brane action)

3. Applications (calculus with instanton).

4. Prototype

of O(1) instantons in two directions: multi instantons/U(1)/recombanitations of

instantons and U(1) instantons intersecting with D-branes/issues of massive ”zero”

modes.

Useful references

R. Blumenhagen, M. Cvetic, P. Langacker, G. Shiu, ”Toward Realistic Intersecting D-

Brane Models”, Ann.Rev.Nucl.Part.Sci. 55 (2005) 71-139, arXiv:hep-th/0502005

R. Blumenhagen, M. Cvetic, S. Kachru, T. Weigand, ”D-brane Instantons in Type

II String Theory”, Invited review to appear in Annu.Rev.Nuc.Part.Sci 2009 59,

arXiv:0902.3251

R. Blumenhagen, M. Cvetic, T. Weigand, ”Spacetime Instanton Corrections in 4D String

Vacua - The Seesaw Mechanism for D-Brane Models”, Nucl.Phys.B771:113-142,2007,

arXiv:hep-th/0609191

L. E. Ibanez, A. M. Uranga, ”Neutrino Majorana Masses from String Theory Instan-

ton Effects”, JHEP0703:052,2007, arXiv:hep- th/0609213

B. Florea, S. Kachru, J. McGreevy, N. Saulina, ”Stringy Instantons and Quiver Gauge

Theories”, JHEP 0705:024,2007, arXiv:hep- th/0610003

1

Spectrum of intersecting D6-branes (CFT)

We consider open strings. Open strings attached to the D-brane satisfy Dirichlet boundary

conditions transversal to the D- brane and Neumann boundary conditions along the D-

brane

∂σXI = 0 at σ = 0, π I = 1, ..., p + 1

XJ = 0 at σ = 0, π J = p + 2, ..., 10

where (σ, τ) denote the world-sheet space and time and XI the spacetime coordinates.

Worldsheet fermions: NS-ψI , R-ψJ . NS sector: Concrete representation of massless

states ↔ Vertex operators. Superconformal ghost sector φ = 0:

VAI (0) = ξI∂zXIeikIXI

(1)

where ξI is the polarization vector in the target space (spin 1 field).

z = eτE+iσ

z = eτE−iσ

with Im z > 0 (only on the upper plane string lives).

Duality trick: We take the whole z-plane. Superconformal ghost (-1) picture:

VAI (−1) = ξIe−φψIeikIXI

V 0AI

= ξIe−φψI (2)

eαΦ has the conformal dimension

[eαΦ] = −α(α+ 2)

2(3)

Similarly

[ψ] =1

2[∂zX] = 1

[VAI ] = 1

VAI = ξIe−φψIeikixi

(Λa ⊗ Λb) (4)

We have N2 massless spin fields (different vertex operators). This is in adjoint represen-

tation of U(N).

2

Orientifold projection

Ω action is σ → −σ. There is an action on worldsheet fermions (−1)F . Z2 involution in

target space M identifies the rigid plane under which the target space has to be invariant.

It is so called O-plane. It acts as a D-brane with negative tension.

Da

D′a

O

Figure 1:

We have to deal with further projection on the states. For Chan-Paton factors Ω(Λa⊗Λb)Ω−1 = −(Λa ⊗ Λb)T

Let us consider string starting on one brane (D6a) and ending on the other (D6b).

CY

×

R3,1

πa

πb

Figure 2:

At each intersection of cycles we want to quantize the theory. πa πb is a topological

number.

πa πb = NaIMIb − MaIN

Ib (5)

πa = NaIAI + MJ

a BJ (6)

AI BJ = δIJ (7)∫

AI

αJ = δIJ (8)

3

∫

BI

βJ = −δIJ (9)

a

b

X1

X2

X3

X4

X5

X6

πθ1 πθ2 πθ3a a

b b

Figure 3:

We have now three Polyakov action, but involved boundary conditions. σ = π on b

cycle and σ = 0 on a cycle. We now involve boundary conditions:

∂σX2I−1 = X2I = 0 at σ = 0

∂σX2I−1 + (tanπθ1)∂σX

2I = 0 at σ = π

X2I − (tanπθI)X2I−1 = 0 at σ = π

where I = 1, 2, 3.

