Superstrings Term 1

Lecturer: Arthur Lipstein

1 Introduction ..... 2
2 Basics ..... 2
2.1 Special Relativity ..... 3
2.2 Field theory ..... 6
2.2.1 Klein-Gordon ..... 6
2.2.2 Maxwell ..... 8
2.3 General Relativity ..... 10
3 Relativistic Point Particle ..... 12
4 Nambu - Goto Action ..... 15
5 Polyakov Action ..... 16
5.1 Symmetries and Gauge-fixing ..... 18
5.2 Equations of motion and boundary conditions ..... 18
5.3 Conserved Charges ..... 20
6 Mode expansion ..... 22
7 Virasoro algebra ..... 26
8 Light cone gauge ..... 28
9 Canonical Quantisation ..... 30
10 Spectrum and Critical Dimension ..... 33
11 CFT: Virasoro from Stress Tensor ..... 36
12 Operator Product Expansion ..... 39
13 Critical Dimension from CFT ..... 43
14 Vertex operators ..... 47
15 String interactions ..... 49

1 Introduction

The main references for this course will be:

Zwiebach: A First Course in String Theory
Becker, Becker, Schwarz: String theory and M-theory
Tong: String Theory (online lecture notes)
Green, Schwarz, Witten: Superstring Theory (Volume 1)
Polchinski: Joe's Little Book of String (Polchinski's online lecture notes)
Srednicki: Quantum Field Theory

This section is based on Zwiebach Chapter 1. The Standard Model describes four fundamental forces (strong, weak, electromagnetic, gravity). The first three give consistent quantum theories, but quantum Einstein gravity is plagued by divergences and is believed to break down at sufficiently small distances/high energies given by

l_{p} = \sqrt{\frac{G ℏ}{c^{3}}} \sim 10^{- 33} c m, m_{p} = \sqrt{\frac{ℏ c}{G}} \sim 10^{19} G e V / c^{2}

where

G =

Newton's constant,

ℏ =

Planck's constant,

c =

speed of light. Planck length

l_{p} =

is the unique length that can be constructed using only powers of

G, ℏ, c

. Similarly Planck mass

m_{p} =

is unique mass that can be constructed in this way.

String theory provides a mathematically consistent theory of quantum gravity (no UV divergences). It also unifies gravity with the other forces. Roughly speaking, gravity comes from fluctuations of closed strings, while gauge theories (like electromagnetism) come from fluctuations of open strings. Since open strings can close to form closed strings, it's not possible to have a theory with only open strings. Hence, gravity is required by string theory.

String theory is also important theoretically because it lead to the discovery of supersymmety and the AdS/CFT correspondence. The latter relates quantum gravity to a lower-dimensional non-gravitational quantum theory. On the other hand making experimentally testable predictions remains a challenge for string theory. This may require a better understanding of its conceptual foundations. The goal of this course is to understand the emergence gravity, its unification with gauge theory, and its good high energy behaviour in the context of bosonic string theory. Supersymmetry will be described next term.

2 Basics

In this section we will review some basic concepts:

Special relativity
Field theory: Klein-Gordon, Maxwell
General relativity

2.1 Special Relativity

The principle of Relativity states that equations of motion should be the same in all frames. Frames are coordinate systems that observers use to measure locations of events in space and time, which are tied to how the observer is moving. There is a special set of frames in which objects follow linear trajectories (at least locally). These are known as inertial frames. Physically, these arise in the absence of gravity and acceleration. Even in the presence of gravity, it is possible for an observer to have a locally inertial frame by going to a state of free-fall. Special Relativity describes motion in inertial frames. General Relativity describes motion in general frames and explains how gravity arises from the curvature of spacetime.

Events are labelled by spacetime coordinates:

x^{μ} = (x^{0}, x^{1}, x^{2}, x^{3})

where

x^{0}

is time and

x^{1, 2, 3}

are spatial directions. Suppose an inertial observer measures two events at

x^{μ}

and

x^{μ} + Δ x^{μ}

. The distance in spacetime between these two events is given by

- Δ s^{2} = - {(Δ x^{0})}^{2} + {(Δ x^{1})}^{2} + {(Δ x^{2})}^{2} + {(Δ x^{3})}^{2}

Now suppose that another inertial observer measures the events to be at

x^{' μ}

and

x^{' μ} + Δ x^{' μ}

. The spacetime distance they measure is the same:

Δ s^{' 2} = Δ s^{2}

Hence

Δ S^{2}

is an invariant interval.

$Δ s^{2} > 0$ time-like
$Δ s^{2} < 0$ spacelike
$Δ s^{2} = 0$

Δ s^{2} = 0

, then the two events can be separated by a light ray. Since the distance between the two events is invariant, the speed of light is the same in all frames.

For events that are infinitesimally close together, we denote the invariant interval as

\begin{aligned} - d s^{2} & = - {(d x^{0})}^{2} + {(d x^{1})}^{2} + {(d x^{2})}^{2} + {(d x^{3})}^{2} \\ = \sum_{μ = 0}^{3} \sum_{ν = 0}^{3} η_{μ v} d x^{μ} d x^{ν} \end{aligned}

where

η_{μ v} = (\begin{array}{cccc} - 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}) Minkowski metric

We will use the Einstein summation convention where repeated indices summed:

- d s^{2} = η_{μ ν} d x^{μ} d x^{ν}

Denote inverse Minkowsi metric with upper indices

η^{μ ν}

. Then

η^{ν ρ} η_{ρ μ} = δ_{μ}^{ν}, where δ_{μ}^{ν} = {\begin{cases} 1 & if μ = ν \\ 0 & if μ \neq ν \end{cases}

We can use the metric to raise and lower indices:

\begin{aligned} b_{μ} \equiv η_{μ ν} b^{ν} \to b^{μ} = η^{μ ν} b_{ν} = η^{μ ν} η_{ν ρ} b^{ρ} = δ_{ρ}^{μ} b^{ρ} = b^{μ} \\ a \cdot b \equiv η_{μ ν} a^{μ} b^{ν} = a^{μ} b_{μ} = a_{μ} b^{μ} = - a^{0} b^{0} + a^{1} b^{1} + a^{2} b^{2} + a^{3} b^{3} \end{aligned}

In general, inertial frames are related by Poincare transformations:

x^{' μ} = L_{ν}^{μ} x^{ν} + T^{μ}

where

L

is a Lorentz transformation and

T

is a translation. This change of coordinatess leaves the metric invariant:

\begin{aligned} d s^{2} = n_{α β} d x^{α} d x^{β} \\ = n_{μ ν} d x^{μ} d x^{ν} \\ = n_{μ ν} \frac{\partial x^{' μ}}{\partial x^{α}} \frac{\partial x^{ν}}{\partial x^{β}} d x^{α} d x^{β} \\ = η_{μ ν} L_{α}^{μ} L_{β}^{ν} d x^{α} d x^{β} \\ \to η_{μ ν} L_{α}^{μ} L_{β}^{ν} = η_{α β} \end{aligned}

In terms of matrices, this can be written as

L^{T} η L = η

. Lorentz transformations can be thought of as rotations in spacetime. For example a rotation in the

x^{1} - x^{2}

plane takes in the form

L = (\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \cos θ & \sin θ & 0 \\ 0 & - \sin θ & \cos θ & 0 \\ 0 & 0 & 0 & 1 \end{array})

and a boost along the

x^{1}

direction is given by

L = (\begin{array}{cccc} γ & γ β & 0 & 0 \\ γ β & γ & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array})

where

β = | \vec{v} | / c, \vec{v}

relative velocity of the observers, and

γ = \frac{1}{\sqrt{1 - β^{2}}}

L

is a Lorentz transformation then so are

L^{- 1}

and

L^{T}

. It follows that

{(L^{- 1})}^{T} η L^{- 1} = η, L η L^{T} = η \to L^{- 1} η {(L^{- 1})}^{T} = η

In terms of Lorentz indices we therefore have

η^{μ ν} {(L^{- 1})}_{μ}^{α} {(L^{- 1})}_{ν}^{β} = η^{α β}

Hence, the Laplacian is Lorentz invariant:

\begin{aligned} η^{μ ν} \frac{\partial}{\partial x^{μ}} \frac{\partial}{\partial x^{' ν}} ϕ & = η^{μ ν} \frac{\partial x^{α}}{\partial x^{μ}} \frac{\partial x^{β}}{\partial x^{' ν}} \frac{\partial}{\partial x^{α}} \frac{\partial}{\partial x^{β}} ϕ \\ = η^{μ ν} {(L^{- 1})}_{μ}^{α} {(L^{- 1})}_{ν}^{β} \frac{\partial}{\partial x^{α}} \frac{\partial}{\partial x^{β}} ϕ \\ = η^{α β} \frac{\partial}{\partial x^{α}} \frac{\partial}{\partial x^{β}} ϕ \end{aligned}

where

ϕ

is some function which is Lorentz invariant, i.e. a scalar field.

