String Theory I - University of Tennessee

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

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〉

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)

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 I - University of Tennessee

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