Let us go to the complexified notation:

ZI = X2I−1 + iX2I =∑

nεZ

αIn−θI

n − θIZ−n+θI +

∑

nεZ

αIn+θI

n + θIZ−n−θI (10)

Note that αI†n+θI

= αIn+θI

and ZI = ZI(θI → −θI).

We can analogously complexify worldsheet fermions:

ΨI = ψ2I−1 + iψ2I =∑

r=n+ 12

ψr−θIz−r− 1

2+θI NS (11)

ΨI = ψ2I−1 + iψ2I =∑

r=n

ψr−θI z−r− 1

2+θI R (12)

We can identify bosons in NS sector.

Twisted bosonic vacuum |σθI 〉.

∂zZI |σθI 〉 ∼ z−1−θI |τθI 〉 (13)

ΨI ∼ eiHI ∼ eiθIHI NS

ΨI ∼ e(i± 12 )HI ∼ ei(θI− 1

2 )HI R

4

The last proportionality is due to the twisted boundary conditions.

0 < θI < 1 for I = 1, 2 and −1 < θ3 ≤ 0.

V−1 = e−φ2∏

I=1

σθI eiθIHI

σ1+θ3ei(1+θ3)H3

V− 12

= uαe−φ

2 sα2∏

I=1

σθI ei(θI− 1

2 )HIσ1+θ3e

i( 12+θ3)H3

with the following conformal weights:

[V−1] =1

2+

2∑

I=1

(θI(1 − θI)

2+

1

2θ2I

)+

(θ3 + 1)(−θ3)2

+(θ3 + 1)2

2

=1

2+

1

2

3∑

I=1

θI +1

2! 1 (14)

[V− 1

2

]=

3

8+ 2

1

8+ 3

1

8= 1 (15)

when∑3

I=1 θI = 0 bosons become massless. If [V 0−1] > 1 then k2

I > 0 and so we have

tachyon, [V 0−1] = 1 then k2

I = 0 and these are massless states, and in the case [V 0−1] < 1 we

have k2I < 0 and these are massive states. [V− 1

2] = 1 means that we always have massless

spacetime fermions.

Now let us consider a system of Na D6a and Nb D6b branes.

Λbβ

Λaα

a

b

Figure 4: D6a-D6b system. Λbβ and Λa

α Chan-Paton factors of b and a branes respectively.α = 1, ...,Na and β = 1, ...,Nb

Na branes have a gauge group U(Na), Nb branes the gauge group U(Nb).

πa πb = NaIMIb − M I

aNbI > 0 (16)

πa πb > 0 (17)

5

b

a′

Figure 5: Intersecting b-brane and a′-brane (the image of a).

a′

a

O

Λaα

Λa′

α

Figure 6: Intersecting a- and a′-branes with O-plane.

In this case we deal with bi-fundamental representation ( a, b).

In this case we are tensoring two fundamental representations

a × a = a + a (18)

a =1

2(πa′ πa + πo πa)

a =1

2(πa′ πa − πo πa)

Let us put in this context D-instantons.

D-instantons

Useful references:

Mirjam Cvetic, Ioannis Papadimitriou, ”Conformal Field Theory Couplings for In-

tersecting D-branes on Orientifolds”, Phys.Rev. D68 (2003) 046001; Erratum-ibid. D70

(2004) 029903, arXiv:hep-th/0303083 .

6

Mirjam Cvetic, Robert Richter, ”Proton decay via dimension-six operators in intersect-

ing D6-brane models”, Nucl.Phys.B762:112-147,2007, arXiv:hep-th/0606001, (Appendix

A).

In Type IIA we will deal with Euclidean D2-branes.

CY

×

R3,1

Ξ

Figure 7:

Introduce instantons to generate couplings absent in perturbative calculation. They

do not preserve U(1)a. a a( a, b) couplings are absent since the U(1) charge is not

preserved. And couplings (10 10) 5H are desired couplings in U(5)GG.

In particular we need neutrino masses to be extremely small. There is a seesaw

mechanism to give masses to neutrinos.