Relativistic Momentum: (from now on we will set

c = 1

and

ℏ = 1

) Massive parties follow timelike trajectories:

- d s^{2} = - {(d x^{0})}^{2} + {(d x^{1})}^{2} + {(d x^{2})}^{2} + {(d x^{3})}^{3} < 0

In this case,

s

is called the proper time. It corresponds to the time elapsed in the rest frame of the particle; where

d x^{1} = d x^{2} = d x^{3} = 0

so,

d s^{2} = {(d x^{0})}^{2}

. We then define the 4 -velocity as

u^{μ} = \frac{d x^{μ}}{d s}

Since

d s

is Lorentz -invariant and

d x^{μ}

is a Lorentz Vector,

u^{μ}

also transforms like a vector under coordinate transformations. More explicitly, it can be written as

\begin{aligned} u^{μ} & = \frac{d x^{0}}{d s} (1, \frac{d x^{1}}{d x^{0}}, \frac{d x^{2}}{d x^{0}}, \frac{d x^{3}}{d x^{0}}) \\ = γ (1, \vec{v}), \vec{v} = \frac{d \vec{x}}{d x^{0}} \end{aligned}

where we noted that

d s^{2} = {(d x^{0})}^{2} (1 - {(d \vec{x} / d x^{0})}^{2})

\begin{aligned} = {(d x^{0})}^{2} (1 - β^{2}) \\ \to \frac{d x^{0}}{d s} = \frac{1}{\sqrt{1 - β^{2}}} = γ \end{aligned}

Note that

u^{μ} u_{ν} = γ^{2} (- 1 + β^{2}) = - 1

Finally, we define the 4-momentum for particle of mass

m

p^{μ} = m u^{μ} \to p^{μ} p_{μ} = m^{2} u^{μ} u_{μ} = - m^{2}

In the non-relativistic limit

β ≪ 1

p^{0} = m + \frac{1}{2} m β^{2} + \dots

which we recognise as the rest energy plus the the kinetic energy. Hence

p^{0} = E

(energy). In general, the energy is related to the mass an spatial momentum as:

p^{2} \equiv p_{μ} p^{μ} = - E^{2} + {\vec{p}}^{2} = - m^{2} \to E^{2} = {\vec{p}}^{2} + m^{2}

It will be convenient later on to use light cone coordinates:

x^{\pm} = \frac{1}{\sqrt{2}} (x^{0} \pm x^{1})

Note that

d x^{+} d x^{-} = \frac{1}{2} (d x^{0} + d x^{1}) (d x^{0} - d x^{1}) = \frac{1}{2} ({(d x^{0})}^{2} - {(d x^{1})}^{2})

Hence, in these coords

\begin{aligned} - d s^{2} & = - 2 d x^{+} d x^{-} + {(d x^{2})}^{2} + {(d x^{3})}^{2} \\ = {\hat{η}}_{μ ν} d {\hat{x}}^{μ} d {\hat{x}}^{ν}, {\hat{x}}^{μ} = (x^{+}, x^{-}, x^{2}, x^{3}) \\ \hat{η} & = (\begin{array}{cccc} 0 & - 1 & 0 & 0 \\ - 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}) \end{aligned}

Similarly, we define

p^{\pm} = \frac{1}{\sqrt{2}} (p^{0} \pm p^{1})

\to p^{2} = - 2 p^{+} p^{-} + {(p^{2})}^{2} + {(p^{3})}^{2} = - m^{2}

2.2 Field theory

2.2.1 Klein-Gordon

This will be based on Zwiebach 10.2-10.4, Srednicki Chapter 3. First we consider scalar (or spin-0) fields. This is described by the Klein-Gordon equation:

η^{μ ν} \partial_{μ} \partial_{ν} ϕ - m^{2} ϕ = 0

where

η^{μ ν} \partial_{μ} \partial_{ν} ϕ = (- \partial_{0}^{2} + {\vec{\nabla}}^{2}) ϕ = \partial^{2} ϕ

We previously showed that the Laplacian is Lorentz invariant we see that the equation of motion is the same in all frames.

Let us Fourier transform expand the scalar field:

ϕ (x) = \int \frac{d^{D} p}{(2 π)^{D}} e^{i p \cdot x} ϕ (p)

where

D

is spacetime dimension, and we recall that

p \cdot x = p_{μ} x^{μ} = - p^{0} x^{0} + \vec{p} \cdot \vec{x}

Plugging the Fourier transform into the equation of motion gives

(\partial^{2} - m^{2}) ϕ (x) = \int \frac{d^{D} p}{(2 π)^{D}} \int^{- e^{i p \cdot x}} (- p^{2} - m^{2}) ϕ (p) = 0 \to (p^{2} + m^{2}) ϕ (p) = 0

For this to hold for arbitrary

ϕ (p)

, the momentum must be "on-shell":

\begin{aligned} p^{2} + m^{2} = 0 & \to {(p^{0})}^{2} - {\vec{p}}^{2} - m^{2} = 0 \\ \to p^{0} = \pm \underset{E_{p}}{\underset{⏟}{\sqrt{{\vec{p}}^{2} + m^{2}}}} \end{aligned}

Hence, for each

\vec{p}

there are two solutions to the equations of motion which can come with arbitrary coefficients

a

and

b

Plugging this solution back into the Fourier expansion then gives

ϕ (x) = \int \frac{d^{D - 1} p}{(2 π)^{D - 1}} \int d p^{0} δ (p^{2} + m^{2}) θ (p^{0}) [a (\vec{p}) e^{- i E_{p} x^{0} + i \vec{p} \cdot x} + b (\vec{p}) e^{i E_{p} x^{0} + i \vec{p} \cdot \vec{x}}]

where we absorbed

1 / 2 π

from measure into definition the coefficients

a, b

. Using

\int d x f (f (x)) = \sum_{i} 1 / | f^{'} (x_{i}) | f (x_{i}) = 0

we finally get

ϕ (x) = \int \frac{d^{D - 1} p}{(2 π)^{D - 1} 2 E_{p}} [a (\vec{p}) e^{- i E_{p} x^{0} + i \vec{p} \cdot \vec{x}} + b (\vec{p}) e^{i E_{p} x^{0} + i \vec{p} \cdot \vec{x}}]

ϕ (x)

is real, it has one propagating degree of freedom. If

ϕ (x)

is complex, it has two real propagating degrees of freedom. Let's take it to be real. Taking the complex conjugate,

\begin{aligned} ϕ^{*} (x) & = \int \frac{d^{D - 1} p}{(2 π)^{D - 1} 2 E_{p}} (a^{*} (\vec{p}) e^{i E_{p} x^{0} - i \vec{p} \cdot \vec{x}} + b^{*} (\vec{p}) e^{- i E_{p} x^{0} - i \vec{p} \cdot \vec{x}}) \\ = \int \frac{d^{D - 1} p}{(2 π)^{D - 1} 2 E_{p}} (a^{*} (- \vec{p}) e^{i E_{p} x^{0} + i \vec{p} \cdot \vec{x}} + b^{*} (- \vec{p}) e^{- i E_{p} x^{\circ} + i \vec{p} \cdot \vec{x}}) \end{aligned}

when we took

\vec{p} \to - \vec{p}

in the second line. From this, we see that

ϕ (x) =

ϕ^{*} (x) ⟷ b (\vec{p}) = a^{*} (- \vec{p})

. Plugging this back in finally gives

ϕ (x) = \int \frac{d^{D - 1} p}{(2 π)^{D - 1} 2 E_{p}} [a (\vec{p}) e^{- i E_{p} x^{0} + i \vec{p} \cdot \vec{x}} + a^{*} (\vec{p}) e^{i E_{p} x^{0} - i \vec{p} \cdot \vec{x}}]

To quantize, we promote

a

and

a^{*}

to creation and annihilation operators

a (\vec{p}), a^{†} (\vec{p}) :

\begin{aligned} [a (\vec{p}), a ({\vec{p}}^{'})] = [a^{†} (\vec{p}), a^{†} ({\vec{p}}^{'})] = 0 \\ [a (\vec{p}), a^{†} ({\vec{p}}^{'})] = (2 π)^{3} 2 E_{p} δ^{3} (\vec{p} - {\vec{p}}^{'}) \end{aligned}

This can be derived using a procedure called "canonical quantization". We wont go into the details, but physically one can think of the quatum field

ϕ (x)

as an infinite set of harmonic oscillations, one for each Fourier mode. For more details, see see section 10.4 of Zwieback and chapter 3 of Srednicki.

vacuum state satifies $a (\vec{p}) | 0 ⟩ = 0$ .
$a^{†} (\vec{p}) | 0 ⟩$ is a single particle state with momentum $\vec{p}$ and energy $E_{p}$ .
$a^{†} ({\vec{p}}_{1}) \dots a^{†} ({\vec{p}}_{n}) | 0 ⟩$ is an $n$ -particle state. The particles have momenta ${\vec{p}}_{1}, \dots, {\vec{p}}_{n}$ and energies $E_{p_{1}, \dots,}, E_{p_{n}}$ .

