String Theory I...String Theory I GEORGE SIOPSIS AND STUDENTS Department of Physics and Astronomy The University of Tennessee Knoxville, TN 37996-1200 U.S.A. e-mail: [email protected]

String Theory I

GEORGE SIOPSIS AND STUDENTS

Department of Physics and AstronomyThe University of TennesseeKnoxville, TN 37996-1200

U.S.A.e-mail: [email protected]

Last update: 2006

ii

Contents

1 A first look at strings 11.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Why Strings? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Point particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 The Mode Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 171.6 Closed strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.7 Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Conformal Field Theory 272.1 Massless scalars in two dimensions . . . . . . . . . . . . . . . . . 272.2 Solution to the Boundary-Value Problem with Green function . 292.3 Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.4 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 332.5 Conformal Invariance . . . . . . . . . . . . . . . . . . . . . . . . . 352.6 Free CFTs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.7 Virasoro Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.8 Mode Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.9 Vertex operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.10 Primary fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492.11 Operator product expansion . . . . . . . . . . . . . . . . . . . . . 502.12 Unitary CFTs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3 BRST Quantization 533.1 Point particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.2 Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3 Mode Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.4 Nilpotency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.5 A note on BRST cohomology . . . . . . . . . . . . . . . . . . . . . 553.6 BRST Cohomology for open strings . . . . . . . . . . . . . . . . . 56

4 Tree-level Amplitudes 594.1 String Interactions . . . . . . . . . . . . . . . . . . . . . . . . . . . 594.2 A Short Course in Scattering Theory . . . . . . . . . . . . . . . . 69

iii

iv CONTENTS

4.3 N-point open-string tree amplitudes . . . . . . . . . . . . . . . . 704.4 Closed Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.5 Moduli . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.6 BRST Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

5 Loop Amplitudes 795.1 One-loop Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . 795.2 String on a Torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.3 The bc system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.4 Vacuum Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 895.5 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.6 Amplitudes on a torus . . . . . . . . . . . . . . . . . . . . . . . . 925.7 Higher Genus Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 96

UNIT 1

A first look at strings

Following “String Theory” by J. Polchinski, Vol.I.

Notes written by students (work still in progress).

For more information contact George Siopsis

[email protected]

1.1 Units

First, we must explain the unit convention we are going to use. Take the fol-lowing two results from Quantum Mechanics and Special Relativity:

E = ~ω (1.1.1)

E = mc2 (1.1.2)

These two equations link energy to frequency and mass through some con-stant of proportionality. The question is, are these constants fundamental innature or created by man? The answer is that they are artificial creations, ex-isting purely because of the units we have chosen to work in. We could easilychoose units such that ~ = c = 1. By doing this, the number of fundamentalunits in the universe is reduced to 1, e.g., energy, all others being related to it:

[energy] = [1/time] = [mass] = [1/length] (1.1.3)

1.2 Why Strings?

Our motivation behind the developement of String Theory is our desire tofind a unified theory of everything. One of the major obstacles that previoustheories have been unable to overcome is the formation of a quantum theoryof gravity, and it is in this respect that String Theory has had notable success

2 UNIT 1. A FIRST LOOK AT STRINGS

(in fact, at present, String theory is the only theory which includes gravita-tional interactions). This leads us to believe that, while String Theory maynot be the final answer, it is certainly a step in the right direction.Let us first discuss the problems one runs into when trying to create a quan-tum theory of gravity using Quantum Field Theory as our guide. Take theHydrogen atom, whose energy levels are given by:

En = −E1

n2, E1 =

~2

2mea0, a0 =

~2

mee2(1.2.1)

where E1 is the ground state energy and n labels the energy levels.Suppose we had only the most basic knowledge of physics: what would weguess the energy of the Hydrogen atom to be? The parameters of the systemare the mass of the electronme, the mass of the protonmp, ~ and the electroncharge e. We may neglect mp as we are interested in the energy levels of theelectron. We might guess the energy to be

E0 = mec2 (1.2.2)

Equations (1.2.1) and (1.2.2) are clearly not the same, but if we take the ratiowe obtain

E1

E0=

e4

~2c2=

(1

137

)2

= α2 (1.2.3)

This is a ratio, so is independent of our choice of units, so α is a fundamen-tal constant that exists in nature independent of our attempts to decribe theworld, and indicates some fundamental physics underlying the situation. Infact α is the fine structure constant and describes the probability for an elec-tromagnetic interaction, e.g. proton - electron scattering (of which hydrogenis a special case in which the scattering results in a bound state).From the diagram of an e-p interaction,

INSERT FIGURE HERE

each vertex contributes a factor e to the amplitude for the interaction, so that

A1 ≡ Amplitude ∼ e2 (1.2.4)

Probability ∼ |Amplitude|2 ∼ e4 ∼ α2 (1.2.5)

Now according to classical analysis, this is the only amplitude we would getfor the interaction, but in quantum mechanics there can be intermediatescattering events that cannot be observed: e.g.,

INSERT FIGURE HERE

contributes o(α2) to the overall amplitude for the interaction. Inserting acomplete set of states, we obtain the amplitude of this second-order processin terms of A1,

A2 ∼∫dE′

E′ |A1|2 ∼ α2

∫dE′

E′ (1.2.6)

1.2 Why Strings? 3

This is logarithmically divergent. However, all higher-order amplitudes havethe same divergence and when we sum the series in α:

Amplitude = () + α() + α2() + ... (1.2.7)

it yields finite expressions for physical quantities.

Let us try it for the gravitational interaction between two point masses, eachof mass M , separated by a distance r. The potential energy is:

V =GM2

r(1.2.8)

which shows upon comparison with electromagnetism that the “charge” ofgravity is eg ∼

√GM . A gravitational “Hydrogen atom” will have energy levels

En ∼ E1

n2, E1 ∼

Me4g2~2

∼ G2M5 (1.2.9)

Comparing with E0 = Mc2, we obtain the ratio

E1

E0∼ G2M4 ∼ e4g (1.2.10)

Immediately we see problems with using this charge to describe the grav-itational interaction, because eg is energy (mass) dependent. The classicalscattering amplitude is

A1 ∼ e2g ∼ GE2 (1.2.11)

where E = Mc2 and the second-order contribution (exchange of two gravi-tons) is

A2 ∼∫dE′

E′ |A1|2 ∼ G2

∫

dE′(E′)3 (1.2.12)

which has a quartic divergence. Worse yet, higher-order amplitudes haveworse divergences, making it impossible to make any sense of the perturba-tive expansion (1.2.7) (non-renormalizability of gravity).

We may see our way to a possible solution by considering the problem of betadecay: Initially it was treated as a three body problem with the proton - neu-tron - electron interaction occuring at one vertex. When the energies of theresultant electrons did not match experiment, the theory was modified to in-clude a fourth particle, the neutrino, and the interaction was ’smeared’ out:the proton and neutron interacted at one vertex, where a W boson was cre-ated, which traveled a short distance before reaching the electron - neutrinovertex.


p

n

e

n

p

e

v

gg

So maybe we can solve our problems with quantum gravity by smearing outthe interactions, so that the objects mediating the force are no longer pointparticles but extended one dimensional objects - strings.

This is the general concept from which we will proceed. It is a difficult task -the gravitational interaction must obey a much larger symmetry than Lorentzinvariance, it must be invariant under completely general co-ordinate trans-formations, and we must of course still be able to describe the weak, strongand electromagnetic interactions.

In this chapter we will take a first look at strings. Initially we examine the com-pletely general equations of motion for a point particle using the method ofleast action, and then apply that method to the case of a general string movingin D dimensions. We will obtain the equations of motion for the string, andthen attempt to quantize it and obtain its energy spectrum. This will highlightsome basic results of string theory, as well as some fundamental difficulties.

1.3 Point particle

We begin by examining the case of a point particle, illustrating the method wewill use for strings. The trajectory of a point particle in D-dimensional spaceis decribed by coordinates Xµ(τ), where τ is a parameter of the particle’s tra-jectory. For a massive particle, τ is its proper time. X0 will be a timelike cordi-

nate, the remaining ~X spanning space. Infinitesimal distances in spacetime:

−dT 2 = ds2 = −(dX0)2 + (d ~X)2 (1.3.1)

1.3 Point particle 5

n

n

a

b

We wish to derive the equation of motion from an action principle. This issimilar to Fermat’s principle of minimizing time along a light ray. For the lightray joining points A and B, this yields Snell’s Law,

n1 sin θ1 = n2 sin θ2 (1.3.2)

Along the trajectory of a free relativistic particle, proper time is maximized.Therefore, trajectories are obtained as extrema of the action

S = m

∫ b

a

dT (1.3.3)

where we multiplied by the mass to obtain a dimensionless quantity (in unitswhere ~ = 1).

x

x

tt

a and b are the fixed start and end points on the trajectory. There is a problemthat when the mass m = 0, since the action is zero. This problem will becleared up a little later.


We can write the action as

S = m

∫ b

a

dτdT

dτ= m

∫ b

a

dτ

√√√√

(dX0

dτ

)2

−(

d ~X

dτ

)2

(1.3.4)

(invariant under reparametrizations τ → τ ′(τ)) from which we can define theLagrangian for the system:

L = m

√√√√

(dX0

dτ

)2

−(

d ~X

dτ

)2

(1.3.5)

Using the convention X = dX/dτ , we write Lagrange’s equations:

d

dτ

(∂L

∂Xµ

)

=∂L

∂Xµ(1.3.6)

For this Lagrangian,∂L

∂Xµ= 0 (1.3.7)

and we obtain the equation of motion for the point particle:

d

dτ

(∂L

∂Xµ

)

=d

dτ

(

Xµ

L

)

= 0 (1.3.8)

To see the physical meaning of this equation, switch from τ to X0 (or set τ =X0 to “fix the gauge”). Then the 3-velocity is

vi =dX i

dX0(1.3.9)

and the equation of motion (1.3.8) reads

uµ = 0 , uµ = γ(1, ~v) , γ =1

L=

1√1 − ~v2

(1.3.10)

i.e., that the acceleration is constant, as expected.We will now look at a better expression for the action: defining an extra fieldη(τ), which at the moment is arbitrary, we write a new Lagrangian:

L =1

2ηXµXµ − 1

2ηm2 (1.3.11)

This has the nice feature that it is still valid for m = 0. Lagrange’s equation forη is

d

dτ

(∂L

∂η

)

= 0 =∂L

∂η(1.3.12)


which implies

− 1

2η2XµXµ − 1

2m2 = 0 ⇒ η =

√

−XµXµ

m2(1.3.13)

Using this equation to eliminate η from the Lagrangian (1.3.11), we get backthe original Lagrangian (1.3.5). Varying Xµ ,we obtain from (1.3.11)

d

dτ

(

Xµ

η

)

= 0 (1.3.14)

which agrees with the previous eq. (1.3.8).We will now examine the meaning of the field η(τ). The trajectory is parame-terized by some co-ordinate of the system in terms of which an infinitesimaldistance along the trajectory, ds, can be expressed. Let us suppose our trajec-tory is along the y axis. Then it would be easiest to parameterize the systemwith the co-ordinate y, and then ds = dy. We could, however, choose the pa-rameter to be θ, the the angle between a line drawn from a fixed point on thex axis at a distance ` to a position on the y axis. Then our new distance wouldbe

ds = `

(1

cos2 θ

)

dθ (1.3.15)

The factor `/cos2θ is our η2. It represents the geometry of the system due toour choice of co-ordinates. In Minkowski space,

ds2 = −γττdτ2 , γττ = η2 (1.3.16)

where γττ is the (single component) metric tensor. Under a reparametriza-tion,

τ → τ ′(τ) , γττ →(dτ

dτ ′

)2

γττ (1.3.17)

i.e. γττ transforms as a tensor (this follows from the invariance of ds2). Wecan see that the action is invariant under this transformation: we have

η′ =dτ

dτ ′η (1.3.18)

and Xµ transforms as

Xµ′=dXµ

dτ ′=

dτ

dτ ′dXµ

dτ(1.3.19)

Thus the Lagrangian transforms as

L′ =dτ ′

dτL (1.3.20)

and the action transforms as

S = m

∫ b

a

dτL = m

∫ b

a

dτ ′L′ (1.3.21)


thus proving its invariance.Now let us form the Hamiltonian for the system: the conjugate momenta toco-ordinate Xµ and the parameter η are

Pµ =∂L

∂Xµ=Xµ

η, Pη =

∂L

∂η= 0 (1.3.22)

Then the Hamiltonian is:

H = PµXµ + Pη η − L = 12η(P

µPµ +m2) (1.3.23)

Here the role of η is that of a Lagrange multiplier; it is not a dynamical vari-able. From Hamilton’s equation:

∂H

∂η= Pη = 0 = P µPµ +m2 (1.3.24)

which is Einstein’s equation for the relativistic energy of a paricle of mass m.Define χ as:

χ =1

2mP µPµ + 1

2m =H

ηm(1.3.25)

Then χ = 0 is a constraint which generates reparametrizations through Pois-son brackets:

δXµ ∼ Xµ , χ =P µ

m, δP µ ∼ P µ , χ = 0 (1.3.26)

We may identifyX0 with time and solve for its conjugate momentum

P0 =

√

~P 2 +m2 (1.3.27)

This is the true Hamiltonian of the system. Equations of motion:

X i =∂P0

∂Pi=P i

P0, Pi = 0 (1.3.28)

same equation as before, if we note vi = X i, 1 − ~v2 = m2/P 20 and therefore,

vi√1 − ~v2

=P i

m(1.3.29)

This system may be quantized by

[Pi , Xj ] = −iδji (1.3.30)

Eigenstates of the Hamiltonian:

H |~k〉 = ω|~k〉 , ω =

√

~k2 +m2 (1.3.31)


Alternatively, we may define light-cone coordinates in spacetime:

X± =1√2(X0 ±X1) ~XT = (X2, . . . , XD−1) (1.3.32)

The Lagrangian reads

L =1

2ηXµXµ − 1

2ηm2 = −1

ηX+X− +

1

2η~X

2

T − 1

2ηm2 (1.3.33)

LetX+ = τ play the role of time; then P+ is the Hamiltonian. X− and ~XT are

the coordinates andP− and ~PT are their conjugate momenta. The Lagrangianbecomes:

L = −1

ηX− +

1

2ηX2i −

1

2ηm2 (1.3.34)

yielding

P− =∂L

∂X−= −1

η(1.3.35)

Pi =∂L

∂X i=

1

ηXi (1.3.36)

The Hamiltonian is

P+ = X−P− + X iPi − L =~P 2T +m2

2P−(1.3.37)

Note that there is no term P+X+ because X+ is not a dynamical variable in

the gauge-fixed theory.

Quantization:

[Pi , Xj ] = −iδji , [P− , X

−] = −i (1.3.38)

Eigenstates of the Hamiltonian:

P+|k−, ~kT 〉 = ω+|k−, ~kT 〉 , ω+ =~k2T +m2

2k−(1.3.39)

To compare with our earlier result, define ω and k1 by

ω+ =1√2

(ω + k1) , k− =1√2

(ω − k1) (1.3.40)

Then ω2 − ~k2 = ω2 − k21 − ~k2

T = 2ω+k− − ~k2T = m2.


1.4 Strings

s=ls=0 x

x

t=2

t=1

Parametrize the string with σ ∈ [0, `]. In analogy to Fermat’s minimization oftime, we will minimize the area of the world-sheet mapped out by the string.To find an expression for the area, pick a point on the worldsheet and drawthe tangent vectors

~tτ =∂ ~X

∂τ= ~X , ~tσ =

∂ ~X

∂σ= ~X ′ (1.4.1)

The infinitesimal parallelepiped with sides ~tτdτ and ~tσdσ has area

dA = |~tτ × ~tσ |dτdσ (1.4.2)

We obtain the total area by integrating over the worldsheet coordinates (τ, σ).The action to be minimized is the Nambu-Goto action

SNG = T

∫

dA (1.4.3)

where T is a constant that makes the action dimensionless (tension of thestring). Using

(~a×~b)2 = ~a2~b2 − (~a ·~b)2 =

∣∣∣∣∣

~a2 ~a ·~b~a ·~b ~b2

∣∣∣∣∣

(1.4.4)

we deduce

SNG = T

∫∫

dτdσ√

− det hab , hab = ∂aXµ∂bXµ (1.4.5)

where the minus sign in the square root is because we are working in Minkowskispace. hab is the two-dimensional metric induced on the worldsheet,

hab =

(

~X2 ~X · ~X ′

~X · ~X ′ ~X ′2

)

(1.4.6)

1.4 Strings 11

and the Lagrangian density is

L =√

− dethab =

√

( ~X · ~X ′)2 − ~X2 ~X ′2 (1.4.7)

where we ignored T (or set T = 1).

σa = (τ, σ) a = 0, 1 (1.4.8)

Lagrange Equation:

∂a∂L

∂(∂aXµ)= 0 ⇒

˙(∂L

∂Xµ

)

+

(∂L

∂Xµ′

)′= 0 (1.4.9)

We have

∂L

∂Xµ=Xµ

~X ′2 −X ′µ~X · ~X ′

L,

∂L

∂Xµ′ =X ′µ~X2 − Xµ

~X · ~X ′

L(1.4.10)

Let us choose the worldsheet coordinates (τ, σ) so that the metric hab be-comes proportional to the two-dimensional Mikowski metric,

hab ∼(

−1 00 1

)

(1.4.11)

Upon comparison with (1.4.6), we deduce the constraints

~X2 + ~X ′2 = 0 , ~X · ~X ′ = 0 (1.4.12)

which may also be cast into the form

( ~X ± ~X ′)2 = 0 (1.4.13)

Using the constraints, the Lagrange eq. (1.4.9) reduces to

∂2Xµ

∂τ2=∂2Xµ

∂σ2(1.4.14)

which is the wave equation. Introducing coordinates

σ± =1√2(τ ± σ) (1.4.15)

so that

∂± =1√2(∂τ ± ∂σ) (1.4.16)

the constraints and the wave equation become, respectively,

(∂± ~X)2 = 0 , ∂+∂−Xµ = 0 (1.4.17)


The general solution to the wave equation is

Xµ = f(σ+) + g(σ−) (1.4.18)

where f and g are arbitrary functions. Let us write out the action explicitlywith the simplifications made above. The parameter τ takes values from -∞to +∞ and σ takes values between 0 and l, the length of the string. Anticipat-ing future results, we write the constant T

T =1

2πα′ (1.4.19)

where α′ is called the Regge slope. Then, from equation (81),

SNG =1

4πα′

∫ +∞

−∞

∫ l

0

dτdσ(X2 − X2) (1.4.20)

=1

4πα′

∫ +∞

−∞

∫ l

0

dτdσ(∂aXµ∂aXµ) (1.4.21)

We now examine the effects of boundary conditions which have yet to betaken into account in the equations of motion. We start off by varying theco-ordinates:

Xµ → Xµ + δXµ (1.4.22)

Starting from equation (93), the variation in the action is

δSNG =1

4πα′

∫ +∞

−∞

∫ l

0

dτdσ (∂aδXµ∂aXµ + ∂aX

µ∂aδXµ) (1.4.23)

=1

2πα′

∫ +∞

−∞

∫ l

0

dτdσ (∂aδXµ∂aXµ) (1.4.24)

And noting the total derivative

∂a(δXµ∂aXµ) = ∂aδX

µ∂aXµ + δXµ∂a∂aXµ (1.4.25)

The last term is just the wave equation, which equals zero, so we are left with:

δSNG =1

2πα′

∫ +∞

−∞

∫ l

0

dτdσ∂a(δXµ∂aXµ) (1.4.26)

=1

2πα′

∫ +∞

−∞dτ[

δXµXµ

]l

0(1.4.27)

We will now introduce two sets of boundary conditions that will get rid of thisterm and leave the equations of motion unchanged at the boundary:Open string (Neumann) boundary conditions, which correspond to there be-ing no forces at the boundary:

1.4 Strings 13

Xµ(σ = 0) = Xµ(σ = l) = 0 (1.4.28)

Closed string boundary conditions, which means there is no boundary andthe string co-ordinates are periodic:

Xµ(σ = 0) = Xµ(σ = l) (1.4.29)

Xµ(σ = 0) = Xµ(σ = l) (1.4.30)

We shall now look at deriving invariant quantities in the theory from symme-tries using Noether’s theorem. We will start with Poincare invariance, whichis invariant under the transformation

Xµ → ΛµνXν + Y µ (1.4.31)

where Λ and Y are constant quantities. We construct the Noether current byapplying this symmetry to the action. Taking the second term, we write thechange in Xµ as

δXµ = Y µ (1.4.32)

The change in the action is found by inserting this into equation (96):

δSNG =1

2πα′

∫ +∞

−∞

∫ l

0

dτdσ∂aYµ∂aXµ (1.4.33)

The Noether current P aµ is defined by

δS =

∫ ∫

dτdσ∂aYµP aµ (1.4.34)

So in this case,P aµ = T∂aXµ (1.4.35)

and∂aP

aµ = T∂a∂

aXµ = 0 (1.4.36)

i.e. the Poincare symmetry has led to a conserved quantity in the Noethercurrent.Now let’s do the same for the variation

δXµ = εΛµνXν (1.4.37)

where ε is a small quantity. The variation in the action is now

δSNG =1

2πα′

∫ +∞

−∞

∫ l

0

dτdσ (Xµ∂aXν −Xν∂aX

µ) Λµν∂aε (1.4.38)

From this we can define the current as

Jµνa = T (Xµ∂aXν −Xν∂aX

µ) (1.4.39)


which is conserved:

∂aJµνa = T (∂aXµ∂aXν−∂aXν∂aX

µ+Xµ∂a∂aXν−Xν∂a∂aX

µ) = 0 (1.4.40)

since the first two terms cancel and the last two terms are the wave equation,which equals zero.

Analogous to electromagnetism, we can define a charge. In EM, the chargeis the integral over a volume of the zeroth (time) component of the current4-vector jµ, so here we define the charge for the current P aµ = (P τµ , P

σµ ) as

Pµ =

∫ l

0

dσP τµ (1.4.41)

P τµ can be seen to be the momentum of the string at a certain point, soPµ, µ >0 is the total momentum of the string, and P0 is the total energy of the string.

Differentiating Pµ with respect to time:

dPµdτ

=

∫ l

0

dσP τµ =

∫ l

0

dσP σµ = P σµ

∣∣∣∣

l

0

= 0 (1.4.42)

where the second equality follows from the wave equation and the last equal-ity form the boundary conditions at 0 and l. This is simply conservation ofmomentum.

Similarly, for the current Jaµν , which we interpret as the angular momentumof the string, we define the charge

Jµν =

∫ l

0

dσJaµν (1.4.43)

and we find

dJµνdτ

= 0 (1.4.44)

So from Poincare invariance we obtain conservation of momentum and an-gular momentum.

We now give two examples to demonstrate the above concepts:

Example 1

1.4 Strings 15

t=0

y

x

t

R

We take a closed string whose initial configuration is a circle centered on thex-y origin with radius R, and whose initial velocity is ~v = 0. Then Xµ =(t, x, y). The solution to the wave equation satisfying these boundary con-ditions is:

x = R cos2πτ

lcos

2πσ

l(1.4.45)

y = R cos2πτ

lsin

2πσ

l(1.4.46)

t =2πR

lτ (1.4.47)

Let’s check the constraints X2 + X2 = 0:

−t2 + x2 + y2 − t2 + x2 + y2 (1.4.48)

= −(

2πR

l

)2

+

(2πR

l

)2

sin2 2πτ

l+

(2πR

l

)2

cos22πτ

l= 0 (1.4.49)

The total energy of the string is given by

P0 =

∫ l

0

dσP τ0 = T

∫ l

0

dσ∂τX0 = T

∫ l

0

dσ∂τ t =2πRT

l

∫ l

0

dσ = 2πRT = E

(1.4.50)The length of the string is 2πR, so T can be identified as energy per unit lengthof the string, i.e. the tension.

Example 2


x

y

Now we consider an open string rotating in the x-y plane. The solution to thewave equation is

x = R cosπτ

lcos

πσ

l(1.4.51)

y = R cosπτ

lsin

πσ

l(1.4.52)

t =πR

lτ (1.4.53)

The speed of each point on the string is given by

~v =

(dx

dt,dy

dt

)

=l

πR

(dx

dτ,dy

dτ

)

= cosπσ

l

(

− sinπτ

l, cos

πτ

l

)

(1.4.54)

From which we see that

~v2 = cos2πσ

l(1.4.55)

Thus at the ends of the string σ = 0, l we see that ~v2 = 1, i.e. the endsof the string travel at the speed of light. This is to be expected, since thestring is massless and there are no forces on the ends of the string (Neumannboundary conditions). The intermediate points on the string don’t travel atthe speed of light because they experience the tension of the string.

The energy of the string is worked out exactly as before, and is found to be:

P0 = TRπ (1.4.56)

Let us now work out the z component of the angular momentum of the string:

1.5 The Mode Expansion 17

Jxy = T

∫ l

0

dσ(x∂ay − y∂ax) (1.4.57)

= T

∫ l

0

dσπR2

l

(

cos2πτ

lcos2

πσ

l+ sin2 πτ

lcos2

πσ

l

)

(1.4.58)

=TπR2

l

∫ l

0

dσ cos2πσ

l= 1

2TπR2 (1.4.59)

From the total energy and angular momentum we can form the quantity

JxyE2

=1

2πT= α′ (1.4.60)

E

J

Regge Slope

2α

Now if we want our strings to correspond to fundamental particles, their an-gular momenta correspond to the spins of the particles, which are either in-teger or half integer. Thus the above relation suggests that if we plot the spinsof particles against their energies squared, we would observe a straight line.This was indeed observed for strongly interacting particles. In fact, string the-ory started out being a theory of the strong interaction, but then QCD camealong. Now string theory become a theory of everything!

1.5 The Mode Expansion

We must first introduce the Hamiltonian formalism for future reference: wehave the Lagrangian

L = 12T (X2 − X2) (1.5.1)

The conjugate momentum to the variable X is


Πµ =∂L

∂Xµ= TXµ (1.5.2)

Then the Hamiltonian is given by

H =

∫ l

0

dσ(ΠµXµ − L) (1.5.3)

= 12T

∫ l

0

dσ

(Π2

T 2+ X2

)

(1.5.4)

= 12T

∫ l

0

dσ(X2 + X2) = 0 (1.5.5)

So this is not a good Hamiltonian. We will find a good one later.Now let us examine the mode expansion for open strings, which obey theNeumann boundary conditions. The general solution to the wave equation isa fourier expansion. Here we write such an expansion as follows:

Xµ = xµ + 2α′π

lpµτ + i

√2α′

∞∑

n=−∞

x6=0

1

nαµne

−πinτl cos

(nπσ

l

)

(1.5.6)

where xµ is a constant and the first mode pµ has been written out explicitly.The constants have been chosen on dimensional groundsThe fact thatXµ must be a real number yields the condition:

αn = (α−n)∗ (1.5.7)

The conjugate momentum toXµ is

Πµ = TXµ = 2α′Tπ

lpµ +

α′T

l

∞∑

n=−∞

x6=0

αµne−πinτ

l cos(nπσ

l

)

(1.5.8)

The centre of mass position of the string is given by

Xµ =1

l

∫ l

0

dσXµ = xµ + 2α′π

lpµτ (1.5.9)

so the centre of mass moves in a straight line. The total momentum is

P µ =

∫ l

0

dσΠµ = 2α′lπ

lpµT = pµ (1.5.10)

In both cases all the harmonic terms vanish on integration.We now move to the light cone gauge mentioned before: we defined the trans-verse co-ordinates X+ and X− and fix the gauge by imposing the condition:

X+ = x+ + 2α′π

lp+τ α+

n = 0 ∀n (1.5.11)


and set

2α′π

lp+ = 1 (1.5.12)

so that

X+ = x+ + τ (1.5.13)

which is the light cone gauge condition from before with an arbitrary con-stant x+.

The centre of mass position can now be written

Xµ = xµ +pµ

p+τ (1.5.14)

The constraint from before:

(X ± X)2 = 0 (1.5.15)

−2(X+ ± X+)(X− ± X−) + (X i ± X i)2

= −((X0 + X1) ± (X0 + X1))((X0 − X1) ± (X0 − X1)) + (X i ± X i)2

= −((X0 ± X0) + (X1 ± X1))((X0 ± X1) − (X1 ± X1)) + (X i ± X i)2

= −(X0 ± X0)2 + (X1 ± X1)2 + (X i ± X i)2 = (Xµ ± Xµ)2 = 0

(1.5.16)

∴ 2(X+ ± X+)(X− ± X−) = (X i ± X i)2 (1.5.17)

∴ 2(X− ± X−) = (X i ± X i)2 (1.5.18)

The mode expansion of the world-sheet fields:

Xµ = xµ + 2α′ π

lpµτ + i

√2α′

∞∑

n=−∞

x6=0

1

nαµne

−πinτl cos

(nπσ

l

)

(1.5.19)

x− ± x− == 2α′π

lp− +

√2α′π

l

∞∑

n=−∞

x6=0

α−n e

−πinl

(τ±σ) (1.5.20)

∴ (xi ± xi)2 =

2α′ π

lpi +

√2α′π

l

∞∑

n=−∞

x6=0

αine−πin

l(τ±σ)

2

(1.5.21)


You can write α−n in terms of αin:

(X i ± X i)2 =pipi

p+p++π2

l22α′

∞∑

n=−∞

n6=0

αi−nαin + 2

π2

l2(2α′)3/2pif(σ, τ) (1.5.22)

Concentrating on the zeroth mode:

2p−

p+=

pipi

(p+)2+

1

α′(p+)2

∞∑

n=−∞

n6=0

αi−nαin (1.5.23)

∴ p− =1

2p+

p

ipi +1

α′

∞∑

n=−∞

n6=0

αi−nαin

= H (1.5.24)

From Einstein’s equation:pµpµ +m2 = 0 (1.5.25)

m2 = −pµpµ = 2p+p− − pipi =1

α′

∞∑

n=−∞

n6=0

αi−nαin (1.5.26)

Now quantizing the system:

[pi, xj

]= −iδij

[Πi(σ), Xj(σ′)

]= −iδijδ(σ − σ′) (1.5.27)

Πµ = TXµ =pµ

l+√

2α′π

lT

∞∑

n=−∞

x6=0

αµne−πinτ

l cos(nπσ

l

)

(1.5.28)

Xµ = xµ + 2α′π

lpµτ + i

√2α′

∞∑

n=−∞

x6=0

1

nαµne

−πinτl cos

(nπσ

l

)

(1.5.29)

[Πi(σ), Xj(σ′)

]=

1

l

[pi, xj

]+i

l

∑

m,n6=0

1

n

[αim, α

jn

]e

−πiτl

(n+m) cos(nπσ

l

)

cos

(mπσ′

l

)

(1.5.30)[αim, α

jn

]= mδijδm+n,0 (1.5.31)

∴[Πi(σ), Xj(σ′)

]= −iδijδ(σ − σ′) (1.5.32)

This is just the commutation relation for the harmonic oscillator operatorswith nonstandard normalization

a =1√mαim a† =

1√mα−m

i (1.5.33)

∴[a†, a

]= 1 (1.5.34)


The state |0, k〉 is defined to be annihilated by the lowering operators and tobe an eigenstate of the center-of-mass momenta

a|0〉 = a|0, k〉 = 0 (1.5.35)

Πm,i|0, k〉 = |0, k〉 (1.5.36)

M2 =1

α′

∞∑

n=−∞

n6=0

∞∑

i=−∞

i6=0

αi−nαin =

1

α′

∞∑

n=−∞

n6=0

∞∑

i=−∞

i6=0

ma†a =1

α′N (1.5.37)

M2|0, k〉 =1

α′

∞∑

n=−∞

n6=0

(D − 2)n

2|0, k〉 =

(D − 2)

2α′

∞∑

n=−∞

n6=0

n

|0, k〉 (1.5.38)

We have to perform the sum:∞∑

n=−∞

n6=0

n (1.5.39)

To perform this sum, we will multiply by the sum by e−2πnε

l and then take thelimit of ε→ 0

∞∑

n=−∞

n6=0

ne−2πnε

l =∂

∂C

∞∑

n=−∞

n6=0

ne−nC =∂

∂C

(1

1 − e−C

)

=e−C

(1 − e−C)2=

1

C2− 1

12

(1.5.40)

where C =2πε

l(1.5.41)

∞∑

n=−∞

n6=0

ne−nC =1

c2− 1

12(1.5.42)

M2 =1

α′

∞∑

n=−∞

n6=0

∞∑

i=−∞

i6=0

αi−nαin =

(D − 2)

2α′

∞∑

n=−∞

n6=0

n = 2p+p− − pipi (1.5.43)

The last equality was obtained using(64)

M2 =(D − 2)

2α′

(1

c2− 1

12

)

=(D − 2)

2α′l2

(2πε)2− (D − 2)

24α′ =(D − 2)

(2πε)2πp+l− (D − 2)

24α′ = 2p+p−−pipi

(1.5.44)

The mass of each state is thus determined in terms of the level of excitation.

M2 =1

α′

(

N − D − 2

24

)

N |0〉 = 0 (1.5.45)


This operator acting on the 0 ket yields:

M2|0〉 = − (D − 2)

24α′ |0〉 (1.5.46)

The mass-squared is negative for D > 2. The state is a tachyon

The lowest excited states of the string are obtained by exciting one of then = 1modes once:

M2(αi−1|0〉) =1

α′

(

1 − D − 2

24

)

(αi−1)|0〉 =1

α′

(26−D

24

)

|0〉 (1.5.47)

Lorentz invariance now requires that this state be massless, so the number ofspacetime dimensions is D = 26

1.6 Closed strings

We must now look at the mode expansion and quantization os closed strings,which are required when looking at string interactions. The procedure is verysimilar to that of open strings, except now we have Dirichlet boundary con-ditions, i.e. Xµ is periodic. The mode expansion is now made up of left andright moving parts:

Xµ = XµR(τ − σ) +Xµ

L(τ + σ) (1.6.1)

such that

XµR = 1

2xµ + α′pµ(τ − σ) + i

√12α

′∞∑

n=−∞

x6=0

1

nαµne

−2πin(τ−σ)l (1.6.2)

XµL = 1

2xµ + α′pµ(τ + σ) + i

√12α

′∞∑

n=−∞

x6=0

1

nαµne

−2πin(τ+σ)l (1.6.3)

The sum of these is periodic. In the sum the integer n is in effect 2n so thereare now twice as many modes as for the open string. Once again, the fact thatXµ is real means that

αµn = (αµ−n)∗ αµn = (αµ−n)∗ (1.6.4)

Quantizing as before:

[αim, α

jn

]=[αim, α

jn

]= mδijδm+n,0 (1.6.5)

[αim, α

jn

]= 0 (1.6.6)

1.6 Closed strings 23

The mass operator is now

M2 = 2p+p−−pipi =2

α′

(

NR +NL +2(D − 2)

24

)

=2

α′

∞∑

n=−∞

n6=0

αi−nαin +

∞∑

n=−∞

n6=0

αi−nαin − 2(D − 2)

24

(1.6.7)There are two symmetries in these expansions: the transformations

τ → τ + constant (1.6.8)

σ → σ + constant (1.6.9)

don’t change the physics of the string. The first symmetry is shared with theopen string, but the spatial translational symmetry is new.The generator of the time translations is H, which we saw before to be

H =

∫ l

0

dσ(X2 + X2) (1.6.10)

and the fact that H = 0 leads to the invariance under time translations. In thecase of closed strings,

H =

∫ l

0

dσ[(XR + XL)2 + (XR − XL)2] (1.6.11)

∼∫ l

0

dσ(X2R + X2

L) = HR +HL (1.6.12)

where we have used the fact that we can write

∂σ ∼ ∂τ (1.6.13)

since τ and σ are interchangeable in the expansions of XR and XL to withina minus sign.This generation of time translations comes from the constraint X2 + X2 = 0,so it is reasonable to suppose that spatial translations come from the otherconstraint X · X = 0. Defining the operator D:

D =

∫ l

0

dσX · X ∼∫ l

0

dσX2 =

∫ l

0

dσ(X2R − X2

L) ∼ NR −NL = 0 (1.6.14)

So

NR = NL (1.6.15)

which common sense tells us must be the case.