νR ab

L

Hu

Figure 8: hDν HuLνR coupling related to U(2)L × U(1)a × U(1)D.

It is hard to tune hDν to be extremely small. Seesaw mechanism introduces direct

mass of right-handed neutrinos.

(hDν < H >)2

MR∼ 10−3Ev

MR ∼ 1011−15Gev

It is away of explaining neutrino masses in seesaw.

7

MRνRνR are perturbatively forbidden. We will not generate them.

If induced by instantons then MR is

MR ∼ e−SclE MsX (19)

where

SclE =

2π

l3sgs

∫

Ξ

Re (Ω3) − i

∫

Ξ

C3 (20)

where Ξ is a three cycle wrapped by an instanton. The suppression factor is the volume

of that cycle. SclE is not invariant under U(1)a.

SclE =

2π

l3s

1

gsVΞ =

2π

α

VΞ

VD(21)

where α is the gauge coupling of the brane, VD is the volume of the cycle wrapped by the

brane. If VΞVD

∼ 1, then e−Scl = e−2πα . Then MR ∼ 1011 Gev.

WZ (Wess-Zumino) action associated with particular Da-brane is

SWZ ∼∫

R3,1×πa

C ∧ eFa ∼∫

(C7 + C5 ∧ Fa + C3 ∧ Fa ∧ Fa) (22)

where C = C7 +C5 +C3 with C7, C5 and C3 seven, five and three forms respectively. First

term of the integral is the key term for tadpole cancellation, the second one describes

gauge bosons associated with U(1) symmetry to be massive.

C5 = CI2αI + D2Iβ

I (23)

C3 = CI0αI + D0Iβ

I (24)

From the second term of WZ action we get

C2 ∧ Fa → −dC2 ∧ Aa = − - dC0 ∧ Aa (25)

C2 = NaICI2 − M I

aD2I (26)

C0 = N0ICI0 − M I

aD0I (27)

C0 = (CI0 , D0I) (28)

We expect in 4 dimensions to get the structure of the type (dC0 − Aa) ∧ (-dC0 − -Aa).

Under a U(1) gauge transformation:

Aa → Aa + dΛ (29)

CI0 → CI

0 + (M Ia )Λ (30)

D0I → D0I + (NaI )Λ (31)

8

If we have orientifolds then

C2 = (NaI − N ′aI)C

I2 − (M I

a − M I′a )D2I (32)

CI0 → CI

0 + Na(MIa − M I′

a )Λ (33)

D0I → D0I + Na(NaI − N ′aI)Λ (34)

if we have Na copies.

Axions transform under U(1), consequently∫

C3 will transform under U(1) as well.

Instantons violate U(1) charge.

Ξ = NΞIAI + MJ

ΞBJ (35)

C3 = C0IαI + D0Iβ

I (36)∫

Ξ

C3 = CI0NΞI − D0IM

IΞ (37)

Under U(1)a:

Im (Scl) → (CI0NΞI − D0IM

IΞ) + N a[NΞI(M

Ia − M I′

a ) − M IΞ(NaI − N ′

aI)]Λ (38)

e−Scl → e−Scl+iQaΞΛ (39)

where

QaΞ = NaΞ (πa − π′a) (40)

Instanton action violates U(1) charge by QaΞ amount. It can be cancelled by charged fields

e−Sclφi.

QaΞ +

∑

i

Qai = 0 ∀a (41)

The dominant contriburions to e−Scl−iSint(M,φ), where M are all the zero modes, φ is

the real matter, come from the minimal volume of the cycle (special Lagrangian)∫

DMe−Scl−iSint(M,φ) = S4Dn.p.(φ) (42)

Now let us get stringy zero modes. Consider strings starting and ending on the same

boundary.

V−1 = XEµ e−φψµ NS (43)

From here we have four bosonic zero modes. XEµ is a polar vector.

In the Ramond sector we have two types of fermionic zero modes associated with

breaking down of supersymmetry.

QαQα (44)

Qα′Qα′ (45)

9

CY

×

R3,1

Ξ

N

D

πa

Figure 9:

D-brane on this background breaks 1/2 of supersymmetry and eliminates Qα′Qα′ part.