2.2.2 Maxwell

This will be based on Zwiebach 3.1,3.3,10.5. Maxwell's equations for electromagnetism:

\begin{aligned} \vec{\nabla} \cdot \vec{E} = ρ \\ \vec{\nabla} \times \vec{B} - \partial_{0} \vec{E} = \vec{J} \\ \vec{\nabla} \cdot \vec{B} = 0 \\ \vec{\nabla} \times \vec{E} + \partial_{0} \vec{B} = 0 \end{aligned}

We can write this in Lorentz-covariant way by introducing field strength and 4-current:

\begin{aligned} F^{μ ν} = (\begin{array}{cccc} 0 & E_{1} & E_{2} & E_{3} \\ - E_{1} & 0 & B_{3} & - B_{2} \\ - E_{2} & - B_{3} & 0 & B_{1} \\ - E_{3} & B_{2} & - B_{1} & 0 \end{array}) \\ J^{μ} = (ρ, \vec{J}) \end{aligned}

Then Maxwell's equations reduce to

\begin{aligned} \partial_{ν} F^{μ ν} = J^{μ} \\ \partial_{μ} F_{ν λ} + \partial_{ν} F_{λ μ} + \partial_{λ} F_{μ ν} = 0 ⟷ \partial_{[μ} F_{ν λ]} = 0 \end{aligned}

Under Lorentz transformations,

F_{μ ν} \to L_{μ}^{α} L_{ν}^{β} F_{α β}, \partial_{μ} \to L_{μ}^{α} \partial_{α}, J_{μ} \to L_{μ}^{α} J_{α}

so the equations have the same form in all inertial frames. Moreover, we see how

\vec{E}

and

\vec{B}

mix into eachother under Lorentz transformations.

The fields can be derived from a vector potential. To see this first note that

\begin{matrix} (1) & \vec{\nabla} \cdot \vec{B} = 0 \to \vec{B} = \vec{\nabla} \times \vec{A} \end{matrix}

Plugging this into second homogeneous Maxwell equation gives

\begin{aligned} 0 & = \vec{\nabla} \times \vec{E} + \partial_{0} \vec{B} \\ (2) & = \vec{\nabla} \times (\vec{E} + \partial_{0} \vec{A}) \to \vec{E} + \partial_{0} \vec{A} = - \vec{\nabla} Φ \to \vec{E} = - \partial_{0} \vec{A} - \vec{\nabla} Φ \end{aligned}

Defining

A^{μ} = (Φ, \vec{A})

, equations (1) and (2) can be written as

F_{μ ν} = \partial_{μ} A_{ν} - \partial_{ν} A_{μ}

Hence, we see that electromagnetism is described by a spin-1 field

A^{μ}

. From this form of

F_{μ ν}

it automatically follows that

\partial_{[μ} F_{ν λ]} = 0

Setting

J^{μ} = 0

, the remaining Maxwell equations can be written as

\begin{aligned} 0 & = \partial_{μ} F^{μ ν} \\ = \partial_{μ} (\partial^{μ} A^{ν} - \partial^{ν} A^{μ}) \\ = \partial^{2} A^{ν} - \partial^{ν} \partial \cdot A \end{aligned}

Note that Maxwell's equations have a gauge symmetry:

A_{μ} (x) \to A_{μ} (x) + \partial_{μ} Λ (x), ie. δ A_{μ} = \partial_{μ} Λ

where

Λ (x)

is an arbitrary function. Under this transformation

F_{μ ν}

is invariant:

\begin{aligned} F_{μ ν} & \to \partial_{μ} (A_{ν} + \partial_{ν} Λ) - \partial_{ν} (A_{μ} - \partial_{μ} Λ) \\ = F_{μ ν} + \partial_{μ} \partial_{ν} Λ - \partial_{ν} \partial_{μ} Λ \\ = F_{μ ν} \end{aligned}

If we Fourier transform to momentum space:

A^{μ} (x) = \int \frac{d^{D} p}{(2 π)^{D}} e^{i p \cdot x} A^{μ} (p)

Then equations of motion are then given by

p^{2} A^{μ} (p) - p^{μ} p \cdot A (p) = 0

In momentum space the gauge transformation is

δ A^{μ} (p) = i p^{μ} Λ (p)

In particular,

δ A^{+} (p) = i p^{+} Λ (p)

. Choosing

Λ = i A^{+} / p^{+} \to δ A^{+} = - A^{+}

, which imposes light-cone gauge:

A^{+} (p) = 0

In this gauge, the

μ = +

component of the equaiton of motion reduces to

p^{+} p \cdot A = 0 \to p \cdot A = 0

Writing out components,

\begin{aligned} 0 = p \cdot A \\ = - p^{+} A^{-} - p^{-} A^{+} + p^{I} A^{I} \\ \to A^{-} = p^{I} A^{I} / p^{\pm} \end{aligned}

where

I = 2, \dots, D - 1

"transverse directions." Setting

μ = I

, the equations of motion then reduce to

p^{2} A^{I} (p) = 0

which are just the equations of motion for

(D - 2)

massless scalars! Hence, we that the Maxwell field has D-2 independent degrees of freedom (one for each transverse degree of freedom).

A^{-}

is not an independent degree of freedom because it can be written in terms of

A^{I}

. We may quantize

A^{I}

as we did the scalar field, writing them in terms of creation and annihilation operators. The quanta of

A^{I}

are called photons. A general single photon state is given by

\sum_{I = 2}^{D - 1} ϵ_{I} {(a^{I} (\vec{p}))}^{†} | 0 ⟩

where

ϵ_{I}

is a polarisation vector.

2.3 General Relativity

This will mostly follow Zwiebach 3.6, 10.6. As mentioned previously, GR describes motion in general frames. Recall that if we start in an inertial frame and perform a coordinate transformation, the metric will become

g_{ρ λ} (x^{'}) = \frac{\partial x^{μ}}{\partial x^{' ρ}} \frac{\partial x^{ν}}{\partial x^{' λ}} η_{μ ν}

where

x^{μ}

are the coordinates of the inertial frame and

x^{' μ}

are the coordinates of the new frame, respectively. If

\partial x^{μ} / \partial x^{' ρ}

is a Lorentz transformation, then

g_{μ ν} = η_{μ ν}

. For more general coordinate transformations,

g_{μ ν}

will not be the Minkowski metric. Physically, this can arise from going to an accelerating frame.

There is another way to have a non-trivial metric: turning on gravity. In this case it will not be possible to transform the metric to Minkowski (except locally) via coordinate transformation. This is because gravity arises from the curvature of spacetime, which can't be set to zero by a coordinate transformation. Minkowski has zero curvature and is therefore a flat spacetime.

Hence, the invariant interval is generally given by

- d s^{2} = g_{μ ν} (x) d x^{μ} d x^{ν}

Under general coordinate transformation (or diffeomorphism)

g_{μ ν}^{'} (x^{'}) = \frac{\partial x^{α}}{\partial x^{' μ}} \frac{\partial x^{β}}{\partial x^{' ν}} g_{α β} (x)

Consider an infinitesimal coordinate transformation:

x^{μ} = x^{μ} - ξ^{μ} (x)

Under this transformation,

\begin{matrix} (3) & δ g_{α β} = ξ^{λ} \partial_{λ} g_{α β} + \partial_{α} ξ^{λ} g_{λ β} + \partial_{β} ξ^{λ} g_{α λ} \end{matrix}

You will show this in the homework. This is the gravitational analogue of a gauge transformation that we saw in Maxwell theory.