Now let us determinant the lowest states in the mass spectrum. For the vac-uum state

M2|0〉 = −4(D − 2)

24α′ |0〉 (1.6.16)

so the mass squared is negative for the vacuum state, as for the open string.The first excited state is |Ωij〉 = αi−1α

j−1|0〉 , where we must remember to keep

NR = NL. We obtain

M2|Ωij〉 =2

α′

(2 − 2(D − 2)

24

)

|Ωij〉 (1.6.17)

As before, we wish M2 = 0 for the first excited state, so again we get D = 26.The situation is a bit more complicated than for the open string case, so let’slook in a bit more detail.The state |Ωij〉 can be split into three parts: a symmetric, traceless part; anantisymmetric part and a scalar part:

|Ωij〉 =

[

12 (|Ωij〉 + |Ωij〉) − 2

D − 2δij |Ωkk〉

]

+ 12 (|Ωij〉 − |Ωij〉) +

1

D − 2δij |Ωkk〉(1.6.18)

We call the three states |Gij〉, |Bij〉, |Φ〉. The symmetric, traceless, spin 2 state|Gij〉 can now be identified with the graviton, which didn’t exist in the openstring theory, so it seems some progress has been made. The spin 0 scalarstate is called the dilaton.Now we impose a further symmetry: invariance under the transformation

σ → −σ (1.6.19)

which is the condition for unoriented strings. This means

XR ↔ XL (1.6.20)

αn ↔ αn (1.6.21)

This condition immediately disallows the antisymmetric state, since underthe transformation,

|Bij〉 → −|Bij〉 (1.6.22)

while |Gij〉 and |Φ〉 remain unchanged.Next let us turn our attention to the fact that we have been working in lightcone gauge, which is not Lorentz invariant. We would like to reassert lorentzinvariance. We can do this by generalizing the commutation identity:

[αim, α

jn

]= mδijδm+n,0 → [αµm, α

νn] = mηµνδm+n,0 (1.6.23)

This gives

1.7 Gauge invariance 25

[α0m, α

0n

]= −mδm+n,0 (1.6.24)

Now let us define a state

|φ〉 = α0−1|0〉 (1.6.25)

The norm of this state is

〈φ|φ〉 = 〈0|α01α

0−1|0〉 = 〈0|[α0

1, α0−1]|0〉 + 〈|α0

−1α01|0〉 = −〈0|0〉 = −1 (1.6.26)

Thus we have a negative norm state, which is physically meaningless. Thistells us there is something wrong with our theory. We will come back to thisin the next couple of chapters, and solve this problem.

1.7 Gauge invariance

Let us examine the open string state

|φ〉 = Aµ(k)αµ−1|0, k〉 (1.7.1)

under the transformation

Aµ → Aµ + kµω(k) (1.7.2)

where ω(k) is an arbitrary function. This is analogous to a gauge transforma-

tion in electromagnetism ~A→ ~A+ ∇ω. The change in |φ〉 is

|δφ〉 = kµωαµ−1|0, k〉 (1.7.3)

and the norm is

〈δφ|δφ〉 = kµkνω2〈0, k|αµ1αν−1|0, k〉 = k2ω2 = 0 (1.7.4)

since k = 0 (the mass is zero, and m2 = kµkν ). Thus the theory has producedgauge invariance! The equivalent state for the closed string is

|φ〉 = gµν αµ−1α

ν1 |0, k〉 (1.7.5)

The gauge transformation is

gµν → gµν + kµων + kνωµ (1.7.6)

This time we find the norm state to be

〈δφ|δφ〉 ∼ k2 = 0 (1.7.7)

Thus gauge invariance holds for the closed string state too. What’s more, wecan identify gµν with the gravitational potential, a sign that general relativitymight be included in the theory.


UNIT 2

Conformal Field Theory

2.1 Massless scalars in two dimensions

We will start by looking at the Polyakov action with one change. The metrichas been replaced with a Euclidean metric δa,b with signature (+,+).

S =T

2

∫

dτdσ(XµXµ +X ′µX ′µ)

where T, Xµ, X ′µ represent the string tension, ∂τXµ, ∂σX

µ respectively.

We can derive the equation of motion by varying the action with respect tothe coordinate X . We find:

δXS = 0 → X ′′µ + Xµ = ∇2Xµ = 0, where ∇2 = ∂2

σ1+ ∂2

σ2

We can define z and z as linear combinations of σ1 and σ2. These will repre-sent the new worldsheet coordinates

z = σ1 + iσ2, z = σ1 − iσ2.

The bar denotes complex conjugate. We can also invert the coordinate trans-formation

σ1 =z + z

2, σ2 =

z − z

2.

Define differentiation:

∂z = ∂ =1

2(∂1 + i∂2) and ∂z = ∂ =

1

2(∂1 − i∂2)

∇2 can be written as 4∂∂, and the volume element is given by

d2z =

∣∣∣∣

1 −11 1

∣∣∣∣dσdτ = 2dσdτ.

28 UNIT 2: Conformal Field Theory

Also define ∫

d2z δ2(z, z) = 1

so that δ(σ1)δ(σ2) = δ2(z, z). The action in complex coordinates is

S = T

∫

d2z∂Xµ∂Xµ .

After varying the action with respect to the coordinate Xµ, we get the equa-tion of motion

∂∂Xµ = 0.

This implies ∂X is a holomorphic function, ie a function of z. Also, ∂X is anantiholomorphic function, ie a function of z.Another useful result is the divergence theorem for complex coordinates. First,let’s look at the divergence theorem for three dimensions i.e., electrostatics.The divergence theorem states that for any well behaved vector field E(x) de-fined within a volume V surrounded by the closed surface S the relation

dS

V

∫

V

d3x ∇ · E =

∮

S

E · n da

holds between the volume integral of the divergence of E and the surface in-tegral of the outwardly directly normal component of E.In 2D:∫

d2z (∂Ez + ∂E z) = i

∮

∂R

(Ezdz −E zdz) where n = (−dz, dz).

We can now write the mode expansions in terms of the complex coordinates.

Xµ(z, z) = xµ − iz − z

2

pµ

p++ i

√

α′

2

∑

n6=0

1

nαµne

2nπiz` + αµne

−2nπiz`

As one can see, this mode expansion can be broken into two pieces (left andright handed).

Xµ(z, z) = XµL(z) +Xµ

R(z)

We see the left handed piece cooresponds to a holomorphic function and theright handed piece to a antiholomorphic function. We will never look at the

2.2 Solution to the Boundary-Value Problem with Green function 29

mode expansion itself. Instead we will always look at various derivatives ofthe mode expansion.

∂XµL =

i

2

pµ

p+− π

`

√2α′

∑

n6=0

αµn exp

[inπz

`

]

,

∂XµR =

i

2

pµ

p+− π

`

√2α′

∑

n6=0

αµn exp

[inπz

`

]

We can absorb the pµ by defining αµ0 a certain way.

Let αµ0 = −i `

2π√

2α′pµ

p+

Now the derivatives on the fields simplify.

∂XµL = −π

`

√2α′

∑

all n

αµn exp

[inπz

`

]

, ∂XµR = −π

`

√2α′

∑

all n

αµn exp

[inπz

`

]

2.2 Solution to the Boundary-Value Problem with

Green function

The solution to the Poisson or Laplace equation in a finite volume V with ei-ther Dirichlet or Neumann boundary conditions on the bounding surface Scan be obtained by means of Greens theorem. In general, we want to solvethe equation,

∇′2G(x,x′) = −4πδ(x − x′)

where G(x,x’) is the potential and the delta function is a point source. Thesolution for G is given as

G(x,x′) =1

|r − r0|+ F(x,x′)

with F satisfying the Laplace equation inside the volume V:

∇′2F (x,x′) = 0

In two dimensions the Poisson equation is given by

∇2G(z, 0; z, 0) = −2πδ(z)δ(z) where ∇2 = ∂∂

= −2πδ2(z, z)

where

G(z, z) = ln |z|2 = ln |z| + ln |z|


This is just the solution for the potential to a line charge in two dimensions.We can prove that G is a solution of the Poisson equation in problem 1.

∂∂G = ∂∂ ln |z|2 = ∂1

z+ ∂

1

z= 2πδ2(z, z), z = 0

Now that we see G is directly related to the potential, we can take the gradientto get the electric field.

Ez = ∂G, Ez = ∂G∫

V

d3x ∇ · E =

∮

S

E · dn = 2πR(1

R) = 2π

2.3 Amplitudes

We are interested in calculating vacuum expectation values

〈0|Xµ1(z1)Xµ2(z2) · · ·Xµn(zn)|0〉.

Let us start with the simplest nontrivial example, the two-point amplitude.

The two-point amplitude

We will focus only on the left-movers.

Xµ = XµL = i

√

α′

2

∑

n6=0

1

nαµn exp

[−2πinz

`

]

.

Aµν = 〈0|Xµ(z)Xν(z′)|0〉

= −α′

2

∑

n,m6=0

1

nmexp

[−2πi(n−m)z

`

]

〈0|αµnανm|0〉

In order evaluate the vacuum expectation value, break the sum into four pieces:m,n > 0; m > 0, n < 0; m < 0, n > 0; m,n < 0. Only the m < 0, n > 0 termssurvive, because the others either kill the vacuum or produce an inner prod-uct of orthogonal states. To evaluate, we express the operators in terms oftheir commutator. This is possible because the second term in the commu-tator kills the vacuum.

[αµn, ανm] = nηµνδm+m,0.

After applying the Kroneker delta our sum reduces to

Aµν =α′

2

∑

n>0

1

nexp

[

−2πin(z − z′)

`

]

.

2.3 Amplitudes 31

This is a sum of the form∑

ωn

n , but only converges for |ω| < 1.

∣∣∣∣exp

[

−2πi(z − z′)

`

]∣∣∣∣=

∣∣∣∣exp

[2π(z − z′)

`

]∣∣∣∣< 1 ⇒ z2 − z′2 < 0, z2 < z′2

So we see the Xs must be time ordered to give a correct result. Therefore,when we calculate any two-point function, we must use the time orderedproduct of the two operators

〈0|X(z)X(z′)|0〉 ⇒ 〈0|T [X(z)X(z′)]|0〉.

We define time ordering as

T [Xµ(z)Xν(z′)] =

Xµ(z)Xν(z′) for z2 < z′2Xν(z)Xµ(z′) for z2 > z′2

,

or in terms of the step function the product becomes

T [Xµ(z)Xν(z′)] = θ(z′2 − z2)Xµ(z)Xν(z′) + θ(z2 − z′2)X

ν(z)Xµ(z′).

Evaluating the sum:

Aµν =α′

2ηµν

∑

n

en

n=α′

2ηµν ln |1 − e−β|

By applying the D’Alembertian to Aµν , we show it is a two-point Green func-tion

∂∂T [Xµ(z)Xν(z′)] = (∂2 + ∂2)T [Xµ(z)Xν(z′)]

= T [∂21X

µXν ] + T [∂1(−δ(z2 − z′2)XµXν + δ(z2 − z′2)X

µXν)]

= T [∂21X

µXν ] + T [∂2δ(z2 − z′2)[Xµ, Xν ]] + T [∂2X

µXν)]

= T [∂21X

µXν ] + δ(z′2 − z2)[∂2Xµ, Xν ] + T [∂2

2Xµ, Xν ]

= T [(∂21X

µ + ∂22X

µ)Xν ] + δ(z′2 − z2)[∂2Xµ, Xν ]

= δ(z′2 − z2)[∂2Xµ, Xν ]

= πα′δ2(z − z′)ηµν

Aµν must be of the form of a Green function, Aµν = ηµνG(z, z′).For future reference Xµ will imply only the holomorphic piece, unless spec-ified otherwise. We may express the time-ordered product in terms of thenormal ordered product minus the singularity.

T [XµXν ] =: XµXν : −α′

2ln |z − z′|ηµν

As z → z′:

Xµ(z)Xν(z′) = −α′

2ηµν ln |z − z′| +

∞∑

k>0

(z − z′)k

k!: Xν∂kX

µ(z′) :


∑

k

1

k!

(

−1

2

α′

2

∫

dzdz′ηµν ln |z − z′| δ

δXµ

δ

δXν

)k

= exp

[

−α′

4

∫


δXµ

δ

δXν

]

Define an operator O such that O = XµXν .

: O := exp

[

−α′

4

∫


δXµ

δ

δXν

]

O

This needs to be inverted.

O = exp

[α′

4

∫


δXµ

δ

δXν

]

: O :

Now O should have no singularities. We have to define the product betweentwo Os. This product will have singularities, unless the product of the twois normal ordered. The product of two time ordered operators, :O1::O2:, hassingularities, whereas the time ordered product, :O1O2:, contains none.

: O1[X ]O2[Y ] := exp

[

−α′

2

∫


δXµ

δ

δXν

]

: O1 :: O2 :

invert

: O1[X ] :: O2[Y ] := exp

[α′

2

∫


δXµ

δ

δY ν

]

: O1O2 :

Example: Let O1 = O2 = ∂Xµ(z)∂Xµ(z) = T (z)

: T (z) :: T (z′) :=: ∂Xµ(z)∂Xµ(z) :: ∂′Xν(z′)∂′Xν(z′) :

There are two possible double contractions and four possible single contrac-tions,

: T (z) :: T (z′) : = 2 ∗ α′

2

2

ηµνηµν(∂∂′ ln |z − z′|)2 − 4 ∗ α

′

2ηµν ln |z − z′| : ∂Xµ∂

′Xν : + : T (z)T (z′) :

=α′2

2

d

(z − z′)4− 2α′

(z − z′)2: T : − α′

z − z′: ∂′T (z′) : + : T (z)T (z′) :

Example: Let O1 =: eik1X(z) :, O2 =: eik2X(z′) :, δδXµ On = iknµOn

: O1O2 : = exp

[α′

2ln |z − z′|ηµν(ik1µ)(ik2ν)

]

: O1 :: O2 :

= exp

[

−α′

2k1 · k2 ln |z − z′|

]

: O1 :: O2 :

= (z − z′)−α′

2 k1·k2 : O1 :: O2 :

⇒: O1 :: O2 : = (z − z′)α′

2 k1·k2 : O1O2 :

= (z − z′)−α′

2 k1·k2 : ei(k1+k2)·X(1 + O(z − z′)) :

In this example we see the time-ordered product for two vertex operators rep-resenting tachyons.

2.4 Noether’s Theorem 33

2.4 Noether’s Theorem

For every symmetry in a theory, there must be some conserved current, ie

∂µjµ = 0 ⇒ Symmetry(S).

We can integrate over the zeroth component of the conserved current to getthe charge.

Q =

∫

d3x∂0j0 → dQ

dt=

∫

R

d3xj0 =

∫

d3x∇ ·~j =

∫

d~s ·~j = 0

Q generates transformations.

δA = iε[Q,A] = iε(QA−AQ) ε << 1

Example: LetQ = HδA = iε[H,A] = A

orQ = ~p~∇A = i[~p,A]

S

S

t

t

t

+

0

−

−

+

t

Look atA(t0):Region R is bounded by [S+, S−].

δA = iε(Q(t+)A(t0) −A(t0)Q(t−))

= iε∫

S+

d3x j0A(t0) −∫

S−

d3x j0A(t0)

= iε

∫

∂R

dSµjµA(t0)

= iε

∫

∂R

dSµT [jµA(t0)]

δA = iε

∫

R

d4x ∂µT [jµA(t0)] ”Ward Identity”


Show the Ward Identity explicitly in two dimensions:

δA =iε

2π

∫

R

d2z ∂aT [jaA(z0)], j = (jz, jz)

=iε

2π

∮

∂R

(dzjz − dzjz)A(z0)

0

∂ R

z

zR

∂aja = ∂zjz + ∂zjz = 0 for special case jz is holomorphic, and jz is antiholo-

morphic.

∮dz

2πjzA(z0) = λ(z0) : jzA(z0) = . . .+

λ(z0)

z − z0+ . . .

δA = ελ+ ελ = −ε(λ− λ)

Example: Define the transformation on X and the current density.

Xµ → xµ + εaµ : jµa =i

α′ ∂aAµ jµz =

i

α′ ∂Xµ

∂jµz = 0 remember X is holomorphic

LetA(z0) =: eik·X :

δA = : eik·(X+εa) :

= (1 + iεk · a)A(z0)

jµzA(z0) =i

α′α′

2ηµν∂ ln |z − z0|(ikν) : A(z0) : +regular terms

= −1

2

kµ

z − z0A(z0)

⇒ λ =1

2kµA(z0) = −λ : Residue

2.5 Conformal Invariance 35

ε(λ+ λ) = ε(kµA(z0))

εaµ(λ− λ) = εaµkµA(z0)

Example: Let us look at translations.

δz = −ε, δXµ = Xµ(z − ε) −Xµ(z)

= −ε∂Xµ

Noether Current

S = − 1

2πα′

∫

∂Xµ∂Xµ

δz = −ε(z, z) δS =1

2πα′

∫

∂(ε∂Xµ)∂Xµ + ∂Xµ∂(ε∂Xµ)

=1

πα′

∫

∂ε(∂Xµ∂Xµ)

2.5 Conformal Invariance

T =1

πα′ : ∂Xµ∂Xµ : ∂T = 0 conserved, T is holomorphic

T =1

πα′ : ∂Xµ∂Xµ : ∂T = 0 T is antiholomorphic

Tττ = − 1

2α′ : X2 + x′2 := Tσσ

Tτσ = Tστ = − 1

α′ X ·X ′ : Traceless T aa = 0.

For an arbitrary function v(z):

j(z) = iv(z)T (z), ∂j = 0.

There are an infinite number of conservation laws. We may calculate the op-erator product expansion for the stress tensor with the fieldXµ

T (z)Xµ(z′) = : ∂Xν∂Xν : Xµ(z′) =1

α′ ηµν∂ ln |z − z′|α′∂Xν(z

′) + ...

=1

z − z′∂Xµ(z′) + ...

T (z)Xµ(z′) =1

z − z′∂Xµ(z′) + ...

From j = iv(z)T (z), ∂aja = 0

∂j = 0 or ∂ j = 0


∂(v(z)T (z)) = 0

v(z)T (z) is a conserved current, where

vTxµ ∼ v∂xµ

z − z′.

Let z transform as δz = z + εV , then

xµ −→ xµ − εv∂xµ

and T (z)A(z′) can be expanded as Laurent series

T (z)A(z′) =a−1

z − z′+

a−2

(z − z′)2+ ...+ regular terms,

then λ is

λ =

∮dz

2πiv(z)T (z)A(z′)

=

∮dz

2πi

[a−1v(z)

z − z′+a−2v(z)

(z − z′)2+ ..

]

= ia−1v(z′) + ia−2∂v(z

′) +i

2!a−3∂

2v(z′) + · · · . ie.

δA = −εa−1v − εa−2∂v −ε

2!a−3∂

2v − · · · ,

= −ε∞∑

n=0

1

n!a−n−1∂

nv.

Therefore, to find a−n, for all n, we need to find how A(z) transforms underthe conformal transformation z → z + εv(z). The simplest case is a scaling:v(z) = z, z → z + δz = (1 + ε)z. Find A(z) that has a simple scaling property(eigenfunction), δA = −hεA, for finite ζ.

A′ = (1 + ε)−hA

= ζ−hA.

or A(z, z) transforms as

A(z, z) → ζ−hζ−hA(z, z′).

For ζ = reiθ ,

A′(z′, z′) → r−(h+h)e−(h−h)θA(z, z′)

h+h is the dimension ofA, and determines the transformation under scaling.h − h determines the transformation under spin. If A is order h, then ∂A isorder h+ 1, i.e.

∂A

∂z=

∂z′

∂z

∂A

∂z′

→ (1 − εz)(1 − hεz)∂A

→ (1 − (h+ 1)εz)∂A.

2.5 Conformal Invariance 37

or∂A, h→ h+ 1

Compare the coefficient of ∂v with the equations for δA. This implies a−2 =hA. For simplicity, let v(z) = 1.Special Case: Do a translation transformation on z, z → z + ε.Then δA =A(z− ε)−A(z) = ε∂A→ a−1 = ∂A. For an arbitary v(z), z → z+ εv(z), δA isgiven by

δA = −hε∂vAA′ = (∂ζ∂z )

−h(∂ζ∂z )−hA

A′ = (1 + ε∂v)−hA.

If δA has only two singularity terms in the form below, call A a primary field,ie.

T (z)A(z′) =∂A

z − z′+

h∂A

(z − z′)2.

Some examples for T (z)A(z′)

T (z)Xµ(z′) = − 1

α′ : ∂Xν∂Xν : Xµ(z′) =1

2ηµν∂ ln(z − z′)∂Xν

=1

z − z′∂Xµ, ie. h = h = 0

T (z)∂2Xµ =1

2ηµν∂∂2 ln(z − z′)∂Xν

=2

(z − z′)3∂Xµ(z)

=2

(z − z′)3[∂Xµ(z′) + (z − z′)∂2X(z′) +

1

2!(z − z′)2∂3Xµ(z′) + ...

=2

(z − z′)3∂Xµ +

2

(z − z′)2∂2Xµ(z′) +

1

z − z′∂∂2Xµ + ...

T (z) : eikX : =1

α′

(α′

2ηµνkν∂ ln(z − z′)

)2

: eik·X : +ηµνkν∂ ln(z − z′) : ∂Xµeik·X :

=α′

4

k2

(z − z′)2: eik·X : +

1

z − z′: ∂eik·X

ForA = eik·X , T (z)A(z′) implies h = α′k2

4 and T (z)A(z′) implies h = α′k2

4 .Therefore

A has weight (h, h) = (α′k2

4 , α′k2

4 ). If a translation is applied to z (z → z + εv),

then eik·X(z) → eik·X(z−εv) and δ(eik·X(z)) = −εv∂eik·X(z), which means thath = 0. We have just shown that h is nonzero, and arrives from normal order-ing (quantum effect).

∂Xeik·X → h = 1 + α′k2

4

∂2Xeik·X → h = 2 + α′k2

4

∂mXeik·X → h = m+ α′k2

4

∂mnXµn ...∂m2Xµ2∂m1Xµ1eik·X → h = mn + ...+m2 +m1 + α′k2

4


It looks like these make a good basis for operators.

Xµ h = 0 h = 0 (0, 0)

∂Xµ h = 1 h = 0 (1, 0)

∂Xµ h = 0 h = 1 (0, 1)

∂2Xµ h = 2 h = 0 (2, 0)

eik·X h = α′k2

4 h = α′k2

4 (α′k2

4 , α′k2

4 )

∂Xeik·X h = 1 + α′k2

4 h = 1 + α′k2

4 (1 + α′k2

4 , 1 + α′k2

4 )...

......

...

T (z) has weight h = 2, but is not a primary field. The TT OPE is given by:

T (z)T (z′) =D

2(z − z′)4+

2

(z − z′)2T (z′) +

2

α′(z − z′): ∂2Xµ∂X

µ :

where ∂2Xµ∂Xµ can be written as 1

2 (∂(∂Xµ∂Xµ)) if T (z) = − 1

α′: ∂Xµ∂Xµ :

+Vµ∂2Xµ

2.6 Free CFTs

In this section we will explore various different conformal field theories. Wemay classify all CFTs by knowing their central charges and three-point func-tions. We will find the central charges for the following free conformal fieldtheories. We will leave the three-point functions for the reader.

Linear dilaton

The TT OPE is given by

T (z)T (z′) ∼ D

2(z − z′)4+ VµV

µα′

2∂2∂′2 ln(z − z′) + ...

=D

2(z − z′)4+

6VµVµα′

2(z − z′)4+ ...

=c

2(z − z′)4+ ... ,

where c = D + 6α′VµV µ. Therefore the central charge for the Linear Dilatontheory can be any number. When the number of dimensions is one or twothe theory is exactly solvable. Also, if we compactify some dimensions V µ canlive in the compact subspace, because there is no need for Lorentz invariancethere.

T (z)Xµ(z′) ∼ ∂ ln(z − z′)∂Xµ + V µα′

2∂2 ln(z − z′) + ...

=1

z − z′∂Xµ +

V µα′

2(z − z′)2, h = 0 and Xµ is not a primary field,

2.6 Free CFTs 39

v(z)T (z)Xµ(z′) ∼ V µα′

2(z − z′)2[v(z′) + (z − z′)∂v + ...] +

v

z − z′∂Xµ + ....

For δXµ = −ε, λ = 12V

µα′∂v + v∂Xµ.

bc theory

Let b and c be anticommuting fields, i.e. spinors.The action can be writen as

S =1

2π

∫

d2zb∂c

The equations of motion are given by: ∂c = 0, ∂b = 0, where b and c areholomorphic. If we let ` = 2π, then we can write b and c as

b(z) = i∑

bneinz , c(z) = i

∑

cneinz,

where

b(z), c(z′) = δ(σ − σ′)equaltime, z2 = z′2 and 0 < σ < 2π or

bm, cn = δm+n,0

〈0|b(z)c(z′)|0〉 =1

1 − ei(z−z′)

∼ 1

z − z”+ regular terms, then

: b(z)c(z′) : = b(z)c(z′) − 1

z − z′

If b has weight hb = λ, then hc = 1 − λ.This is known since the action hasweight 0 and the volume element has weight (−1,−1). From the transforma-tion δz = ε(z), b will change to

b′ =

(∂z′

∂z

)λ

b(z − ε)

= (1 − λ∂ε)(b− ε∂b), then

δb = −λ∂b− ε∂b, and

δc = −(1 − λ)∂c− ε∂c

δS =

∫

∂ε((∂b)c− λ∂(bc)),

where (∂b)c− λ∂(bc) is the Noether current in this case, then T is

T = : (∂b)c : −λ∂(: bc :), and

c(z)b(z′) =1

z − z′+ ...

b(z)c(z′) =1

z − z′+ ...


T (z)T (z′) = −(

1

z − z′

)2

− 2λ∂

(1

z − z′∂

1

z − z′

)

+ λ2∂∂′1

(z − z′)2

=−1 + 6λ− 6λ2

2(z − z′)4+ ...

=c

2(z − z′)4

c = −2 + 12λ− 12λ2 = 1 − 3(2λ− 1)2.

From the Linear Dilaton theory c = D + 6αV 2. Let the charges from the twotheories be equal, ie. D+ 6αV 2 = 1− 3(2λ− 1)2 and solve for V . For the casewhere D = 1, one will obtain

V =1√2π

(2λ− 1).

Can the bosons be equivalent to the fermions? We will see later. Let us explorea special case, λ = 1

2 . We find V = 0, c = 1 and, b, c can be writen as a linearcombination of scalar fields Ψ1 and Ψ2

b(z) =1√2(Ψ1 + iΨ2), c(z) =

1√2(Ψ1 − iΨ2).

The action may be expressed in terms of the Ψs

S =1

4π

∫

d2z(Ψ1∂Ψ1 + Ψ2∂Ψ2),

with a stress tensor

T = −1

2Ψi∂Ψi, i = 1, 2.

Another interesting case is for λ = 2 and V=0. Then the central charge, cbecomes c = −26 from c = 1 − 3(2λ − 1)2. This is the result obtained inchapter 1.

βγ theory

The next example, the bosonic case, let β and γ be commuting scalar fields.The βγ action is given by:

S =1√2π

∫

d2zβ∂γ,

where∂β = ∂γ = 0

The procedure is the same as for the spinor case.

β(z)γ(z′) =1

z − z′+ ...

γ(z)β(z′) = − 1

z − z′+ ...

c = −1 + 3(2λ− 1)2

2.7 Virasoro Algebra 41

Note that if λ = 32 , then c = 11. If we combine the central charges for the four

theories, the number of physical dimensions reduces from twenty-six to ten.

Xµ D (b, c) −26Ψµ d

2 (β, γ) 11

D +d

2− 26 + 11 = 0 =⇒ D = 10.

From

XµL(z) = xµ − α′

2pµz + i

√

α′

2

∑ 1

nαµne

inz

〈XL(z), XL(z′)〉 = ln(1 − ei(z−z′))

∼ ln(z − z′) + ...

2.7 Virasoro Algebra

The worldsheet of a free closed string moving through space-time looks like acylinder whose radius may fluctuate depending on the excitation mode. Wecan map this cylinder to the complex plane. This map would be equivalentto sqeezing one end of the string so it looks like a cone.Then smash the coneinto a plane. Now the worldsheet coordinates can be expressed as complexcoordinates, where r represents τ and the phase, θ, represents the position onthe string, σ (z 7→ z, cylinder 7→ plane). For closed strings λ = 2π, 0 < σ < 2π

z = e−iz

timeslices

time

2π0

const.

z z

time

Instead of expanding in Fourier modes, do a Laurent Expansion.

Xµ ∼∑ 1

nαµnz

−n

What happens to T (z):

T (z) =1

α′ : ∂Xµ∂Xµ :

=∑

Tmeimz

?=

∑

Tmz−m


T (z) doesn’t transform as easily as guessed. Subtract the term with centralcharge to make it a primary field (tensor). Then transform each coordinate.

T (z)T (z′) =c

2(z − z′)4+

2

(z − z′)2T (z′) +

1

z − z′∂T (z′) + . . .

multiply both sides by v(z)

v(z)T (z)T (z′) = v(z)[. . .]

δT = −ελ = εc

2

1

3!∂3v − 2ε∂vT − εv∂T

z → z + εv(z) z → z = e−iz

T (z) = (∂z

∂z)2T (z) +

c

12z, z,

where z, z is known as a Schwarzian derivative, defined as

z, z =2z′′′z′ − 3(z′′)2

2(z′)2,

which equals 1/2 for our example.

T = −z2T +c

24

We can invert this and solve for z,

T (z) = −z−2T +c

24z−2

=∑

Lmz−m−2, Lm = −Tm +

c

24δm,0

Invert the equation and solve for L.

Lm =

∮dz

2πizm+1T

The Hamiltonian∫ 2π

0dσ2πT can now be written in terms ofL. Including left and

right movers, the Hamiltonian is

H = L0 + L0 −c+ c

24,

and the number operator is N = L0 − L0. We can see that ∂Lm = 0 since∂[zm+1T ] = 0. This implies all Lms generate symmetries.

2.7 Virasoro Algebra 43

Commutators

Recall the Ward identity for any operator A : δA = iε[Q,A] implies Q can bewritten as an integral of the current, j. Q =

∮dz2πi j(z) The OPE of T with some

primary field of weight h is given by:

T (z)A(z′) =h

(z − z′)2A(z′) +

∂A(z′)

z − z′+ . . .

Look at the variation of A:

δA = −εh∂vA− εv∂A

choose v = zm+1 j = zm+1T Q =∮j ∼ Lm

⇒ δA = iε[Lm, A]

[Q,A] = h(m+ 1)zm + Zm+1∂A = [Lm, A]

expand A in a Laurent expansion:

A = (∂z

∂z)−h

∑

Amz−m

=∑

Amz−m−h

This is the expansion for a primary field in the z coordinates. Look at thecommutator of L with A in these cordinates.

[Lm, An] = h(m+ 1)An+m − (n+m+ h)An+m

= [(h− 1)m− n]Am+n

We got an algebra from an OPE. We know the algebra, but what are the repre-sentations of the algebra.

δT = expected +c

12∂3v

[Lm, T ] = expected +c

12(m+ 1)m(m− 1)zm−2

[Lm, Ln] = expected +c

12(m3 −m)δm+n,0

= (m− n)Lm+n +c

12(m3 −m)δm+n,0.

This is the Virasoro algebra. Let us focus on the special case m = 0. Theaction of the Hamiltonian L0:

[L0, An] = −nAn


even if An = Ln. If

L0|ψ〉 = E|ψ〉, |ψ′〉 = Ln|ψ〉

L0|ψ′〉 = [L0, Ln]|ψ〉 + LnL0|ψ〉= −nLn|ψ〉 +ELn|ψ〉= (E − n)Ln|ψ〉= (E − n)|ψ′〉

For n > 0 ⇒ Ln|ψ〉 has lower energy then |ψ〉. Is the spectrum of L0 un-bounded? This needs to be fixed.Let us look at the n = −1, 0, 1 Virasoro subalgebra. Is this analogous to theraising and lowering operators for angular momentum.

[L0, L1] = −L1, [L0, L−1] = L−1, [L1, L−1] = 2L0

We notice that this closed algebra is independent of c, therefore for every CFTwe should get this subalgebra.This is a Lie algebra SL(2,R) and is not compact.Look at Quantum Mechanics:

[L0, L+] = L+, [L0, L−] = −L−, [L+, L−] = 2L0

This is close to the algebra above, but not the same.The difference is actuallyvery important!

2.8 Mode Expansions

Free Scalars

Now that we know how to transform to these z coordinates, we can look at thesame calculations and look for similarities.The mode expansion is given by,

XµL = xµ − i

α′

2pµ ln z + i

√

α′

2

∑

m>0

1

mαµmz

−m.

Increasing timec

c

Increasing time

2.8 Mode Expansions 45

The two-point function which has to be radially ordered is given by,

〈XµL(z)Xν

L(z′)〉 =α′

2

∑ 1

m(z′

z)ηµν |z′| < |z| → time ordering...i.e. radial ordering

=α′

2ln |1 − z′

z|ηµν .

The normal ordered product is given by,

: XX := XX =α′

2ln |z − z′| in z picture.

Now we can compare our definition for :: to switching a and a† around.

: XµL(z)Xν

L(z′) : = : (X+L +X−

L )µ(X+L +X−

L )ν :

= XµL(z)Xν

L(z′) + [X(−)(z′), X(+)(z)]

= XµL(z)Xν

L(z′) − α′

2

∑

n

1

nηµν

z′

z

n

[xµ,−iα′

2pµ ln z] = XX +

α′

2ηµν ln |z − z′|.

From normal ordering

Xµ(z)Xν(z′) =α′

2ηµν ln(z − z′)+ : Xµ(z)Xν(z′) :,

the operator product can be writen as product = singularity + normal orderingproduct From

Lm =

∮dz

2πizm+1T (z), charge

where zm+1T (z) is a conserved current and

∂Xµ = −i√

α′

2

∞∑

m=−∞αµmz

−m−1

P µ =

√

2

α′αµ0

Xµ = xµ + pµ ln z + i∑

m6=0

1

mαµmz

−m

T (z) =1

α′ : ∂Xµ∂Xµ :

=1

2

∑

n1,n2

z−n1−1z−n2−1 : αµn1αn2µ :

Lm =1

2

∑

n

: αµm−nαnµ :


The Hamiltonian (L0) is given by:

L0 =α′

4p2 +

∞∑

n=1

αµ−nαnµ

What about the equal-time commutator? That means |z| = |z ′|. To calcu-late Xµ(z)Xν(z′), we need to approach |z| → |z′| from |z| > |z′|. To calculateXν(z)Xµ(z′), we need to approach |z| → |z′| from |z| < |z′|. Thus, the com-mutation relations are

[Xµ(σ),Πν (σ′)] = 2πiδ(σ − σ′)

[Xµ(σ), Xν(σ′)] = 0, where X = XL(z) + XR(z)

[XµL(z), Xν

L(z′)] = XµL(z)Xν

L(z′) −XνL(z′)Xµ

L(z)

=α′

2ηµν ln(z − z′) − α′

2ηµν ln(z − z′)

=πi

2α′ηµν

d

dz(step function)

=πi

2α′ηµνδ(z − z′),

where differentiating a step function, a delta function is obtained.N.B. The commutator [XL, XL] 6= 0 means XL is not a coordinate. It is acombination of a coordinate and momentum.Another interesting example

eAeB = e[A,B]eBeA, then

: eik1X1 :: eik2X2 : = e±iπα′

2 k1·k2 : eik2X2 :: eik1X1 :,

For the special case D = 1, Xµ = α′

2 Ψ, kµ1 = ±√

2α′

= kµ2 and let O1 = e±iψ

and O2 = e±iψ, then

O1O2 = e±iπO2O1 = −O2O1, (Quantum group!)

or

O1O2,O2O1 = 0

ψ(z)ψ(z′) ∼ ln(z − z′), then

: eiψ(z) :: e−iψ(z′) : ∼ e− ln(z−z′) : eiψ(z)e−iψ(z′) :

∼ 1

z − z′: eiψ(z)e−iψ(z′) :

: eiψ(z) :: eiψ(z′) : ∼ eln(z−z′) : eiψ(z)eiψ(z′) :

∼ (z − z′) : eiψ(z)eiψ(z′) : .

Example: bcCFT

2.9 Vertex operators 47

We can write b and c in terms of ψ: b =: eiψ : and c =: eiψ :. The stress tensoris given as

Tψ =: ∂ψ∂ψ : +V ∂2ψ.

From b(z) =∑bmz

−m−λ and c(z) =∑cmz

−m−1+λ, we may calculate theOPE, b(z)c(z′) ∼ 1

z−z′ . The anticommutator between b and c is

bm, cn = δm+n,0.

bn and cn are annihilation operators for n > 0. For the zero modes, m,n = 0the anticommutator is

b0, c0 = 1

If we let |Ψ〉 be a null state of b, ie. b0|ψ〉 = 0 and c0|ψ〉 = |χ〉, then

c0|χ〉 = c0c0|ψ〉 = 0, bm, bn = cm, cn = 0

b0|χ〉 = b0c0|ψ〉 = b0, c0|ψ〉 = |ψ〉, then

〈ψ|ψ〉 = 〈χ|b0b0|χ〉 = 0

〈χ|χ〉 = 〈ψ|b0b0|ψ〉 = 0, and

〈ψ|χ〉 6= 0.

So |ψ〉 and |χ〉 are independent vacua. Hilbert space: act on |ψ〉, |χ〉 with thecreation operators b−n, c−n, n > 0. By convention we will group b0 and c0with creation and annihilation operators respectively. Then |ψ〉 is the vac-uum, |ψ〉 = |0〉.N.B. Define 〈0| = 〈χ| and 〈ψ|χ〉 = 〈χ|ψ〉 = 1.