V θα

−1/2 = θαe−φ

2 SαeP

i12 iHi (46)

V τ−1/2 = ταe

−φ2 Sαe−i 1

2

Pi Hi (47)

1√3

∑

i

Hi = H (48)

θα and τ α are two goldstinos. In order to have only θα mode for N = 1 supersymmetry

we put one instanton on O-plane.

θα 1 (49)

τ α 0 (50)

These are O(1) instantons. There are additional zero modes if the cycles are not rigid.

These are undesired zero modes. So, we will look only for O(1) rigid instantons.

I : Ξ− Da (51)

We have two bosonic twist fields∏2

i=1 σi1/2 in complexified coordinates. In the NS sector

we have integer moded fermionic modes and in R sectod half-integer moded modes.

In NS sector:

V−1 = e−φ2∏

i=1

σi1/2S

αVint (52)

V−1 = e−φ2∏

i=1

σi1/2S

αVint (53)

[V−1] =1

2+ 2

12(1 − 1

2)

2+ 2

1

2(1

2)2 + [Vint] = 1 + [Vint] (54)

We get massive modes. Bosonic massless modes are absent.

10

In R sector:

V−1/2 = e−φ2

2∏

i=1

σi1/2Vint (55)

θ1 θ2

θ3

Figure 10:

θ1, θ2 > 0, θ3 < 0 then

Vint =2∏

i=1

σθI ei(θI− 1

2 )HIσHθ3ei( 1

2+θ3)H3 (56)

[V− 12] = 1 has massless zero modes. We denote by λa the state V− 1

2. It is charged under

the brane, carries an index a, and it is in the fundamental representation [−1E , a].

λa = |Ξ πa| + |Ξ π′a| (57)

Number of zero modes is equal to the number of intersections of instanton and brane.

Ξ πa > 0 [−1E , a] (58)

Ξ πa < 0 [1E, a (59)

Ξ π′a > 0 [−1E , a] (60)

Ξ π′a < 0 [1E, a] (61)

(62)

The charge that zero modes carry is given by

Qaλ = Na(Ξ πa − Ξ π′a) (63)

∫DXµ

E

∫Dθα

∫ ∏

I,i

2λiIe

−Scl−Sint(λ,θα,Φi) = S4Dn.p.(φ

i,ψiα) (64)

where Φ = φ+ θαψα.

11

λia

φab

a b

Ξλi

a

φab

λjb

λjb

Figure 11: Yukawa coupling φλλ

Sint = Xk(Ii)(Jj)φ

kλiIλ

jJXk

(Ii)(Jj)θαψk

αλiIλ

jJ (65)

each λiI appears exactly ones.

S4Dn.p. =

∫

n.p.

DXµ

∫D2θe−Scl

∏

k,Ii,Jj

XkIi,JjΦ

k (66)

νR = (−1a, +1b) (67)

νRνR coupling has deficit of charge (−2a, +2b). We have to insure that

Ξ πa = 2 (68)

Ξ πb = −2 (69)

First equation means two copies of modes of charge (−1E , 1a), and the second one two

copies of modes of charge (+1E ,−1b). Ξ (SM) = 0 and all their images of (π′a, π′b).

aΞ

a′

10

λa

λa

Figure 12:

We choose U(5)a × U(1)b. We want to introduce instanton that will create the exact

number of zero modes which is 5. Single zero mode already carries charge 5.

λa = (+1E, a) (70)

λb = (−1E, a) (71)

12

Ξ πa = −1 (72)

Ξ πb = +1 (73)

aΞ

a′

10

λa

λa

aΞ

a′

10

λa

λa

aΞ

10

λa

λb

b

+ +

Figure 13:

More general instantons

• Gauge instantons. There are more zero modes and more interactions among zero

modes. There is a contribution of instantons that do not seat on the O-plane.

• U(1) instantons. U(1) × U(1) → O(1). There is a large number of zero modes.

• Instantons far away from O-plane, but intersecting with many zero modes. Also τ

modes.

Mass terms are (θ1 + θ2)µµ′ + (−θ3)µ′µ. Susy term: −θ3(µµ′ + µµ′ + wαwα). We have

also coupling to τ : τα(wαµ + µwα).

13