In the presence of gravity, the metric becomes dynamical and its equation of motion is the Einstein equation which states that the curvature of spacetime can be derived from the distribution of matter and energy. The Einstein equations are invariant under diffeomorphisms. We will not write them out in general but instead consider small fluctuations around Minkowski (i.e. weak gravity):

g_{μ ν} (x) = η_{μ ν} + h_{μ ν} (x)

In the absence of matter and energy, the linearised equations of motion for

h_{μ ν}

are

\partial^{2} h^{μ ν} - \partial_{α} (\partial^{μ} h^{ν α} + \partial^{ν} h^{μ α}) + \partial^{μ} \partial^{ν} h = 0

where

h^{μ ν} = η^{μ α} η^{ν β} h_{α β}

and

h = η^{μ ν} h_{μ ν}

If we expand (3) to linear order in

h_{μ ν}

and

ξ_{μ}

we get

δ h_{μ ν} = \partial_{μ} ξ_{ν} + \partial_{ν} ξ_{μ}

The equations of motion are invariant under this variation. To see this, let's Fourier transform to momentum space, which amounts to the replacement

\partial_{μ} \to

+ i p_{μ}

. The equations of motion then become

S^{μ ν} = p^{2} h^{μ ν} - p_{α} (p^{μ} h^{ν α} + p^{ν} h^{μ α}) + p^{μ} p^{ν} h = 0

and the diffeomorphism becomes

δ h^{μ ν} = i p^{μ} ξ^{ν} (p) + i p^{ν} ξ^{μ} (p)

Under this, the equations of motion have the following variation:

\begin{aligned} δ S^{μ ν} = i p^{2} (p^{μ} ξ^{ν} + p^{ν} ξ^{μ}) & - i p_{α} p^{μ} (p^{ν} ξ^{α} + p^{α} ξ^{ν}) \\ - i p_{α} p^{ν} (p^{μ} ξ^{α} + p^{α} ξ^{μ}) + 2 i p^{μ} p^{ν} p \cdot ξ \end{aligned}

The terms cancel in pairs.

We can use this gauge symmetry to set all components of the metric with a + component to zero:

\begin{aligned} δ h^{+ +} = 2 i p^{+} ξ^{+} = - h^{+ +} \to ξ^{+} = i h^{+ +} / 2 p^{+} \\ δ h^{+ -} = i p^{+} ξ^{-} + i p^{-} ξ^{+} = - h^{+ -} \to ξ^{-} = (i h^{+ -} - p^{-} ξ^{+}) / p^{+} \\ δ h^{+ I} = i p^{+} ξ^{I} + i p^{I} ξ^{+} = - h^{+ I} \to ξ^{I} = (i h^{+ I} - p^{I} ξ^{+}) / p^{I}, I \in 2, \dots, D - 1 \end{aligned}

For this choice of

ξ^{μ}

, we get light-cone gauge:

h^{+ +} = h^{+ -} = h^{+ I} = 0

For this gauge choice, the equations of motion reduce to (HW)

\begin{matrix} p^{2} h^{I J} = 0, h^{I I} = 0 \\ h^{I -} = \frac{1}{p^{+}} p_{J} h^{I J}, h^{- -} = \frac{1}{p^{+}} p_{I} h^{- I} \end{matrix}

Note that

h^{I -}, h^{- -}

are determined by

h^{I J}

so are not independent degrees of freedom.

Hence,

h^{I J}

can be treated like massless scalars, and

h^{- I}

and

h^{- -}

are determined from these. Since

h^{I J}

is a symmetric traceless

(D - 2) \times (D - 2)

matrix, the number of degrees of freedom in the gravitational field is

(\binom{D - 2}{2}) + (D - 2) - 1 = \frac{1}{2} (D - 2) (D - 3) + D - 3 = \frac{1}{2} D (D - 3)

For

D = 4

, we see that gravity has 2 degrees of freedom. After quantisation, we can once again write

h^{I J}

in terms of creation and annihilation operators. The quanta of the gravitational field are called gravitons. One-graviton states are given by

\sum_{I, J = 2}^{D - 1} ϵ_{I J} {(a^{I J} (\vec{p}))}^{†} | 0 ⟩

where

ϵ_{I J}

is a symmetric traceless tensor.

3 Relativistic Point Particle

This will primarily follow Tong Section 1.1. In non-relativistic quantum mechanics, spatial coordinates

X^{i}

are operators white

X^{0}

is a label. On the other hand, in relativity space and time are on equal footing so we have two choices: to demote spatial coordinates to labels and introduce fields giving QFT, or to promote time to an operator. The latter approach is less convenient in practice for the study of point particles, but generalises more straightforwardly to string theory.

Alter promoting

X^{0}

to and operator, we must introduce a new parameter

τ

to parameterize the world line. One then integrates over all paths between initial and final spacetime points. The action should be Lorentz-invariant and extremised by the classical trajectory. The natural candidate is the proper time of the particle's trajectory:

S = - m \int d s

where

\begin{aligned} d s^{2} & = - d X^{μ} d X^{ν} η_{μ ν} \\ = - {\dot{X}}^{μ} {\dot{X}}^{ν} η_{μ ν} d τ^{2}, {\dot{X}}^{μ} = \partial_{τ} X^{μ} \end{aligned}

X^{μ} (τ)

, encodes the embedding of the world line into the "target space," which we take to be Minkowski space.

Hence the action for a relativistic point particle is

S = - m \int d σ \sqrt{{\dot{X}}^{μ} {\dot{X}}^{v} η_{μ v}}

This is manifestly invariant under Poincare transformations of target space:

X^{μ} \to L_{ν}^{μ} X^{ν} + T^{μ}

It also has "reparameterisation invariance":

τ \to \tilde{τ} (τ)

Indeed, we see that

\begin{aligned} S & = - m \int d \tilde{τ} \sqrt{- \partial_{\tilde{τ}} X^{μ} \partial_{\tilde{τ}} X^{ν} η_{μ ν}} \\ = - m \int d τ \frac{\partial \tilde{τ}}{\partial τ} \sqrt{- (\frac{\partial τ}{\partial \tilde{τ}} \partial_{τ} X^{μ}) (\frac{\partial τ}{\partial \tilde{τ}} \partial_{τ} X^{ν}) η_{μ ν}} \\ = - m \int d τ \sqrt{- \partial_{τ} X^{μ} \partial_{τ} X^{ν} η_{μ ν}} \end{aligned}

We can use this symmetry to set

\tilde{τ} = X^{0} (τ)

(this is know as static gauge), so that

S = \int d τ \sqrt{1 - {\dot{\vec{X}}}^{2}}

where we relabeled

\tilde{τ}

τ

. Lorentz-invariance is no longer manifest. Let's expand the static gauge action in the non-relativistic limit:

\begin{aligned} S = \int d τ \sqrt{1 - {\dot{\vec{X}}}^{2}} \\ = - m \int d τ (1 - \frac{1}{2} {\dot{\vec{X}}}^{2} + \dots) \end{aligned}

where ... indicates higher order terms in

\dot{\vec{X}}

. The first term is the rest-mass energy and the second term is the non-relativistic kinetic energy.

We will now compute the equations of motion and quantize using the manifestly Lorentz-invanaut action. Let's compute the equations of motion by extremising the action. Consider the variation

X^{μ} (τ) \to X^{μ} (τ) + δ X^{μ} (τ)

\begin{aligned} δ S = \int d τ δ L ({\dot{X}}^{μ} (τ)) \\ = \int d τ \frac{\partial L}{\partial {\dot{X}}^{μ}} δ {\dot{X}}^{μ} = \int d τ [\partial_{τ} (\frac{\partial L}{\partial {\dot{X}}^{μ}} δ X^{μ}) - \partial_{τ} (\frac{\partial L}{\partial {\dot{X}}^{μ}}) δ X^{μ}] \\ = {\frac{\partial L}{\partial {\dot{X}}^{μ}} δ X^{μ} |}_{τ = τ_{i}}^{τ = τ_{f}} - \int d τ \partial_{τ} (\frac{\partial L}{\partial {\dot{X}}^{μ}}) δ X^{μ} \\ = - \int d τ \partial_{τ} (\frac{\partial L}{\partial {\dot{X}}^{μ}}) δ X^{μ} \end{aligned}

where we take

δ X^{μ} (τ_{i}) = δ X^{μ} (τ_{f}) = 0

. Hence, we find that

δ S = 0 \to \partial_{τ} Π_{μ} = 0

where

Π_{μ} = \frac{\partial L}{\partial {\dot{X}}^{μ}}

is the canonical momentum.

In the present case

L = - m \sqrt{- {\dot{X}}^{2}}, {\dot{X}}^{2} = {\dot{X}}^{μ} {\dot{X}}^{ν} η_{μ ν}

. Hence,

Π_{μ} = \frac{\partial L}{\partial {\dot{X}}^{μ}} = - \frac{m}{2} \frac{1}{\sqrt{- {\dot{X}}^{2}}} (- 2 {\dot{X}}_{μ}) = + \frac{m {\dot{X}}_{μ}}{\sqrt{- {\dot{X}}^{2}}} = m \frac{d X_{μ}}{d s}

Note the

Π^{2} = - m^{2}

. The equations of motion are

0 = \partial_{τ} Π_{μ} = m \frac{d}{d τ} (\frac{d X^{μ}}{d s}) \to \underset{\frac{d}{d s}}{\underset{⏟}{\frac{1}{\sqrt{- {\dot{X}}^{2}}} \frac{d}{d τ}}} (\frac{d X^{μ}}{d s}) = \frac{d^{2} X_{μ}}{d s^{2}} = 0

Hence, particles follow linear trajectories which satisfy the mass shell constraint, as expected.