2.9 Vertex operators

Vertex operators are one-to-one corresponding to their states, I ' |0 > for anoperator ∂Xµ and momentum operator pµ

∂Xµ(z) = −i√

α′

2

∑

m

αµmz−m−1

pµ|0〉 = 0,

pµ|0; k〉 = kµ|0; k〉, then

∂Xµ(z)|0〉 = −i√

α′

2

∞∑

m=1

αµ−mz−m−1|0〉

= −i√

α′

2(αµ−1 + αµ−2z + αµ−3z

2 + ...)|0〉 letting z → 0

∂Xµ(0) = −i√

α′

2αµ−1|0〉


z

0

∞

From |A〉 ' A(z) or |A〉 = A(0)|0〉, ∂Xµ operating at a bra is

〈0|∂Xµ(z) = −i√

α′

2〈0|

∞∑

m=1

αµ−mz−m−1, letting z → ∞

〈0|∂Xµ(∞) = −i√

α′

2〈0|αµ−1,

then the product can be consider as

〈0|︸︷︷︸

time at +∞

time order stuffed |0〉︸︷︷︸

time at −∞

From the equation of ∂Xµ, αµ1 is

αµ−1 =

∮dz

2πiz−1∂Xµ(z)︸︷︷︸

conservedcurrent

,

z → e−iz

αµ−2|0〉 also can be obtained from

∂2Xµ(z)|0〉 = −i√

α′

2(αµ−2 + 2αµ−3z + ...)|0〉, for any αµ−m

∂mXµ(0) = −i√

α′

2((m− 1)!αµ−m +O(z))|0〉, letting z → 0

αµ−m(0) ' i

√

2

α′1

(m− 1)!∂mXµ(z).

αµ−mαν−n|0〉 can be obtained from

: ∂mXµ∂νXν : |0〉 = −α′

2(m− 1)!(n− 1)!αµ−mα

ν−n + ...

︸︷︷︸

go to

0 in the infinite past|0〉

2.10 Primary fields 49

Some examples of vertex operators are

eik·X |0〉 = |0; k〉limz→0

: eik·X(z) : |0〉 = eik·X |0〉, then

: ∂Xm1∂Xm2 ...eik·X : ' α−m1α−m2 ....|0; k〉

2.10 Primary fields

A primary fieldA, A → |A〉, A(0)|0〉 = |A〉, with a state |m1 +m2 + ...; k > canbe writen as

|A〉 = α−m1µ1αµ2

−m2...|0; k〉where

A(z) =1

(n1 − 1)!

1

(n2 − 1)!... : ∂n1Xµ1∂n2Xµ2 ...eik·X : .

If we let λ = 2 for the bc theory:

b−m|ψ〉 −→ 1

(m− 2)!∂m−2b, b0|ψ〉 = 0

c−m|ψ〉 −→ 1

(m+ 1)!∂m+1c

Tbc(z) = : (∂b)c : −λ∂ : (bc) :,

where

b(z) =∑

bmz−m−λ

c(z) =∑

cmz−m+λ

The Virasoro operator in this case is

Lbcm =

∮dz

2πizm+1Tbc

=∑

n

−(n+ λ)bncm−n + λ(m+ 1)bncn−m + aδn,0

=∑

n

(mλ− n) : bncn−m : +aδm,0

Tbc(z)Tbc(z′) ∼ c

2(z − z′)4+

2

(z − z′)2Tbc +

1

z − z′∂Tbc

where c = 1 − 3(2λ− 1)2, and the commutation relations for L are given by:

[Lbcm, L

bcn

]= (n−m)Lbcn+m +

c

12(m3 − n)δn+m,0


For m = 1 and n = −1, the commutator is

[Lbc1 , L

bc−1

]|Ψ〉 = 2Lbc0 |Ψ〉

L1L−1|Ψ〉 − L−1L1|Ψ〉 = λb0c1(1 − λ)b−1c0|Ψ〉= λ(1 − λ)c1b−1|Ψ〉= λ(1 − λ)c1, b−1|Ψ〉= λ(1 − λ)|Ψ〉

2.11 Operator product expansion

Consider the commutator betweenLm and A

[Lm, A] = zm+1∂A+ h(m+ 1)zmA, where

A(z) =∑

Anz−n−h

For m = 0,

[L0, A] = z∂A+ hA

[L0, An] = −nAnas z → 0, [L0, A(0)] = hA(0)

L0|A〉 = [L0, A(0)] |0〉 +A(0)L0|0〉, L0|0〉 = 0,

= hA(0)|0〉 = h|A〉as z → 0, [Lm, A] = form > 0, then

Lm|A〉 = 0,m > 0,

where L0 is bounded from below.

2.12 Unitary CFTs

Define an inner product 〈...|...〉 such that L†m = L−m. We want an inner prod-

uct of positive norm.〈ψ|Lm|χ〉 = 〈L†

mψ|χ〉Example: For Xµ: 〈0; k|0; k′〉 = 2πδ(k − k′).[αµm, α

νn] = ηµνδm+n,0, αµ†m = αµ−m.

‖ αµ−1|0; k〉 ‖2< 0 for µ = 0 and ‖ αµ−1|0; k〉 ‖2> 0 for µ = i 6= 0.This can be corrected by letting Φ → X ′. This conformal field theory is uni-tary and its action is

S =1

2πα′

∫

d2z∂φ∂φ.

Theorem: For highest weightA, h ≥ 0,.Proof: 2h〈A|A〉 = 〈A| [L1, L−1] |A〉 =‖ L−1|A〉 ‖2≥ 0.

2.12 Unitary CFTs 51

Corollary: Any eigenstate of L0 has h ≥ 0 (it has energy ≥ h h.w.s. energy).Theorem: c > 0: c

12 (m3 − m) = 〈A| [Lm, L−m] |A〉 − 2m〈A|L0|A〉 ≥ 0 forL0|A〉 = 0.And we have gotten what we wanted; a positive norm for the highest weightstate.


UNIT 3

BRST Quantization

BRS&T: Becchi-Ronet-Stora & Tyutin.

3.1 Point particle

Recall

S =1

2

∫

dτ

(1

ηXµXµ − ηm2

)

=

∫

dτ (pµxµ − ηχ)

where

χ =1

2(pµpµ +m2).

The constraint χ = 0 generates the transformation

δXµ = εXµ, χ = εpµ, δpµ = 0.

Quantization: |~k〉, H = p0 =√

~p2 +m2, H |~k〉 =√k2 +m2|~k〉. where

ω =√

k2 +m2

is the dispersion relation.

Sexier approach: Let εbe anticommuting, say ε→ εc, where εc are commutingand anticommuting respectively. Promote c to coordinate status. Let b be itsconjugate momentum, so

Sbc =

∫

dτbc.

The action S′ = S + Sbc =∫dτ(pµx

µ + bc) is invariant under

δBXµ = εcpmu, δBp

µ = 0

δBb = −ε(χ−m2)

54 UNIT 3: BRST Quantization

Check:

δS′ =

∫

dτ[ε(cpµ)pµ − ε(χ−m2)c

]=

∫

dτεd

dτ(1

2c(pµpµ +m2)) = 0.

Generated by QB = cχ Nilpotent:

Q2B =

1

2QB, QB = 0

Quantization: b, c = 1, [Pµ, Xν ] = −iηµν . The b, c theory is much like b0, c0

in strings. States |ψ〉, |χ〉

b0|ψ〉 = 0, c0|χ〉, c0|ψ〉 = |χ〉, b0|χ〉 = |ψ〉.

Include momentum, |k〉 ⊗ |ψ〉 = |k, ψ〉 (pµ|k〉 = kµ|k〉)

QB|k, ψ〉 =1

2(k2 +m2)|k, χ〉, QB |k, χ〉 = 0,

where |k, ψ〉 is closed for k2 +m2 = 0 and |k, χ〉 is closed. Set of closed states|k, ψ〉, k2 +m2 = 0 is a set of physical states, in agreement with the analysisabove (gauge fixed).

3.2 Strings

Recall S = 12πα′

∫d2z∂Xµ∂Xµ. Constraint T = 1

α′∂Xµ∂Xµ = 0 (and similarly

for T ) generates conformal transformations

δXµ = εv∂Xµ

Now make v anticommuting, v → c, then b conjugate momentum bc systemwe already studied.

Sbc =1

2π

∫

d2zb∂c.

Let us guess that S′ = S + Sbc is invariant under the transformations

δBXµ = iσc∂Xµ, δBb = iεT...

This does not quite work. We need T → T +Tbc and then δBc 6= 0. The correcttransformations are

δBXµ = iσc∂Xµ, δBb = iε(T + Tbc), δBc = iεc∂c.

Then δS′ = 0. The corresponding Noether current is

jB = cT +1

2: cTbc : +

3

2∂2c.

Require: jB have weight h = 1, so the chargeQB =∮

dz2πijB has h = 0 (confor-

mally invariant scalar operator). If we look at the cT part in jB we see hT = 2therefore hc = −1. Therefore the bc system must have λ = 2 and hb = 2.

3.3 Mode Expansion 55

3.3 Mode Expansion

QB =∑

n

cnL−n+1

2

∑

m,n

(m−n) : cmcnb−m−n : −c0 =∑

n

: cn

(

L−n +1

2Lbc−n − δn0

)

:

where the minus sign in front of c0 comes from l(1−λ)2 = −1 and is in disagree-

ment between mode and conformal normal ordering.

3.4 Nilpotency

Q2B =

1

2QB, QB =

1

2

∑([LTOTm , LTOTn ] − (m− n)LTOTm+n

)c−mc−n

where LTOTm = Lm + Lbcm − δm,0. The right hand side is 112 (D + cbc)(m

3 −m)where cbc = 1 − 3(2λ− 1)2 = −26. Our BRST charge is then

Q2B ∼ 1

12(D − 26) = 0

if and only if D = 26. Conversely, suppose Q2B = 0. DefineLTOTm = QB , bm.

Then

[LTOTm , LTOTn ] = [LTOTm , QB, bm] = Qb, [Lm, bn]= QB, (m− n)bm+n = (m− n)Lm+n.

Physical states are annihilated by QB, (QB |phys〉 = 0). Note that |phys〉 +QB|ψ〉 is also physical. they represent the same system (likeAµ and Aµ + ∂µψin QED). Therefore, |phys〉=equivalence class |A〉+QB |ψ〉. Cohomology of theconformal groupNote:

(〈A| + 〈ψ|QB)(|B〉 +QB|ψ〉) = 〈A|B〉

for physical |A〉, |B〉 (QB |A〉 = QB |B〉 = 0) where we assume Q†B = QB. In

particular, 〈ψ|QB |B〉 = 0 therefore 〈ψ|QB = 0.

3.5 A note on BRST cohomology

Given a group with symmetry group G generated by the algebra

[Li, Lj ] = ifkijLk.

Introduce ghosts bi, ci such that

ci, bj = δij , ci, cj = bi, bj = 0.


Define the BRST charge

QB = ciLi −i

2fkijc

icjbk

= ci(

Li +1

2Lbci

)

, Lbci = −ifkijcjbk

Q2B =

1

2QB, QB = icicjfkijLk − ifklmc

lcmci, bkLi −1

2fkijf

mkl c

icjclbm = 0.

due to the Jacobi identity

[[Li, Lj ], Lk] + [[lj , Lk], Li] + [[Lk, Li], Lj ] = 0

ifmij [Lm, Lk] + ifmjk[Lm, Li] + ikmki [Lm, Lj ] = 0

−fmij f lmkLl − fmjkflmiLl − fmkif

lmjLl = 0

For strings, [Lm, Ln] = (m− n)Lm+n and fkmn = (m− n)δk,m+n cm = c−m, so

QB = c−mLm − 1

2(m− n)c−mc−nbm+n.

3.6 BRST Cohomology for open strings

Open strings are easier than closed strings, but they are entirely similar. Weintroduce a vacuum |ψ〉 such that b0|ψ〉 = 0 and |χ〉 = c0|ψ〉. Then 〈ψ|ψ〉 = 0,but 〈χ|ψ〉 6= 0 so, we define the inner product by 〈ψ|c0|ψ〉.Let |ψ〉 ⊗ |k〉 = |ψ; k〉, 〈k|k′〉 = (2π)DδD(k − k′). Physical states will be con-structed from |ψ〉, so b0|phys〉 = 0. Then L0|phys〉 = QB , b0|phys〉 = 0.

L0 = α′p2 +∑

n

nb−ncn +∑

n

αµ−nαnµ − 1

H = |ψ〉, b0|ψ〉 = 0, L0|ψ〉 = 0QB |ψ〉 = |Z〉, b0|Z〉 = L0|ψ〉 = 0, L0|Z〉 = [L0, QB]|ψ〉 = 0.

Therefore QB : H → H. In H, |k〉 is specified by |~k〉, because k0 is given in

terms of ~k through L0 = 0. Therefore we can define the inner product

〈~k|~k′〉 = 2k0(2π)D−1δD−1(~k − ~k′).

which is a Lorentz invariant defintion.Example: |ψ;~k〉

L0|ψ;~k〉 = (α′p2 − 1)|ψ;~k〉 = 0 ⇒ k2 =1

α′ .

QB |ψ;~k〉 = 0, |ψ;~k〉 6= QB |Z〉

3.6 BRST Cohomology for open strings 57

Therefore |ψ;~k〉 are all the cohomology classes. Same as in the light-conequantization.

Example: |ψ〉 =(

Aµ(~k)αµ−1 + β(~k)b−1 + γ(~k)c−1

)

|ψ;~k〉

〈ψ|ψ〉 = (A∗µA

µ∗ + β∗γ + γ∗β)〈ψ;~k|ψ;~k〉

There are 26 positive-norm states: Ai, β = γ (αi−1|ψ;~k〉, (b−1 + c−1)|ψ;~k〉), 2

negative-norm states: A0, β = −γ, (α0−1|ψ;~k〉, (b−1 − c−1)|ψ;~k〉).

QB |ψ〉 = 0 ⇒ (c−1k · α1 + c1k · α−1)|ψ〉 = 0 ⇒ (kµAµc−1 + βkµα

µ−1)|ψ;~k〉 = 0

Therefore k · A = 0 and β = 0. This gets rid of negative-norm states k0A0 6= 0for all k0 6= 0 and β = γ = 0 is the other negative-norm state. 26 states remain:2 have zero-norm:

kµαµ−1|ψ;~k〉, c−1|ψ;~k〉.

They are orthogonal to all physical states 〈...|ψ〉 = 0.

c−1|ψ;~k〉 is exact.

Proof: Let |Z〉 = Aµαµ−1|ψ;~k〉, k · A 6= 0. Then QB |Z〉 = k · Ac−1|ψ;~k〉. There-

fore c−1|ψ;~k〉 = 1k·AQB |Z〉.

k · α−1|ψ;~k〉 is exact.

Proof: Let |Z〉 = b−1|ψ;~k〉, then QB |Z〉 = k · α−1|ψ;~k〉. In each BRST coho-molgy class there is a gauege equivalence Aµ = Aµ + αkµ.

No-Ghost Theorem

In the light-cone gauge, we considered the space H⊥ is a Hilbert space. No-Ghost Theorem: BRST cohomology is isomorphic to H⊥.Definition: α±

m = 1√2(α0m ± α1

m).

Therefore[α+m, α

−n ] = −mδm+n,0, [α+

m, α+n ] = [α−

m, α−n ] = 0.

Q1 is the part of Q proportional to the α−−m oscillators.

Q1 = −√

2α′k+∑

m

a+−mcm.

Lm = α+0 α

−m +

∑

n

α+nα

−m−n +

1

2

∑

n

αinαim−n, Q =

∑

m

L−mcm + ...

Q21 = 0.

Definition: R = 1√2α′k+

∑

m a+−mbm

S = Q1, R =∑

m

(mb−mcm +mc−mbm − α+−mα

−m − α−

−mα+m)


N.B.:[Q1, S] = [Q, Q,R] = Q1RQ1 −Q1RQ1 = 0

Theorem: |ψ〉 ∈ Large(S) ∪Kernel(Q1) ⇒ |ψ〉 is exact.Proof: |ψ〉 = S|Z〉, Q1|ψ〉 = Q1|Z〉 = 0. Therefore, |ψ〉 = Q,R|Z〉 =Q1R|Z〉.Corollary: Cohomology of |ψ〉 non-trivial only if |ψ〉 ∈ ker(S), i.e., S|ψ〉 = 0.Now |ψ〉 ∈ H⊥ ⇒ Q1|ψ〉 = 0 (trivial from the definition of Q1). From the defi-nition ofS in terms of the oscillators, S|ψ〉 = 0 only if |ψ〉 has noα±

−m, b−m, c−mexcitations, therefore |ψ〉 ∈ H⊥. Therefore theQ1 cohomology is H⊥.The no-ghost theorem for Q1: Let us go back toQB .Define U = QB − Q1, R = QB , R − S, therefore QB, R = S + U . Map|ψ〉 ∈ ker(S) 7→ |Z〉 = (1−S−1U+S−1US−1U+...)|ψ〉. S−1 makes sense for allU |ψ〉 contains α±

−m excitations therefore, U |ψ〉 /∈ ker(S) Clearly, (S + U)|ψ〉 =S|ψ〉 = 0. This establishes an isomorphism

ker(S) ∼= ker(S + U).

We can show the cohomology of QB ∼= ker(S + U) just like we showed co-homology of Q1

∼= ker(S). Therefore, coh(QB) ∼= ker(S + U) ∼= ker(S) ∼=coh(Q1) ∼= H⊥. Q.E.D.Inner products: 〈Z1|Z2〉 = 〈ψ1|ψ2ψ〉 (positive definite).

UNIT 4

Tree-level Amplitudes

4.1 String Interactions

In particle theory, we need to introduce a multi-particle space (Fock space)where creation and annhilation are possible. In string theory, the tools wehave developed for one string are sufficient for the description of multi-stringstates and interactions! The entire quantum theory of strings is based onthese tools!

Example of particle interactions

k

p1

p2

−e

+e +e

−e

p4

p3

γ

k’

ep

4

p3

p2

+e

−e−e

p1

γ

+

1/k2: inverse of the Klein-Gordon operator φ = 0, −1 ∼ 1/k2 There is apole at k2 ∼ 0, e.g., β-decay

60 UNIT 4: Tree-level Amplitudes

n

W

e−

ν

p+

Amplitude∼ 1k2−m2

W

, pole at k2 = m2W , resonance.

Strings

The interaction consists of strings joining and splitting. Where do they join?This is a stupid question. It depends on the time slicing. Therefore this is afuzzy interaction. Moreover, the shape (geometry) of the surface is not im-portant, only the topology is important. There is one diagram for all tree dia-grams.

4.1 String Interactions 61

Example

Interaction point

⟩

|A⟩Incoming string

Outgoing string

(arbitrary)

+∞

−∞

t

|B

There is an arbitrary interaction point. The amplitude is constructed by join-ing two semi-infinite cylinders. Map the cylinders to a plane:

C|A⟩

|B⟩z

C

cylinder → C⋃∞ = S2 (sphere). This is done through stereographic pro-

jection (sphere=fat cylinder).


P’

N

S

O

P

Amplitude: sphere with states (operator insertions) at North and South poles.Notice the equivalence of the two poles (clinder z → 1

z ).

Open Strings

Make a strip by cutting the cylinder in half along the axis.

z

We then map the strip to the upper-half plane which can then be mapped tothe unit circle via the mapping z → z−i

z+i .

80

Each string is a semi-infinite cylinder (or strip), which is mapped to a disk.When we put two on a sphere, they were simply represented by insertion ofA(z) at z = 0, z = ∞.Guess: For scattering of N strings we can do the same, i.e., on a sphere selectpoints z1, z2, ..., zN and insert operators Ai(zi). Then the amplitude is

A ∼ 〈0|A1(z1)A2(z2)...AN (zN )|0〉.


NowA(0) is equivalent to∮

Cdz2πiA(z).

For conformal invariance, we require that all Ai have dimension hi = 1 sothat

∫dzAi(z) have zero dimension (conformally). Then we should define

Amp ∼∫

dz1...dzN 〈0|A1(z1)...|0〉.

In fact, the measure should read∫d2z1...d

2zN , but we will not be writing thez piece explicitly. The proper dimension of Ai(z, z) should be hi = 1, hi = 1.In general,

A(z) ∼: ∂m1X∂m2X...eik·X :,

where h = m1 +m2 + ...α′k2 = 1. We shall work with the simplest caseA(z) =eik·X , k2 = 1

α′. The rest is similar.

Complication: The amplitude is conformally invariant: z → z + εv(z) wherev(z) is analytic. v(z) should be analytic everywhere in C ∪ ∞. We need tocheck that the transformation is analytic at infinity. So let z 7→ 1

z = z′.

δz′ = − 1

z2δz = −ε 1

z2v(z) = −εz′2v

(1

z′

)

.

therefore v(z) = a+bz+cz2 so that z′2v(

1z′

)is analytic. This is a six-parameter

family of transformations. It includes SO(3) (rotation group). Special Cases:• z 7→ z + εa generated by L−1. Recall [Lm, A] = zm+1∂A + h(m + 1)zmAwhere h = 1 for BRST invariance. So [L−1, A] = ∂A − 1

zA i.e., L−1 generatestranslations in z.Finite transformation: z 7→ z + a,

A(z) → eaL−1A(z)e−aL−1 = A(z + a).

•z 7→ z + εbz = (1 + εb)z generated by L0.

[L0, A] = z∂A+A.

A(z) → ebL0A(z)e−bL0 = A(ebz).

Finite transformation: z → ebz.•z 7→ z + εcz2 generated by L1.

[L1, A] = z2∂A+ zA.

Finite transformations: z → z1−cz = z′

A(z) → ecL1A(z)e−cL1 = A

(z

1 − cz

)

.

Combination of all three: z 7→ az+bcz+d , ad − bc = 1 defines the group SL(2,C)

whose algebra is

[L1, L−1] = 2L0, [L1, L0] = L1, [L−1, L0] = −L−1.


This is a closed algebra (no constant term) and is common in all conformalfield theories.How come a matrix entered acting on a number z? Answer: Consider thevector

(z1, z2)

(a bc d

)(z1z2

)

=

(az1 + dz2cz1 + dz2

)

.

Let z = z1/z2. Then

z 7→ az1 + bz2cz1 + dz2

=az + b

cz + d.

For open strings: the Real axis is a boundary, and the group of symmetriesbecomes SL(2,R), a, b, c, d ∈ R. Then under z → az+b

cz+d , ∂ is invariant. Theupper-half plane maps to itself.Amplitude for open strings:

z z zz1 2 3 4

z

Amp ∼ 〈V (z1)V (z2)...V (zN )〉, zi ∈ R,

where the product is time ordered and thus the zis are ordered. How do weintegrate over zi? Due to SL(2,R) symmetry, we have redundency, so naiveintegral would be proportional to the volume of SL(2,R) which is infinite! Weneed to fix the gauge by choosing three points. Easiest to fix them to (0, 1,∞).This is an arbitrary choice, but all choices are equivalent by the SL(2,R) sym-metry. We will integrate over the rest of the parameters.

Example 1: Three tachyons

Consider three tachyons, Vi(z) =: eiki·X(z) : The amplitude is given by

A ∼ 〈0|V1(z1)V2(z2)V3(z3)|0〉,

where

Xµ(z) = xµ − iα′

2pµ ln |z|2 + i

√

α′

2

∑

m6=0

1

maµm(z−m + z−m

).

Since z ∈ R, Xµ reduces to

Xµ(z) = xµ − iα′pµ ln |z| + i√

2α′∑

m6=0

1

maµmz

−m.


Since we can fix three points, let us choose (z1 = ∞, z2 = 1, z3 = 0), then

V3(z3 = 0)|0〉 = |0; k3〉, 〈0|V1(z1 = ∞) = 〈0;−k1|.

The amplitude becomes

A ∼ 〈0;−k1|V2(z2 = 1)|0; k3〉 = 〈0;−k1|0; k2 + k3〉 = δD(k1 + k2 + k3).

One can derive this for arbitrary z1, z2, z3, due to the SL(2,R) symmetry.

Example 2: Two tachyons and one vector

Consider two tachyons, Vi(z) =: eiki·X(z) : and a vector,Vj(z) =: Aµ∂Xµeikj ·X(z) :,

where k2j = 0. We may act the vertex operators on the vacuum states

A ∼ 〈0;−k1|V2(1)Aµ∂Xµ|0; k3〉 ∼ 〈0;−k1|eik2·xe

α′

2 a1·k2Aµαµ−1|0; k3〉

∼√

2α′A · k2δD(k1 + k2 + k3). (4.1.1)

A is transverse to it’s momentum (A · k3 = 0) therefore, the amplitude is

A ∼√

α′

2A · (k2 − k1)δ

D(k1 + k2 + k3),

where the dot product represents the coupling of the electromagnetic poten-tial to the charged scalar. We may check the gauge invariance of the ampli-tude. Using the gauge transformation Aµ → Aµ + ωkµ3 , the amplitude be-comes

δ(A) ∼ k3 · (k2 − k1) = k22 − k2

1 = 0.

Example 3: Four tachyons

This is the first nontrivial amplitude. Due to the SL(2,R) symmetry, we mayfix three operators. Now we have an extra operator we can not fix. We mustintegrate over its parameter. After we operate vertex operators on the vacuumstates, the amplitude is given by

A ∼ 〈0;−k1| : eik2·X(1) :: eik3·X(z) : |0; k3〉.

This is a time-ordered product and we must integrate over z from [0, 1]. Theamplitude becomes

A ∼∫ 1

0

dz〈0;−k1| : eik2·X(1) :: eik3·X(z) : |0; k4〉.

Using the mode expansion of Xµ,

A ∼∫ 1

0

dz〈0;−k1|eik2·xe√

2α′ m>0 k2·αm/meik3·xei

α′

2 k3·p ln |z|e√

2α′ n>0 k3·α−n/n|0; k4〉.


Using the Hausdorff formula, eAeB = e[A,B]eBeA,

A ∼∫ 1

0

dz〈0;−k1|z2α′k3·k4e−2α′k2·k3 zm/m|0; k3 + k4〉

∼∫ 1

0

dzz2α′k3·k4(1 − z)−2α′k2·k3δD(k1 + k2 + k3 + k4)

Define the Mandelstam variables

θk

k

k

k

1

3

2

4

s = (k1 +k2)2 = (k3 +k4)

2 = −2k3 ·k4−2

α′ , t = −(k2 +k3)2, u = −(k2 +k4)

2,

and

s+ t+ u = − 4

α′ .

The amplitude expressed in terms of Mandelstam variables becomes

A ∼∫ ∞

0

dzz−α′s−2(1 − z)−α

′t−2δD(k1 + k2 + k3 + k4) ∼ B(−α′s− 1, α′t− 1),

where B is the Euler-beta function with the property

B(x, y) =Γ(y)Γ(y)

Γ(x+ y).

This is known as the Veneziano amplitude. Note, there are poles at −α′s−1 =0 and α′t− 1 = 0. Let us focus on the first pole (−α′s− 1 = 0).

A ∼ Γ(−α′s− 1) =Γ(−α′s)

α′s+ 1+ ... (Γ(x+ 1) = xΓ(x))

∼ − 1

−α′s+ 1+ ...

The pole is due to an intermediate tachyon (s = −1/α′). Unitarity requiresThis checks, since The next pole is at α′s = 0.

Γ(−α′s− 1) = −Γ(−α′s)

α′s+ 1=

Γ(−α′s+ 1)

(α′s+ 1)(α′s)=

1

α′s+ ...

The amplitude becomes

A ∼ 1

α′s

Γ(−α′t− 1)

Γ(−α′t− 2)=

Γ(−α′t− 2)

α′s+ ... =

u− t

2s+ ...


where we used the condition s+ t+ u = −4/α′.Check unitarity: The amplitude is gauge invariant. Summing over the polar-izations

∑εµεν = ηµν gives the amplitude

A ∼ α′ (k1 − k2)(k3 − k4)

2k2=u− t

2s.

All the poles in α′s : α′s = −1, 0, 1, 2, ... which are the masses of the openstring states. (α′s2 = N − 1 from L0 − 1 = 0). Curious Result: same structureof poles we obtain for α′t, since the amplitude is symmetric in s and t. Thiswould also be true of a field theory amplitude.Alternate derivation of the poles: It is instructive to find the poles without per-forming the integral for two reasons. (a) We can not always do the integral. (b)We can see what type of world-sheet contributes to the pole (physical picturefor an effective field theory).Let z → 0

A ∼∫

0

dzz−α′s−2 + ... =

z−α′s=1

−α′s− 1

∣∣∣∣∣0

+ ... = − 1

α′s+ 1+ analytic.

Taylor expansion:

A ∼∫

0

z−α′s−2(1 − z)−α

′t−2 =

∫

0

dzz−α′s−2(1 + (α′t+ 2)z + ...),

where the first and second terms in the expansion represent the α′s = −1 andα′s = 0 poles respectively. The other poles are acquired through higher orderterms in the expansion. Poles in α′t are obtained from z → 1.

3 k 4

0 81z

k k k1 2

There is no reason to restrict∫dz to

∫ 1

0 dz. We would like to extend the integral

to∫∞∞ dz. The integral becomes

∫ 0

−∞ +∫ 1

0+∫∞1

.∫ 0

−∞: ordering (k1 + k2 + k3 + k4) which is∫ 1

0 with k2 ↔ k1. The effect is

switching t and u. This can be seen through the transformation z 7→ 1 − 1z

which maps (0, 1) 7→ (−∞, 0). Therefore, if∫ 1

0= I(s, t) then

∫ 0

−∞ = I(t, u).

Similarly,∫∞1 = I(s, u). Therefore, the integral becomes

∫ ∞

−∞= I(s, t) + I(s, u) + I(t, u).

Now the amplitude is completely symmetric in s, t, u.


BRST invariance

If V (z) has weight h = 1, then∫dzV (z) has weight h = 0. It is BRST invariant.

Let us check this.

[Q, V (z)] =∑

c−n[Ln, V (z)] =∑

c−n(zn+1∂V (z) + (n+ 1)znV (z))

= c(z)∂V (z) + ∂c(z)V (z) = ∂(c(z)V (z))

Therefore

[Q,

∫

V ] =

∫

∂(c(z)V (z)) = 0.

What happens with the three V s that we fixed? To turn them into h = 0 oper-ators, we multiply them by c(z). Then c(z)V (z) has the weight h = 0.

Q, cV = Q, cV − c[Q, V ] = c∂cV − cc∂V − c∂cV = 0.

Now in the amplitude, we have three c(z)s, zi = 0, 1,∞. The amplitude mustbe defined with respect to the SL(2,R) invariant vacuum. Recall:

b0|ψ〉 = 0, |χ〉 = c0|ψ〉.

Lbcm =∑

n

(2m− n) : bncm−n : −δm,0

So,

Lbc0 =∑

n

n : b−ncn : −1, Lbc1 =∑

n

(2−n) : bnc−n :, Lbc−1 =∑

n

(−2−n) : bnc−n−1 : .

The operators act on the states

Lbc0 |ψ〉 = −|ψ〉, Lbc1 |ψ〉 = 0, Lbc−1|ψ〉 = b−1|χ〉,

So |psi〉 is not invariant. Let |0〉 = b−1|ψ〉.

[Lbc0 , b−1] = b−1, [Lbc1 , b−1] = 2b0, [Lbc−1, b−1] = 0.

Therefore,

Lbc0 |0〉 = b−1|ψ〉−b−1|ψ〉 = 0, Lbc−1|0〉 = b−1b−1|ψ〉 = 0, Lbc−1|0〉 = b−1b−1|χ〉 = 0.

So, |0〉 is SL(2,R) invariant.The ghost contribution is

〈0|c(∞)c(1)c(0)|0〉, c(z) =∑

n

cnz−n+1.

c(0)|0〉 = c1|0〉 = |ψ〉, 〈0|c(∞) = 〈ψ|, ψ|c(1)|ψ〉 = 〈ψ|c0|ψ〉 = 1.

4.2 A Short Course in Scattering Theory 69

High Energy

θk

k

k

k

1

3

2

4

kµ1 = (E/2, ~p), kµ2 = (E/2,−~p), kµ3 = (−E/2,−~p′), kµ4 = (−E/2, ~p′).

where(E2

)2 − ~p2 = m2, |~p′| = p. The Mandelstam variables become

s = −(k1+k2)2 = E2, t = −(k1+k3)

2 = (4m2−E2) sin2 θ

2, u = −(k1+k4)

2 = (4m2−E2) cos2θ

2.

The high energy limit is equivalent to the small angle limit, where s→ 0 and tis fixed. The gamma function is approximated by

Γ(x) ∼ xxe−x√

2π

x.

The amplitude is

A ≈ Γ(−α′s− 1)Γ(−α′t− 1)

Γ(−α′s− α′t− 2≈ s−α

′s−1

s−α′s−α′t−2eα

′t+1Γ(−α′t−1) ∼ sα′t+1Γ(−α′t−1).

This is the Regge behavior. At the poles α′t − 1 ∼ −n, the amplitude goes asA ∼ sn which is the exchange of a particle of spin n.For a fixed angle,θ=fixed: s, t→ ∞, s/t = fixed. The amplitude becomes

A ∼ s−α′s−1t−α

′t−1

(s+ t)−α′s−α′t∼ s−α

′st−α′t

uα′u∼ e−α

′(s ln s+t ln t+u lnu)

≈ e−α′(s ln(s/s)+t ln(t/s)+u ln(u/s))

≈ e−α′s( t

sln t

s+ u

sln u

s

≈ e−α′s(− sin2 θ

2 ln sin2 θ2−cos2 θ

2 ln cos2 θ2 )

≈ e−Cs, C > 0.

unlike in field theory, where the amplitude goes as A ∼ s−n. Therefore theunderlying smooth extended object of size

√α′.

4.2 A Short Course in Scattering Theory

We define the 〈in| state in the real infinite past (t→ −∞), and the |out〉 state inthe infinite future (t → ∞). These states are both described by free particles.There is an isomorphism

|in〉 = S|out〉, S = limt→∞

eiHt/~..


To conserve probabilities, S must be unitary, S†S = 1 (c.f. unitarity of evolu-tion operator, U = eiHt/~). The transition probability (S = I + iT ) is

|〈i−∞|f∞〉|2 = |〈i|T |f〉|2,

where |i > and |f〉 represent states in the same Hilbert space. We will discardthe I because it represents |i〉 → |i〉 (forward scattering i.e., along the beam:undetectable).Unitarity

S†S = I = I + i(T − T †) + T †T.

Therefore

〈i|T |f〉 − 〈i|T †|f〉∗ = i〈i|T †T |f〉.

Insert complete sets of physical states

〈i|T |f〉 − 〈i|T †|f〉∗ = i∑

n

〈i|T †|n〉〈n|T |f〉,

2Im < i|T |f〉 =∑

n

< i|T |n〉〈f |T |n〉∗. (4.2.1)

Viewed as a function of s, 〈i|T |f〉 has poles in s. Away from the pole, 〈i|T |f〉 isreal, so the left hand side vanishes.Near the pole, we obtain a behavior ∼ 1

s+m2 (pole at s = −m2). to find theimaginary part, first regulate the amplitude

1

s+m2→ 1

s+m2 + iε

for small ε. Then

Im1

s+m2→ Im

1

s+m2 + iε=

−ε(s+m2)2 + ε2

= −πδ(s+m2).

Therefore, for

INSERT FIGURE HERE

the imaginary part is

INSERT FIGURE HERE

This is in agreement with unitarity.

4.3 N-point open-string tree amplitudes

Amp ∼ 〈: eik1·X(z1) : ... : ek2·X(zn) :〉 = A(z1, ..., zn)

4.3 N-point open-string tree amplitudes 71

Consider∂1A(z1, ..., zn) ∼ 〈∂z1 : eik1·X(z1) : ... : ek2·X(zn) :〉

To evaluate this, consider the OPE

ik · ∂X(z) : eik1·X(z1) :=α′k2

1

2(z − z1)eik1·X(z1) : +∂1 : eik1·X(z1) : +...

So, first replace ∂1 : eik1·X(z1) : by ∂X(z) : eik1·X(z1) : in Amp and define

fµ(z) = 〈∂Xµ(z) : eik1·X(z1) : ... : eikn·X(zn) :〉

The singularity structure of fµ(z) can be deduced from OPEs

z

1 z3z2 zn...

C

z

∂X(z) : eik1·X(z1) := − iα′kµ12(z − z1)

eik1·X(z1) : +...

Therefore,

fµ(z) = − iα′

2A(z)

n∑

i=1

kµ1z − zi

+ ...

Behavior at z → ∞ : z′ = 1z

∂Xµ =∂z′

∂z∂′Xµ = − 1

z2∂′Xµ

which implies

fµ(z) = − 1

z2〈∂′Xµ + ...〉.

Therefore, as z → ∞, fµ(z) ∼ 1z2 → 0 (〈∂′Xµ : ... :〉 analytic at ∞) Therefore

the holomorphic piece vanishes and

fµ(z) = − iα′

2A

n∑

i=1

kµiz − zi

.

Now define a contour C surrounding all zi’s. There are two ways to evaluate

the contour integral,∮

dz2πif(z). Cauchy ⇒

∮dz2πif

µ(z) = − iα′

2 A∑ni=1

kµi

z−zi, or

in the transformed coordinate z′=1z , C encircles z′ = 0 where fµ(z) is analytic.

Therefore∮

dz

2πifµ(z) = 0 ⇒

n∑

i=1

kµi = 0


The momentum is conserved.Now consider ik · f and compare with the OPE

ik · ∂x(z) : eik·X (z1) :=α′k2

1

2(z − z1): eik1·X(z1) : +∂1 : eik1·X(z1) : +...

which implies

ik · f =α′

2

k21

z − z1A+

α′

2

∑

i6=1

k1 · kiz − zi

A

Therefore

∂1A =α′

2

k21

z − z1A+

α′

2

∑

i6=1

k1 · kiz − zi

A.