To quantize, promote

Π_{μ}

to the operator

Π_{μ} = - i \partial / \partial X^{μ}

. Then we have the canonical quantisation condition

[X^{μ} (τ), Π_{ν} (τ^{'})] = i δ (τ - τ^{'}) δ_{ν}^{μ}

Moreover the constraint

Π^{2} + m^{2} = 0

becomes

(- \frac{\partial}{\partial X^{μ}} \frac{\partial}{\partial X^{ν}} η^{μ ν} + m^{2}) Ψ (x) = 0

where

Ψ

is a wavefunction. Hence we recover the Klein-Gordon equation we saw previously for scalar fields. Note that

Ψ (x)

is not a field, however. In this formalism we can only define a single-particle state which is described by the wavefunction. In contrast, in QFT we can construct multiparticle states. Similarly, when we generalise this approach to string theory, we will only be able to construct single-panticle states. To define multi-particle states in string theory, we need string field theory, which is beyond the scope of this course.

4 Nambu - Goto Action

This will follow Tong section 1.2 and BBS section 2.2. Whereas a particle sweeps out a world line in spacetime, a string sweeps out a

2 d

surface called a world sheet. Points on the worldsheet are parameterised by two worldsheet coordinates

σ^{α} = (σ^{0}, σ^{1}) = (τ, σ) . X^{μ} (τ, σ)

describes the embedding of the world sheet into spacetime.

open strings: $σ \in [0, π]$
closed strings: $σ \in [0, 2 π)$ (periodic)

Note that BBS takes

σ \in [0, π)

but we will follow the conventions of Tong.

Whereas the action for a relativistic point particle is proportional to the length of the worldline, the action of a string should be proportional to the area of the worldsheet. Hence, the classical string motion extremizes the area. To compute the area of the

2 d

surface swept out by the string, first compute the induced metric on the surface:

γ_{α β} = \frac{\partial X^{μ}}{\partial σ^{α}} \frac{\partial X^{ν}}{\partial σ^{β}} η_{μ ν}

The area of the surface is then given by

Area = \int d^{2} σ \sqrt{- \det γ}

Noting that

γ_{α β} = (\begin{array}{cc} {\dot{x}}^{2} & \dot{X} \cdot X^{'} \\ \dot{X} \cdot x^{'} & X^{' 2} \end{array})

, where

{\dot{X}}^{μ} = \partial_{τ} X^{μ}, X^{' μ} = \partial_{σ} X^{μ}

we see that

\begin{matrix} (4) & Area = \int d σ d τ \sqrt{{(\dot{X} \cdot X^{'})}^{2} - {\dot{X}}^{2} X^{' 2}} \end{matrix}

Let's verify this is the area when the target space in Euclidean

(R^{D})

rather than Minkowski space. Consider an infinitesimal region in

σ, τ

space and let's compute the area of the corresponding region in the target space. The vectors tangent to the boundary of the region in the target space are:

d {\vec{l}}_{1} = \frac{\partial \vec{X}}{\partial σ} d σ, d {\vec{l}}_{2} = \frac{\partial \vec{X}}{\partial τ} d τ

If the angle between the two vectors is

θ

, then the area is

\begin{aligned} d (Area) = | d \vec{l_{1}} | | d {\vec{l}}_{2} | \sin θ = \sqrt{d {\vec{l}}_{1}^{2} d {\vec{l}}_{2}^{2} (1 - \cos^{2} θ)} = \sqrt{d {\vec{l}}_{1}^{2} d {\vec{l}}_{2}^{2} - {(d \vec{l_{1}} \cdot d {\vec{l}}_{2})}^{2}} \\ = \sqrt{{\dot{X}}^{2} X^{' 2} - {(\dot{X} \cdot X^{'})}^{2}} d σ d τ \end{aligned}

which indeed takes the form of (4).

After this discussion we now present the Nambu-Goto action for a relativistic string:

S = - T \int d σ d τ \sqrt{{(\dot{X} \cdot X^{'})}^{2} - {\dot{X}}^{2} X^{' 2}}

where

T

is the string tension. To understand physical interpretation of

T

, let's take the non-relativistic limit. Using reparameteristion invariance

σ^{α} \to

σ^{' α} (τ, σ)

(which you will prove in HW) we may choose "static gauge":

X^{0} = τ, X^{1} = σ

Note that in this gauge

0 < σ < L

, where

L

is the spatial length of the string. Then

\begin{aligned} \det γ & = \det (\begin{array}{cc} {\dot{X}}^{2} & \dot{X} \cdot X^{'} \\ \dot{X} \cdot X^{'} & X^{' 2} \end{array}) \\ = \det (\begin{array}{cc} - 1 + {({\dot{X}}^{I})}^{2} & {\dot{X}}^{I} X^{' I} \\ {\dot{X}}^{I} X^{' I} & 1 + {(X^{' I})}^{2} \end{array}) \\ = (- 1 + {({\dot{X}}^{I})}^{2}) (1 + {(X^{' I})}^{2}) + \dots \\ = - 1 + {({\dot{X}}^{I})}^{2} - {(X^{' I})}^{2} + \dots \end{aligned}

where

I \in {2, \dots, D - 1}

labels the transverse directions and ... indicate higherorder terms in

{\dot{X}}^{I}

X^{' I}

, which we neglect in non-redativistic limit. In this limit,

\begin{aligned} S = - T \int d σ d τ \sqrt{- \det γ} \\ \sim - T \int d σ d τ \sqrt{1 - {({\dot{X}}^{I})}^{2} + {(X^{' I})}^{2}} \\ \sim T \int d τ d σ [- 1 + \frac{1}{2} {({\dot{X}}^{I})}^{2} - \frac{1}{2} {(X^{' I})}^{2}] \end{aligned}

Let's compute the rest energy:

S_{rest} = - T \cdot \int d τ d σ = - m \int d τ, where m = L T

This is the rest mass contribution to the action of a point-particle with mass

L T

. Hence,

T

is mass per unit length and has dimensions ( length )

^{- 2}

(in units where

ℏ = c = 1)

. It is conventional to define

T = \frac{1}{2 π α^{'}}

, where

α^{'}

is known as "Regge slope" for reasons we will see later. It is also conventional to define

α^{'} = \frac{1}{2} l_{s}^{2}

where

l_{s}

is the "string length".

5 Polyakov Action

This will mainly be based on BBS pg

26, 27, 30, 31

. Although the Nambu-Goto action has a nice physical interpretation as the area of a string worldsheet, the
equations of motion are complicated and it is hard to quantize because of the square root. In practice, we use the Polyakov action:

S = - \frac{T}{2} \int d^{2} σ \sqrt{- h} h^{α β} \partial_{α} X^{μ} \partial_{β} X^{ν} η_{μ ν}

where

α \in {0, 1}, h_{α β} (τ, σ)

is the worldsheet metric,

\begin{aligned} h = \det h_{α β} \\ h^{α β} = {(h^{- 1})}^{α β} \end{aligned}

Let us show that this is classically equivalent to the NG action. First compute the equation of motion for

h^{α β}

δ S = - \frac{T}{2} \int d^{2} σ [δ (\sqrt{- h}) h^{α β} \partial_{α} X \cdot \partial_{β} X + \sqrt{- h} δ h^{α β} \partial_{α} X \cdot \partial_{β} X]

Noting the

δ \sqrt{- h} = - \frac{1}{2} \sqrt{- h} h_{α β} δ h^{α β}

(which you will provie on HW)

\begin{aligned} δ S & = - \frac{T}{2} \int d^{2} σ [- \frac{1}{2} \sqrt{- h} h_{α β} δ h^{α β} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X + \sqrt{- h} δ h^{α β} \partial_{α} X \cdot \partial_{β} X] \\ = - \frac{T}{2} \int d^{2} σ \sqrt{- h} δ h^{α β} [\partial_{α} X \cdot \partial_{β} X - \frac{1}{2} h_{α β} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X] \\ = - \frac{T}{2} \int d^{2} σ \sqrt{- h} δ h^{α β} T_{α β} \end{aligned}

where

T_{α β}

is the "stress tensor". Hence action extremised

(δ S = 0)

when

T_{α β} = 0

In this case, we have

\partial_{α} X \cdot \partial_{β} X = \frac{1}{2} h_{α β} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X

Taking determinant of both sides (recall they are

2 \times 2

matrices):

\det (\partial_{α} X \cdot \partial_{β} X) = h {(\frac{1}{2} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X)}^{2}

Multiplying by

(- 1)

and taking square rot:

\sqrt{- \det (\partial_{α} X \cdot \partial_{β} X)} = \frac{1}{2} \sqrt{- h} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X

Recalling the Nambu-Goto action, we see that

\begin{aligned} S & = - T \int d^{2} σ \sqrt{- \det (\partial_{α} X \partial_{β} X)} \\ = - \frac{T}{2} \int d^{2} σ \sqrt{- h} h^{γ} \partial_{γ} X \cdot \partial_{δ} X \end{aligned}

as claimed.

5.1 Symmetries and Gauge-fixing

The Polyakov action has the following symmeties:

Poincare transformations of target space: $X^{μ} \to X^{μ} = L_{ν}^{μ} X^{ν} + T^{μ}$
Reparameterisations:

σ^{α} \to σ^{' α} (τ, σ), h_{α β} (τ, σ) \to h_{α β}^{'} (τ^{'}, σ^{'}) = \frac{\partial σ^{γ}}{\partial σ^{' α}} \frac{\partial σ^{δ}}{\partial σ^{' β}} h_{γ δ} (τ, σ)

Check:

\begin{aligned} S & = - \frac{T}{2} \int d^{2} σ^{'} \sqrt{- h^{'}} h^{' α β} \frac{\partial X^{μ}}{\partial σ^{' α}} \frac{\partial X^{ν}}{\partial σ^{' β}} η_{μ ν} \\ = - \frac{T}{2} \int d^{2} σ \det (\frac{\partial σ^{' α}}{\partial σ^{β}}) \sqrt{- \det (\frac{\partial σ^{γ}}{\partial σ^{α}} \frac{\partial σ^{δ}}{\partial σ^{' β}} h_{γ δ})} h^{α β} \frac{\partial X^{μ}}{\partial σ^{α}} \frac{\partial X^{ν}}{\partial σ^{β}} η_{μ ν} \\ = - \frac{T}{2} \int d^{2} σ \sqrt{- h} h^{α β} \frac{\partial X^{μ}}{\partial σ^{α}} \frac{\partial X^{ν}}{\partial σ^{β}} η_{μ ν} \end{aligned}

where we noted that

\sqrt{- \det (\frac{\partial σ^{γ}}{\partial σ^{' α}} \frac{\partial σ^{δ}}{\partial σ^{' β}} h_{γ δ})} = \det (\frac{\partial σ^{α}}{\partial σ^{' β}}) \sqrt{- h}

Weyl transformations: $h_{α β} \to e^{ϕ (τ, σ)} h_{α β}$

Check:

\sqrt{- h} h^{α β} \to \sqrt{- e^{2 ϕ}} h e^{- ϕ} h^{α β} = \sqrt{- h} h^{α β}

Note that

h_{α β}

is a symmetric

2 \times 2

matrix so has three independent components. Since we have two reparameterisations (one for each

σ^{α}

) we can fix two components of the metric, laving only one unfixed. Let's choose it to be conformally fiat:

h_{α β} = e^{- ϕ} η_{α β}, n_{α β} = (\begin{array}{cc} - 1 & 0 \\ 0 & 1 \end{array})

Using Weyl invariance, we can then fix

ϕ = 0

to give

h_{α β} = η_{α β}

This is called "conformal gauge." In this gauge, the Polyakov action takes a very simple form:

S = - \frac{T}{2} \int d^{2} σ n^{α β} \partial_{α} X \cdot \partial_{β} X = \frac{T}{2} \int d^{2} σ ({\dot{X}}^{2} - X^{' 2})

5.2 Equations of motion and boundary conditions

This will be based on BBS pg 31-33. We already found the equations of motion for

h_{α β}

, which imply vanishing of stress tensor. In conformal gauge, this reduces to

\begin{aligned} 0 = T_{01} = T_{10} = \dot{X} \cdot X^{'} \\ 0 = T_{00} = T_{11} = \frac{1}{2} ({\dot{X}}^{2} + X^{' 2}) \end{aligned}

Adding and subtracting list constraint equivalently gives

{(\dot{X} \pm X^{'})}^{2} = 0

These are called the "Virasoro constraints."

Next, let's compute equations of motion for

X^{μ}

. Varying the action gives:

\begin{aligned} δ S & = - \frac{T}{2} \int d^{2} σ 2 \partial_{α} X^{μ} (\partial_{β} δ X_{μ}) η^{α β} \\ = - T \int d^{2} σ η^{α β} \partial_{β} (\partial_{α} X_{μ} δ X^{μ}) + T \int d^{2} σ (\partial^{2} X^{μ}) δ X_{μ} \end{aligned}

Neglecting boundary terms, we find the equations motion:

\partial^{2} X^{μ} = {\ddot{X}}^{μ} - X^{'' μ} = 0

Now let's look at boundary terms more carefully.

\begin{aligned} δ S_{b d r y} = - T \int d τ d σ [- \partial_{τ} ({\dot{X}}_{μ} δ X^{μ}) + \partial_{σ} (X_{μ}^{'} δ X^{μ})] \\ = - T \int d τ d σ \partial_{σ} (X_{μ}^{'} δ X^{μ}) \end{aligned}

where we discarded boundary terms along the time direction.

Closed string:

δ S_{b d r y} = - T \int d τ [{X_{μ}^{'} δ X^{μ} |}_{σ = 2 π} - {X_{μ}^{'} δ X^{μ} |}_{σ = 0}] = 0

since

X^{μ} (τ, σ) = X^{μ} (τ, σ + 2 π)

Open string:

δ S_{b d r y} = - T \int d τ [{X_{μ}^{'} δ X^{μ} |}_{σ = π} - {X_{μ}^{'} δ X^{μ} |}_{σ = 0}]

In this case, boundary terms must vanish individually. This can be achieved with Dirichlet or Neumann boundary conditions:

Dirichlet: $δ X^{μ} = 0$ at $σ = 0$ or $π$

In other words,

X^{μ}

is fixed to a constant at

0, π

. This breaks Poincare invariance.

Neumann: $X^{μ} = 0$ at $σ = 0$ or $π$

This preserves Poincare invariance. We will only consider Neumann boundary conditions. Dirichlet boundary conditions will be considered next term.

5.3 Conserved Charges

This will be based on BBS pg 37,38, Srednicki Chapter 22. As shown earlier, Poincare transformations are symmetries of the string worldsheet theory:

X^{μ} \to X^{' μ} = L_{ν}^{μ} X^{ν} + T^{μ}

For infinitesimal Lorentz transformations

L^{μ}_{ν} = δ_{ν}^{μ} + δ ω^{μ}_{ν}

, we have

X^{' μ} = X^{μ} + δ ω_{ν}^{μ} X^{v} + T^{μ} \to δ X^{μ} = δ ω_{ν}^{μ} X^{v} + T^{μ}

This symmetry implies that momentum and angular momentum of the string are conserved. The relation between symmetries and conserved quantities is given by Nether's theorem. Let's prove it for a general scalar field theory and then apply it to the string worldsheet theory.

Noether's theorem

Consider a set of scalar fields

ϕ_{a} (x)

, with Lagrangian

L (ϕ_{a}, \partial_{μ} ϕ_{a})

. If

L

is invariant under the infinitesimal change

δ ϕ_{a} (x)

, then the following current is classically conserved:

j^{μ} = \frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} δ ϕ_{a}

This is known as the "Noether current."

Corollary

The following quantity

Q = \underset{spatial intigral at fixed time}{\underset{⏟}{\int d^{D - 1} X j^{- 0} (x)}}

is constant in time. This is known as the "Noether charge."

Proof

First let's compute the equations of motion. Consider an infinitesinal variation

δ ϕ_{a}

(not necessarily a symmetry). Classical equations of motion determined by
demanding that

δ S = 0

\begin{aligned} 0 & = δ S = \int d^{D} x δ L \\ = \int d^{D} x [\frac{\partial L}{\partial ϕ_{a}} δ ϕ_{a} + \frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} δ \partial_{μ} ϕ_{a}] \\ = \int d^{D} x [\frac{\partial L}{\partial ϕ_{a}} δ ϕ_{a} + \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} δ ϕ_{a}) - \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})}) δ ϕ_{a}] \\ = \int d^{D} x [\frac{\partial L}{\partial ϕ_{a}} - \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})})] δ ϕ_{a} \\ \to \frac{\partial L}{\partial ϕ_{a}} - \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})}) = 0 "Euler-Lagrange equations" \end{aligned}

where we discarded boundary terms that come from integrating total derivatives in the third line.

Now consider a symmetry variation

δ ϕ_{a}

. By definition, this leaves

L

invariant:

0 = δ L = \frac{\partial L}{\partial ϕ_{a}} δ ϕ_{a} + \frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} \partial_{μ} δ ϕ_{a} = \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})}) δ ϕ_{a} + \frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} \partial_{μ} δ ϕ_{a} = \partial_{μ} (\frac{\partial L}{\partial (\partial_{μ} ϕ_{a})} δ ϕ_{a})

where we used the Euler Lagrange equations to obtain the third equality. Hence,

\partial_{μ} j^{μ} = 0

. This, proves Noether's theorem.