Therefore

∂1 lnA =α′

2

∑

i6=1

k1 · kiz1 − zi

A.

Repeating for other points,

∂j lnA =α′

2

∑

i6=j

kj · kizi− zj

A.

By integrating we obtain lnA =∑

i<j ln |zi − zj |ki·kj + const where we addedthe z piece. Therefore,

A ∝∏

i<j

|zi − zj |α′ki·kj .

For open strings, α′ → 2α′, so

A ∝∏

i<j

|zi − zj |2α′ki·kj .

SL(2, R) Invariance

z → z′ =az + b

cz + d, czz′ + dz′ = az + b→ z =

dz′ − b

a− cz′, ad− bc = 1.

Therefore,

zi − zj =dz′i − b

a− cz′i− dzj − b

a− cz′j=

z′i − z′j(a− cz′i)(a− cz′j)

. (4.3.1)

Therefore,

A ∝∏

i<j

|zi−zj |2α′ki·kj =

∏

i<j

|z′i−z′j |2α′ki·kj

∏

(a−cz′i)2α′k2

i , k2i =

1

α′ . (4.3.2)

4.4 Closed Strings 73

dzi =dzi

(a− cz′i)2.

If we let zj → zi in (4.3.1), we find that the amplitude is invariant underSL(2, R) transformations. The measure is given by

∏

dzi =∏

dz′i∏

(a− cz′i)−2,

however, the last factor cancels with the overall factor in (4.3.2).

4.4 Closed Strings

For open strings we found four tachyons,

Aopen ∼∫ ∞

−∞dz z2α′k3·k4(1 − z)2α

′k2·k3δD(k1 + k2 + k3 + k4)

=

∫ 0

−∞+

∫ 1

0

+

∫ ∞

1

where

∫ 1

0

= I(s, t) =

∫ 1

0

dz z−α′s−2(1 − z)−a

′t−2δD(k1 + k2 + k3 + k4)

∫ 0

−∞= I(t, u),

∫ ∞

1

= I(s, u), z ∈ R.

For closed strings, z is the entire C and we need to multiply the holomorphicand anti-holomorphic pieces, so

Aclosed ∼∫

d2z |z|−α′s/2−4|1 − z|−α′t/2−4.

Note: s→ s/4 is due to the different expansion of theXµs. The tachyon massis m2 = − 4

α′, whereas for the open string it is, m2 = − 1

α′.

To calculate the amplitude for the closed string, treat z and z as independentvariables and deform the contour of integration until it coincides with the realaxis. Then z, z ∈ R. We must take care with the branch cuts.There are three cases.(i) z < 0: Contour for z:

1

z

0


C has branch cuts on the same side and therefore contributes nothing.

(ii) z > 1:

There is no contribution for the same reason as in (i).

(iii) 0 < z < 1:

C

z

10

Aclosed ∼∮

dz z−α′s/4−2(1 − z)−α

′t/4−2 ×∫ 1

0

dz z−α′s/4−2(1 − z)−α

′t/4−2

Contribution from the upper side of C is

∫ ∞

1

dη|η|−α′s/4−2e−iπ(α′t/4+2)|1 − η|−α′t/4−2 × I(s/4, t/4).

The lower side gives

∫ ∞

1

dη |η|−α′s/4−2e+iπ(α′t/4+2)|1 − η|−α′t/4−2 × I(s/4, t/4).

Therefore the amplitude for the closed string is

Aclosed ∼ sinπα′t

4I(t/4, u/4)I(s/4, t/4).

This can be cast in a symmmetric form by using the transformation proper-ties of the Gamma function

Γ(x)Γ(1 − x) =π

sin(πx), for

−α′t

4− 1

So, since s+ t+ u = 4m2 = −16/α′

Γ(−α′t/4− 1)Γ(2 + α′t/4) =π

sin(α′tπ/4).

Therefore the amplitude is given by

Aclosed ∼ πΓ(−α′s/4− 1)Γ(−α′t/4− 1)Γ(−α′u/4− 1)

Γ(−α′s/4− α′t/4− 1)Γ(−α′t/4− α′u/4− 1)Γ(−α′u/4− α′s/4− 1).

4.5 Moduli 75

4.5 Moduli

Build closed-string four-point amplitude as follows. In the z-plane, drill holes.This will represent the diagram on the left with amputated legs. Now attachthe legs by telescopically collapsing each semi-infinite tube to a disc:

′z

Next, patch the discs on the z-plane. This produces a sphere with four punc-tures. There will be regions of overlap where z ′ = f(z).

of disk

∂ of hole in z−plane

z

∂

By conformal transformations, I can fix three points (due to SL(2,C) symme-try). The fourth point cannot be fixed. Call it z. Punctured spheres withtwo different z’s, are not related by a conformal transformation. There areinequivalent surfaces.They are parametrized by two parameters, z1 and z1 ∈ C. These parametersare called moduli and their space, moduli space (although it should be calledmodulus space) (c.f. vector space). They are also called Teichmuller parame-ters. They label conformally inequivalent surfaces. To calculate amplitudes,we need to integrate over the moduli.


E.g., the four-point amplitude, 〈V1(∞)V2(1)V3(z1)V4(0)〉need to integrate overz1 →

∫d2z1. In general, N-point amplitudes integrate

∫d2z1...d

2zN−3 at z =∞, 1, 0 we specified V ∼ cc : eik·X :. We can do the same for the unfixed V ’sto put them all on equal footing.Thus, let Vi = cc : eiki·X :, ∀i. Since we introduced an extra c, c, we need tocompensate for it with a b, b insertion.To do this work as follows. Shift z1 → z1 + δz1. This is implemented in thez′-plane by a coordinate transformation

z′ → z′ + δz1vz(z′, z′).

where vz is of course not conformal (depends on z as well as z). Introduce theBeltrami differential.

ψ = ∂zvz

There is a similar differential for the complex conjugate

ψ = ∂zvz

If vz represents a conformal transformation, then ψ, ψ = 0. Thus ψ encodesinformation about conformally inequivalent surfaces.We will insert 1

2π

∫d2z′(pψ + bψ)× anti-holomorphic in the amplitude. We

integrate over the patch that we will use so

Amp ∼∫

d2z1〈V1V2V3V4

(1

2π

∫

d2z′(pψ + bψ) × (anti)

)

〉

Since ∂zb = 0, the integral is ∼∫d2z′

(

∂z(bvz) + ∂z(bv

z))

. Therefore it can be

written as (divergence theorem)

B1 =1

2πi

∮

C

(

dz′bvz − dz′bvz)

where C is in the overlap region of z and z′.

C

z

Explicitly,

vz =∂z′

∂z1.

4.6 BRST Invariance 77

In the overlap region, z = z′ + z1, so vz + 1 = 0, therefore vz = −1. Therefore

1

2πi

∮

C= dz′bvz = −b−1, b(z) =

∑

bnz−n−2, c =

∑

cnz−n+1.

∫

dz1b−1V3 =

∫

dz1b−1cc : eik3·X(z1) := c : eik3·X(z1)

where

b−1c =

∮

Cdz′b(z′)c(z1) = 1.

∫

dz1b−1c : eik3·X(z1) :=: eik3·X(z1) :

so the b-insertions kill cc from V3 and the amplitude is as before.

4.6 BRST Invariance

QB, B1 =1

2πi

∮

c

dz′(Tvz − dzT vz

)

Recall

T (z′)V3(z1) =h

(z′ − z1)2+

1

z′ − z1∂V3 + ..., h = 0!

Therefore QB, B1V3 ∼∮dz′∂V3 = 0 unless the moduli space has ∂ (not true

here, but argument is general and sometimes ∂ 6= 0).


UNIT 5

Loop Amplitudes

5.1 One-loop Amplitudes

For closed string amplitudes, we consider a sphere which has six isometries(parametrized by three complex numbers a, b, c ∼ SL(2,C)), so we had to putfour punctures to get a modulus (conformally inequivalent surfaces).

For loop amplitudes (containing virtual strings - quantum mechanical cor-rections, only present due to Heisenberg’s uncertainty principle) we need toconsider higher-genus Riemann surfaces (fortunately they have all been clas-sified).

, ...,

We will start with the torus. Unlike the sphere, there exist tori that are con-formally inequivalent, without any punctures. So we need to study the torusby itself first. What does it represent? A virtual string that lives in the vac-uum. Or nothing creating a pair of strings which then annihilate to producenothing again.

Are we about to study nothing? You bet! There is energy associated with noth-ing and this energy is observable if gravity is present! It is the CosmologicalConstant.

First let us study the geometry

80 UNIT 5: Loop Amplitudes

0

z

ϕ2ϕ1

λ1

λ1

λ2

λ1+λ2

z ≈ z + λ1 ≈ z + λ2 ∴ z ≈ z + nλ1 +mλ2.

Different choices of λ1, λ2 lead to conformally inequivalent surfaces.

Example: z → ζz (rescaling) ⇒ λ1 → 1ζλ1, λ2 → 1

ζλ2 However τ = λ1

λ2=invariant!

Now change λ1, λ2 thusly:

(λ′2λ′1

)

=

(a bc d

)(λ2

λ1

)

, a, b, c, d,∈ Z, ad− bc = 1

Then

z ≈ z + n′λ′1 +m′λ′2

≈ z + (n′ m′)

(a bc d

)(λ2

λ1

)

≈ z + nλ2 +mλ1

where(

nm

)

=

(a bc d

)(n′

m′

)

Therefore it is the same torus. But λ′1, λ′2 → τ ′ =

λ′

2

λ′

1= aτ+b

cτ+d .

Conclusion: τ, tau′ define the same torus. τ up to SL(2,Z) transformationsuniquely labels conformally inequivalent tori. Therefore τ is a modulus.

We need to integrate over τ . Find the integration region. SL(2,Z) is generatedby two transformations (S, T ).

T : τ ′ = τ + 1, S : τ ′ =1

τ

5.1 One-loop Amplitudes 81

1

λ2

λ′2

z

λ

0

1

λ2

z

−λ2

λ

0

In the τ-plane, T divides it into inequivalent regions (mapping one regioninto the adjacent region). So concentrate on − 1

2 ≤ Re(τ) ≤ 12 .

−½ ½

τ

The Imτ axis, τ = it, acting with S: it′ = − 1it , so t′ = 1

t mapping (1,∞) ↔(0, 1). the point i is fixed and so the entire arc τ = eiθ.

τ

C

0F

F1

C′

−1 −½ ½ 1

The region above the unit circle and in between |Re(τ)| = 12 is “irreducible”,

called the fundamental region,F0.


S maps τ = 12 + it onto τ ′ = − 1

12 +it

=− 1

2+it14+t2

, ∞ → 0, t =√

32 → − 1

2 + i√

32

So F0 → F1.

−1 −½ 1

τ

½All Fundamental regions are equivalent. Integration should be over one fun-damental region (does not matter which one).

5.2 String on a Torus

Set λ1 = 2π (without l.o.g.). If τ = it, then

INSERT FIGURE HERE

This is a cylinder (closed string propogating for time t and then coming backto where it started from). In general, τ = τ1 + iτ2, to t = τ2. τ1 representsan angle (0 ≤ τ1 ≤ 2π). The string is twisted by τ1 before it gets identifiedwith the initial string. This is the “cylinder” picture. We can map it onto the“sphere” picture as before, z = e−iz .

1

z

z ≈ z + 2π, trivial

z ≈ z + 2πτ

z ≈ ze−2πiτ , |z| ≈ |z|e2πτ2 : unit circle ≈ circle of radius e2πτ2 with twist τ1.

5.2 String on a Torus 83

Quantum Mechanics

Start with a state |n〉, evolve for “time” 2πτ , |n〉 → |n(τ)〉 = eiH(2πτ)|n〉. For theleft-movers,H = L0 − D

24 .

Transition amplitude: An = 〈n|n(τ)〉 = 〈n|ei(L0−D/24)(2πτ)|n〉 Define Z(τ) =∑

nAn(τ) =∑

n〈n|ei(L0−D/24)(2πτ)|n〉. If n is an eigenfunction of the “Hamil-tonian”, (L0 −D/24)|n〉 = En|n〉 then Z(τ) =

∑e2πτiEn = Tr (e2πτi(L0−D/24)).

We need to include the right-movers. Define

Z(τ) = Tr (e2πiτ(L0−D/24)e−2πiτ(L0−D/24)).

If q = e2πiτ , then

Z(τ) = Tr (q(L0−D/24)q(L0−D/24))

Let us calculate it. Recall ...

L0 + L0 =α′

2p2 +N + N, L0 − L0 = N − N

ThereforeZ(τ) = (qq)−D/24Tr (e−2πτ2(α

′/2p2)qN qN )

where N =∑

n αµ−nαµn, N =

∑

n αµ−naµn. For each n and µ, αµ−nαµn has

eigenvalues nNµn ∈ N0, where Nµn is the occupation number.Therefore

Tr qN qN =∏

µn

∑

N,N

(qnNµn qnNµn) =∏

µ,n

( ∞∑

N=0

qnNµn

)

∞∑

N=0

qnNµn

Each sum is a geometric series, therefore

Tr qN qN =

∞∏

n=1

(1 − qn)−2D.

To calculate Tr e−2πτ2(α′/2p2), make the space finite and Euclidean. Then pµ

has discrete eigenvalues (n/L), where L is the size of the box.

∑

n

f(n/L) = L∑

1L

f(2πn/L) = L

∫dpx2π

f(px).

Repeat for other dimensions and we get

Tr → LD∫

dDp

(2π)Df(p)

The t-componentE = p0ψip0 to make the integral finite, so

Tr e−2πτ2(α′p2/2) = iLD

∫dDp

(2π)De−2πτ2(α

′p2/2)


= iLD(∫

dp

2πe−2πτ2(α

′p2/2)

)2

= iLD(

2π√

α′τ2)−D

Putting everything together,

Z(τ) = iLD

(

2π√

α′τ2|q1/24∞∏

n=1

(1 − qn)|2)−D

.

We now need to check the modular invariance. τ → τ + 1 ⇒ q → q, obviousinvariance! τ → −1/τ is not at all obvious! In order to show the invariance wemay use the powerful machinery developed by Jacobi centuries ago, knownas the Jacobi-Theta functions, Θ.

Θ-functions

These are functions with nice modular properties.Definition:

ϑ(ν, τ) =

∞∑

n=−∞eπin

2τ+2πinν =

∞∑

n=−∞qn

2/2zn, q = e2πiτ , z2πiν

The Jacobi-Theta functions satisfy the heat equation

i

π

∂2ϑ

∂ν2+ 4

∂ϑ

∂τ= 0.

Periodicity properties:

ϑ(ν + 1, τ) = ϑ(ν, τ)

ϑ(ν + τ, τ) = e−πiτ−2πiνϑ(ν, τ) (5.2.1)

These uniquely define ϑ up to a multiplicative constant.Check: ν → ν + 1 ⇒ z → z ∴ ϑ(ν + 1, τ) = ϑ(ν, τ). ν → ν + τ ⇒ z → qz

ϑ(ν + τ, τ) =∑

qn2/2znqn = q−1/4

∑

q(n+1)2/2zn

= q−1/4z−1∑

q(n+1)2/2zn+1

= q−1/4z−1ϑ(ν, τ)

= e−πiτ−2πiνϑ(ν, τ)

Equations (5.2.1) admit a different form of solution (which must be the sameby uniqueness).

ϑ(ν, τ) =

∞∏

m=1

(1 − qm)(1 + zqm−1/2)(1 + z−1qm−1/2)


Check: ν → ν + 1 is trivial to check ν → ν + τ : z → qz therefore

∏

→∏

×1 + z−1q−1/2

1 + zq1/2= (∏

) × z−1q−1/2

as before. The normalization constants also match. Check the limit q → 0. ϑis usually called ϑ3. A general ϑ function is given by

ϑ

[ab

]

(ν, τ) =∞∑

n=−∞eπi(n+a)2τ+2πi(n+a)(ν+b) = eπia

2τ+2πia(ν+b)ϑ3(ν+aτ+b, τ)

where

ϑ3(ν, τ) = ϑ

[00

]

(ν, τ).

Interesting ϑ:

ϑ1 = −ϑ[

1/21/2

]

(ν, τ) = i∞∑

n=−∞(−)nq(n−1/2)2/2zn−1/2

= 2eiπτ/4 sinπν∞∏

m=1

(1 − qm)(1 − zqm)(1 − z−1qm)

The product forms are useful to find zeros of ϑ3.

z = −q±(m−1/2) ⇒ e2πiν = eπiτ(2m−1)+πi

Therefore

ν =1

2(τ + 1), ν + 1, ν + 2, ...

= n1 + n2τ, ν + τ, ν + 2τ, ...

Modular Transformations

T : τ → τ + 1:

ϑ3(ν, τ) =∑

eπin2τ+πin2+2πinν

=∑

eπin2τ+2πin(ν+1/2)+πin2−πin

= eπin2τ+2πin(ν+1/2), eπin(n−1) = e2πik = 1

= ϑ3(ν + 1/2, τ)

S : τ → − 1τ : This transformation is more difficult. First, convert the sum to

an integral:

ϑ3(ν, τ) =∑

qn2/2zn =

∫ ∞

−∞dxqx

2/2zx∑

n

δ(x− n)


and∑

n

δ(x− n) =

∞∑

m=−∞e2πixm.

Proof: If x ∈ Z, then clearly∑→ ∞. If x /∈ Z, then

∞∑

m=−∞e2πixm =

1

1 − e2πix+

1

1 − e−2πix− 1

= 2Re(1− e2πix)−1 − 1 = 2Ree−πix

−2i sin(πx)− 1 = 0

Also,∫ 1/2

−1/2

∞∑

m=−∞e2πixm =

∑

m6=0

1

2πime2πix

∣∣∣∣∣∣

1/2

−1/2

+ 1

If m 6= 0, then e2πimx∣∣1/2

−1/2∼ sinπm = 0, therefore

∫ 1/2

−1/2= 1. By periodicity,

∑e2πixm =

∑

n δ(x− n). Therefore

ϑ3(ν, τ) =∑

m

∫ ∞

−∞dxqx

2/2zxe2πixm

=∑

m

∫ ∞

−∞dxeπiτx

2

e2πiνxe2πimx

=∑

m

∫ ∞

−∞dxeπiτ(x+(ν+m)/τ)2−πi(ν+m)2/τ

=1√−iτ

e−iπν2/τ∑

m

e−πim2/τ−2πiνm/τ

=1√−iτ

e−πiν2/τϑ3(ν/τ | − 1/τ)

Therefore

ϑ3

(ν

τ| − 1

τ

)

=√−iτeπiν2/τϑ3(ν|τ).

Similarly for ϑ1, we obtain

ϑ1(ν|τ + 1) = eπi/4ϑ1(ν|τ)

ϑ1

(ν

τ| − 1

τ

)

= −√−iτeπiν2/τϑ1(ν|τ)

ϑ1 can be related to the partition function Z(τ) as follows. Recall

ϑ1(ν|τ) = 2eπiτ/4 sinπν

∞∏

m=1

(1 − qm)(1 − zqm)(1 − z−1qm).


To get Z(τ), we can not just set z = 1, because then sinπν = 0, so ϑ1 = 0. Weneed to differentiate with respect to ν first.

∂νϑ1(ν|τ) = 2πeπiτ/4(

cosπν∏

(...) + sinπν∂ν∏

(...))

Now set ν = 0 and the second term vanishes. Therefore

∂νϑ1(ν|τ) = 2πeπiτ/4

[ ∞∏

m=1

(1 − qm)

]3

= 2π

[

q1/24∞∏

m=1

(1 − qm)

]3

.

Notice the appearance of 1/24 in the exponent! It is needed for the modularinvariance! Therefore

η(τ) = q1/24∞∏

m=1

(1 − qm) =

(∂νϑ1(ν|τ)

2π

)3

where η(τ) is the Dedekind η-function.

Modular Properties of the Dedekind function

τ → τ + 1

η(τ + 1) = eπ1/12η(τ)

1

τ∂νϑ1

(

0| − 1

τ

)

= −√−iτ∂νϑ1(0|τ)

∂νϑ1

(

0| − 1

τ

)

= (−iτ)3/2∂νϑ1(0|τ)

η

(

−1

τ

)

=√−iτη(τ)

So we see that ϑ and η are not invariant under modular transformations. Wereally only care about the partition function, which depends on η. Let uscheck how the partition function transforms under the modular transforma-tions from knowing how η transforms. The partition function is

Z(τ) = iLD(

1

2π√α′τ2

|η(τ)|−2

)D

Under τ → τ + 1, τ2 does not change, nor does |η(τ)|. Under τ → − 1τ , τ2 →

τ2|τ |2 , so 1√

τ2→ |τ |√

τ2and |η(τ)|−2 → 1

|τ | |η(τ)|−2

Therefore 1√τ2|η(τ)|−2 is modular invariant and so is Z(τ).


5.3 The bc system

Recall L0 =∑n : b−ncn : −1. This is in the “sphere” picture. To go back to the

cylinder picture, use

L0 → L0 +c

12

where c = 13. Therefore

L0 =∑

n : b−ncn : +1

12

which is the generator of the t-translations for left-movers.

Zbc = q1/2q1/2Tr qL0 qL0

To calculate Tr qL0 , start with b−1c1 + c−1b1. b−1c1 which has eigenvectors

|0〉 = |ψ〉, |1〉 = b−1|ψ〉

with respective eigenvalues 0, 1.We can show (b−1c1)

2 is a projection operator by using the bc algebra,

b−m, cn = δ0,m+n, bm, bn = cm, cn = 0.

(b−1c1)2 = b−1c1b−1c1 = b−1(1 − b−1c1)c1 = b−1c1.

Therefore, the eigenvalues 0, 1 are the only eigenvalues. Now

Tr |ψ〉〈ψ| = 〈χ|ψ〉 = 〈ψ|c0|ψ〉

is the only sensible definition. Define 〈0| = 〈ψ|, 〈1| = 〈ψ|c1.

Tr |1〉〈1| = 〈1|1〉 = 〈ψ|c1c1b−1|ψ〉 = −1.

The eigenvectors of c−1b1 are |ψ〉 and c−1|ψ〉 so overall, the b−1c1+c−1b1 eigen-values of |ψ〉, b−1|ψ〉, c−1|ψ〉, c−1b−1|ψ〉 are 0, 1, 1, 2 respectively. Therefore

Tr qb−1c1+c−1b1 = q0 − q1 − q1 + q2 = (1 − q)2.

Similarly, we obtain

Tr qn(b−ncn+c−nbn) = (1 − qn)2.

ThereforeTr qL0 =

∏

(1 − qn)2,

and

Zbc(τ) = (qq)1/2∞∏

n=1

|1 − qn|4 = |η(τ)|4.

5.4 Vacuum Energy 89

5.4 Vacuum Energy

CombineZ(τ) for Xµ with Zbc(τ) to get the total partition function.

Ztotal(τ) = iLD(

1

2π√α′τ2

|eta(τ)|−1

)D

|η(τ)|4 = iLD(2π√

α′τ2)−D|η(τ)|−2(D−2).

So D → D − 2 which is good since only the transverse modes should con-tribute. This is not modular invariant. However, note that we have a confor-mal transformation left: rigid translations (similar to the sphere where we hadSL(2,C) invariance). There is a one to one correspondence between pointson the torus and translations. Therefore the number of translations is pro-portional to the volume of the torus.

2π

22πτ = (2π)2τ2

We need to average over translations, so divide by the volume of the group((2π)2τ2). We also have a reflection symmetry, so we need to multiply by 1/2.If we have vertex operators, we need to fix one of the positions (c.f. a spherewhere we fixed three positions). Finally the one-loop vacuum string ampli-tude is

Zphysical = iLD∫

F0

dτdτ

2(2π)2τ2(2π√

α′τ2)−D|η(τ)|−2(D−2).

This is the cosmological constant. Why? C.f. with the point particle.

E =√

~p2 +m2.

On a circle of length ` the temperature is T ∼ 1/`. The partition function isgiven by

Z(`) = LD∫

dDp

(2π)De−χ

`

, χ =1

2(−E2 + ~p2 +m2) =

1

2(p2 +m2)

= iLD (2π`)−D/2 e−m2`/2. (5.4.1)

What information can we gain by evaluating the vacuum amplitude for thestring? The vacuum amplitude defined earlier is given by,

Z =

∫ ∞

0

dl

2πZ(l).


where

Z(l) = LD∫

dDp

(2π)De−χl = iLD

(1√2πl

)D

e−m2l/2

There is an ultraviolet divergence when l = 0. Cut off the integration regionat ε and later let ε→ 0. Apply a Wick rotation (p0 → ip0) and integrate over p0.

Z(l) = i

∫ ∞

−∞

dp0

2πexp[− l

2(p2

0 + ~p2 +m2)]

∼ 1√lexp[− l

2(~p2 +m2)]

Now do the integral over l:

Z =

∫ ∞

ε

dl

l

1√lexp[− l

2(~p2 +m2)]

∼√

~p2 +m2

∫ ∞

ε

dx

x3/2e−x/2

∼√

~p2 +m2 ∗ constant

Z(l) ∼ LD∫

dD−1~p

(2π)D−1

√

~p2 +m2

∼ LD∫

dD−1~p

(2π)D−1E(~p)

This is equivalent to integrating over all possible modes of the string. We candefine the energy density as Λ.

Λ =Z

LD∼∫

dD−1~p

(2π)D−1E(~p) “Cosmological Constant”

This implies the cosmological constant is the sum of the vacuum energiesfrom each of the states of the string.How does the Cosmological constant play a role in one loop correction of thevacuum?

Λ ∼∫

F0

dτ dτ

τ2(2π√

α′τ2)−D |η(τ)|−2(D−2)

Notice that there is no ultraviolet divergence, because τ is not allowed to ap-proach zero (|τ | ≥ 1 in F0). One the other hand, if τ2 → ∞ (very long torus)

|q| = |e2πiτ | = e−2πτ2 → 0

so we may expand

|η(τ)|−2 = |q|−1/12|1 − q + ...|−2 = eπτ2/6(1 + 4Req + ...).

5.5 Thermodynamics 91

The dominant contribution∫ ∞ dτ2

τ2(√

4π2α′τ2)−26e4πτ2

c.f. with (2πτ2)−D/2 exp(−m2`/2). Since ` = 2πα′τ2 the square of the mass

becomes m2 = −4/α′ and we have a tachyon! This diverges due to m2 <0. Other terms in the sum are due to other modes with m2 ≥ 0, so they allconverge.

5.5 Thermodynamics

In this section we look at various thermodynamic properties of the partitionfunction.

τ1 = 0 τ = iτ2 q = e−2πiτ2

In the limits above what can we find out about the partition function?

Z(τ2) = |q|−D/12Tr(

qL0 qL0

)

= e−πDτ2/6Tr(e−2πτ2H

): H = L0 + L0

= e−πDτ2/6∑

n

e−2πτ2En :1

T= 2πτ2

High-temperature limit: τ2 → 0. Then q → 1 and Z(τ2) is dominated by high-weight states. By modular invariance, τ2 → −1/τ2, it is related to Z(1/τ2).Small-temperature limit: τ2 → ∞ ⇒ q → q′ = e−2π/τ2 → 0

Z(τ2) =∑

n

e−En/T ' e−E0/T +∑

n6=0

e−En/T + ...

Only theE0 = 0 will survive in the sum.

Z

(1

τ2

)

= e− πD

6τ2

τ2 → 0 : Z(τ2) = Z

(1

τ2

)

= e− πD

6τ2

Entropy

Z =∑

E

ρ(E)e−E/T saddle point approximation

There exists a stationary exponent when

d(ln ρ−E/T ) = 0.

Define the entropy as S = ln ρ. Therefore dE = TdS.

Z =∑

eS−E/T d(S − E/T) = 0 ⇒ dS =dE

T


Free energy

F = −T lnZ = −T(

S − E

T

)

= E − TS

The entropy can be found from the Free energy, S = − ∂F∂T . Comparing with

Z(τ2) we see there is extra factor of e−D/12T . Modular invariance impliesZe−D/12T is invariant. Therefore

Z(τ)e−D/12T = Z

((2π)2

T

)

e−π2DT/12

Z(τ) = eD/12Te−π2DT/12Z

((2π)2

T

)

Z(1/T ) is a slowly varying function in the saddle point approximation. Whenwe go from statistical mechanics to thermodynamics Z

(1T

)is ignored.

lnZ(T ) =D

12T− π2DT

12

F = −D

12+π2DT 2

12=D

12(π2T 2 − 1), S = −∂F

∂T=Dπ2T

6,

E = F + TS =D

12(π2T 2 + 1) S ≤ πE.

At high temperature the energy is proportional to the square of the temper-ature (E ∝ T 2). This represents the “Casimir energy in 2-D”, which in four-dimensions is given by E ∝ T 4. We may express S in terms of E

D

12π2T 2 = E − D

12,

S2 =

(Dπ2

6

)2

T 2 =D2

26π4 12

π2D

(

E − D

12

)

= 4Dπ2

12

(

E − D

12

)

,

S = 2π

√

D

12

(

E − D

12

)

“Cardy”

or S ≤ πE which is the same as the black hole entropy, where Bekenstein’sequation is S ≤ SB ∼ E. If the equation is generalized to 4-D, it gives theFriedman-Robertson-Walker universe equation for the Hubble constant.

5.6 Amplitudes on a torus

ForM tachyons, Vi =: eiki·X(zi) together with the right movers, the amplitudecan be writen as

5.6 Amplitudes on a torus 93

12

M

3

A(z1, z2, ..., zM ) = Tr V1 .... VMq(L0−D/24) × (c.c.)

where

XL = x+α′

2pz + i

√

α′

2

∑

n>0

1

nαne

−inz, [x, p] = i

At the moment, let us omit the complex conjugate. We can insert it whenneeded. There are two parts to the amplitude: the p-integral and the N-modes.

∫dDp

(2π)D〈p|e−πτ2α′p2eik1·x/2eiα

′k1·pz/2 · · · eikm·x/2eiα′kM ·pzM/2|p〉.

using XL = x + α′

2 pz + i√

α′

2

∑

n>01nαne

−inz. To evaluate the integral, com-

mute all eik1·x/2 through. When they hit |p〉, they change |p〉 → |p+ k1 + . . .+kM 〉. Then 〈p|p+ k1 + . . .+ kM 〉 = δD(k1 + . . .+ kM ) (conservation of momen-tum).As we push the factors through, the amplitude becomes

∫dDp

(2π)Deiπτp

2

ei(k1z1+...+kMzM )·pδD(k1 + . . .+ kM )

This is a Gaussian integral. We need to shift the momentum p→ p−Im(k1z1+. . .+ kMzM )/πτ2. The amplitude becomes

A ∼ (2π√

α′τ2)−D/2δD(k1+...+kM )e−α

′/πτ2[Im(k1z1+...+kM zM )]2∏

i<j

e−α′/2ki·kj(zi−zj)×c.c.

Using conservation of momentum, we may also write

(k1z1 + . . .+ kMzM )2 = −∑

i<j

ki · kj(z1 − zj)2.

Next, do the oscillators. We can do each oscillator separately, so fix µ, n anddo α−nµ , αµn. The trace is

T µn = Tr

M∏

i=1

exp

(1

nkiµα

µ−n

)

exp

(

− 1

nkiµα

µne

−niz)

qαµ−nαnµ

where there is no sum on µ!


To evaluate the trace we need to insert the identity I =∑

E |E〉〈E| and thenuse Tr A =

∑

E〈E|A|E〉. For a complete set of states, choose

|κ〉 = e1nκ·α−n |0〉,

where κ ∈ C. The identity operator is

I =1

nπ

∫

d2κe−|κ|2/n|κ〉〈κ|.

To show this, consider an eigenstate of the number operator N =∑α−nαn

(dropped the µ to avoid confusion).

|`〉 =1√`!

(α−n)`√n

|0〉, 〈`|`〉 = 1.

Then

〈`|κ〉 =1

`!〈`|(κ

n)`α`−n|0〉,

=(κ

n

)` 1√`!,

〈`|I |`′〉 =1

nπ

1√

(`!)2

∫

d2κ e−|κ|2/n( κ

n

)` (κ

n

)`′

For ` 6= `′, 〈`|I |`′〉 = 0 which can be shown by using polar coordinates. For` = `′

〈`|I |`〉 =1

nπα′

∫

d2ke−|κ|2/n( |κ|n

)2`

= 1,

which implies< `|I |`′ >= δ``′ .

Also qN |κ〉 = |κqn〉.PROOF:

qNα−nq−N = α−n + ln q[N,α−n] +

1

2(ln q)2[N, [N, [α−n]]

= α−n + n ln qα−n +n2

2(ln q)2α−n + ...

= en ln qα−n = qnα−n

qN (α−n)`q−N ) = qn`(α−n)`

qNe1nκα−nq−N = e

1nqnκα−n ,

where we used the Glauber identity (eAeB = eBeAe[A,B]). Therefore

qN |κ〉 = |κqN 〉.

5.6 Amplitudes on a torus 95

The trace becomes

T µn =1

nπ

∫

d2κe−|κ|2/n〈κ|m∏

i=1

e1nki·α−ne

nizie−

1nki·αne

−niz

e1nqnκα−n |0〉.

As we push exp(− 1nk1αne

−niz) through, the commutators contribute to anew factor given by

exp

(

− 1

nk1q

nκe−niz1)

.

After moving the exponentials over to act on the states, the commutators con-tribute an overall factor of

∏

i<j

exp

(

− 1

nki · kjeni(zi−zj)

)

,∏

i

exp

(

− 1

nkiq

nκe−nizi

)

Finally push exp(

1n xαn

)through. We get a factor exp

(1nkiκe

nizi)

from each

vertex and exp(

1nq

n|κ|2)

from |qnκ〉. Therefore

T µn =1

nπ

∫

d2κe−(1−qn)|κ|2/n∏

i<j

exp

(

− 1

nki · kjeni(zi−zj)

)∏

i

exp

(

− 1

nkiq

nκe−nizi

)

This is a Gaussian integral: 1π

∫d2κe−c|κ|

2

eaκ+bκ = 1ce

−ab/c.Therefore

T µn ∼ 1

1 − qn

∏

i<j

e−1nki·kje

−ni(zi−zj )

exp

− (∑kie

nizi)(∑kie

−nizi)qn

n(1 − qn)

.

Use∑

i<j

ki · kj = −1

2

∑

ki2 = −M.

∏

n,µ

T µn =

∞∏

m=1

(1 − qm)−D∏

i<j

e−ki·kjemi(zi−zj + qme−mi(zi−zj) − 2qm

m(1 − qm)

where we see the whole second product is a new contribution. If we use theidentity

∞∑

m=1

1

m

xm

1 − ym= −

∞∑

n=0

ln(1 − xyn)

we find

∏

n,µ

T µn =

∞∏

m=1

(1 − qm)−D∏

i<j

e−ki·kj ∞

n=0 ln(1−qnei(zi−zj ))+ln(1−qn+1e−i(zi−zj))−2 ln(1−q)

=∞∏

m=1

(1 − qm)−D∏

i<j

(1 − qei(zi−zj))α′ki·kj

∞∏

n=1

[(1 − qnei(zi−zj))(1 − qne−i(zi−zj))

(1 − qn)2

]α′ki·kj

.


The first factor∏

i<j(1−qei(zi−zj))α′ki·kj is combines with

∏

i<j e−α/2(zi−zj)ki·kj

to give∏

i<j

sin(zi − zj)α′ki·kj/2.

Then

sin(zi − zj)

∞∏

n=1

[(1 − qnei(zi−zj))(1 − qne−i(zi−zj))

(1 − qn)2

]α′ki·kj

∼ϑ1

(zi−zj

2π |τ)

∂νϑ1(0|τ),

where

ϑ1(ν|τ) = 2eπiτ/4 sinπν∏

(1 − qm)(1 − zqm)(1 − z−1qm).

As we recall, ϑ1 has nice modular properties, given by

ϑ1(ν|τ + 1) = eiπ/4ϑ1(ν|τ),

ϑ1

(ν

τ| − 1

τ

)

= −i√−iτeπiν2/τϑ1(ν|τ).

Let us compare the amplitude to the sphere amplitude. The M-point ampli-tude for the sphere was

A = 〈V1...VM 〉 ∼∏

i<j

|zi − zj |α′kk·kj .

For small |zi − zj |, we may make the approximation

ϑ1

(zi − zj

2π|τ)

∼ sinπν ∼ zi − zj .

So, the torus is similar to the sphere at short distances.

5.7 Higher Genus Surfaces

All two-dimensional surfaces may be classified by their genus. These surfacesare classified the number of handles they posess. There is no classification forsurfaces of dimension greater than two. The genus of a sphere and torus arezero and one respectively.

......

.....

5.7 Higher Genus Surfaces 97

Plumbing Fixture

This is a method of creating higher genus surfaces from lower genus surfaces.

2

1

R2

R1

R2

z1 z

R

Coordinate patches z1 and z2 may belong to different or same surface. Cut adisk of radius R1 around the origin of each. Then glue annular regions R1 <|zi| < R2. (i = 1, 2) by identifying

R2 R1R1 R 2

z1 z2

Example 1: z1 and z2 on the same plane. Add a handle to create a torus.