Now integrate

\partial_{μ} j^{μ}

over

(D - 1)

-dimensional space:

\begin{aligned} 0 & = \int d^{D - 1} \partial_{μ} j^{μ} \\ = \int d^{D - 1} (\partial_{0} j^{0} + \underset{⏟}{\vec{\nabla} \cdot \vec{j}}) \end{aligned}

Discarding boundary terms from integrating the total derivative finally gives

0 = \partial_{0} \underset{Q}{\underset{⏟}{\int d^{D - 1} x j^{0} (x)}}

Hence,

\partial_{0} Q = 0

Now let's apply Nether's theorem to worldsheet theory. We can think of

X^{μ} (τ, σ)

as 2d scalar fields with Lagrangian

L = - \frac{T}{2} η^{α β} \partial_{α} X^{μ} \partial_{β} X^{ν} η_{μ ν}

For an infinitesimal traslation of the target space

δ X^{μ} = b^{μ}

, the Noether current is

j^{α} = \frac{\partial L}{\partial (\partial_{α} X^{μ})} \underset{b^{μ}}{\underset{⏟}{δ X^{μ}}} = - T \partial^{α} X^{μ} b_{μ}

Since

b_{μ}

is arbitrary, we have a conserved current for each component of

b_{μ}

j_{α}^{μ} = T \partial_{α} X^{μ}

(where we have lowered

α

index and inutiplied by -1 ). The corresponding Noether charges are

p^{μ} \equiv \int d σ j_{0}^{μ} = T \int d σ {\dot{X}}^{μ}

This encodes the momentum of the string.

Similarly, for an infinitesimal Lorentz transformation

δ X^{μ} = δ ω^{μ}_{ν} X^{ν}

\begin{aligned} j^{α} & = \frac{\partial L}{\partial (\partial_{α} X^{μ})} δ X^{μ} = - T \partial^{α} X_{μ} δ ω_{ν}^{μ} X^{ν} \\ = \frac{T}{2} (X^{μ} \partial^{α} X^{ν} - X^{ν} \partial^{α} X^{μ}) δ ω_{μ ν} \end{aligned}

where we recall that

δ ω_{μ ν} = - δ ω_{ν μ}

. We have a Noether current for each independent component of

δ ω_{μ ν}

j_{α}^{μ ν} = T (X^{μ} \partial_{α} X^{ν} - X^{ν} \partial_{α} X^{μ})

where we multiplied by 2 . The corresponding Noether charges are

J^{μ ν} = \int d σ j_{0}^{μ ν} = T \int d σ (X^{μ} {\dot{X}}^{ν} - X^{ν} {\dot{X}}^{μ})

This encodes angular momentum of string.

6 Mode expansion

This will be based on Tong section 1.4, pg 52,53, and BBS pg 33-35,37,39,40. Recall the equations of motion

\partial_{α} \partial^{α} X^{μ} = 0

Let us introduce the lightcone cords:

σ^{\pm} = τ \pm σ \to τ = \frac{1}{2} (σ^{+} + σ^{-}), σ =

\frac{1}{2} (σ^{+} - σ^{-})

. We then find

\begin{aligned} \partial_{+} = (\partial_{+} τ) \partial_{τ} + (\partial_{+} σ) \partial_{σ} = \frac{1}{2} (\partial_{τ} + \partial_{σ}) \\ \partial_{-} = (\partial_{-} τ) \partial_{τ} + (\partial_{-} σ) \partial_{σ} = \frac{1}{2} (\partial_{τ} - \partial_{σ}) \\ \to \partial_{+} \partial_{-} X^{μ} = \frac{1}{4} (\partial_{τ} + \partial_{σ}) (\partial_{τ} - \partial_{σ}) = \frac{1}{4} ({\ddot{X}}^{μ} - X^{''}) = 0 \\ \to X^{μ} (τ, σ) = X_{L}^{μ} (σ^{+}) + X_{R}^{μ} (σ^{-}) \end{aligned}

Hence, the general solution can be written as a sum of left and right-movers.

For closed strings,

X^{μ} (τ, σ) = X^{μ} (τ, σ + 2 π)

. The most general periodic solution can be expanded in Fourier modes:

\begin{aligned} X_{L}^{μ} (σ^{+}) = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{+} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} {\tilde{α}}_{n}^{μ} e^{- i n σ^{+}} \\ X_{R}^{μ} (σ^{-}) = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{-} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} e^{- i n σ^{-}} \end{aligned}

A few notes about the mode expansion:

Recall that $α^{'} = \frac{1}{2} l_{s}^{2}$ , as required by dimensional analysis. Normalisation of $1 / n$ chosen for later convenience.
$X_{L}^{μ}, X_{R}^{μ}$ not periodic because of terms linear in sigma $^{\pm}$ , but their sum gives $\frac{1}{2} α^{'} p^{μ} (σ^{+} + σ^{-}) = α^{'} p^{μ} τ$ , which is periodic.
$p^{μ}$ is the momentum:

T \int d σ {\dot{X}}^{μ} (τ, σ) = \frac{1}{2 π α^{'}} (2 π α^{'} p^{μ}) = p^{μ}

where we noted that oscillatory terms integrate to zero.

$x_{0}^{μ} + α^{'} p^{μ} τ$ is the center of mass of the string:

⟨ X^{μ} ⟩ = \frac{1}{2 π} \int d σ X^{μ} (τ, σ) = x_{0}^{μ} + α^{'} p^{μ} τ

Reality of $X_{L, R}$ implies

α_{- n}^{μ} = {(α_{n}^{μ})}^{*}, {\tilde{α}}_{- n}^{μ} = {({\tilde{α}}_{n}^{μ})}^{*}

Indeed,

\begin{aligned} X_{L}^{*} (σ^{+}) & = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{t} - i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} {({\tilde{α}}_{n}^{μ})}^{*} e^{i n σ^{+}} \\ = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{+} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} {({\tilde{α}}_{- n}^{μ})}^{*} e^{- i n σ^{+}} \end{aligned}

where we took

n \to - n

in the second line. This is equal to

X^{μ} (σ^{+})

iff

{(α_{- n}^{μ})}^{*} =

α_{n}^{μ}

. Similar story for

X_{R}

The solution must also satisfy the Virasoro constraints:

{(\dot{X} \pm X^{'})}^{2} = 0

Recall that these came from vanishing of the stress tensor:

T_{α β} = \partial_{α} X \cdot \partial_{β} X - \frac{1}{2} h_{α β} h^{γ δ} \partial_{γ} X \cdot \partial_{δ} X

In conformal gauge with lightcone coordinates,

h_{α β} = (\begin{array}{cc} h_{+ +} & h_{+ -} \\ h_{- +} & h_{- -} \end{array}) = - \frac{1}{2} (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}) \to h^{α β} = - 2 (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})

To see this, note that

d s^{2} = - d τ^{2} + d σ^{2} = - d σ^{+} d σ^{-} = - \frac{1}{2} (d σ^{+} d σ^{-} + d σ^{-} d σ^{+})

Hence, the vanishing of the energy momentum tensor becomes

\begin{aligned} T_{+ +} = \partial_{+} X \cdot \partial_{+} X = \frac{1}{4} {(\dot{X} + {\dot{X}}^{'})}^{2} = 0 \\ T_{-} = \partial_{-} X \cdot \partial_{-} X = \frac{1}{4} {(\dot{X} - X^{'})}^{2} = 0 \end{aligned}

while

T_{+ -} = T_{- +} = \partial_{+} X \cdot \partial_{-} X - \frac{1}{2} h_{+ -} (h^{+ -} + h^{- +}) \partial_{+} X \cdot \partial_{-} X = 0

T_{+ -} =

T_{- +}

trivially vanishes.