Example 2: z1 and z2 on two tori.

Plumbing is a reversible process. One can cut handles and then patch disksto create a lower genus surface. Let us look at an interesting case, q → 0. Thenfor fixedR1,R2 → 0, so an annulus maps to a semi-infinite cylinder, an objectwith physical meaning!

Examples:

equivalent to

equivalent to

This is the boundary of the moduli space. Recall that violations to the BRSTsymmetry may also come from the boundary of the moduli space, so pinchedhandles are important.

Amplitudes

We will start with a trivial example: two spheres connected by a cylinder.

A1 = 〈0|V1...VM |E〉 = 〈0|V1...VMVE |0〉A2 = 〈E|VM+1...VN |0〉 = 〈0|V ∗

EVM1 ...VN |0〉

where

|E〉 = VE |0〉 and I =∑

E

|E〉〈E|

A =∑

E

A1A2 = 〈0|V1...VMVM+1VN |0〉 = 〈0|V1...VN |0〉

This is just a N-point amplitude on a sphere, obviously since S2 ∪ S2 = S2

Next, let us move to an example which is nontrivial, but known.


1z z2

If q = 1, then the identification: z1z2 = 1 which is the two patches on thesphere around zero and infinity respectively. We might as well choose z1 =0, z2 = ∞. Then the amplitude,A, becomes

A = 〈E|V1...VM |E〉 = 〈0|V ∗EV1...VMVE |0〉

and the trace of the amplitude becomes

Tr A =∑

E

A = Tr V1...VM .

Now let us have z1z2 = q, need z2 → z2q, a conformal transformation underwhich

VE →(∂z′2∂z2

)h(∂z′2∂z2

)h

VE = qhqhVE .

E is the set of h’s (weights), where

L0|E〉 = L|E〉, L0|E〉 = L|E〉.

Therefore ∑

E

A = Tr(

V1...VM qL0 qL0

)

, for q 6= 1.

Generalize ...

A =∑

E

〈V1...VM |E〉〈E|VM+1 ...VN |0〉qhqh

to obtain an amplitude of a higher genus surface.

Divergences

We have already seen that there are no ultra-violet divergences that plaguequantum field theories (short distance effects- breakdown of theory, i.e., newphysics at short distances). String theory is the ultimate theory, no matterhow short the distance.Example: The torusPuzzle: as vertices come together on a torus, we get a singularity. Is this ashort distance effect? No! It is like the case of a sphere (topology plays no


role). Recall the singularity in 〈V1(∞)V2(1)V3(z)V4(0)〉 as z → 1. We obtainpoles, which after a conformal transformation seem to come from a particlepropagating for a long time.

LONG

2 1

43

That is a long distance effect, i.e., an “infrared divergence” which is expected.z → 1 means the distance becomes small on the worldsheet (bogus?). A longintermediate state is a spacetime concept (REAL!). Similarly, on a torus

pinching

equivalent byconformal transformation

Special Case I

All vertices but one come together.

i2

1

k1k

kpoles ~

2+mConservation of momentum implies k = k1, but k2

1 = −m2, so 1/(k2 +m2) =1/0, a singularity (infrared).We get rid of it in quantum field theory as follows. Let k2 6= −m2. Then

= δ +δ2

k2 +m2+

δ3

(k2 +m2)2+ . . .

=δ

1 − δk2+m2

=δ(k2 +m2)

k2 +m2 − δ

Notice that the pole has been shifted by δ, so the mass is corrected by quan-tum effects: δm2 = −δ. This means that we should have started with a parti-cle of massm2 − δ and notm2. It was a poor perturbative expansion (singular


perturbation theory). It is the same in string theory. We got 1/0 because weasked the wrong question.Note: Massless particles receive no quantum corrections as required by gaugeinvariance.

Special Case II

All vertices come together

k

So the pole from the massless modes goes as 1/k2 = 1/0! (Bad)Again, we are asking the wrong question. To calculate the contribution ofthe massless modes, after we insert

∑

E |E〉〈E|, we pick E = 0. That is aninsertion of ∂Xµ∂Xµ. If we regulate the integral, e.g., by cutting the length ofthe connecting cylinder by Lmax, we obtain

Amp ∼ C

∫

d2z∂Xµ∂Xµ

where C is an infinite constant.Now this can also come from a perturbation in the action

∫d2z∂Xµ∂Xµ which

tells us that the metric in spacetime, instead of being flat,Gµν = ηµν (Lorentz)should be Gµν = (1 − C)ηµν . The flat background is not a good zeroth orderapproximation because of the gravitational effects of the string.

String Theory II

GEORGE SIOPSIS AND STUDENTS

Department of Physics and AstronomyThe University of TennesseeKnoxville, TN 37996-1200

U.S.A.e-mail: [email protected]

Last update: 2006

ii

Contents

6 Compactification and Duality 16.1 The Kaluza-Klein Mechanism . . . . . . . . . . . . . . . . . . . . 16.2 Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36.3 Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 56.4 Vertex Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86.5 Amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96.6 Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106.7 R =

√α′ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

6.8 Away fromR =√α′ . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6.9 T-duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

7 Superstrings 257.1 Bosons and fermions . . . . . . . . . . . . . . . . . . . . . . . . . 257.2 The ghosts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.3 Mode Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . 297.4 Open Strings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337.5 The Ramond (R) sector . . . . . . . . . . . . . . . . . . . . . . . . 387.6 Superstring Theories . . . . . . . . . . . . . . . . . . . . . . . . . 40

8 Heterotic Strings 518.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518.2 The spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

9 Low Energy Physics 579.1 Type IIA Superstring . . . . . . . . . . . . . . . . . . . . . . . . . . 579.2 Supergravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

10 D-Branes 6110.1 T-duality (again) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6110.2 D-branes at angles . . . . . . . . . . . . . . . . . . . . . . . . . . 6410.3 Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 6610.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

iii

iv CONTENTS

UNIT 6

Compactification and Duality

6.1 The Kaluza-Klein Mechanism

Before we introduce the Kaluza-Klein mechanism, let us briefly review elec-tromagnetism and gauge symmetry. The field strength is given by

Fµν = ∂µAν − ∂νAµ,

and the action

S =

∫

d4x

(

−1

4FµνF

µν +AµJµ

)

.

leads to the Maxwell equations

∂µFµν = Jν .

Gauge invariance in the theory implies Aµ → Aµ − ∂µλ provided the currentis conserved (∂µJ

µ = 0). The conserved charge is given by

Q =

∫

d3xJ0,dQ

dt= 0.

Suppose Jµ is due to a scalar field Φ which has massm. If we forget about thecharge for the moment, let pµ be the momentum of the particle Φ represents(Φ represents no particle that anybody has observed). Einstein tells us

pµpµ = −m2

Quantize the system: pµ → i∂µ, so ∂µ∂µΦ = m2Φ (Klein-Gordon equation).

This is obtained from the action

S =1

2

∫

d4x ∂µΦ∗∂µΦ +m2|Φ|2

2 UNIT 6. COMPACTIFICATION AND DUALITY

The conserved current: jµ = Φ∗∂µΦ− c.c. , ∂µJµ = 0. If Φ has charge, we

need to couple Jµ to Aµ. This is done by pµ → pµ − qAµ, or ∂µ → ∂µ + iqAµ,where q is the charge. Therefore the action for a charged scalar field, Φ is

S =1

2

∫

d2x(

(∂µ − iqAµ)Φ∗(∂µ + iqAµ)Φ +m2|Φ|2)

comparing with∫

d4x AµJµ, we obtain

Jµ = − iq2

(Φ∗∂µΦ− Φ∂µΦ∗).

Gauge invariance: Φ→ eiqλΦ, Aµ → Aµ−∂µλ. We have (∂µ+igAµ)Φψeigλ(∂µ+

igAµ)Φ So the action and the currents are gauge invariant. Since λ is real, |Φ|is invariant, so Φ moves on a circle in the complex plane as λ changes. λ rep-resents an angle in this picture.Kaluza-Kleins suggestion was to take this picture literally and assume the thee&m is nothing but the effect of an extra (fifth) dimension. How? Let us see...Imagine a five-dimensional manifold in which the fifth dimension is a circleof radius R. Choose the coordinates

xµ = (t, x, y, z, u), u ≡ u+ 2πR.

The line element is

ds2 = GMNdxMdxN = Gµνdx

µdxν + 2Gµudxµdu+Guudu

2.

SupposeGMN is independent of u (invariance under u translation). ParametrizeGMN as follows: Aµ = Gµu/Guu, gµν = Gµν −GuuAµAν . Then

ds2 = gµνdxµdxν +Guu(du+Aµdx

µ)2

Reparametrizaions u → u + λ(xµ) implies Aµ → Aµ − ∂µλ, i.e., gauge trans-formations!Let pµ be the momentum conjugate to u. Then eiapµf(u) = f(u+a) (pµ gener-ates translations). Since f(u+2πR) = f(u), we need e2πiRpµ = 1, so pµ = n/R(momentum is quantized). A general f(u) may be expanded in momentumeigenstates (einu/pR).

f(u) =

∞∑

n=−∞ane

inu/R.

For a field φ(xµ) = φ(xµ, u), we have

φ(xµ) =

∞∑

n=−∞an(x

µ)einu/R.

The wave equation ∂µ∂µφ = 0 (massless φ) becomes

∂µ∂µΦn −

n2

R2Φn = 0

6.2 Strings 3

i.e., an(xµ) represents a massive field of massm = n/R from a four-dimensional

point of view. Einstein’s equation correspondingly reads pµpµ = −n2/R2.

At energies E 1/R, we only see the n = 0 mode of α0(xµ). At high energies

(early universe), we see more modes.What about charge? To isolate the effects of Aµ, letGuu = 1 and gµν = ηµν .From φ =

∑

an(xµ)einu/R and u→ u+ λ(xµ) we obtain an(x

µ)→ einλ/Ran.Comparing with φ → eiqλφ for a field φ of charge q, we see an indication thatq = u/R for the mode an. So, q = m; the mass of an.To see the gauge invariance in full swing, go back to the wave equation for anand put back in the curvature of space-time:

Dµ∂µ(aneinu/R) = 0

where

DMvN = ∂MvN − ΓLMNvL, DMvM = ∂MvM −GMNΓLMNvL,

and the Christophel symbol expressed in terms of the metric may be writtenas

ΓLMN =1

2GLP (∂MGPN + ∂NGMP − ∂PGMN ) .

A short calculation reveals

(∂µ +Aµ∂u)2(ane

inu/R) = 0

∂µ + in

RAµ)an =

n2

R2an

which confirms that an has charge q = n/R. Moreover, the Maxwell equationscome from the Einstein action

S ∼∫

d5x√−GR(5),

where

G = detGMN = −1 R(5) = −1

4FµνF

µν .

So the Einstein action reduces to electromagnetism! If we put back four-dimensional curvature (gµν 6= ηµν), then we get G = detgµν and R(5) =R(4) − 1

4FµνFµν which is four-dimensional gravity and electromagnetism!

6.2 Strings

Consider a closed string moving along u (as well as x, y, z, ...), described bythe function V (σ, τ). The action is (concentrate on U )

S =1

2πα′

∫

d2z ∂U∂U


and we still demand U ≡ U + 2πR. As before, center-of-mass momentum isquantized: p = n/R, n ∈ Z. Recall the mode expansion without compactifi-cation:

Xµ(σ, τ) = xµ+2πα′pµτ

`+i

√

α′

2

∑

m6=0

1

m

(

αµme−2πim(σ+τ)/` + αµme

2πim(σ−τ)/R)

,

which can also be written in terms of z = e2πi(σ+τ)/`, z = e−2πi(σ−τ)/` as

Xµ(z, z) = xµ + 2πα′pµτ

`+ i

√

α′

2

∑

m6=0

1

m

(

αµmz−m + αµmz

−m) ,

with derivatives

∂Xµ(z, z) = −iα′

2pµz−1 − i

√

α′

2

∑

m6=0

αµmz−m−1,

∂Xµ(z, z) = −iα′

2pµz−1 − i

√

α′

2

∑

m6=0

αµmz−m−1.

Notice that the momentum is p = 12πα′

∮

(dz∂X − dz∂X) and if we go aroundthe string once, we obtain

∮

(dz∂X + dz∂X) = 0. (6.2.1)

With u ≡ u + 2πR (compactified), (6.2.1) is no longer necessarily true. Whenwe go around the string, u can change by a multiple of 2πR, i.e.,

U(σ + `) = U(σ) + 2πRw, w ∈ Z.

This allows a solution of the form U(σ, τ) = 2πwR σ` . Since z/z = e4πiσ/` ⇒

2πσ/` = −i/2 ln(z/z), so

U(σ, τ) = − i2wR ln

(z

z

)

has to be added. We obtain

U(z, z) = u− iα′

2

n

Rln |z|2 − i

2wR ln

(z

z

)

+ i

√

α′

2

∑

m6=0

1

m(αmz

−m + αmz−m),

= u− iα′

2

(

n

R+wR

α′

)

ln z − iα′

2

(

n

R− wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

m(αmz

−m + αmz−m),

with the derivatives

∂U(z, z) = −iα′

2pLz

−1 − i√

α′

2

∑

m6=0

αmz−m−1,

∂U(z, z) = −iα′

2pRz

−1 − i√

α′

2

∑

m6=0

αmz−m−1,

6.3 Partition Function 5

where PLR

= nR ± wR

α′ . Notice that

1

2πα′

∮

(dz∂X − dz∂X) =1

2(pL + pR) = p =

n

R,

and1

2πα′

∮

(dz∂X + dz∂X) =1

2(pL − pR) =

wR

α′ 6= 0.

We may express the Virasoro generators in terms of the left and right-handedmomenta.

L0 =α′p2

L

4+

∞∑

n=1

α−nαn, L0 =α′p2

R

4+

∞∑

n=1

α−nαn

These are the same as in the uncompactified case, except now pL 6= pR.

6.3 Partition Function

First, let us recall the uncompactified case

Z = Tr (qL0−1/24qL0−1/24), q = e2πiτ ,

= Tr (qq)α′p2/4−1/24

∏

n

( ∞∑

Nn=0

qnNn

)

∞∑

Nn=0

qnNn

,

=(

Tr e−πτ2α′p2)

∣

∣

∣

∣

∣

∏

n

(1− qn)−1

∣

∣

∣

∣

∣

2

(qq)−1/24,

= |η(τ)|−2V

∫

dp

2πe−πτ2α

′p2 ,

= 6|η(τ)|−2 V

2π

1√α′τ2

,

which is invariant under modular transformations (τ → τ + 1, τ → −1/τ ).

Notice that for Xµ = xµ − iα′

2 pµ ln |z|2, which is a solution of the wave equa-

tion,we have ∂Xµ = −iα′

2 pµz−1, ∂Xµ = −iα′

2 pµz−1 and the action is

S =1

2πα′

∫

d2 z∂Xµ∂Xµ =

α′

8πp2

∫

d2z

|z|2 , z = ei(σ1+iσ2),

=α′

8πp22VTorus, (VTorus = 2π(2πτ2))

= α′p2τ2π.

Thereforee−S = e−πτ2α

′p2 ,

which is the factor whose trace contributes to the partition function.


Partition Function for Compactified Space

Z = Tr (qL0−1/24qL0−1/24),

= |η(τ)|−2∑

n,w

qα′p2L/4qα

′p2R/4, pLR

=n

R± wR

α′

= |η(τ)|−2∑

n,w

exp

[

−πτ2α′(

n2

R2+w2R2

α′2

)

+ 2πiτ1nw

]

.

Use the Poisson resummation formula:

∞∑

n=−∞e−πan

2+2πibn =1√a

∞∑

m=−∞e−π(m−b)2/a.

Therefore the partition function for the compactified case is

Z = |η(τ)|−2 R√τ2α′

∑

m,w

exp

[

−πR2

α′τ2

(

(m− τ1w)2 + τ22w

2)

]

,

= |η(τ)|−2 V

2π√α′τ2

∑

m,w

exp

(

−πR2

α′τ2|m− τw|2

)

,

which is the same as the uncompactified case. Modular invariance implies

τ → τ + 1 ⇔ m→ m+ w,

τ → −1

τ⇔ m→ −w, w → m.

Notice that the solution with the correct boundary conditions satisfying

U(σ1 + 2π, σ2) = U(σ1, σ2) + 2πwR,

U(σ1 + 2πτ1, σ2 + 2πτ2) = U(σ1, σ2) + 2πmR.

can be written as

U(σ1, σ2) =wR

τ2(τ2σ

1 − τ1σ2) +mR

τ2σ2

where

∂1U = wR, ∂2U =R

τ2(m− wτ1).

The action is given by

S =1

4πα′

∫

dσ1dσ2

[

(∂1U)2 + (∂2U)2]

=1

4πα′ 2π2πτ2R2

τ22

|m− wτ |2 + ...


Thermodynamics

From thermodynamics we know the partition function is

Z =∑

E

e−βE, β =1

T

and T is the temperature.

Let |E〉 be the eigenstate of the Hammiltonian with eigenvalue E. Then

Z =∑

E

〈E|e−βH |E〉 = Tr e−βH .

This is a special case of the string partition function (β ∈ R). To calculate this,insert the complete sets

Z =∑

E,x,y

〈E|x〉〈x|e−βH |y〉〈y|E〉

Letβ → it~, then 〈x|e−βH |y〉 → 〈x|e−iHt/~|y〉 = 〈x(t)|y(0)〉 a correlator (Greensfunction) of the Schrodinger equation. Suppose t is small. Then insert 1 =∑

p |p〉〈p|, where |p〉 is an eigenstate of the momentum (H = H(p, q)).

〈x(t)|y(0)〉 =∑

p

〈x|p〉e−iHt/~〈p|y〉,

=

∫

dp e−ipxe−iHt/~eipy.

For x− y ' −qt, so

〈x(t)|y(0)〉 =

∫

dp ei(qp−H)t =

∫

dp eiS , S =

∫ t

0

dt′ (q −H)

The dominant contribution is from the stationary point, ∂S∂p = 0. This is at

q = ∂H∂p , which is the Hamilton-Jacobi equation. Then

〈x(t)|y(0)〉 = eiS .

This integrates for finite t. Then

Z =∑

e,x,y

eiS(x,y)ψ∗E(x)ψE(y) =

∑

x

eiS(x,x)

(sum over all possible closed paths) and we used the orthogonality of |x〉 suchthat 〈x|y〉 = δ(x − y) and assumed the energy eigenstates formed a completeset.


6.4 Vertex Operators

Recall

U(z, z) = u−iα′

2

(

n

r+wR

α′

)

ln z−iα′

2

(

n

r− wR

α′

)

ln z+i

√

α′

2

∑

m6=0

1

m

(

αmz−m + αmz

−1)

Momenta pL, pR are different. Their eigenvalues are

pL =n

r+wR

α′ , pR =n

r− wR

α′ .

In the uncompactified case, u commutes with p = 12 (pL + pR), [u, p] = i. Here

pL, pR are independent operators, so u should alsp consist of two indepen-dent operators, u = uL + uR, such that

[uL, pL] = [uR, pR] = i.

Thus U may be broken into holomorphic (UL) and antiholomorphic (UR)pieces as

UL(z) = uL − iα′

2pL ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m,

UR(z) = uR − iα′

2pR ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m.

The operator product expansions are

UL(z)UL(0) ∼ −α′

2ln z, UR(z)UR(0) ∼ −α

′

2ln z, UL(z)UR(0) ∼ 0.

The vertex operator also splits into a holomorphic and antiholomorphic pieces

VL(z) =: ei(nr+wRα′ )UL(z) :, VR(z) =: ei(

nr−wRα′ )UR(z) :

c.f., the uncompactified case,

VL(z) =: eik·XL(z) :, VR(z) =: eik·XR(z) :

The OPE for these vertex operators is

V kL (z)V k′

L (0) = : eik·XL(z) :: eik′·XL(0) :,

∼ eik·k′ α′

2ln z : ei(k+k

′)XL(0) :,

= zα′2k·k′ : V k+k

′

L (0) : .

This has a branch cut. If you let z go around the circle C once, it picks up a

factor eπiα′k·k′ (z

α′2k·k′ = e

α′2k·k′ ln z → e

α′2k·k′(ln |z|+iArgz)). On the other hand,

6.5 Amplitudes 9

z−α′2k·k′ picks up a factor e−πiα

′k·k′ . The two factors cancel each other, so theOPE of the full vertex operator is single valued: (V = VLVR)

V k(z, z)V k′(0, 0) ∼ |z|α′k·k′/2V k+k

′(0, 0)

In the compactified case, let kL = nr + wR

α′ , kR = nr − wR

α′ . Then, similar to theuncompactified case, we find

V kL(z)V k′L(0) ∼ |z|α′kL·k′L/2V

kL+k′LL (0) V kR (z)V k

′R(0) ∼ |z|α′kR·k′R/2V kR+k′R(0)

As z goes around a circle C, we obtain factors e−πiα′kL·k′L , e−πiα

′kR·k′R . Thetotal factor for the full vertex is eiπα

′(kL·k′L−kR·k′R) = e2πi(nw′+n′w) = 1, so the

full vertex is ok.Subtlety: If instead of going around, consider points z1, z2 and interchangethem and let (kL, kR) ↔ (k′L, k

′R) i.e., consider the commutator of two ver-

tices. This is equivalent to letting z ↔ −z, which introduces a factor (−1)α′2kL·k′L =

eiπα′2kL·k′L in the left part and eiπ

α′2kR·k′R in the right part. Overall,

eiπα′(kL·k′L−kR·k′R) = e2πi(nw

′+n′w) = ±1

So if nw′+n′w is odd, the vertices anticommute! To remedy this, we will definethe vertex as

V kL,kR = CkL,kr (p) : ei(kLUl+kRUR) :

where CkL,kR is known as a cocyle. One possible choice is all choices areequivalent.

CkL,kr (p) = eiπα′2

(kL−kR)r, p = /2(pL + pR)

It satisfiesCk(p)Ck′ (p) = Ck+k′ (p), k = (kL, kR).

When we commute two vertices, we pick up the factor

eiπα′2

(kL−kR) k′2 e−iπ

α′2

(k′L−k′R)k2 = eπi(nw

′−n′w)

Thus the overall factor is now

eπi(nw′−n′w)eπi(nw

′+n′w) = e2πinw′= 1.

Thus the overall factor is one, so the vertices always commute.

6.5 Amplitudes

Recall in the uncompactified case,

A = 〈V1(z1)V2(z2) + ...+ VN (zN )〉 ∼ 2πδ(k1 + k2 + ...+ kN )∏

i<j

|zi − zj |α′ki·kj

(6.5.1)


where the vertex operators are given by

Vi(zi) =: eiki·X(zi) : .

The delta function in (6.5.1) comes from the zero mode, eiki·(x−iα′2p ln |z|2). Ex-

plicitly

〈eik1·(x−iα′

2p ln |z|2)...eikN ·(x−iα′

2p ln |z|2)〉 = 〈eik1·(x−iα

′2p ln |z|2)...eikN−1·(x−iα

′2p ln |z|2)|kN 〉

∼ 〈0|k1 + k2 + ...+ kN 〉∼ δ(k1 + k2 + ...+ kN )

Each eiki·x shifts the state |k〉 → |k + ki〉. In the compactified case, we get thesame holomorphic and antiholomorphic except momenta are not different:

∏

i<j

|zi − zj |α′ki·kj →

∏

i<j

(zi − zj)α′kLikLj/2(zi − zj)α

′kRikRj/2.

The zero modes contribute

δ(kL1+kL2+...+kLN)δ(kR1+kR2+...+kRN) ∼ δ(n1+n2+...+nN)δ(w1+w2+...+wN ).

Cocycles give additional± signs.

6.6 Spectrum

Recall in the uncompactified case:

L0 =α′p2

4+N, N =

∞∑

n=1

α−nαn.

In the compactified case, p→ pL inL0 and p→ pR in L0. A string will travel in(t, x, y, z, ..., u) i.e., the 26 dimensional space-time with one-dimension com-pactified. Then

L0 =α′(p2 + p2

L)

4+N, p2 = pµp

µ

and

N =

∞∑

n=1

αµ−nαnµ +

∞∑

n=1

α−nαn,

and similarly for L0. The Mass-shell condition states

(L0 − 1)|phys〉 = (L0 − 1)|phys〉 = (L0 − L0)|phys〉.

The Hamiltonian is a constraint. We define the mass by m2 = kµkµ, where kµ

is the eigenvalue of pµ. Then

L0 − 1 = 0⇒ α′

4(−m2 + k2

L) +N − 1 = 0⇒ m2 = k2L +

4

α′ (N − 1)

6.6 Spectrum 11

From the antiholomorphic piece we get

m2 = k2R +

4

α′ (N − 1)

Subtract the two and we find

N −N =α′

4(k2L = k2

R) = nw

If we add the two conditions we find

m2 =2

α′ (N + N − 2) +k2L + k2

R

2=n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2)

We now explicitly see the Kaluza-Klein mass term and the winding potentialenergy.

Massless States

n = w = 0, N = N = 1, same as in the noncompactified theory. They are:

αµ−1αν−1|0; k〉, αµ−1α−1|0; k〉, α−1α

µ−1|0; k〉, α−1α−1|0; k〉.

The states are represented by the (graviton and antisymmetric tensor), vec-tors, and scalar particles respectively. Recall in Kaluza-Klein field theory, wehad gµν , Aµ, Guu. Our gravition corresponds to gµν , the scalar to Guu, butwe have two vectors instead of one! Which vector corresponds to Aµ? Toanswer this question consider three-point amplitudes of two tachyons and avector. The tachyons have momenta k1, k2 and the vector has momentum k.A tachyon is described by the vertex

: eikLUl+ikRUR+ik·X :⇔ state |0, k, kL, kR〉

The two vectors are

|B〉 = Bµ(k)αµ−1α−1|0; k〉, |C〉 = Cµ(k)α

µ−1α−1|0; k〉

The amplitude for |B〉 is

A ∼ Bµ〈0; k1, k1L, k1R| : eik2LUL+ik2RUR+ik2·X(1) : αµ−1α−1|0; k〉

The relevent parts of U, Xµ are:

UL = uL − iα′

2pL ln z + ...

UR = rR − iα′

2pR ln z + i

√

α′

2α1z

−1 + ...

Xµ = xµ − iα′

2pµ ln |z|2 + i

√

α′

2αµ1 z

−1 + ... where z = 1


So

A ∼ Bµα′

2〈0; k1 + k2, k1L + k2L, k1R + k2R|α1k2Rα

ν1k2να

µ−1α−1|0; k〉

∼ α′k2 · Bk2Rδ(k1 + k2 + k)δ(k1L + k2L)δ(k1R + k2R)

∼ α′(k2 − k3) ·Bk2Rδ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2)

where we used k · B = 0 (gauge invariance coming from Q|B〉 = 0 or ob-serve that if B ∝ k, |B〉 is a null state. Notice that the two tachyons haveopposite quantum numbers n,w which makes sense because they annihilateeach other. Alternatively, if k2 is outgoing, the two tachyons have the samequantum numbers (scattering of a single tachyon). The diagram for the othervector C is

A′ ∼ (k2 − k3) · Bk2Lδ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2).

i.e., k2R → k2L, Bµ = Cµ. Notice that if the sum (corresponding to |B〉 + |C〉)is

A+A′ ∼ α′(k2 − k3) ·Bn

Rδ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2)

i.e., the strength of the interaction is proportional to the change. Therefore,the photon is

Aµ(k)(αµ−1α−1 + αµ−1α−1)|0; k〉.

The other vectorA′µ(k)(α

µ−1α−1 − αµ−1α−1)|0; k〉 leads to the amplitude

A−A′ ∼ (k2 − k3) ·BwR

α′ δ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2)

so it couples to a different charge: the winding number (magnetic?) This isabsent in particle theory and is only a string effect.

6.7 R =

√α′

When R =√α′, m2 = 0 implies n2

α′ + w2

α′ + 2a′ (N + N − 2) = 0, N −N = nw.

Apart from n = w = 0, N = N = 1, we have the following possibilities

• n = w = ±1, in which case we have N + N = 1, N − N = +1, soN = 0, N = 1.

• n = −w = ±1, in which case we have N + N = 1, N − N = −1, soN = 1, N = 0.

• n = ±2, w = 0, in which case we have N + N = 0, N − N = 0, soN = N = 0.

• n = 0, w = ±2, in which case we have N + N = 0, N − N = 0, soN = N = 0.

6.7 R =√α′ 13

The first and second possibilities are new vectors and they are charged! Thisis reminiscent of the Weak interactions where we haveW± (charged vectors).Together with a mixture of γ and Z0, they form a triplet, which is a repre-sentation of SU(2). Similarly, matter comes in doublets (e, νe), (u, d) whichalso transform under SU(2), which is a gauge symmetry, much like electro-magnetism, where the photon and matter transform under U(1). (W± alsohave mass, but that is only because they ate a Higgs). Recall in E&M (matterrepresented by a scalar, e.g., tachyon - both fiction)

S =

∫

d4x

[

−1

4FµνF

µν + |DµΦ|2 +m2|Φ|2]

,

where

Dµ = ∂µ + iqAµ, Fµν = ∂µAν − ∂νAµ.Gauge invariance: Φ → eiqλΦ, Aµ → Aµ − ∂µλ based on the gauge groupU(1).To extend this to weak interactions, where we have three vectors,A1

µ, A2µ, A

3µ,

we view them as a vector in an abstract space (three-dimensional). Rotationsin this space should be independent of the physics. In other words, the action,S, must be invariant under such rotations. We might guess that we shouldhave

−1

4

3∑

i=1

F iµνFiµν

(a Weak field for each vector). We also need∫

d4x∑

i AiµJ

iµ, where J iµ is

made of Aiµ. These requirements severly restrict the form of the action. Itturns out that the fields must be defined by

F iµν = ∂µAiν − ∂νAiµ − εijkAjµAkν

and the action is

S = −1

4

∫

d4x∑

i

F iµνFiµν

and leads to nonlinear Maxwell equations given by

∂µFiµν − εijkAjµF kµν = 0.

Gauge transformations: Aiµ → Aiµ − ∂µλi − εijkAjµλ

k (infinitesimal). Finite

transformations: Introduce Pauli matrices, σi. Define the matrix field

Aµ =1

2Aiµσ

i

where σi represents the Pauli spins matrices which obey the algebra

[σi, σj ] = 2iεijkσk


The field strength is

Fµν =1

2F iµνσ

i

S = ei2λiσi ∈ SU(2), SS† = I

For λ 1 we may expand S

S ' 1 +i

2λjσj

Let us see how Aµ = 12A

iµσ

i transforms under a finite transformation

Aµ → S (Aµ − i∂µ)S†

→ (1 +i

2λiσi)(Aµ − i∂µ)(1−

i

2λjσj)

→ Aµ −1

2σj∂µλ

j +i

4λi[σi, σj ]Ajµ +O(λ2)

→ Aµ −1

2σj∂µλ

j − 1

2εijkσiλjAkµ

Aiµ → Aiµ − ∂µλi − εijkλjAkµ (6.7.1)

For the matter fields, we need to define DµΦ. Φ is a doublet, (u, d) or (e, νe),etc. (not quite, because of the spin), so define

Dµ = ∂µ + iAµ.

whereAµ is a matrix. Notice that we have no degree of freedom in introducinga charge q, because

Φ→ SΦ, Aµ → S(Aµ − i∂µ)S†,

so

DµΦ → ∂µ(SΦ) + iA′µSΦ,

= S(∂µΦ + S†∂µSΦ + iS†A′µSΦ),

= SDµΦ. (6.7.2)

This would have worked nicely had we chosen

Φ→ S′Φ, S′ = ei2qλiσi ,

because iS′†A′µS

′ 6= Aµ−S†∂µS. This is only true in the Abelian case, S = eiqλ,because all “matrices” commute.We found two vectors that coupled to charges pL and pR (or n and w). Thecharge operator (which measures the charge of a state) is then the momen-tum, and there are two of them. The corresponding (conserved) currents gen-erate gauge symmetries, both being U(1), so the gauge group is U(1)× U(1).

6.7 R =√α′ 15

The two currents are ∂U and ∂U . To see their action, consider a state of charge(n,w) or (kL, kr), e.g., V =: eikLU+ikRU :. The OPEs are given by

∂U(z)V (0, 0) ∼ ∂UL(z) : eikLUL :

∼ ikL∂

(

−α′

2ln z

)

: eikLUL(0) :

∼ −ikLα′

2· 1zV (0, 0)

∂U(z)V (0, 0) ∼ −ikRα′

2· 1zV (0, 0)

The charges are

QL =1

2πi

∮

C

dz∂U(z) QR = − 1

2πi

∮

C

dz∂U(z).

so

[QL, V ] = −iα′

2kLV [QR, V ] = −iα

′

2kRV

The electric charge is Q ∼ QL +QR. These are generators of gauge transfor-mations, indeed

δV = −λ[QL, V ] = −iα′

2kLλV

so V → (1 + iα′

2 kLλ)V which is the infinitesimal of V → eiα′2kLλV , or for the

state |V 〉 = V (0), |V 〉 → eiα′2kLλ|V 〉 (similarly for QR).

At the special radius (R =√α′), we have vectors which are charged. They

must combine with the two photons, just like the X± combine with the neu-tral vector in weak interactions (must because we know of no other consistenttheory that has charged vectors).The vectors are

1© = αµ−1|0; k;±1,±1〉, 2© = αµ−1|0; k;±1,±1〉

1© corresponds to the vertex : αµ−1|0, k,−1,−1〉 ∼ : ∂Xµeik·Xe−i2√α′ UL(z)

:(since n = w, kR = 0)

2© corresponds to the vertex : αµ−1|0, k,−1,+1〉 ∼ : ∂Xµeik·Xe−i2√α′ UR(z)

:(since n = −w, kL = 0)Just like with the two photons, they lead to conserved currents.

j±(z) =: eikLUL(z), j±(z) =: eikRUR(z)

where

kL =n

R+wR

α′ = ± 2√α′, kR = ± 2√

α′.

The OPEs are given by


j+(z)j+(0) ∼ : ei 2√

α′ UL(z) :: ei 2√

α′ UL(0) :

∼ e−4

α′ ln(z)−α′2 : e

i 4√α′ UL(0)

∼ z2 : ei 4√

α′ UL(0) : ∼ 0

j−(z)j−(0) ∼ 0

j+(z)j−(0) ∼ z−2 : ei 2√

α′ UL(z)e−i 2√

α′ UL(0) :

= z−2

[

1 + z i2√α′ ∂UL + z2( ) + ...

]

∼ 1

z2+ i

2√α′∂UL

1

z

Define j3 = i√α′ ∂U (the photon!). The OPEs may be expressed in terms of the

photon.

j+(z)j−(0) ∼ 1

z2+

2

zj3(z) + ...

j3(z)j+(0) ∼ − 2

α′

(

−α′

2∂ ln z

)

j+(0) ∼ 1

zj+(0)

j3(z)j−(0) ∼ −1

zj−(0)

Be defining j1 = 12 (j+ + j−), j2 = 1

2 (j+ − j−), we may write all the OPEs inthe form

ja(z)jb(0) ∼ 1

2z2δab + iεabcjc(0).

The corresponding charges

Qa =

∮

dz

2πija(z)

satisfy an SU(2) algebra

[Qa, Qb] = iεabcQc

so the gauge group is SU(2)×SU(2) (which is enlarged fromU(1)×U(1)). Youcan think of this as an abstract three-dimensional space (in fact, two) in whichparticles are free to rotate. U(1) is then a subgroup of SU(2) correspondingto rotations around the z-axis. In fact, the symmetry of the theory has aninfinite number of generators (much like T (z) generated an infinite numberof symmetries through its modes, Ln). Expand

ja(z) =∑

jamz−m−1

6.8 Away fromR =√α′ 17

so that the charges Qa are the zero modes,Qa = ja0 . Then the current algebrais an affine Lie algebra (Kac-Moody)

[jaµ, jbν ] =

m

2δm+n,0δ

ab + iεabcjcm+n.

The constant tells us that ja is not a tensor. It can be shown that a generalalgebra has km

2 δm+n,0δab constant term, k ∈ N (level), so in our case k = 1.

6.8 Away from R =

√α′

Once we realize there is a symmetric at the special radius R =√α′, you can

not ignore it when you move away fromR =√α′. This is becauseR is dynam-

ical and we have already seen that there is a string mode, Φ(k)α−1α−1|0; k〉which mixes withGuu and changesR, much like the graviton gµν(k)(α

µ−1α

ν−1+

αν−1αµ−1)|0; k〉 changes the background metric (in the uncompactified case).

So what happens to the SU(2) symmetry as we move away from R =√α′?

Recall weak interactions...A scalar called the Higgs moves from the unstable symmetric point to a stableminimum of the potential (Mexican hat) So |Φ| goes from 0 to a value 〈|Φ|〉 ∼ν. Φ can settle into any minimum and all positions are equivalent. Howevereach position breaks the symmetry. It is similar to the SUN-EARTH system.The underlying physics (Newton’s law) is rotationally invariant, but the orbitof the Earth is not (it is an ellipse, even as a circle there is an axis that breaksthe symmetry).Recall the action for a scalar.

S =

∫

d4x(

|DµΦ|2 + V (Φ))

.

It contains a term quadratic in the vectorsAiµ, which gives rise to a mass termafter the shift |Φ| → |Φ|+ ν.Back to strings...Massless scalars ∀R : ηµν α

µ−1α

ν−1|0; k〉 (dilaton) α−1α−1|0; k〉 (Guu|). The latter

changes R and corresponds to the vertex

: ∂U(z)∂U(z)eik·X :,

or: j3(z)j3(z)eik·X : .

AtR =√α′, we have additional scalars

1© : ∂UeikLULeik·X : (c.f. vector : ∂Xµeik·XeikLUL :) or : j±j3eik·X :

2© : ∂UeikRUReik·X : (c.f. vector : ∂Xµeik·XeikRUR :) or : j3j±eik·X :

3© : n = ±2, w = 0⇒ kL = kR =n

R= ± 2√

α′, vertex : eiklULeikRURek·X = j±j±eik·X

4© : n = 0, w = ±2⇒ kL = −kR =wR

α′ = ± 2√α′ , vertex : eiklULeikRURek·X = j±j±eik·X


Putting everything together, we have

: ja(z)jb(z)eik·X :

transforming as a (3,3) of SU(2) × SU(2). This is the Higgs (not a doublet,unlike weak interactions). At the symmetric point, all scalars are massless.Away from the symmetric point, all except j3j3 get masses

m2 =

(

R2 − α′

Rα′

)2

,

using

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2)

breaking the SU(2) × SU(2) symmetry, down to U(1) × U(1) (with one (two)massless vectors).

Unlike with weak interactions, we can get arbitrarily close to the symmetricpoint by varying R→

√α′. This shows that the potential contains a flat direc-

tion (it costs nothing to move along this direction) - not a Mexican hat.

Conclusion

There is a underlying gauge symmetry SU(2)× SU(2) in string theory whichis not present in particle theory (KK). Strings see space-time in an unusualway-not yet understood.

6.9 T-duality

Recall

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2).

As R → ∞ , n = 0 states becomes infinitely massive and decouple. w = 0states go to a continuum of small masses so we get ordinary particle theoryat an uncompactified extra dimension. As R → 0, w = 0 states becomes in-finitely massive and decouples. n = 0 states so to a continuum, so this is sim-ilar to theR→∞ limit, i.e., we still have an extra uncompactified dimension,even though it has been shrunk to 0! This behavior generalizes to a symmetryunder

R→ R′ =α′

R.

Spectra are identical if we just interchange (n ↔ w). THM: The theories at R

and R′ = α′

R are identical.

6.9 T-duality 19

Proof: Let us start with theR theory. RecallU = UL+UR. DefineZ = UL−UR.Z has the same OPE’s as U , because they all come in pairs, so the signs cancel(e.g., ZRZR = (−UR)(−U(R)). However,

UL = uL − iα′

2

(

n

R+wR

α′

)

ln z...

UR = uR − iα′

2

(

n

R− wR

α′

)

ln z...

so

Z = uL − uR − iα′

2

[(

n

R+wR

α′

)

ln z −(

n

R− wR

α′

)

ln z

]

+ ...

(6.9.1)

and theR′ theory has

Z = uL + uR − iα′

2

[(

n

R+wR

α′

)

ln z +

(

n

R− wR

α′

)

ln z

]

+ ...

so

U = uL + uR − iα′

2

[(

w

R+nR

α′

)

ln z −(

w

R− nR

α′

)

ln z

]

+ ...

which has the same momentum as Z if we interchange (w ↔ n). QED

The self-dual point is R = R′, which is the special point we discussed before.The set of inequivalent theories lies in the internal [α′,∞), so there is a “min”R =

√α′ from the string point of view.

T-duality, R ↔ α′

R is a Z2 symmetry. It is part of SU(2)× SU(2). Indeed, note

that, if δR = R − Rmin, then for small δR, δR′ = R′ +√α′ = α′

R −√α′ =√

α′

R (√α′ −R) ' −δR since (R '

√α′). δR = j3j3, so reversing its sign means,

e.g., in terms of the first SU(2), a reflection in the 12-plane.

However, to get δR → −δR, we may also rotate (generated by j1) around the1-axis by π. This is an SU(2) transformation. Thus the points

√α′ + δR and√

α′ − δR are gauge equivalent.

R =

√α′

k∼= R = k

√α′

Recall

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2).

For R = k√α′, we have massless scalars withN = N = 0,

m2 =n2

k2α′ +

(

wk2

α′

)2

− 4

α′ = 0⇒ n = ±2k, w = 0.


In the T-dual theory, R =√α′/k, these scalars have n = 0, w = ±2k (w ↔ n).

Therefore, kL = wRα′ = ± 2

α′ = −kR, and the vertex operators are

: eikLULeikRUReik·X :=: e±i2

α′ ULe∓i2

α′ UReik·X :=: j±j∓eik·X :

which is part of the setjajbeik·X :

at the symmetric pointR =√α′! How come? (Total of three: j3j3, j+j−, jij+)

To see the connection you must think of twists. Why? Because you can! Letus start with k = 2 for simplicity. We wish to compare r =

√α′ andR =

√α′/2

(equivalent to R = 2√α′). Consider the expansions

UL = uL − iα′

2

(

n

R+wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m,

UR = uR − iα′

2

(

n

R− wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m.

For

R =√α′ : UL = uL − i

√

α′

2(n+ w) ln z + ...

R =

√α′

2: UL = uL − i

√

α′

2(n+

w

2) ln z + ...

To mimic R =√α′/2 at R =

√α′, we need to (a) n → 2n, (b) w → w/2. (a)

is easier so let us try it first. We need to restrict n to even numbers. This is arestriction on the Hilbert space.c.f. Harmonic Oscillator: H = p2/2m + 1/2 mω2x2. H has eigenvalues (n +1/2)~ω. We can restrict wavefunctions to evenfunctions. This is consistent,because H is even (H commutes with the parity operator). Then H has theeigenvalues (2n+ 1/2)~ω. Given Ψ(x), we can construct the even function

ψeven =1

2(1 + P )ψ(x) =

1

2(ψ(x) + ψ(−x)),

where 1/2(1 + P ) is a projection operator.For strings, the restriction on the Hilbert space (even n) is consistent. Indeen,consider two even-n states,

V =: eikLULeikRUR :, V ′ =: eik′LULeik

′RUR :

The OPE gives

V (z, z)V (z′, z′) ∼ z α′

2kLk

′Le

α′2kRk

′RVkL+kL′ ,kR+k′

R.

The operator similar to parity is (−1)n where n ∼ charge ∼momentum, so itacts just like parity, U → −U .

6.9 T-duality 21

Having completed (a), we turn to (b). This is harder. We need to allow half-integer winding numbers. How can a closed string wind half-way? Answer:Fold the circle U ≡ U + 2πR by identifying point U ≡ −U . There are two fixedpoints: 0 and πR = −πR. This creates a singular “manifold” known as anorbifold. Now the string can wind half-way, say from −πR/2 to πR/2 (thesetwo points are identified, so the string is closed).In general, the ends of the string can be at opposite points, i.e., U(σ + 2π) =−U(σ). We are allowed to impose anti-periodic boundary conditions! Doesthe theory make sense? There is no a priori guarantee that it will, but alas,it does (also note, p = 0, string cannot move away from the fixed point, son = 0). Of course, we need to restrict the Hilbert space again to “even par-ity” states. Again, this is a consistent truncation and the resulting theory isidentical to the theory on a R =

√α′/2 circle.

The truncated theory is called “twisted”. Now let us compare the masslessscalars. In the original R =

√α′ theory we had 9 (3× 3) scalars, : jajbeiκ·X :.

These are indeed the massless scalars atR =√α′/2. The above generalizes to

∀k ∈ N.

Open Strings

Just like closed strings, open strings have a quantized momentum in the com-pact dimensions, p = n

R . However, there is no winding for open strings, sow = 0 (just like KK particles). The mass formula is

m2 =n2

R2+

1

α′ (N − 1).

As R → 0, n 6= 0 states become infinitely massive and decouple. Thus thecompact dimensions disappears. That would be considered normal behavior,were it not for the fact that open strings cannot help but create and interactwith closed strings. The latter exhibit weird behavior (the compact dimensiondoes not disappear as R→ 0). So how does one reconcile the two pictures? Itis easier to think in terms of the R → ∞ limit and we have already seen thatthis is possible with closed strings because of T-duality. Recall that the theoryatR′ = α′/R is equivalent to the theory atR if written in terms ofZ = UL−URinstead of UL + UR.Recall the expansion

U(z, z) = u− iα′p ln |z|2 + i

√

α′

2

∑

m6=0

1

mαm(z−m + z−m).

For compact U , p = n/R.

z = eπi(σ+τ)/`, z = e−πi(σ−τ)/`

We will set ` = π for simplicity. So

∂σUL = ∂σz∂U = iz∂UL, ∂τUL = iz∂UL,∂σUR = ∂σ z∂UR = −iz∂UR, ∂τUR = iz∂UR,


so

∂σZ = iz∂U+iz∂U, ∂τU = iz∂U + iz∂U = ∂σZ

so

Z(σ = π)− Z(σ = 0) =

∫ π

0

dσ∂σZ =

∫ π

0

dσ∂τU =

∫ π

0

dσ∂τ (2α′pτ)

= 2α′pπ = 2α′πn

R= 2πnR′

In other words, the ends of the string lie at the same point in the compactdimension (in terms of the dual coordinate). Including the noncompact di-mensions, this implies that end-points lie on a hyperplane (D-brane).

Notice that translation invariance is broken, or equivalently, the momentumin the compact dimension is not conserved. Since p = n

R and in the dulatheory n ↔ w, this is equivalent to the non-conservation of winding numberin the small R theory. That is obvious. A wound closed string can break intotwo open strings.

Consider massless modes. These are as in the uncompactified case, i.e., thephoton: αµ−1|0; k〉. We need to split it into uncompactified αµ−1|0; k〉 and com-pactified, α−1|0; k〉 components. Corresponding vertices

: ∂τXµeik·XeinU/R : (σ = 0) : ∂τUe

ik·XeinU/R :

but n = 0 for the massless and k2 = 0. The former is a photon tangent to theD-brane. The letter can be written as

: ∂σZeik·X : .

Just like the gravition gµν∂Xµ∂Xνeik·X : contributes to the background and

curves it, the vertexA∂σZeik·X shifts the position of the D-brane z → z+A∂σZ

(∂σz is perpendicular to the D-brane). Therefore, the D-brane is a dynamicalobject. Its fluctuations are described by open strings attached to it.

The D-brane is our Universe! Notice that the photon (and other particles)are confined to the D-brane. No wonder we never wander off into the extradimension(s). On the other hand, gravity has to be present in the extra di-mension, because gravity creates space.

This D-brane fills space. We can imagine more compact dimensions and havep non-compact dimensions. Then we have a Dp-brane.

Scattering

Let us compare open and closed strings. We will need to mix them in order todescribe scattering by D-branes. Recall the operator product expansions forclosed strings

X(z, z)X(0, 0) ∼ −α′

2ln |z|2 + ...

6.9 T-duality 23

We also have∂∂(X(z, z)X(0, 0)) ∼ −πα′δ2(z, z)

so G(z, z) = X(z, z)X(0, 0) is a Green function. ln |z|2 satisfies the boundarycondition of periodicity in σ : z = ei(σ+τ), so σ → σ + 2π ⇒ z → z.It is the electrostatic potential of the uniformly charged straight line. Openstrings, z = ei(σ+τ), but not 0 ≤ σ ≤ π, so z is on the upper-half plane. Theboundary is the real axis and that is where the vertex operators are. The Greenfunction (and the OPE) is found in two steps. First we need to satisfy Neu-mann boundary conditions, ∂σX = 0.This translates into ∂nX = 0 (normal to boundary vanishes), which can besatisfied by adding an image charge at z. This is different from electrostatics,where the tanget needs to vanish, requiring an opposite charge for the image.Thus,

G(z, z; z′, z′) = −α′

2ln |z − z′|2 − α′

2ln |z − z′|2 (6.9.2)

When both z and z′ approach the boundary (i.e., they become real), we obtain

G(z, z; z′z′) = −α′ ln |z − z′|2 = −2α′ ln |z − z′|.

This shows that the OPE for open strings ought to be

X(z)X(0) ∼ −2α′ ln |z|.

Recall the amplitudes:Closed strings:

An = 〈: eik1·X(z1, z1)eik2·X(z2, z2)...e

ikn·X(zn, zn) :〉.

View this as a function of z1 and differentiate. We obtain

∂z1An = 〈: ik1∂Xeik1·X(z1, z1)e

ik2·X(z2, z2)...eikn·X(zn, zn) :〉.

By making use of the OPE ∂X(z) : eik·X (0) :∼ −ik α′

21z eik·X (0) we obtain

∂z1An =α′

2An∑

i

k1kiz1 − zi

.

Integrating ...

An ∝∏

i<j

|zi − zj |α′kikj

which is the holomorphic and antiholomorphic pieces multiplied together.For open strings, a similar argument yields An ∝

∏

i<j |zi − zj |2α′kikj . For

closed string emission from a D-brane, we need to use (6.9.2). (See (6.2.33)Polchinski)


UNIT 7

Superstrings

7.1 Bosons and fermions

Bosonic strings have the action

S =1

2πα′

∫

d2z∂Xµ∂Xµ.

We wish to build a theory that has supersymmetry (SUSY). Why? It turns outthat this is the only (known) way of obtaining a consistent theory.

For SUSY, each boson (commuting field), must have a fermionic (anticom-muting) counterpart. We have already seen anticommuting fields. We calledthem b, c. Recall the b, c action

Sbc =1

2π

∫

d2zb∂c.

and their OPEs are

b(z)c(0) ∼ 1

z.

The wave equation was given by ∂b = ∂c = 0, i.e., b and c are purely holomor-phic. The energy-momentum tensor is

T =: (∂b)c : −λ∂(: bc :)

where we assume the weights hb = λ, hc = 1 − λ. The OPE for the energy-momentum tensor is

T (z)T (0) ∼ c

2z4+

2

z2T (0) +

1

z∂T (0)

where c = −3(2λ − 1)2 + 1. Earlier we required λ = 2, so c = −26 (henceD = 26 for the bosonic string) in order to do BRST quantization properly

26 UNIT 7: Superstrings

(Q2BRST = 0). A more symmetric choice is λ = 1

2 . Then hb = hc = 12 and c = 1.

Define

b =1√2(ψ1 + iψ2), c =

1√2(ψ1 − iψ2).

Then the action is

S =1

2π

∫

d2zb∂c =1

4π

∫

d2z(ψ1∂ψ1 + ψ2∂ψ2).

The stress-energy tensor written in terms if the new fields may be expressedas

T (z) = −1

2ψ1∂ψ1 −

1

2ψ2∂ψ2.

The system splits into two identical copies. Since c = 1, for the two together,each system has c = 1

2 .Pick one such system, ψ = ψ1, say. Make D copies of it, ψ → ψµ (µ =0, 1, ..., D− 1) and let us try ψµ as a SUSY partner of Xµ.The stress-energ tensor is given by

T = − 1

α′ ∂Xµ∂Xµ −

1

2ψµ∂ψµ.

The TT OPE becomes

T (z)T (0) ∼ (3D/2)

2z4+

2

z2T (0) +

1

z∂T (0)

where we used

Xµ(z, z)Xν(0, 0) ∼ −α′

2ηµν ln |z|2, ψµ(z)ψν(0) ∼ 1

zηµν .

T (z) is a conserved current that generates conformal transformations whichare symmetries of the theory (in fact v(z)T (z) is conserved for arbitrary v(z),leading to an infinite number of symmetries). The new theory (Xµ, ψµ) haseven more symmetries! Let us define a supercurrent as

TF = i

√

α′

2ψµ(z)∂Xµ(z)

Any η(z)TF (z) is conserved and generates a symmetry mixingXµ and ψµ (su-perconformal transformation) - ηmust be anticommuting so that ηTF is com-muting. To see this consider

TF (z)Xµ(0, 0) ∼ −i√

α′

2

α′

2

1

zψµ(0) = −i

√

α′

2

1

zψµ(0)

TF (z)ψµ(0) ∼ i

√

2

α′1

z∂Xµ(0, 0)

7.1 Bosons and fermions 27

So

δXµ = −iε∮

dz

2πiη(z)TF (z)Xµ(0, 0) = −

√

α′

2εηψµ(0)

δψµ = −iε∮

dz

2πiη(z)TF (z)ψµ(0) = −

√

2

α′ εη∂Xµ(0, 0)

The other OPEs are given by

T (z)TF (0) ∼ 2

(

− 1

a′

)

(

i

√

α′

2

)

(

α′

2∂2 ln |z|

)

∂Xµ(z)ψµ(0) +

(

−1

2

)

(

i

√

α′

2

)

(

∂1

z

)

ψµ(z)∂Xµ(0)

+

(

−1

2

)

(

i

√

α′

2

)

1

z∂ψµ(z)∂X

µ(0)

T (z)TF (0) ∼ 3

2z2TF (0) +

1

z∂TF (0)

TF (z)TF (0) ∼(

i

√

α′

2

)21

z

(

α′

2∂2 ln |z|

)

D +

(

i

√

α′

2

)21

z∂Xµ∂Xµ

+

(

i

√

α′

2

)2(

α′

2∂2 ln |z|

)

ψµ(z)ψµ(0)

∼ D

z3+

2

zT (0).

The first OPE shows that TF has weight h = 3/2. There is a correspondingconstruction for the anti-holomorphic operators. Since ψµ is holomorphic,we need to add a new anti-holomorphic fermionic field ψµ(z) with the action

S =1

4π

∫

d2zψµ∂ψµ.

The wave equation is given by

∂ψµ = 0,

so, indeed ψµ is anti-holomorphic. They OPE is

ψµ(z)ψν(0) ∼ 1

zηµν .

The stress-energy tensors are

T = −1

2ψµ∂ψµ, TF = i

√

α′

2ψµ∂Xµ.

The OPEs are similar to the OPEs of their holomorphic counterparts. Noticethat the central charge for this theory is c = 3D/2. This is now a superconfor-mal theory (N = 1, N = 1 where N, N counts the number of TF , TF ’s). Otherexamples


7.2 The ghosts

Recall,

Sbc =1

2π

∫

d2zb∂c, T = (∂b)c− λ∂(bc), b(z)c(0) ∼ 1

z.

The weights and central charge for the bc system are

hb = λ, hc = 1− λ, cbc = −3(2λ− 1)2 + 1.

Since (b, c) are anti-commuting fields, their partners will have to be commut-ing. We have already met them. They are the (β, γ) fields with action

Sβγ =1

2π

∫

d2zβ∂γ,

which is the same action as the bc action. Let hβ = λ′, hγ = 1 − λ′. Thecombined system will have SUSY if we can find a TF that mixes b, c with β, γ.Such a TF will most likely contain a (∂β)c (c.f. (∂b)c in Tbc and (∂β)γ in Tβγ ,

Tβγ = (∂β)γ − λ′∂(βγ).

Since h = 3/2 for TF , we need 1 + λ′ + 1 − λ = 3/2, i.e., λ′ = λ − 1/2. Thecentral charge is

cβγ = 3(2λ′ − 1)2 − 1 = 3(2λ− 2)2 − 1.

The central charge for the combination of the two systems becomes

ctotal = cbc + cβγ = −3(2λ− 1)2 + 3(2λ′ − 2)2 = 3(3− 4λ).

For the special (interesting) case λ = 2, in which cbc = −26 (hence d = 26 forbosonic strings), we have ctotal = 3(3−4×2) = −15. If we combine this systemwith the (Xµ, ψµ, ψµ), for which c = 3D/2 and demand ctotal = 0, we need3D/2− 15 = 0⇒ D = 10. Therefore superstrings must live in 10-dimensions.

Linear Dilaton

Recall

T (z) = − 1

α′ ∂Xµ∂Xµ + Vµ∂

2Xµ,

where Vµ is a fixed vector (breaking translational invariance). The centralcharge for this theory is

c = D + 6α′V µVµ.

By adding the fermion ψµ, with T = − 12ψ

µ∂ψµ and c = D/2, we obtain

c =3D

2+ 6α′V µVµ,

and

TF = i

√

2

α′ψµ∂Xµ − i

√2α′Vµ∂ψ

µ.


7.3 Mode Expansions

Let us do closed strings first. Recall the expansion

∂Xµ(z) = −i√

α′

2

∑

m

αµmz−m−1,

where αµ0 =√

α′

2 pµ and [αµm, α

νn] = mηµνδm+n,0. Xµ obeys periodic boundary

conditions. We could have imposed anti-periodic boundary conditions onXµ, and we did so with U (the compactified coordinate) and got an orbifold,but this breaks translational invariance. That is ok for dimensions we cannotsee (e.g., compactified), but not for the four dimensions that describe ourspace-time. ψµ and ψµ on the other hand have no such concerns (also notethe absence of a spin-statistics theorem in two-dimensions), so we have twopossibilities.

• anti-periodic boundary conditions (Neveu-Schwarz (NS)):ψµ(σ+2π) =−ψµ(σ).

• periodic boundary conditions (Ramond (R)): ψµ(σ + 2π) = ψµ(σ).

These have two distinct Hilbert spaces (sectors). There are also two Hilbertspaces for ψµ, so in all there are four Hilbert spaces (sectors): NS-NS, R-NS,NS-R, R-R.Let us first describe ψµ in NS. ψµ is a function of σ + τ . When expandingin Fourier modes, because of anti-periodicity, only the terms e−i(2m+1)(σ+τ)/2

contribute (since σ → σ+2π ⇒ e−i(2m+1)(σ+τ)/2 → e−πi(2m+1)e−i(2m+1)(σ+τ)/2)Define r = m+ 1/2 ∈ Z + 1/2, then

ψµ(σ + τ) =√i∑

r∈Z+ 12

ψµr e−ir(σ+τ)

where the factor of√i was introduced for convenience. Transforming to the

z-picture, z = ei(σ+τ), we have

ψµ(z) =

(

∂w

∂z

)h

ψµ(σ + τ)

=1√izψµ(σ + τ)

=∑

r∈Z+ 12

ψµr z−r− 1

2

which is a Laurent expansion. We saw the same in terms of the Xµ field. Weobtain anti-commutation relations of the ψµ fields by analyzing the OPE

ψµ(z)ψν(0) ∼ 1

zηµν .


The anti-commutation relations are

ψµr , ψνs = ηµνδr+s,0.

We find similar results for the right-moving sector

ψµ(z) =∑

r∈Z+ 12

ψµr z−r−1

2 , ∂Xµ(z) = −i√

α′

2

∑

m∈Z

αµmz−m−1,

and the anti-commutation relations are

ψµr , ψνs = ηµνδr+s,0.

The stress-energy tensor is

T (z) =∑

m∈Z

Lmz−m−2, h = 2.

The OPE gives the Virasoro algebra with central extension

[Lm, Ln] = (m− n)Lm+n +c

12m(m− 1)(m+ 1)δm+n,0.

In terms of the OPEs we find

TF (z)TF (0) ∼ 3

2z2TF (0) +

1

z∂TF (0).

We may expand TF (z) in terms of modes

TF (z) =∑

r∈Z+ 12

Grz−r− 3

2 .

Recall

[Lm, Gr] = ((h− 1)m− r)Gr+m = (1

2m− r)Gr+m.

Finally

TF (z)TF (z′) ∼ D

(z − z′)3 +2

z − z′T (z′),3D

2= c, so D =

2c

3.

Find the anit-commutator Gr, Gs in two steps. First

Gr =

∮

dz

2πirr+

12TF (z),

and∮

dz

2πizr+

12 TF (z)TF (z′) =

∮

dz

2πizr+

12

D

(z − z′)3 + 2z′r+12 T (z′)


f(z) = zr+12 , f ′(z) =

(

r +1

2

)

rr−12 , f ′′(z) =

(

r2 − 1

4

)

zr−32 ,

so∮

dz

2πizr+

12

D

(z − z′)3 =D

2

(

r2 − 1

4

)

z′r−32

Second step: apply∮

dz′

2πiz′s+ i

2 to isolateGs:

Gr, Gs =D

2

(

r2 − 1

4

)∮

dz′

2πiz′r+s−1 + 2

∮

dz′

2πiz′r+s+1T (z′)

= 2Lr+s +D

2

(

r2 − 1

4

)

δr+s,0

= 2Lr+s +c

12

(

4r2 − 1)

δr+s,0

The algebra of (Lm, Gr) closes, as expected: NS algebra. Next, let us study themode expansion: using

∂Xµ = −i√

α′

2

∑

m∈Z

αµmz−m−1, ψµ =

∑

r∈Z+ 12

ψµr z−r−1

2 ,

and

T (z) = − 1


1

2ψµ∂ψµ = − 1


1

4(ψµ∂ψµ − (∂ψµ)ψµ)

we have

Lm =

∮

dz

2πizm+1T (z) =

1

2

∑

n,n′

∮

dz

2πiαµnαn′µz

−n−n′−m−1 +1

4

∑

r,r′

∮

dz

2πiψµr ψr′µ(r − r′)z−r−r

′+m−1

=1

2

∑

n∈Z

αµm−nαnµ +1

4

∑

r∈Z+ 12

(2r −m)ψµm−rψrµ.

TF (z) = i

√

2

α′ψµ∂Xµ ⇒ Gr =

∮

dz

2πizr+

12TF (z) =

∑

n,r′

∮

dz

2πiαµnαr′µz

−n+r+r′−1 =∑

n∈Z

αµnψr−n µ.

Normal ordering: No question in Gr, ∀ r and Lm, ∀m 6= 0. Potential problemwith L0. After normal ordering, we get L0 + a where a is a constant to bedetermined. To determine a, look at [L+, L−1] = 2L0. We have L1|0〉 = 0, so〈0|[L+, L−1]|0〉 = 〈0[L+1L−1]|0〉 = ||L−1|0〉||2, because L+

−1 = L1.Now L−1|0〉 = 1

2

∑

αµ−1−nαnµ|0〉 + 14

∑

(2r + 1)ψµ−1−rψrµ|0〉 There are non-vanishing terms only if−n− 1 < 0 n < 0, i.e., 0 < n < −1 which is impossible!Also 0 < r < −1, which implies r = 1/2, but then 2r+1 = 0, so it also vanishes.Therefore

L−1|0〉 = 0, ||L−1|0〉||2 = 0,

so〈0|2L0|0〉 = 2a = 0 ⇒ a = 0.

In the above, we used αµn|0〉 = ψµr |0〉, n, r > 0, and the hermicity property,(αµ−n)

† = αµn, (ψµ−r)† = ψµr .


The ghosts

The ghost system (b, c;β, γ) is a superconformal system on its own right. It isopposite to (Xµ, ψµ) in that the role of Xµ is played by the fermionic (b, c).So b, c, obey periodic boundary conditions (necessary due to definition ofQBRST). Then (β, γ) may obey periodic (R) or anti-periodic (NS) boundaryconditions. Let us do NS first. Recall

hb = λ, hc = 1− λ, hβ = λ′, hγ = 1− λ′, λ′ = λ− 1

2.

We are interested in the λ = 2 case, in order to couple this system to the(Xµ, ψµ) system. Then

hb = 2, hc = 1− 1, hβ =3

2, hγ = −1

2,

and the expansions are

b =∑

m∈Z

bmz−m−2, c =

∑

m∈Z

cmz−m+1, β =

∑

r∈Z+ 12

βrz−r− 3

2 , γ =∑

r∈Z+ 12

γrz−r+ 1

2 .

From the operator product expansions, we get standard (anti) commutators

bm, cn = δm+n,0, [γr, βs] = δr+s,0.

bm, cm, βr, γr are all annihilation operators for r,m > 0. Recall the subtletywith the zero modes b0, c0, satisfying b0, c0 = 1. We have two choices for thevacuum. Choose c0|0〉 = 0. The conformal generators are

Lm =

∮

dz

2πizm+1T (z),

T (z) = (∂b)c− λ∂(bc) + (∂β)γ − λ′∂(βγ)

= (∂b)c− 2∂(bc) + (∂β)γ − 3

2∂(βγ)

=∑

n,n′

(−n′ − 2)bn′z−n′−3cnz

−n+1 − 2(−n− n′ − 1)bn′cnz−n−n′−2

+∑

r,r′

(

−r′ − 3

2

)

βr′z−r′− 5

2 γrz−r+1

2 − 3

2(−r − r′ − 1)βr′γrz

−r−r′−2

=∑

n,n′

(n′ + 2n)bn′ccz−n−n′−2 +

1

2

∑

r,r′

(3r + r′)βr′γrz−r−r′−2

So

Lm =

∮

dz

2πizm+1T (z) =

∑

n

(m+ n)bm−ncn +1

2

∑

r

(m+ 2r)βm−rγr

7.4 Open Strings 33

The SUSY generators are

Gr =

∮

dz

2πizr+

12TF (z), TF (z) = −1

2(∂β)c+ λ′∂(βc)− 2bγ

where

TF (z) =∑

s,n

−1

2

(

−s− 3

2

)

βsz−s− 5

2 cnz−n+1 +

3

2

(

−s− n− 1

2

)

βscnz−s−n− 3

2 − 2bnγsz−n−s− 3

2

=∑

s,n

−1

2(2s+ 3n)βscnz

−n−s− 32 − 3bnγsz

−n−s− 32 .

So

Gr =

∮

dz

2πizr+

12TF (z) = −

∑

n

1

2(2r + n)βr−ncn + 2bnγr−n.

Normal ordering: again, only L0 has a problem; should be L0 + a. To find a,consider [L1, L−1] = 2L0.

L−1 =∑

n

(n− 1)b−1−ncn +1

2

∑

r

(2r − 1)β−1−rγr.

When applied to the ground state, |0〉, only the terms n = −1, r = −1/2contribute (recall c0|0〉 = 0), soL−1|0〉 = −2b0c−1|0〉−β−1/2γ−1/2|0〉. Similarly,we obtain 〈0|c−1 = 〈0|(2b1c0 + β1/2γ1/2). So

〈0|[L1, L−1]|0〉 = 〈0|L1L−1|0〉 = −2〈0|(b1c0b0c−1−β 12γ 1

2β− 1

2γ− 1

2|0〉 = −2+1 = −1.

So 2a = 〈0|2L0|0〉 = −1, so a = −1/2 (−1 from the bc and 1/2 from the βγ).

7.4 Open Strings

Open strings do not have independent ocsillators αµn, αµn. Instead, αµn = αµn.

Thus,

∂Xµ(z) = −i√

α′

2

∑

m

αµmz−m−1, ∂Xµ(z) = −i

√

α′

2

∑

m

αµmz−m−1.

where αµ0 =√

2α′pµ (c.f. αµ0 =√

α′

2 pµ for closed strings). Similarly for the

ψµ’s:

ψµ(z) =∑

r

ψµr z−r−1

2 , ψµ(z) =∑

r

ψµr z−r− 1

2 .


The spectrum

For a physical state, |ψ〉, we demand

Ln|ψ〉 = Gr|ψ〉 = 0, for r, n > 0.

Also, L−n|ψ〉, Gr|ψ〉 are orthogonal to all physical states |ψ′〉 : 〈ψ′|L−n|ψ〉 =〈ψ|Ln|ψ′〉 = 0, and similarly for Gr|ψ〉. They are in the equivalence class ofzero. Check also L−n|ψ〉 is null: ||L−n|ψ〉||2 = 0. Physical states also obey theconstraint

(

L0 −1

2

)

|ψ〉 = 0

i.e., the Hamiltonian H = L0 − 1/2 = 0 (vanishes).We build the Hilbert space by applying αµ−n, ψ

µ−r oscillators only (no ghost

modes- the lead to states in the same equivalence classes as above) to theground state.

H =1

2

∑

n∈Z

: αµ−nαnµ : +1

2

∑

r

r : ψµ−rψrµ : −1

2

plus the ghost oscillators, but they do not contribute. Since α0 =√

2α′pµ foropen strings, we have

H = α′p2 +N − 1

2, N =

∞∑

n=1

αµ−nαnµ +

∞∑

r= 12

rψµ−rψrµ.

The lowest state: |0; k〉 for which N = 0, so α′k2 − 1/2 = 0, so m2 = −k2 =−1/2α′, a tachyon!So we still have a tachyon. This was to be expected, because we took thebosonic theory and enlarged it therefore we should expect the new SUSY the-ory to contain all the states of the bosonic theory and more.The next state: |1〉

Aµ(k)ψµ

− 12

|0; k〉

has N = 12 . We see that this is a massless state since

α′k2 +1

2− 1

2= 0, ⇒ m2 = −k2 = 0.

Also, Gr =∑

n αµnψr−nµ, so when G1/2 acts on our state, only the n = 0 term

contributes. So

G 12|1〉 = αµ0Aµ(k)|0; k〉 =

√2α′k ·A|0; k〉 = 0, ⇒ k · A = 0,

i.e., transverse polarization. Also note that this is a null state:

G− 12|0; k〉 = αµ0ψ− 1

2µ|0; k〉 =

√2α′k · ψ− 1

2|0; k〉

7.4 Open Strings 35

i.e., the state with longitudinal polarization is null (and orthogonal to all phys-ical states).Thus the massless state is aD−2 = 8 dimensional vector. It transforms underthe group SO(8). For closed strings, the situation is similar.TheH = 0 constraint translates into L0 = L0 = 0, i.e.,

α′

4p2 +N − 1

2=α′

4p2 + N − 1

2= 0

Notice the difference in α′p2 → α′

4 p2, which is due to the different definitions

of αµ0 between closed and open string.

The lowest state: |0; k〉with α′

4 k2 = 1

2 , som2 = −k2 = − 2α′ which is a tachyon!

The next level: Aµνψµ−1/2ψ

ν−1/2|0; k〉, with α′

4 k2 = 0, i.e., m2 = 0. This de-

omposes into a scalar, an antisymmetric tensor, and a traceless symmetrictensor:

Aµν =1

D − 2Aρρηµν +

1

2(Aµν −Aνµ) +

1

2(Aµν + Aνµ +

2

D − 2Aρρηµν .

SectionGetting rid of the tachyon Comparing the tachyon with the masslessstates, there is a clear difference: the tachyon has one less fermionic excita-tion then the massless states. If we select the states with as odd number offermionic excitations, that will get rid of the tachyon. This is similar to theharmonic oscillator, where we could select, e.g., all the odd states and stillhave a perfectly well defined physical system.The operator that did the trick there was P (parity) which commuted withthe Hamiltonian and could therefore be simultaneously diagonalized with it.Here we need to find an operator that has two eigenvalues and commuteswith all generators of space-time symmetries (not just the Hamiltonian). Thespace-time symmetries from the Lorentz group (Poincare group rather, but

Lorentz suffices). Let us review briefly. The angular momentum ~L = ~r × ~p. Interms of components we have

Lx = ypz − zpy, Ly = zpx − xpz , Lz = xpy − ypx.

where x and p obey the commutation relations [xi, pj ] = δij .

Define the antisymmetric tensorLij = xipj−xjpi, thenLi = 12εijkLjk . An an-

tisymmetric tensor is a vector in three-dimensions. Not so in four-dimensions.So generalize Lij → Lµν = xµpν − xνpµ, [xµ, pν ] = iηµν which includes time.~L generates rotations:

δxi = − i2ωkl[Lkl, xi] = ωijxj

where ωij is an anti-symmetric tensor. In terms of the vector ~ω we have δ~x =~ω × ~x. This generalizes to Lµν : δxµ = ωµνx

ν . For e.g., L01, we have δt =ω01x, δx = −ω01t, a boost! L0i is a boost in the xi-direction. The algebra of


these Lorentz generators is

[Lµν , Lρσ] = [xµpν − xνpµ, xρpσ − xσpρ]= i(ηνρLµσ − ηµρLνσ − ηνσLµρ + ηµσLνρ)

Lie algebra of SO(3, 1), or in D-dimensions, SO(D − 1, 1). Introduce spinors:we need to add a piece to Lµν that will rotate the spinor (or boost it). Callthis piece Σµν . It needs to satisfy the same SO(D − 1, 1) algebra and willcommute with Lµν by constuction (since Lµν involves space-time and Σµνinvolves fermionic operators).Guess:

Σµν = −i∑

r

ψµr ψν−r = − i

2

∑

r

[ψµr , ψν−r].

Then the algebra is

[Σµν ,Σρσ ] = −1

4

(

∑

r

[ψµr , ψν−r],

∑

s

[ψρs , ψσ−s]

)

= −(

∑

r

ψµr ψν−r,∑

s

ψρs , ψσ−s

)

= i(ηνρΣµσ − ηµρΣνσ − ηνσΣµρ + ηµσΣνρ)

where we used ψµr , ψνs = ηµνδr+s,0.Σµν generates Lorentz transformations on the fermionic fields ψµ(z). No-tice that in D = 10, there are five operators that commute with each other:Σ01, Σ23, Σ45, Σ67, Σ89 (trivial - they contain different ψµr modes). They canbe simultaneously diagonalized. How do they act? Let us be specific and con-sider Σ23. It acts on ψ2

r , ψ3r as follows:

[

Σ23, ψ2r

]

= −i∑

s

[ψ2sψ

3−s, ψ

2r ] = −

∑

s

ψ2s , ψ

2rψ3

−s = iψ3r

[

Σ23, ψ3r

]

= −[Σ32, ψ3r ] = −iψ2

r

Eigenstates: ψ2r + iψ3

r , ψ2r − iψ3

r .

[

Σ23, ψ2r + iψ3

r

]

= ψ2r + iψ3

r eigenvalue : +1[

Σ23, ψ3r − iψ3

r

]

= −ψ2r + iψ3

r eigenvalue : −1

Consider a finite transformation (rotation) U(θ) = eiθΣ23

. Then U(θ)(ψ2r +

iψ3r)U

†(θ) = eiθ(ψ2r + iψ3

r).Proof:

Σ23(ψ2r + iψ3

r) = (ψ2r + iψ3

r)(1+Σ23)⇒ (Σ23)n(ψ2r + iψ3

r = (ψ2r + iψ3

r)(1+Σ23)n

⇒ U(θ)(ψ2r + iψ3

r)U†(θ) = (ψ2

r + iψ3r)e

iθ(1+Σ23) = eiθ(ψ2r + iψ3

r)U(θ).

7.4 Open Strings 37

Similarly, U(θ)(ψ2r + iψ3

r)U†(θ) = e−iθ(ψ2

r + iψ3r). In particular, for θ = π,

the action of U(π) on both ψ2r ± iψ3

r is the same. Therefore U(π)ψ2,3r U †(π) =

eiπψ2,3r = −ψ2,3

r , i.e., U(π) and ψ2,3r anti-commute!

On the other handU(π) commutes with all otherψµr , µ 6= 2, 3. ThusU(π) onlyhas two eigenvalues, ±1, like parity! If a state has an even number of ψ2

−r, 3’s

(r > 0), then it belongs to eigenvalue +1 - with an odd number of ψ2,3−r ’s, it has

U(π) = −1. E.g.:

ψ2−r|0〉 : U(π)ψ2

−r |0〉 = −ψ2−rU(π)|0〉 = −ψ2

−r|0〉 : (−1)

U(π)ψ2−r1ψ

3−r2 |0〉 = −ψ

2−r1U(π)ψ3

−r2 |0〉 = ψ2−r1ψ

3−r3 |0〉 (+1)

etc.We can do the same with all other Σ’s. Thus we have

U1(π) = eπΣ12

, U2(π) = eiπΣ23

, U3(π) = eiπΣ45

, U4(π) = eiπΣ67

, U5(π) = eiπΣ89

.

Notice that U1(π) has no i in the exponential. This is because ψ0r , ψ

0s =

−δr+s,0. The product

U1(π)U2(π)...U5(π) = eiπ(−iΣ01+Σ23+Σ45+Σ67+Σ89) = eiπF .

This anti-commutes with allψµr . F is a fermion number operator. eiF will playthe role of parity in the harmonic oscillator case. Correction: eiπFVgh will. Vgh

is the ghost contribution. Since there are no ghost oscillators, all it does is acton the vacuum: Vgh|0〉 = −|0〉. Thus restrict Hilbert space to eigenstates ofeiπFVgh of eigenvalue +1 (invariant states). This gets rid of the tachyon, foreiπFVgh|0; k〉 = −|0; k〉 but keeps all massless states ψµ−1/2|0; k〉.

Consistent truncation

Since eiπF is made of Lorentz generators it is guaranteed to be conserved bythe OPEs of vertex operators. So even states will produce even states whenthey interact with other even states.Thus, we now have a consistent string theory without a tachyon! Or do we?We still need to check modular invariance. TheXµ part of the partition func-tion is modular invariant by itself,

ZX(τ) =

(

1

2π√α′τ2|η(q)|−2

)D

, η(q) = q1/24∞∏

n=1

(1− qn), q = e2πiτ .

the fermionic part of the partition function is similarly calculated. The resultis a Jacobi-theta function. But, alas, it is not modular invariant. This can beseen without doing any calculation as follows.Before we demanded ψµ(σ + 2π) = −ψµ(σ) (anti-periodic boundary condi-tions). On a torus, we demand ψµ(z + 2π) = −ψµ(z) and also ψµ(z + 2πτ) =−ψµ(z). But then,

ψµ(z + 2π(τ + 1)) = −ψµ(z + 2πτ) = +ψµ(z).


Therefore the transformation τ → τ + 1 changes the boundary conditionsto periodic! Therefore τ → τ + 1 is not a symmetry of the theory. Our the-ory is not modular invariant. The above argument also shows how to fix thetheory. We need to include (somehow) the sector in which ψµ obeys periodicboundary conditions. That is the Ramond sector and we study it next.

7.5 The Ramond (R) sector

The R-sector can only exist in two-dimensions, because there is no spin-statisticstheorem there. The mode expansion is

ψµ(z) =∑

n∈Z

ψµnz−n− 1

2

indices are integers, since ψµ(z) is periodic. The expansion has a factor ofz−1/2, because the weight of ψµ is h = 1/2. Therefore this is not a Laurentexpansion and has a branch cut. We still have

ψµm, ψνn = ηµνδm+n,0

as in the NS-sector. We also have the same algebra for Lm, Gr (note it is nowGm, m ∈ Z).

Normal ordering

We only have a problem with L0. Using [L1, L−1] = 2L0, we have

2〈0|L0|0〉 = 〈0|L+1L−1|0〉,

L−1|0〉 =(

1

2

∑

n

αµ−1−nαnµ +1

4

∑

n

(2n+ 1)ψµ−1−nψnµ

)

|0〉

For the α’s we need −1 − n, n < 0, so −1 < n < 0, which is impossible. Forthe ψ’s, we need−1− n, n ≤ 0, so−1 ≤ n ≤ 0, so n = 0, or n = −1. Therefore

L−1|0〉 =1

4

(

−ψµ0ψ−1µ + ψµ−1ψ0µ

)

|0〉 = 1

2ψµ−1ψ0µ|0〉.

Therefore

〈0|L1L−1|0〉 =1

4〈0|ψ0νψ

ν1ψ

µ−1ψ0µ|0〉,

=1

4〈0|ψµ0ψ0µ|0〉

=1

8〈0|ψµ0 , ψ0µ|0〉

=D

8

Therefore 〈0|L0|0〉 = D/16 = a (i.e., L0 =: L0 : −D/16).

7.5 The Ramond (R) sector 39

The ghosts

b =∑

m∈Z

bmz−m−2, c =

∑

m∈Z

cmz−m+1, β =

∑

r∈Z+ 12

βrz−r−3

2 , γ =∑

r∈Z+ 12

γrz−r+ 1

2 .

where β, γ are not Laurent expansions. The algebras are

bm, cn = δm+n,0, [γm, βn] = δm+n,0,

which are the same as before, but in addition, the zero modes: [γ0, β0] = 1,i.e., γ0, β0 are creation and annihilation operators respectively. This define|0〉 by bm|0〉 = 0, m > 0, βm|0〉 = 0, m ≥ 0 and cm|0〉 == gm|0〉 = 0 form > 0.

Normal ordering

L0 again has a problem. We can solve as we did before.

L−1|0〉 =

(

∑

n

(n− 1)b−1n−1cn +1

2

∑

n

(2n− 1)β−n−1γn

)

|0〉

=

(

−b−1c0 −1

2β−1γ0

)

|0〉

There is only one possibility since −1 < n ≤ 0, so

〈0|L1L−1|0〉 = −〈0|b0c1b−1c0|0〉 −1

4〈0|β0γ1β−1γ0|0〉

= −1− 1

4= −5

4

and

〈0|L0|0〉 =1

2〈0|L1L−1|0〉 = −

5

8= a.

The spectrum

First observe that the defintion |0〉 is ambiguous. Indeed |0〉 is defined byψµm|0〉, m > 0. But then ψµ0 |0〉 is as good as |0〉, for ψνmψ

ν0 |0〉 = −ψν0ψµm|0〉 =

0, m > 0. the ground state is then a representation of the algebra of the zeromodes, ψµ0 , ψν0 = ηµν (Clifford - Dirac algebra). |0〉 therefore is a spinor.Instead of one spin, here we have five, because we are in ten-dimensions.The spin operators are Σ01,Σ23,Σ45,Σ67,Σ89. They commute with each otherso they can be simultaneously diagonalized. We can then define a basis ofground states |s1, s2, s3, s4, s5〉 where si = ±1/2(i = 1, 2, 3, 4, 5). We will usethe notation ~s = (s1, s2, s3, s4, s5). There are 25 = 32 such states (c.f. 22 = 4states in the four-dimensional Dirac spinor). All states built from |~s〉 have


integer +1/2 spin, because ψµ−m has spin one (eigenstate of SXXXX+1 witheigenvalue +1). to be contrasted with NS-sector where all states have integerspin. Thus, the inclusion of the R-sector is important, because we need allspins to describe Nature.

The Hamiltonian (L0) has const. D/16 − 5/8 = 10/16 − 5/8 = 0, so H =α′p2 +N (c.f. H = α′p2 +N − 1/2 in the NS-sector)

The lowest level: N = 0, soH = 0 andm2 = −p2 = 0. There is no tachyon! Thelowest states, |0; k〉 are massless! Non-trivial constraint: G0|~s; k〉 = 0. Relevantpiece: G0 =

√2α′pµψ

µ0 , so kµψ

µ0 |~s; k〉 = 0 which is the Dirac equation (γµ =

1√2ψµ0 , then kµψ

µ|~s; k〉 = 0). Notice also that the algebra G0, G0 = 2L0, i.e.,

G20 = L0. G0 is the square root of the Hamiltonian!

This is just like in the Dirac case. It is also a generic feature of a SUSY theory:the Hamiltonian can be written as the square of a SUSY charge.

Notice that this also implies that the ground state has zero eigenvalue, be-cause G0|0〉 = 0, which makes it very hard to have a finite cosmological con-stant in a SUSY theory. In terms of the fields, the contribution of the bosonalways exactly cancels the contribution of the fermions (due to SUSY boson↔ fermion) and we get zero vacuum expectation energy (cosmological con-stant). The R-sector can also be split into two eigenspaces of eiπF with eigen-values±1. The ground states belongs to +1.

7.6 Superstring Theories

We may now combine the NS and R-sectors to form a consistent superstringtheory. We need to have analycity in the OPEs (which is not guaranteed inthe R-sector, due to branch cuts in the expansions of the fields). This severelyconstrains the possibilities (we also do not want a tachyon) to ...

IIA : (NS+, NS+) (R+, NS+) (NS+, R−) (R+, R−)IIB : (NS+, NS+) (R+, NS+) (NS+, R+) (R+, R+)IIA′ : (NS+, NS+) (R−, NS+) (NS+, R+) (R−, R+)IIB′ : (NS+, NS+) (R−, NS+) (NS+, R−) (R−, R−)

It can be shown that IIA′ is the same as IIA (also, similarly, IIB′ is the sameas IIB)

Proof: Transform X9 → −X9, ψ9 → −ψ9, ψ9 → −ψ9. Then eiπS89 → e−iπS

89

(same eigenvalue), but S89|0〉 → −S89|0〉, so the sign is reversed in the R-sector (S89 annihilates the NS vacuum (no zero modes), so no change there).Therefore this transformation maps R+→ R− and vice versa. QED

Open Strings

Only one possibilty: type I: NS+,R+. The projection of eigenspaces of eiπF and

eiπF is known as the Gliozzi-Scherk-Olive (GSO) projection.

7.6 Superstring Theories 41

The resulting theories turn out to have space-time SUSY and obey the spin-statisics theorem (which has to be obeyed for D > 2). The fact that space-time SUSY and the spin-statistics theorem emerge is rather unexpected. Onewould expect that these two should be evident from the start - built in formal-ism. This fact remains elusive.

Modular Invariance

We have already seen that modular invariance for the NS-NS sector alonecannot possibly work. Now we have a multitude of sectors and a hope thatmodular transformations will map one onto others and somehow the combi-nation will be invariant. Let us start with the NS-sector. Only the NS+ subsec-tor appears. The partition function for the Xµ’s is the same as before and wehave already established it is modular invariance, so we will concentrate onthe ψµ’s.

The partition function is as always

ZNS+ = Tr (qH ), q = e2πiτ .

If |ψ〉 is in NS+, then eiπF |ψ〉 = |ψ〉. To find such a |ψ〉, we can start with anarbitrary state |ψ′〉 and project onto the eigenspace of eπiF of eigenvalue +1.The projection operator is

P =1

2(1 + eiπF ), P 2 = P.

Also, eiπFP |ψ′〉 = P |ψ′〉, so eigenvalue +1. Thus, to compute the Tr NS(PA) =12Tr NSA+ 1

2Tr NS(eiπFA). First trace: for each µ, we have the creation oper-ators ψµ−r, r > 0 where of course r ∈ Z + 1

2 . A state can have 0 or 1 ψµ−r, since(ψµ−r)

2 = 0 (fermionic mode). So for fixed r, µ we get a factor q0 + qr = 1 + qr

(since N = 0, r) the rest of H has already been considered in theXµ part).

Varying r, we get a product

∏

r>0

(1 + qr) =

∞∏

m=1

(1 + qm−1/2).

Varying µ, we get eight copies of this product (because only the transverse µ’scontribute and there are 10− 2 = 8 of them). Thus

Tr NSqH =

(

q−1/48∞∏

m=1

(1− qm− 12 )

)8

.

NB the factor of q−1/48 which comes from the new tensor transformation ofT (stress-energy “tensor”) as we go from z to σ + τ (z = ei(σ+τ)) c.f. in thebosonic case we got q−1/24, double becuase for a boson c = 1 whereas for a


fermion c = 1/2. We can write this partiion function in terms of the Jacobiϑ-function. Recall ... (z = e2πiν , q = e2πiτ )

ϑ00(ν, τ) =

∞∏

m=1

(1− qm)(1 + zqm−1/2)(1 + z−1qm−1/2)

ϑ01(ν, τ) =∞∏

m=1

(1− qm)(1− zqm−1/2)(1− z−1qm−1/2)

ϑ10(ν, τ) = 2eπiτ/4 cosπν∏

m=1

(1− qm)(1 + zqm)(1 + z−1qm)

ϑ11(ν, τ) = −2eπiτ/4 sinπν∏

m=1

(1− qm)(1− zqm)(1− z−1qm)

For ν = 0, z = 1, so

ϑ00(ν, τ) =

∞∏

m=1

(1− qm)(1 + qm−1/2)(1 + qm−1/2)

ϑ01(ν, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)(1− qm−1/2)

ϑ10(ν, τ) = 2q1/8∏

m=1

(1− qm)(1 + qm)(1 + qm)

ϑ11(ν, τ) = −2q1/8 sinπ0∏

m=1

(1− qm)(1− qm)(1− qm) = 0!

Also η(τ) = q1/24∏∞m=1(1− qm). Thus,

ϑ00(0, τ) = q−1/24η(τ)

( ∞∏

m=1

(1 + qm−1/2)

)2

.

So

Tr NSqH =

(

ϑ00(0, τ)

η(τ)

)

.

Next, let us do Tr NSeπiF qH . Let us fix µ and r again. If the state has zero

ψµ−r’s then F = 0 and N = 0. So we get q0 = 1. If the state has one ψµ−r, theneiπF = −1 (recall eiπFψµ−re

−πiF = −ψµ−r) and N = r, so we get−qr.So this case differs from the previous one by a mere sign change which impliesthatϑ00 → ϑ01. Moreover, the ground state |0〉has eigenvalue−1 (eiπF |0〉NS =−|0〉NS), so we get an overall “-” sign. Thus

Tr NSeiπF qH = −

(

ϑ01(0, τ)

η(τ)

)4

.

Similarly, in the Ramond sector,

ZR+ = Tr R+PqH =

1

2

(

Tr RqH + Tr Re

πiF qH)

.


The creation modes are ψµ−m ,m > 0 and we need to take special care of thezero modes ψµ0 .Fix µ andm > 0. Then, for zeroψµ−m’s, we obtain q0 = 1 and for one ψm−mu, weobtain qm, so overall, 1 + qm. The ground state has energy H = 1/16 (normalordering constant we obtained earlier, so 1 + qm → q1/16(1 + qm). Varyingµ, m we obtain the product

Tr RqH =

(

q116 q−

148

∞∏

m=1

(1 + qm)

)8

× “0”

where “0” is the contribution of the zero-modes. Recall the ground state |~s〉.Consider the Σ23 spin, for example. There are two states | ↑〉, | ↓〉, with spin± 1

2 respectively. Each contributes 1, so overall 1 + 1 = 2. we have four inde-pendent such states (since we have eight transverse dimensions Σ01 does notproduce independent physical states). Therefore the overall factor “0” = 24

c.f. with

ϑ01(0, τ) = 2q1/8∞∏

m=1

(1− qm)(1 + qm)2 = 2q1/8q−1/24η(τ)

( ∞∏

m=1

(1 + qm)

)2

Tr RqH =

2q1/8q−1/24

( ∞∏

m=1

(1 + qm)

)2

4

= −(

ϑ10(0, τ)

η(τ)

)4

where the minus sign comes from space-time spin-statistics (ghosts). Finally,Tr Re

iπF qH gives a similar product, but with two changes

• 1 + qm → 1− qm (- form eiπF , as in NS-sector

• | ↑〉 and | ↓〉 have opposite eigenvalues, contributing 1− 1 = 0!

Therefore

Tr R = qH = −(

ϑ10(0, τ)

η(τ)

)4

= 0.

Putting everything together, the partition function for ψµ in NS+ and R+ sec-tor is

Zψ(τ) = Tr NS+qH + Tr R+q

H

=1

2Tr NSq

H + Tr NS1

2eπiF qH +

1

2Tr Rq

H +1

2Tr Re

πiF qH

=1

2(η(τ)4(

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 + ϑ11(0, τ)

4)

.

This is complicated combination of Jacobi-theta functions, yet not only is itmodular invariant, but it vanishes identically! This should not be too surpris-ing, since we have space-time SUSY and so the cosmological constant should


vanish. This fact was known to Jacobi himself, for he proved the “abstruseidentity”

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 = 0,

and we have already seen ϑ11(0, τ)4 = 0.

Of course the total Z is a product of Zψ and Z→ = Z∗ψ in the case of IIB and

(Z ′ψ)∗ =

1

2η(τ)4(

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 − ϑ11(0, τ)

4)

which of course Z ′ψ = Zψ.

Modular invariance of type-I

Type-I is an open string theory. Instead of a torus, we have a cylinder.The cylinder can easily be deduced from the torus. Recall for the torus

Z =

∫

F0

dτdτ

4τ2Z(τ), Z(τ) = Tr qH , q = e2πiτ

where F0 is the fundamental region and dτdτ4τ2

is a modular invariant mea-

sure on the torus (τ2(2π)2 is the volume of the torus = volume of the groupof translations). The cylinder defines a more honest partition function, be-cause τ → t ∈ R and Z(t) = Tr qH , q = e−2πt, i.e., τ = iτ2, τ2 = t, and

Z =

∫ ∞

0

dt

2tTr e−2πtL0 .

Notice that thee is no fundamental region, so we have potential divergencesfrom both limits t → ∞ and t → 0. t → ∞ is usually associated with theIR region (long-distance, low energy). t → 0 is associated with UV diver-gences (short-distances - high energies). In closed strings, there is no t → 0limit, for it is cut by the restriction to the fundamental region F0. In the openstring case, it is there. But does open string theory have UV divergences? Thatwould make it as bad as field (particle) theory. To answer this, concentrate onXµ, µ = 0, 1, .., D − 1. The partition function (easily deduced from the torus)is

Z(t) = Tr e−2πtL0 = iV(√

8π2α′t)−D

(η(it))−(D−2)

c.f. on torus:

Z(t) = Tr e−2πiτL0e−2πiτL0 = iV(

√

4π2α′τ2)−D

|η(it)|−2(D−2)

where

η(it) = e−πt/12∞∏

m=1

(1− e−2πmt).


LetD = 26. In the t→∞ limit, we may expand

(η(it))−24 = e2πt∞∏

m=1

(1− e−2πmt)−24 = e2πt(1 + 24e−2πt + ...)

= e2πt + 24 + ...

Each term in the expansion comes from a certain mass level. The first termis from the tachyon, and diverges, because m2 < 0. The second term is fromthe massless modes (24 transverse photons). Again, it diverges, but only log-arithmically. This is expected and is similar to field theory. These divergencescancel in physical quantities.Now look at t → 0. This appears to be a high energy effect, but it is not! Thecylinder becomes very thin and it looks like a closed string is being created,propogating and disappearing again (NB: t does not represent a physical dis-tance). So t → 0 is still an IR effect (long-distance). To show this, use themodular property, η(−1/τ) =

√iτη(τ). For τ = it, we get η(i/t) =

√tη(it), so

η(it) =1√tη

(

i

t

)

.

Change variables to s = πt . Then, apart from constants

Z ∼∫ ∞

0

dt

tt−13η(it)−24 =

∫ ∞

0

dt

t2η

(

i

t

)−24

∼∫ ∞

0

dsη

(

is

π

)−24

.

t→ 0 is obtained by expanding in large s,

η

(

is

π

)−24

= e2s + 24 + ...

(same expansion as before). The first term is from the tachyon (pathological).The second term is from the massless modes. The propagator for them is 1/k2

and since k2 = −m2 = 0, we have 1/0 =∞. The pole is due to the propagatorfor a long time (on-shell).Let us return to type-I. In this case d = 10, so for theXµ’s, we have

ZX(t) = iV (8π2α′t)−5η(it)−8.

Moreover, there is a subtlety: define the world-sheet parity Ω by

Ω : Xµ(σ)→ Xµ(π − σ).

In terms of modes, Ωαmn uΩ−1 = (−1)nαµn (recall Xµ(σ) ∼ ∑αµne

−inσ). Ob-viously, Ω2 = 1, sp Ω has two eigenvalues, ±1. We need to restrict to the+1 eigenspace for consistency of the theory. This is easily implemented: weneed to keep the states with and even number of αµ−n modes. [NB: In bosonictheory, this would give garbage, for it would exclude the photon! Here, thephoton is ψµ−1/2|0; k〉, so it has 0 (even!) αµ−n’s.


The various partition functions are not affected by the presence of Ω, but weget an extra factor of 1/2 from the projection 1

2 (1 + Ω). Thus

Z =

∫ ∞

0

dt

2t

1

2

1

2ZX(t)Zψ(t)

whereZψ(t) = ϑ00(0, it)

4 − ϑ01(0, it)4 − ϑ10(0, it)

4 − ϑ11(0, it)4

and the two factors of 1/2 come from Ω and the GSO projections respectively.To study the divergences (even though Zψ = 0! - we still need to study the,otherwise Zψ = 0 is a∞−∞ = 0 statement; also these divergences appear(and did not cancel) in other amplitudes) Define s = π/t. Then

ZX(t) = iV

8π(8π2α′)5

∫ ∞

0

ds η

(

is

π

)−8

Zψ

(π

s

)

.

Modular properties

η(it) =1√tη(i/t), ϑ00(0, it) =

1√tϑ00(0, i/t).

Separate NS and R. Then in the NS-sector

ZNS(t) = iV

8π(8π2α′)5

∫ ∞

0

ds η

(

is

π

)−12(

ϑ00(0, is/π)4 − ϑ10(0, is/π)4)

.

To leading order, η(

isπ

)−12= q−1/2 = es and

ϑ10(0, is/π)4 = 24q1/2 = 24es, ϑ00(0, is/π)4 = 1 + ...

So

ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + o(e−2s)).

Notice that the tachyon has disappeared, but of course, we still have the di-vergence from the sixteen massless modes, as expected. What can we do?Well, the cylinder is not the only possibility. We also have the Mobius stripand the Klein bottle.

The Mobius Strip

Same as the cylinder, but we twist before we identify the ends. In other words,Xµ(ω, 2πt) = Xµ(π − σ, 0) = ΩXµ(σ, 0)Ω−1. The partition function is verysimilar to the cylinder. The only difference is the insertion of the parity op-erator, Ω. Thus ZMobius = Tr (qL0Ω). The action of Ω is simple. If a state hasan even (odd) number of αµ−n’s, Ω = +1(−1). Thus, (1− qm)−1 is replaced by(1− (−)mqm)−1, q = e=2πt and so

η(it) = e−πt/12∞∏

m=1

(1− e−2πmt)⇒ e−πt/12∞∏

m=1

(1− (−)me−2πmt)


which can be written in terms of

ϑ00(0, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)2.

As follows: let τ = 2it,

ϑ00(0, 2it) =

∞∏

m=1

(1− qm)(∏

(1 + e−2πt(2m−1)))2

=

∞∏

m=1

(1− qm)−1[

∏

(1 + e−2πt(2m))(1 + e−2πt(2m−1))]2

= e−πt/161

η(2it)

[

∏

(1− (−)me−2πtm)]2

soe−πt/12

∏

(1− (−)me−2πtm) =√

ϑ00(0, 2it)η(2it)

replaces η(it). zero modes are still the same , so... Recall the cylinder

ZX = iV

(

1√8π2α′t

)D

η(it)−(D−2)

The partition function for the Mobius strip is

ZX = iV

(

1√8π2α′t

)D

(ϑ00(0, 2it)η(it))−(D−2)/2

.

Next, do the ψ’s. Easier to work in the R-sector (only one contribution)

Zψ,R = Tr ΩqNψ = −24

[

q1/16q−1/48∞∏

m=1

(1 + (−)mqm)

]8

which can be written in terms of Jacobi-theta functions as follows:

ϑ01(0, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)

ϑ10(0, τ) = 2q1/8∞∏

m=1

(1− qm)(1 + qm)2

so

ϑ01(0, τ)ϑ10(0, τ) = 2q1/8∞∏

m=1

(1− qm)∏

m

(1 + (−)(√q)m)2,

so let q = e−4πt.

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2= 2eπt/3e−πt/2

∏

m

(1 + (−)(√q)m)2


so

Zψ,R = −(

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2

)4

.

AtD = 10

ZR =

∫

0

dt

2t

1

2

1

2ZXZψ,R = iV

∫

0

dt

8t(8π2α′t)−5

(

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2

)4

.

To study the t→ 0 limit, switch the variable to s = π/t. Then

ZR = iV8

(8π2α′)5

∫ ∞

0

ds

(

ϑ01(0, 2is/π)ϑ10(0, 2is/π)

η3(2is/π)ϑ00(2is/π)

)4

for small s, we have

ϑ01(0, 2is/π) ' 1 + ...,

ϑ01(0, 2is/π) ' 2q1/8 + ...

ϑ00(0, 2is/π) ' 1 + ...,

η(2is/π) ' q1/24 + ...

soϑ01ϑ10

η3ϑ00=

2q1/8

q1/8= 2 + ...⇒

(

ϑ01ϑ10

η3ϑ00

)4

= 24 + ... = 16 + ...

no tachyon, and sixteen massless modes contributing, as expected. c.f. forthe cylinder,

ZR = −ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + ...).

of opposite sign, but they do not cancel!

The Klein Bottle

Even though a bottle looks more appropriate for closed strings, and ampli-tudes are defined in terms of closed string modes, the Klein bottle contributesto open strings.DEFINITION: Consider a torus with τ = it. we identify the sides σ = 0 andσ = 2π and obtain a cylinder, but just like with the Mobius strip, we identifythe sides τ = 0 and τ = 2πτ by twisting them first

Xµ(σ, 0) = Xµ(−σ, 2πt) = ΩXµ(σ, 2πt)Ω−1

The partition function is given by

ZX = Tr Ωe−2πtL0e−2πtL0


In this case, ΩαµnΩ−1 = −αµn (unlike for open strings, whereαµn → −αµn) There-

fore, the diagonal elements if Ω have exactly the same αµn’s as αµn’s.So the trace is effectively over theαµn’s only, which explains why this is an openstring amplitude.For the diagonal elements of Ω we have Ω = +1 (even total # of αµn, α

µn.) and

Ł0 = L0, soZX = Tr e−4πtLoΩ

∣

∣

Ω=1

which is the same as open string partition function, but with q2 instead of q(or 2t instead of t)

ZX = iV (4πα′t)−D/2(η(2it))−(D−2)

Note the first factor has a 4 rather than an 8 due to the closed string.The partition function for theψµ’s is obtained similarly. The result is the sameas the open string (cylinder) again, but with t→ 2t.

ZNSψ =1

(η(2it))4[

ϑ400(2it)− ϑ4

10(2it)]

and ZNSψ = −ZRψ . Overall

ZNS =

∫ ∞

0

dt

2t

1

2

1

2ZXZ

NSψ = iV

∫ ∞

0

dt

8t(4π2α′t)−s(η(2it))−12

[

ϑ400(2it)− ϑ4

10(2it)]

andZR = −ZNS. The study of the t→ − limit can be copied from the cylinderwith an extra 210 factor

ZNS = i210V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

Again there is no tachyon, but alas, Zcylinder +Zmobius +Zklein still has a non-vanishing divergence. What do we do? We need to introduce Chan-Patonfactors!Chan-Paton factors were first introduced in QCD, where the string was madeof glue. They attached quarks at the ends of the string which carried indiceslabeling color.In the present setting, we will introduce them because we can. They do notspoil Lorentz invariance, because they live at the ends of the string. They areuseful because they give us extra degrees of freedom, which are needed todescribe gauge interactions.e.g. E&M: Kaluza-Klein added an extra index throughout the string (didn’tknow about strings, but that is what they effectively did.) (X0, ..., X3, X4): X4

was the extra-dimension. This spoiled Lorentz invariance, but that was ok,because we only care about Lorentz invariance in four dimensions. The ex-tra dimension gave us a gauge group (U(1)) corresponding to a photon. Moredimensions give us more complicated gauge groups and extra degrees of free-dom.


With Chan-Paton factors, the gauge group does not come from extra dimen-sions, but from extra degrees of freedom at the ends of the string (open ofcourse). Yet another innovation of string theory!So all states now carry two more indices |0〉 → |0, ij〉, so, e.g. we now haven2 tachyons or photons, if i, j = 1, 2, ..., n. Thus, the photon can be the weakboson multiplet (W±, Z0), or the gluon.How does this effect the partition function? For the cylinder, all n2 states con-tribute equally, so Z is multiplied by n2. For the Mobius strip ,because of thetwist, i needs to be identified with j, and there are n possibilities, ZMobius →nZMobius.For the Klein bottle, we have no indices, because we have closed strings, soZKlein → ZKlein.Overall, the partition function is now

Z = n2Zcylinder + nZMobius + ZKlein.

Recall for the R-sector

Zcylinder = −i V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

ZMobius = i26V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

ZKlein = −i 210V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

Z = −i(n− 25)2V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

We obtain a finite answer if and only if n = 25 = 32. This implies that out ofall possible gauge groups, type I string theory makes a unique choice: SO(32).This was a crucial discovery that led to the explosion of interest in string the-ory.

UNIT 8

Heterotic Strings

8.1 Introduction

The Heterotic string was introduced in 1985 by the “string quartet” (GrossHarvey, Martinec and Rohm).

Basic idea: left and right movers need not be in the same theory. not even inthe same dimension!

Thus, take the holomorphic part as bosonic (∂Xµ(z), µ = 0, 1, ..., 25) and theanti-holomorphic part as a superstring (∂Xµ(z), ψµ(z), µ = 0, 1, ..., 9).

Then the holomorphic piece has c = 26, so we need to throw in the bc ghostswith c = −26 in order to have a vanishing central charge. The anti-holomorphicpiece has a central charge c = 15, so we need couple it to the superconformalghosts (b,c and γ, β) with central charge, c = −15.

It is convenient to split ∂Xµ into ∂Xµ, µ = 0, 1, ..., 9, which will then combinewith the anti-holomorphic piece toXµ(z, z) and the rest ∂Xµ, µ = 10, 11, ..., 25have c = 16 and have no anti-holomorphic partners. We may replace themwith ψA(z), A = 1, 2, ..., 32 which have the same central charge (c=32/2=16).The two theories are the same even though it may not be obvious.

Now we can write the action

S =

∫

d2z

(

1

2πα′ ∂Xµ∂Xµ + ψµ∂ψµ + ∂A∂ψA

)

, µ = 0, 1, ..., 9, A = 1, 2, ..., 32.

so µ is a space-time index and A is an internal index which represents thegauge degrees of freedom. Think of ψA as a 32 dimensional vector rotatedin an abstract space. The gauge group is then the group of rotations, SO(32),which is large enough (too large!) to accommodate all interactions we see inNature.

Notice that SO(32) is also the gauge group in type-I theory, yet they are differ-ent, for in type-I, SO(32) is at the ends of the strings, whereas in the heterotictheory, SO(32) is along the entire string.

52 UNIT 8: Heterotic Strings

Modern wisdom holds that type-I and heterotic only look different. Deep in-side they are different manifestations of the same theory. The operator prod-uct expansions are as the usual

Xµ(z, z)Xν(0, 0) ∼ −ηµν α′

2ln |z|2

ψµ(z)∂ν(0) ∼ ηµν1

z

ψA(z)ψB(0) ∼ δAB1

z← Euclidean signature!

Energy momentum tensor:

T (z) = − 1

α∂Xµ∂Xµ −

1

2ψA∂ψA , T (z) = − 1

α∂Xµ∂Xµ −

1

2ψA∂ψA.

SUSY currents:

TF = i

√

2

α′ ψµ∂Xµ, , TF = 0

so this theory has N = 0, N = 1 SUSY (world-sheet)To build the Hilbert space, we need to specify the boundary conditions onψA and ψµ. ψµ is as before, leading to NS and R sectors - we may apply GSOprojection to split the sectors in NS± and R±.ψA is tricky. It is not restricted by Lorentz invariance, because A is an in-ternal index. We may only require that T (z) be periodic, so ψA(σ + 2π, τ) =OABψA(σ, τ) where O is a 32×32 orthogonal matrix. This leaves the quadraticform ψA∂ψA and T (z) invariant. A host of possibilities, but only two work!Possibility 1

ψA(σ + 2π, τ) = ±ψA(σ, τ)

same sign from all components. This is easy and is the same as before. Definethe fermion number operator F = Σ12 + Σ34 + ... + Σ31 32 (we now have 16spins). The GSO projection is onto eigenspaces of eiπF .We will select eiπF = +1, thus restricting to NS+, R+ forψA. Partition FunctionRecall for ψµ:

Zψ =1

2

[

(

ϑ00(0, τ)

η(τ)

)4

−(

ϑ10(0, τ)

η(τ)

)4

−(

ϑ01(0, τ)

η(τ)

)4

±(

ϑ11(0, τ)

η(τ)

)4]

= 0

which vanished by the abstruse identity.In our case, instead of 4=8/2, we have 16=32/2 (no time-like coordinate there-fore all components contribute) Answer:

Zψ =1

2

[

(

ϑ00(0, τ)

η(τ)

)16

+

(

ϑ10(0, τ)

η(τ)

)16

+

(

ϑ01(0, τ)

η(τ)

)16

±(

ϑ11(0, τ)

η(τ)

)16]

Where the first “+” is due to the ghosts (β, γ) from which are absent in thiscase. The second “+” sign is due to space-time statistics, ψA is a scalar.

8.2 The spectrum 53

This partition function is multiplied byZψ, which vanishes, but it is still usefulto demonstrate the modular invariance of ZψZ

∗ψ

.

Under τ → − 1τ

ϑ01(0,−1/τ) =√−iτϑ00(0, τ)

ϑ01(0,−1/τ) =√−iτϑ10(0, τ)

ϑ10(0,−1/τ) =√−iτϑ01(0, τ)

η(−1/τ) =√−iτη(τ)

It is obvious that both Zψ and Zψ are invariant since ϑ10 ↔ ϑ01.Under τ → τ + 1

ϑ00(0, τ + 1) = ϑ01(0, τ)

ϑ01(0, τ + 1) = ϑ00(0, τ)

ϑ10(0, τ + 1) = eiπ/4ϑ10(0, τ)

η(τ + 1) = eiπ/12η(τ)

Zψ → e−16πi/12Zψ = e2πi/3Zψ and Zψ → −e−4πi/12Zψ = e2πi/3Zψ. Thereforethe product ZψZ

∗ψ

is invariant under modular transformations.

8.2 The spectrum

Constraints: L0 = L0 = 0. Do L0 first (anti-holomorphic part). This is thesame as before.

NS : L0 =α′

4p2 + N − 1

2= 0.

Lowest state: |0, k〉 has no N = 0, so m2 = −k2 = −2/α′, a tachyon. This has

eiπF |0, k〉 = −|0, k〉, where the minus sign is due to the (β, γ) ghosts. Thereforeit is not a NS+.Next level: ψµ− 1

2

|0, k〉 has N = 12 , so m2 = −k2 = 0. It is a vector transforming

under SO(8). It has eiπF ψµ− 12

|0, k〉 = ψµ− 12

|0, k〉 ∈ NS+.

R-Sector L0 = α′

4 p2 +N + a

a can be deduced as follows: recall in bosonic theory a = −1. This is becausea = −(D − 2)/24. Here D = 10, so we get a = −1/3. In the NS sector, weget a = −1/2, because the fermions contributed−1/6(1/2 = 1/3 + 1/6)-eachfermion contributes −1/48. Now we have 32 fermions, so they contribute−32/48 = −2/3. Overall a = −1/3− 2/3 = −1 in NS.In the R sector, we get a = 0, so fermions contribute 1/24 each. Overall, a =−1/3+ 32/24 = +1 in the R sector which implies all modes are massive in theR sector.In NS, the lowest state has m2 = −k2 = − 4

a′ , which represents a tachyon! Ithas eiπF |0, k〉 = |0, k〉, so we keep it.


The next level has N = 1/2 : ψA−1/2|0, k〉, so m2 = k2 = − 2α′ , which is another

tachyon, but eiπFψA−1/2|0, k〉 = −ψA−1/2|0, k〉, so we must reject it.

The next level hasN = 1 : m2 = k2 = 0, massless! There are two possibilities:

Aµ(k)αµ−1|0; k〉 , BABψ

A1/2ψ

B−1/2|0; k〉

Both possibilities have the correct GSO projection i.e., eiπF = +1, so we keepthem.Aµ(k) represents a photon with 8 transverse polarizations. BAB is an anti-symmetric 32× 32 matrix with 32×31

2 = 496 components.Summary

m2 NS+ R+ NS+ R+−4/α′ |0; k〉 − − −

0 Aµ, BAB − ψµ−1/2|0; k〉 |~s; k〉

Closed strings states must have L0 = L0, so the tachyon in NS+ is rejected,because there is no tachyon in NS+ or R+. At the massless level we have

Aµναµ−1ψ

ν−1/2|0; k〉, Aµ,~sαµ−1|~s; k〉

whereAµν may be decomposed via a scalar, antisymmetric and traceless sym-metric tensors as 8×8 = 64 = 1+28+35. Aµ~s is the supersymmetric partner toAµν and may be decomposed into 8+56 irreducible representations of SO(8).We also have

B(~s)ABψ

A−1/2ψ

B−1/2|~s; k〉, BµABψA−1/2ψ

B−1/2ψµ −1/2|0; k〉

so BµAB represents a gauge boson with 496 components. c.f. gluon has Aµi ,where i = 1, 2, ..., 8. So SO(32) is a gauge symmetry in space-time, just likeSU(3) is a gauge symmetry for QCD.Comparing with SO(32) type-I theory, we see big differences. In type-I theorySO(32) resides at the ends of the strings and the spectrum does not match thatof the heterotic string (except at the massless level). Yet, these two theoriesare one and the same (to be proved)!Possibility #2

Divide ψA(z) into two groups, ψA, A = 1, 2, ..., 16, ψB , B = 17, 18, ..., 32. Thisis possible because

ψA(σ + 2π) = −ψA(σ) (NS1) ψA(σ + 2π) = +ψA(σ) (R1)ψB(σ + 2π) = −ψB(σ) (NS2) ψB(σ + 2π) = +ψB(σ) (R2)

Total of four possibilities: NS1 +NS2, NS1 +R2, R1 +NS2, R1 +R2. Thereare two GSO projections because we have two fermion number operators. Wewill restrict to eiπF1 = eiπF2 = +1.Partition Function

Zψ = ZψAZψB = Z2ψA ,

8.2 The spectrum 55

because ZψA = ZψB . ZψA is the same as Zψ we derived in possibility #1, ex-cept 16→ 8 (we have half as many ψAs now). Therefore

ZψA =1

2

[

(

ϑ00

η

)8

+

(

ϑ10

η

)8

+

(

ϑ01

η

)8

∓(

ϑ11

η

)8]

.

Modular invariance: τ → −1/τ leave the partition function invariant (trivial).τ → τ + 1 takes ϑ00 ↔ ϑ01, ϑ10 → eiπ/4ϑ10, so ϑ8

10 → ϑ810. So only η → eiπ/12η,

i.e., η−8 → e−2πi/3η−8, and the partition transforms as ZψA → e−2πi/3ZψA ⇒Z2ψA → e−4πi/3Z2

ψA = e2πi/3Z2ψA . The additional factor cancels when we mul-

tiple by Z∗ψ→ e−2πi/3Z∗

ψ.

Gauge Group: Obviously, this theory has SO(16) × SO(16) symmetry. How-ever, this is only a subgroup os the full gauge group, which is E8 ×E8 (excep-tional group).To summarize: there exist only two possibilities for heterotic strings, withgauge groups SO(32) and E8 ×E8, respectively.


UNIT 9

Low Energy Physics

9.1 Type IIA Superstring

Sectors: (Ns+,NS+), (R+,NS+), (NS+,R-), (R+,R-).(NS+, NS+): massless states: Aµνψ

µ−1/2ψ

ν−1/2|0; k〉 Aµν is decomposed into a

scalar, antisymmetric field and traceless symmetric field: 8× 8 = 1 + 28 + 35The scalar field represents the dilaton. We do not see it in Nature and it isbelieved to have settled into it’s ground state value and not affect dynamicsany further. We will set it to zero to avoid complications in an already com-plicated discussion (it should be set to a constant, but we can always tweakcouplings, etc., so setting it to zero will be fine). Let Bµν be the antisymmet-ric tensor and gµν the traceless symmetric tensor (graviton). The dynamics ofgµν is described by the Einstein action

Sg =1

4πG10

∫

d10x√−g R,

where G10 is the ten-dimensional Newton’s constant. This can be derivedfrom string theory tree-level amplitudes and loop amplitudes introduce cor-rections. Bµν has field strength

Hµνρ = ∂µBνρ − ∂νBµρ + ∂ρBµν .

Hµνρ is totally antisymmetric in it’s indices. The action is given by

SB = − 1

8πG

∫

d10x√−gHµνρHµνρ,

where indices are raised and lowered by gµν .

(R+,R-): massless states: |~s; k〉 ⊗ |~s′; k〉, 8× 8 of them.Recall ψµ0 |~s; k〉 is also a ground state (annihilated by all ψµr , r > 0).

States decompose into Cµψµ0 |0〉 and Cµνρψ

[µ,0 ψ

[ν,0 ψ

ρ]]0 |0〉 where we antisym-

metrize over all indices. There are 8Cµ (transverse µ) and 56Cµνρ (8+56 = 64)

58 UNIT 9: Low Energy Physics

so they span the ground states. The action is given by

SR = − 1

8πG10

∫

d10x√−g

(

FµνFµν + Fµνρσ F

µνρσ)

where Fµν = ∂µCν − ∂νCµ is the field strength of Cµ.

Fµνρσ = Fµνρσ −1

4(CµHνρσ + CνHρσµ + CρHσµν + CσHµνρ)

where Fµνρσ = ∂µCνρσ + ... (add terms such that Fµνρσ is completely antisym-metric) and is the field strength of Cµνρ.There is one more contribution to the action that does not involve the metric(topological). This is a Chern-Simons term given by

SCS = − 1

8πG10

∫

d10xεµ1µ2...µ10Bµ1µ2Fµ3µ4µ5µ6

Fµ7µ8µ9µ10

The total action is the sum of all the actions and is given by

S = Sg + SB + SR + SCS

There is a fermionic counterpart which we will not discuss.

9.2 Supergravity

Let us compare with supergravity (SUGRA). Unfortunately, SUGRA lives in 11dimensions, yet it looks so much like the type-II superstring, that is hard toignore. It turns out that (modern wisdom holds) we really live in eleven di-mensions and ten dimensional strings are really an eleven dimensional the-ory! What theory? Nobody knows ... M-Theory.We have seen this problem with dimensions before. We compactified one di-mension a la Kaluza-Klein, then let R → 0 and the extra dimension wouldnot go away. Same here. We will compactify the eleventh dimension to get10D superstrings, but the eleventh dimension will remain lurking in the back-ground.Action:

SSUGRA11 =1

4πG11

∫

d11x√−G

(

R(11) − 1

2FMNQRF

MNQR

)

− 1

24πG11

∫

d11x εM1M2...M11AM1M2M3FM4M5M6M7

FM8M9M10M11

where FMNQR is the field strength ofAMNQ (FMNQR = ∂MANQR+...), whereF is completely antisymmetric. The last term is a Chern-Simons term and isgauge invariant

δAMNQ = ∂MλNQ + ...

9.2 Supergravity 59

even though it doesn’t look like it.We need to reduce the dimension from eleven to ten to compare with super-strings. We will do that a la Kaluza-Klein. Assume nothing depends on theeleventh coordinate and call it “u”.The metric: ds2 = GMN (xµ)dxµdxν , M,N = 0, 1, ..., 10, µ = 0, 1, .., 9 We maydecompose the metric as such

ds2 = Gµνdxµdxν + 2Gµudx

µdu+Guudu2

LetGuu = 1 for simplicity (fixes the size of the extra dimension, which can berescaled later).Introduce vector Amu = Gµu and metric gµν = Gµν − AµAν . Then we maywrite ds2 = gµνdx

µdxν + (du + Aµdxµ)2. The potential AMNQ may also be

grouped into Aµνρ and Aµν = Aµνu (no other components exist, becauseAMNQ is antisymmetric, so we can not have two u indices). So now the fieldcontent becomes (from the 10D perspective) gµν , Aµ, Aµν , Aµνρ, very similarto the type-II superstring.Futhermore,

R(11) = R(10) − 1

2FµνF

µν

where Fµν = ∂µAν − ∂νAµ.

FMNQRFMNQR = FµνρF

µνρ + Fµνρσ Fµνρσ

whereFµνρ = ∂µAµρ+... , Fµνρσ = Fµνρσ−AµFνρσ+... andFµνρσ = ∂µAνρσ+...and

1

6εM1M2...M11AM1M2M3

FM4M5M6M7FM8M9M10M11

= εµ1µ2...µ10Aµ1µ2Fµ3µ4µ5µ6

Fµ7µ8µ9µ10

= εµ1µ2...µ10Aµ1µ2µ3Fµ4µ5µ6

Fµ7µ8µ9µ10+ total derivative

The last equality can easily be verified by integrating by parts.Also

√−G =

√−g.If the eleventh dimension has length 2πR, then the action becomes

SSUGRA11 =1

4πG10

∫

d10x√−g

(

R(10) − 1

2FµνF

µν − 1

2FµνρF

µνρ − 1

2Fµνρσ F

µνρσ

−1

2εµ1µ2...µ10Aµ1µ2

Fµ3µ4µ5µ6Fµ7µ8µ9µ10

)

where G10 = 2πRG11 is the 10D Newton’s constant and we have rescaledAµν → 1√

2πRAµν , Aµνρ → 1√

2πRAµνρ.

This action is identical to the one obtained from type-IIA superstring if weidentify

gµν → gµν , Aµ → Cµ, Aµν → Bµν , Aµνρ → Cµνρ.

60 UNIT 9: Low Energy Physics

UNIT 10

D-Branes

10.1 T-duality (again)

Consider type-II theories. We have

IIA : (NS+, NS+) (R+, NS+) (NS+, R−) (R+, R−)IIB : (NS+, NS+) (R+, NS+) (NS+, R+) (R+, R+)

Compactify the 10th dimension on a circle of radius R in IIA, say. As weshowed in the bosonic theory (argument is the same) the theory atR is iden-

tical to the theory at R′ = α′

R (T-duality).

To show this, we started with the coordinate X9 = U = UL(z) + UR(z) andintroduced the coordinate z = UL(z)− UR(z). The resulting theory is at R′ =α′

R . In other words, the parity transformation on the right-moving part (only!).

X9R(z)→ −X9

R(z)

relates the theory atRwith the theory atR′ = α′

R . Because of the superconfor-mal invariance, this parity transformation is also applied tp the superpartner,ψ9(z)

ψ9(z)→ −ψ9(z).

This, in particular, reverses the chirality of the states in the antiholomorphicpart, so R- ↔ R+. Therefore IIA ↔ IIB, because that is the only difference

between the two theories. Therefore IIA at R is equivalent to IIV at R′ = α′

R .

In particular, the IIA R-R fields, Cµ, Cµνλ are mapped onto the IIB R-R fields,C, Cµν , Cµνρσ as follows:

C9 → C, Cµ → Cµ9, Cµν9 → Cµν , Cµνλ → Cµνλ9.

Of course, e.g., Cµνρσ is obtained from Cµνρσ9 in IIA, but Cµνρσ9 is not an in-dependent field (can be expressed in terms of Cµ, Cµνλ) 8 + 56 = 64.

62 UNIT 10: D-Branes

Type-I Strings If we compactify the 10th dimension, X9(z, z), then the theory

in the R → 0 limit is mapped onto a T-dual theory at R = α′

R → ∞ whichcontains a D-brane.Recall the argument, in theR′ theory, the 10th dimension is

Z(z, z) = X9L(z)−X9

R(z), ∂σZ = ∂τX9

so

Z(σ = π)− Z(σ = 0) =

∫ π

0

dσ∂σZ =

∫ π

0

dσ∂τX9 =

∫ π

0

∂τ (2α′pτ)

= 2α′pπ = 2α′ n

Rπ = 2πnR′ = 0.

Translation invariance is broken in the T-dual theory. Massless modes (sameas in uncompactified theory)

NS : Aµψµ−1/2|k〉, Aψ

9−1/2|k〉, R : |~s; k〉

where Aµ represents a photon tangent to the brane. The second state shiftsthe position of the brane making it a dynamical object. (A is a function of k→ its F.T. is a function of Xµ, µ = 0, 1, ..., 8).Even though the translation invariance is broken, the original theory has 32supersymmetries! Of these, only half are broken. Thus the brane is a super-symmetric object with 16 supersymmetries! This large amount of symmetryimplies the existence of conserved charges. What are they?Our brane has 8+1 dimensions, so its volume element couples to the R-R po-tential, Cµ1µ2,...,µ9

(dV ∼ εµ1µ2...µ9dxµ1 ...dxµ9 ).

Recall familiar examples:

• A point charge qmoving along a trajectory xµ(τ) has the action q∫

dτvµAµ =∫

dτjµAµ = q∫

dxµAµ. The charge q is conserved.

• The magnetic flux: Φ =∫

~B · d~s , ~B = ∇ × ~A Define a field strength:Fij = ∂iAj − ∂jAi. Then Bi = 1

2εijkFjk , so Φ =

∫

FjkdSjk where

dSjk = 12εijkdSi is the surface element. This is the magnetic charge,

i.e., 0. Similarly, for the electric charge field, ΦE =∫

F0idΣ0i ∝ q.

For the R-R charge on the D8 brane, we have

Q ∝∫

dxµ1 ...dxµ9Cµ1...µ9

If we dualize two more dimension, the brane becomes a 6+1 dimensional ob-ject, (D6-brane). Two more gives D4, two more gives D2 and two more gives aD0 brane which represents a point particle. The charges are

∫

dxµCµ,∫

dxµ1dxµ2 ...Cµ1µ2...

which are the R-R fields in type IIA theory! On the other hand, the D(2p+1)-branes couple to Cµ1µ2

, Cµ1µ2µ3µ4, etc., which are the potentials in the type-

IIB theory!

10.1 T-duality (again) 63

Not all these potentials are independent. Consider, e.g., D0-brane coupledto Cµ. The D0-brane is a point particle (with strings attached- hairy) withcharge q which is the source of Cµ and field strength Fµν = ∂µCν − ∂νCµ. q isan electric charge. The flux

∫

FµνdΣµν ∝ q (Gauss’ Law).

In four dimensional electromagnetism we may define the dual of Fµν as Fµ =12εµνρσF

ρσ which interchanges ~E ↔ ~B. Then the electric charges become

magnetic charges. One may define a vector potential Aµ corresponding to

Fµν and describe electromagnetics in terms of Aµ instead of Aµ. Aµ can notbe defined globally, since the magnetic flux around a charge is no longer zero,but it can be defined in patches, or almost everywhere apart from the string(Dirac string). If we include both electric and magnetic charges, then noaction can be defined, yet the theory still makes sense. The existence of amonopole leads to quantization of the electric charge (Dirac).Proof: Consider a point particle moving from ~x1 → ~x2. Its wavefunctionchanges ψ(~x1) → ψ(~x2). If I want to compare ψ(~x1) and ψ(~x2), then I willdefine the quantity ψ(~x2) ∗ ψ(~x1). In the limit ~x2 → ~x1 (closed path) weobtain |ψ(~x1)|2. Gauge invariance: ψ(~x) → eiqλ(x)ψ(~x), so ψ∗(~x2)ψ(~x1) →eiq(λ(x2)−λ(x1))ψ∗(~x2)ψ(~x1) This is not a gauge-invariant oblect. To make it

gauge-invariant, multiply by eiq~A·d~, ~A→ ~A−∇λ, so δeiq

~A·d~ = e−iq(λ(~x1)−λ(~x2)),

so ψ∗(~x2)eiq

~A·d~ψ(~x1) is gauge-invariant (physical)!

Go around a loop: we have eiq ~A·d~|ψ(~x1)|2. By Stoke’s theorem,∮

C~A · d~ =

∫

S~B · d~s (flux through S).

If the path shrinks to zero, then∮

C~A · d~=

∫

S~B · d~s = 0.

In the presence of a magnetic monopole,∫

S~B · d~s = m, the magnetic charge,

so eiqS~B·d~s = eiqm. We must have eiqm = 1, therefore qm = 2πn, i.e., q is

quantized even if only one magnetic monopole exists in the entire Universe.Returning to D-branes, the Cµ potential on the D0-brane has field strengthFµν whose dual is εµ1µ2...µ10Fµ9µ10

(8 indices). It corresponds tp a potentialwith seven indices, Cµ1µ2...µ7

which resides on a D6-brane.Thus the D0 electric charge is a source for the same field for which the D6-branes magnetic charge is a source. More generally, the electric Dp-branecharge and the magnetic D(6-p)-brane charge are sources for the same field.Action for D0-branes electromagnetism:

S = −1

2

∫

d10x√−gFµνF µν + q

∫

dxµAµ

The potential between two points (D0-branes) is a Coulomb potential (in10D)

V (y) ∝ q2

y7

In momentum space, this is obtained from the propagator − ik2 where kµ is

the momentum of the exchanged boson (photon). Then

V (y) = −i∫

d10kei~k·~y q

2

k2= −i 15V

32π4

q2

y7


where V =∫

dω.With D-branes, the potential comes from the exchange of closed strings. Thismay also be viewed as an open string with ends at y = 0 and y = y movingaround a loop. We already know how to calculate it.The answer is

Z =

∫ ∞

0

dt

2tZ(t), Z(t) = Tr e−2πtL0 .

Recall our result earlier

ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + o(e−2s)), s =π

t.

Now we have 9 dimensions (16 compactified), so s = 92 . Also there is no

integral over spatial momenta, only the energy D0-branes have world-lines,

so the contribution from 0-modes(

8π2α′t)−D/2 →

(

8π2α′t)−1/2

, therefore,

there is an additional factor(

8π2α′t)−(1−D)/2 →

(

8π2α′t)9/2

.An extra factor of 4 = 2 × 2 ( 2 from XXXXX and no need to average overorientations). Finally, since Z(σ = π)−Z(σ = 0) = y, the expansion contains

an extra termZ = y σπ + ...which gives an extra contribution toL0 = y2

4π2α′ + ....

Therefore the extra factor is given by e−2πt y2

4π2α′ = e−ty2/2πα′

. The partitionfunction becomes

Z → iV (4× 16)

8π(8π2α′)5

∫ ∞

0

πdt

t2(8π2α′t)9/2e−ty

2/2πα′

= iV (2π)(4π2α′)315

32π4

1

y7

This is compared to the potential V (y) = −i 15V32π4

q2

y7 . In fact it is the R-sector

ZR = V (y), but ZR = −ZNS, so q2 = 2π(4π2α′)3.This generalizes to Dp-branes: (8π2α′t)9/2 → (8π2α′t)(9−p)/2. The potentialgeneralizes to Vp(y) ∼ 1

y7−p and the charges become q2p = 2π(4π2α′)3−p.

For the D6-brane, q26 = 2π(4π2α′)3 , so q6q0 = 2π, the Dirac quantization condi-

tion with n = 1! In general, qpq6−p = 2π, confirming that the Dp-brane andD(6-p)-brane act as electric and magnetic sources for the same field.

10.2 D-branes at angles

So far we have considered similar D-branes separated by a distance y. Theseare parallel D-branes. More generally, we can have a Dp-brane and a Dp’-brane along different subspaces and they may even intersect e.g., a D8-braneobtained by dualizing X9 and a D8-brane from X8 dualization. These twobranes have the space (X1, X2, ..., X7) common. One may be obtained by ro-tating the other by 90 in the (X8, X9) plane. An open string may stretch be-tween these two branes. Then itsX9 coordinate will obey Dirichlet boundary

10.2 D-branes at angles 65

conditions at one end and Neumann boundary conditions at the other. TheX8 coordinate is reversed: Neumann boundary conditions at one end andDirichlet at the other. Thus the modes expansions will be different. Recall forNeumann boundary conditions on both ends (set τ = 0 for simplicity)

XµNN(σ) = xµ + i

√2α′

∑ 1

nαµn cos(nσ).

Check ∂σXNN6µ = 0 at σ = 0, π.For Dirichlet boundary conditions on both ends,

XµDD(σ) =

yσ

π− i√

2α′∑ 1

nαµn sin(nσ)

where y is the separation of the two (parallel in the µ-direction) branes. CheckXµDD(0) = 0, Xµ

DD(π) = y. XµNN is split into holomorphic and antiholomor-

phic pieces as such

XµL =

1

2xµ + i

√

α′

2

∑ 1

nαµne

inσ

XµR =

1

2xµ + i

√

α′

2

∑ 1

nαµne

−inσ

XDD is split as XµDD = Xµ

L − XµR (dual!) For DN-b.c., i.e., Xµ

DN (σ = 0) =0 , ∂σX

µDN (σ = π) = 0, we obtain

XµDN (σ) = −

√2α′

∑

r∈Z+1/2

αµrr

sin(rσ).

For ND-boundary conditions, we haveXµND(σ) = i

√2α′∑

r

Z+1/2αµrr cos(rσ).

The superpartners ψµ and ψµ are similar.Generalize to general angles. Suppose that there is an angle φ between thebranes and consider strings stretched between the two. DefineZ = X8 + iX9

(the brane at X9 = 0 is not rotated-no loss of generality). At σ = 0, X9 = 0and ∂σX

8 = 0, so Im(Z) = 0, Re(Z) = 0. At σ = π, the brane is rotated by φ,so Z → eiφZ, so Im(Ze−iφ = ∂σ Re(Ze−iφ) = 0.We may expand in terms of the modes

Z =

√

2

α′

∑

r∈Z+φπ

αrreirσ +

√

2

α′

∑

r∈Z−φπ

α+r

re−irσ

αr and α+r are independent, because they involve α8

r and α9r (αr = α8

r + α9r).

At σ = 0: Z ∼ O(αr+α+r ), so, Im(Z) = 0, and ∂σZ ∼ i(αr−α+

r ), so, Re(∂σZ) =0.At σ = π: Z ∼ (αr + α+

r )eiφ, so, Im(Ze−iφ) = 0, and ∂σZ ∼ i(αr − α+r )eiφ, so,

Re(∂σZe−iφ) = 0.


10.3 Partition Function

The partition function has contributions from both the αr’s and α+r ’s. It is

easy to see that for q = e−2πt

Z = qa∏

r∈Z+φπ

(1− qr)−1∏

r∈Z−φπ

(1− qr)−1,

= qa∞∏

m=0

(1− qm+ φπ )−1

∞∏

m=1

(1− qm−φπ )−1,

where a is the Casimir energy (normal ordering constant in L0 =: L0 : −a).Recall a = −1/24 for a boson, because a = 1

2

∑∞n=1 n = 1

2ζ(1) = −1/24. Herethe sum becomes

1

2

∑

r∈Z−φπ

r =1

2

∞∑

m=1

(

m− φ

π

)

=1

2

[

1

24− 1

8

(

2φ

π− 1

)2]

.

To prove this, look at the twisted sum problem (Polchinski 2.9.19) done lastsemester.Also,

1

2

∑

r∈Z+φπ

r>0

r =1

2

∞∑

m=0

(

m+φ

π

)

=1

2

[

1

24− 1

8

(

2

(

1− φ

π

)

− 1

)2]

=1

2

[

1

24− 1

8

(

1− 2φ

π

)2]

,

which is the same as before. So, a = 124 − 1

8

(

1− 2φπ

)2

. Therefore,

Z = qa(1− z)−1

[ ∞∏

m=1

(1− zqm)(1− z−1qm)

]−1

, z = qφ/π = e−2φt = e2πiν

This can be expressed in terms of

ϑ11(ν, it) = −2q1/8 sinπν

∞∏

m=1

(1− qm)(1− zqm)(1− z−1qm),

and

η(it) = q1/24∞∏

m=1

(1− qm).

Indeed,

η(it)

ϑ11(ν, it)= −1

2q−1/24−1/8−a 1

sinπν

[

∏

(1− zqm)(1− z−1qm)]−1

= −1

2q1/24−1/8−a 1− z

sinπνZ

= iqφ2/2π2

Z


Therefore

Z = −1eφ

2t/πη(it)

ϑ11(ν, it).

Similarly for the fermion, we obtain

Z =ϑab(ν, it)

eφ2t/πη(it),

for a, b = 0, 1 (NS-NS, NS-R, etc.)Notice that the bosonic Z diverges as ν → 0, i.e., φ → 0. In this limit thetwo branes become parallel to each other, and the string is free to move alongthem, i.e., it has an additional (continuous) momentum, whose trace givesTr qL0 ∼ V√

8πα′t, where V is the volume of the dimension along the brane.

Therefore,

Z = qa1√

8π2α′t

∞∏

m=1

(1− qm)−2, a = 1/24− 1/8 = −1/12

=V√

8π2α′t(η(it))−2.

The fermionic partition functions Zab do not change. Suppose as φ→ 0, bothbranes are in the X8 direction. Now take the dual of X8. Since we have Neu-mann boundary conditions in X8 (Dirichlet in X9), in the dual, we will haveDirichlet in X8. So in the dual picture, the two branes will become distinctpoints separated by a distinct y.If originally we had Dp-branes, we end up with D(p-1) branes in the dualspace. Open strings are stretched between the two branes. Thus, instead of

a continuous momentum, we now have a contribution y2

4π2α′ in L0, therefore

e−ty2/2πα′

, , y2 = y28 + y2

9 , in general. The partition function is

Z = qae−ty2/2πα′

∞∏

m=1

(1− qm)−2 = e−ty2/2πα′

(η(it))−2

Example: Consider two D4-branes at an angle φ1 in the 23-plane, φ2 in the45-plane, φ3 in the 67-plane, φ4 in the 89-plane and separated by a ditancey in the 1-directioin. In each plane, we obtain a partition function for thefermions:

Zab(φi, it) =ϑab(νi, it)

eφ26t/πη(it)

, , ν=iφit/pi, i = 1, 2, 3, 4.

Putting them together, the fermionic partition function is

Zf =1

2

[

4∏

i=1

ϑ00(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ10(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ01(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ11(νi, it)

eφ2i t/πη(it)

]

.


Generalizing our earlier result, when φi = 0 = νi,

Zψ =1

2η4(it)

(

ϑ400(0, iτ)− ϑ4

10(0, iτ)− ϑ401(0, iτ)− ϑ4

11(0, iτ))

.

Earlier we used the abtruse identity to show Zψ = 0. Now, we shall use thegeneralization of the abstruse identity:

∞∏

m=1

ϑ400(0, iτ)−

∞∏

m=1

ϑ410(0, iτ)−

∞∏

m=1

ϑ401(0, iτ)−

∞∏

m=1

ϑ411(0, iτ) = 2

∞∏

m=1

ϑ11(ν′i, it)

ν′i = iφ′it/π, φ′1 = 12 (φ1 + φ2 + φ3 + φ4), φ′2 = 1

2 (φ1 + φ2 − φ3 − φ4)φ′3 = 1

2 (φ1 − φ2 + φ3 − φ4), φ′4 = 12 (φ1 − φ2 − φ3 + φ4)

Notice∑4

i=1 φ′2i =

∑4i=1 φ

2i , so

∏4i=1 e

φ2i t/π =

∏4i=1 e

φ′2i tπ and

Zf =

∏4i=1 ϑ11(ν

′i, it)e

−φ′2i t/π

η4(it).

Bosons: Recall in the 89-plane

Zboson = −ieφ2t/πη(it)

ϑ11(ν, it)

so in the 234...9 direction

Zboson = η4(it)

4∏

i=1

eφ2i t/π

ϑ11(νi, it).

In the 0(time)-direction, we have a continuous distribution, so Z ∼ 1√8π2α′t

.

In the 1-direction, we have branes separated by a distance y, so L0 = y2

4π2α′ +

..., so Z1 ∼ e−ty2/2πα′

.Multiplying everything, the partition function becomes (potential)

V = −∫ ∞

0

dt

t

1√8π2α′t

e−ty2/2πα′

4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)

This is a complicated function of y. We will calculate it for large distances. If yis large, the dominant contribution to the integral comes from small t (due to

the ety2/2πα′

factor). If we set t = 0 in the ϑ-function, we obtain a constant andthe integral diverges. We will calculate the force, which is a physical quantityand define the potential on the integral, V = −

∫

Fdy.

F = −dVdy

= −y∫ ∞

0

dt

πα′e−ty

2/2πα′

√8π2α′t

4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)

10.4 Scattering 69

From

ϑ11(ν, it) = −2q1/8 sinπν

4∏

m=1

(1− qm)(1− zqm)(1− z−1qm),

andϑ11(−iν/t, i/t) = −i

√teπν

2/tϑ11(ν, it),

we obtain4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)=

4∏

i=1

ϑ11(−iν′i/t, i/t)ϑ11(−iνi/t, i/t)

.

As t→ 0, q = e−2π/t → 0, so Π→ ∏4i=1

sin iπν′i/t

sin iπνi/t

νi = iφt/π → iπνi/t = −φi, Π =

4∏

i=1

sinφ′isinφi

.

So

F ∼ Cy∫ ∞

0

dt√te−ty

2/2πα′, y →∞ const. :

1

πα′√

8π2α′

4∏

i=1

sinφ′isinφi

.

and the potential is V ∼ Cy.

10.4 Scattering

How do you make a D-brane move? Simple. Motion in e.g., the 1-directionis motion in Minkowski space (X0, X1) just like a rotation in Euclidean space(X8, X9) we studied above.

(

X0

X1

)

→(

cosh ζ sinh ζsinh ζ cosh ζ

)(

X0

X1

)

,

(

X8

X9

)

→(

cos ζ sin ζ− sin ζ cos ζ

)(

X8

X9

)

.

whereX1 = vX0 and the speed (v) is defined by the rapidity (ζ) as v = tanh ζ.The rapidity is related to the velocity via

cosh ζ =1√

1− v2, sinh ζ =

v√1− v2

.

Consider two parallel Dp-branes moving with relative velocity v in the X1-direction and separated by a distance y in the z-direction (branes are perpen-dicular to bot X1 and X2). In the 01-plane (Minkowski), we may copy ourearlier result with the substitution φ = −iζ: The bosonic part of the partitionfunction is

Zbosonic (01) = −ieφ2t/πη(it)

ϑ11(ν, it), φ = −iζ, ν = iφt/π − ζt/π.


The fermionic part is

Zab =ϑab(ν, it)

eφ2t/πη(it).

In the rest of the direction, the D-branes are parallel, so all other angles arezero. Therefore, the fermionic piece is

Zf =1

2η4(it)e−φ

2t/π[

ϑ00(ν, it)ϑ300(0, it)− ϑ10(ν, it)ϑ

310(0, it)

−ϑ01(ν, it)ϑ301(0, it)− ϑ11(ν, it)ϑ

311(0, it)

]

This may be computed by applying the generalized abstruse identity. We have

φ1 = φ, φ2 = φ3 = φ4 = 0,

so

φ′1 = φ′2 = φ′3 = φ′4 =1

2φ

and therefore

Zf =1

2η4(it)e−φ

2t/πϑ411(

1

2φ, it).

The bosonic piece in the other directions (X2, X3, ..., X9 total of eight ... sixof which are transverse) is

Z bosonic2,3,...,9

= Vp

(

1√8π2α′t

)p

e−ty2/2πα′

(η(it))−6

Therefore the partition function is

Z = −iVp∫ ∞

0

dt

t(8π2α′t)−p/2e−ty

2/2πα′ ϑ411(ν/2, it)

ϑ11(ν, it)(η(it))−9, ν = ζt/π

As the branes move the distance changes to r2 = y2+v2τ2. The potential maybe extracted from

Z = −1

∫ ∞

−∞dτV [r(τ), v].

We easily obtain

V (r, v) = i2Vpv

(√

8π2α′)p+1

∫ ∞

0

dt t(5−p)/2e−tr2/2πα′ ϑ4

11(iζ/2π, i/t)

η9(i/t)ϑ11(iζ/π, i/t),

where we used the modular properties of the ϑ and η functions.Note: as v → 0, u→ 0, so V → 0.Since

ϑ11(ν, it) = −2q1/8 sinπν∏

(1−qm)(1−zqm)(1−z−1qm), η(it) = q1/24∏

(1−qm)

10.4 Scattering 71

we have, as v → 0, ν → 0, Z → 1.

ϑ411(iζ/2π, i/t)

η9(i/t)ϑ11(iζ/π, i/t)=

8i sinh4(ζ/2)

sinh(ζ)+ ... =

1

2v3 + ..., ζ → v

So

V (r, v) = − 2Vpv4

(√

8π2α′)p+1

∫ ∞

0

dt t(5−p)/2e−tr2/2πα′

+ o(v6)

∼ − v4

r7−pVp

α′p−3

Problem: as r → 0, V → ∞! How can string theory claim finiteness at shortdistances (r is real distance - not bogus!)?

Answer: Let r → 0 before expanding in v. r only appears in e−tr2/2πα′

. If werescale t→ t/r2, the r → 0 corresponds to large t. If t is large in ϑ, η, then

ϑ411

η9ϑ11→ sinh4

(

vt4

)

sinh(vt), ζ ∼ v

From the exponential, t ∼ 2πα′/r2 dominates. ut ∼ 2πα′u/r2, so in the limitthat r → 0, ut becomes large and the integral oscillates rapidly. OScillation ona scale ut ∼ 1, i.e., 2πα′u ∼ r2, i.e., r ∼

√α′v. This is the effective scale probed

by the brane: r0 =√α′v. A slow brane (v → 0) probes scales smaller than the

string scale! Moreover, we obtain an uncertainty in the position

δx ≥√α′v.

The time it takes for this scattering process is

δt ∼ δx/v

Therefore,

δxδt ≥ δx

v

√α′v ' α′v

v= α′.

A new uncertainty principle! It implies that coordinates do not commute! Itseems that Nature is described by noncommutative geometry. What can thispossibly mean??Consider two branes separated by a distance y. Strings ending on the samebrane have a massless mode each, so we have two massless modes. A stringstretched between the two branes has

L0 =y2

4π2α′ + ...

This extra term makes L0 > 0 i.e., there are no massless modes. At low ener-gies, we only see two massless particles from the two branes. However, wheny → 0, the stretched string develops a massless mode, and there are two of


them. So when the two branes coincide, there are four massless modes. Thesefour modes can be grouped into a matrix Xij in an obvious notation.Each Xij is the position of the brane! When we develop a particle theory weneed to treat the position of the brane as a 2 × 2 matrix. More generally, ndistinct branes have n massless modes. The particle theory is just n copiesof the same theory. When all n branes coincide, we have n2 massless modes.Each massless mode corresponds to a symmetry of the theory (U(1)). With n2

massless states the symmetry is enhanced to U(n) (n2 generators).Familiar ExamplesPhoton: Fµν = ∂µAν − ∂νAµ, U(1) symmetry (Aµ → Aµ + ∂µλ).3 Photons: F iµν = ∂µA

iν − ∂νAiµ, U(1)3 symmetry.

Weak Bosons: DemandSU(2) symmetry which has three generators, soF iµν 6=∂µA

iν = ∂νA

iµ. There is a correction, to obey the enhanced symmetry Fµν =

∂µAν − ∂νAµ + [Aµ, Aν ] (Aµ is a matrix ... Aµ = Aiµσi)Gluons: Demand SU(3) - eight gluons (32 − 1).

Fµν = ∂µAν − ∂νAµ + [Aµ, Aν ], Aµ = Aiµλi

where λi represent the Gell-Mann matrices. The action is given by

S ∼∫

d4x Tr FµνFµν .

If we only had eight copies of electromagnetism, we would have

S ∼∫

d4x Tr F iµνFµνi .

Now we have interactions between gluons - enhanced symmetry (gluons andweak bosons, unlike photons have charge).Potential: SetAµ =constant, then

Tr FµνFµν ∼ Tr [Aµ, Aν ]

2.

Back to D-branes: Xµ is like aµ (that can be made precise - see Polchinski8.6). So the enhanced symmetry contains a potential

V ∼ Tr [Xµ, Xν ]2

where µ, ν run over that dimension transverse to the branes. Expand aroundXµ = 0 in a Taylor series. There are no linear or quadratic terms in Xµ, sothere is no mass term (which would come from V (φ) = V (0) + V ′(0)φ +12V

′′(0)φ2/m2 + ...)So we have kn2 massless modes, where k is the number of transverse dimen-sions. Also, V = 0 if and only if all [Xm, Xn] = 0, i.e., all Xm commute. Thiscan be accomplished if we make them all diagonal. There are n diagonal el-ements, each corresponding to one of the D-branes. Thus this potential cor-rectly describes n coincident non-interacting free D-branes.

Documents

String Theory I...String Theory I GEORGE SIOPSIS AND STUDENTS Department of Physics and Astronomy The University of Tennessee Knoxville, TN 37996-1200 U.S.A. e-mail: [email protected]