Let us compute Fourier expansion of

T_{\pm \pm}

. First note that

\begin{aligned} \partial_{-} X^{μ} & = \partial_{-} (X_{L}^{μ} (σ^{+}) + X_{R}^{μ} (σ^{-})) \\ = \partial_{-} X_{R}^{μ} (σ^{-}) \\ = \partial_{-} (x_{0}^{μ} + \frac{α^{'}}{2} p^{μ} σ^{-} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} e^{- i n σ^{-}}) \\ = \frac{α^{'}}{2} p^{μ} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} (- i n e^{- i n σ^{-}}) \\ = \sqrt{\frac{α^{'}}{2}} \sum_{n} α_{n}^{μ} e^{- i n σ^{-}}, α_{0}^{μ} = \sqrt{\frac{α^{'}}{2}} p^{μ} \end{aligned}

Hence,

\begin{aligned} T_{-} = {(\partial_{-} X^{μ})}^{2} & = \frac{α^{'}}{2} \sum_{n, m} α_{n} \cdot α_{m} e^{- i (n + m) σ^{-}} \\ = \frac{α^{'}}{2} \sum_{n, m} α_{n - m} \cdot α_{m} e^{- i n σ^{-}}, taking n \to n - m \\ = α^{'} \sum_{n} L_{n} e^{- i n σ^{-}}, L_{n} = \frac{1}{2} \sum_{m} α_{n - m} \cdot α_{m} \end{aligned}

Similarly,

T_{+ +} = α^{'} \sum_{n} {\tilde{L}}_{n} e^{- i n σ^{ϕ}}

where

{\tilde{L}}_{n} = \frac{1}{2} \sum_{m}^{n} {\hat{α}}_{n - m} \cdot {\tilde{α}}_{m}, {\tilde{α}}_{0}^{μ} = α_{0}^{μ}

The Virasoro constraints

T_{+ +} = T_{- -} = 0

then imply

L_{n} = {\tilde{L}}_{n} = 0

Let's look of the

n = 0

constraints more closely:

\begin{aligned} L_{0} & = \frac{1}{2} \sum_{m} α_{- m} \cdot α_{m} = \frac{1}{2} α_{0}^{2} + \frac{1}{2} \sum_{m < 0} α_{- m} α_{m} + \frac{1}{2} \sum_{m > 0} α_{- m} \cdot α_{m} \\ = \frac{α^{'}}{4} p^{2} + \frac{1}{2} \sum_{m < 0} α_{m} \cdot α_{- m} + \frac{1}{2} \sum_{m > 0} α_{- m} \cdot α_{m} \\ = - \frac{α^{'}}{4} M^{2} + \sum_{m > 0} α_{- m} \cdot α_{m} \end{aligned}

where we replaced

α_{- m} \cdot α_{m}

with

α_{m} \cdot α_{- m}

in the second term of the second line. Note that this only holds classically. In quantum theory

[α_{m}^{μ}, α_{- m}^{μ}] \neq 0

leading to an ordering ambiguity. Similarly we find

{\tilde{L}}_{0} = - \frac{α^{'}}{4} M^{2} + \sum_{m > 0} {\tilde{α}}_{- m} \cdot {\tilde{α}}_{m}

Hence

L_{0} = {\tilde{L}}_{0} = 0 \to

\begin{aligned} M^{2} & = \frac{4}{α^{'}} \sum_{n > 0} α_{- n} \cdot α_{n} = \frac{4}{α^{'}} \sum_{n > 0} {\tilde{α}}_{- n} \cdot {\tilde{α}}_{n} \\ = \frac{2}{α^{'}} \sum_{n > 0} (α_{- n} \cdot α_{n} + {\tilde{α}}_{- n} \cdot {\tilde{α}}_{n}) \end{aligned}

The first line implies the constraint

\sum_{n > 0} α_{- n} \cdot α_{n} = \sum_{n > 0}^{n} {\tilde{α}}_{- n} \cdot {\tilde{α}}_{n}

which is known as "level matching."

We can follow a similar procedure for open strings with Neumann boundary conditions. This will be a HW exercise, so let's just summarize the results:

Mode expansion: $X^{μ} (τ, σ) = x_{0}^{μ} + 2 α^{'} p^{μ} τ + 2 i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} \cos (n σ) e^{- i n τ}$

Note that there is only one set of oscillators. There is also a factor of 2 difference in the

p^{μ}

term compared to closed string.

Virasoro constraints:

0 = T_{\pm t} = α^{'} \sum_{n} L_{n} e^{- i n σ^{\pm}} \to L_{n} = 0

where

L_{n} = \frac{1}{2} \sum_{m} α_{n - m} \cdot α_{m}, α_{0}^{μ} = \sqrt{2 α^{'}} p^{μ}

Now there is only one set of Virasoro modes.

Mass formula:

L_{0} = 0 \to M^{2} = \frac{1}{α^{'}} \sum_{n > 0} α_{- n} \cdot α_{n}

7 Virasoro algebra

This will be based on BBS pg 40,41,58-62. After choosing conformal gauge

h_{α β} = n_{α β}

, their is still a residual reparamaterisation symmetry. Let

δ σ^{α} = - ξ^{α}

be an infinitesimal reparametrisation and

Λ

be an infinitesimal parameter for a Weyl rescaling. The residual symmetries obey

δ h^{α β} = \partial^{α} ξ^{β} + \partial^{β} ξ^{α} = Λ η^{α β}

This is known as the "conformal killing equation" and the solutions are known as conformal Killing vectors. In other words, we are free to make infinitesimal diffeomorphisms that correspond to Weyl transformations. These are known as "conformal transformations." Conformal transformations change distances, but preserve angles. Theories invariant under conformal transformations are known as "conformal field theories." The string worldsheet theory is a 2d CFT.

D > 2

spacetime dimensions, there are

(\binom{D + 2}{2})

solutions, which generate

S O (D, 2)

corresponding to the "conformal group" in

D

-dimensional Minkowski space. In

D

-dimensional Euclidean space, the conformal group is

S O (D + 1, 1)

. In

D = 2

, the conformal killing equation has an infinite number of solutions, so the conformal group is infinite-dimensional.

Let's verify this. In lightone cords

σ^{\pm} = σ^{0} \pm σ^{1}, ξ^{\pm} = ξ^{0} \pm ξ^{1}

the conformal Killing equations becomes

\begin{aligned} 2 \partial_{+} ξ_{+} = Λ n_{+ +} = 0 \\ \partial_{+} ξ_{-} + \partial_{-} ξ_{+} = Λ n_{+ -} = - \frac{Λ}{2} \\ 2 \partial_{-} ξ_{-} = Λ n_{- -} = 0 \end{aligned}

where we recall that in light cone cords,

(\begin{array}{ll} η_{+ +} & η_{+ -} \\ η_{- +} & η_{- -} \end{array}) = - \frac{1}{2} (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})

This is solved by

ξ_{+} (τ, σ) = f_{+} (σ^{-}), ξ_{-} (τ, σ) = f_{-} (σ^{+})

, where

f_{\pm}

are arbitrary functions respecting boundary conditions of worldsheet. Raising indices,

ξ^{\pm} =

f^{\pm} (σ^{\pm})

. For closed strings a complete basis of solutions is given by

f_{n}^{\pm} = e^{i n σ^{\pm}}, n \in Z

which are periodic in

σ \to σ + 2 π

. We may then define an infinite set of generators

V_{n}^{\pm} = e^{i n σ^{\pm}} \frac{\partial}{\partial σ^{\pm}} "Virasoro generators"

which generate the diffeomorphisms

δ σ^{\pm} = ϵ_{n}^{\pm} V_{n}^{\pm} (σ^{\pm}) = ϵ_{n}^{\pm} e^{i n σ^{\pm}}

, where

ϵ^{\pm}

are infinitesimal parameters.

Let's Wick-rotate

τ \to - i τ

so the worldsheet becomes Euclidean and then map it to complex plane:

z = e^{i σ^{-}} = e^{i (- i τ - σ)} = e^{τ - i σ}, \bar{z} = e^{i σ^{+}} = e^{i (- i τ + σ)} = e^{τ + i σ}

Under this mapping, the cyclinder gets mapped to a plane. The infinite past on the cylinder corresponds to the origin of the plane. We also see that conformal transformations correspond to analytic functions on the complex plane:

z \to w (z), \bar{z} \to \bar{w} (\bar{z})

The Virasoro generators become

\begin{aligned} V_{n}^{+} & = e^{i n σ^{+}} \frac{\partial}{\partial σ^{+}} = {\bar{z}}^{n} \frac{\partial \bar{z}}{\partial σ^{+}} \frac{\partial}{\partial \bar{z}} = i {\bar{z}}^{n + 1} \partial_{\bar{z}} \equiv - i {\bar{l}}_{n} \\ V_{n}^{-} & = e^{i n σ^{-}} \frac{\partial}{\partial σ^{-}} = i z^{n + 1} \partial_{z} \equiv - i l_{n} \end{aligned}

l_{n} = - z^{n + 1} \partial_{z}, {\bar{l}}_{n} = - {\bar{z}}^{n + 1} {\bar{\partial}}_{z}

These generate infinitesimal conformal transformations:

δ z = ϵ_{n} l_{n} (z) = - ϵ_{n} z^{n + 1}, δ \bar{z} = {\bar{ϵ}}_{n} {\bar{l}}_{n} (\bar{z}) = - {\bar{ϵ}}_{n} {\bar{z}}^{n + 1}

and obey the following algebra:

[l_{n}, l_{n}] = (m - n) l_{m + n}, [{\bar{l}}_{m}, {\bar{l}}_{n}] = (m - n) {\bar{l}}_{m + n}, [l_{m}, {\bar{l}}_{n}] = 0

which corresponds to two copies of the "classical Virasoro algebra." We will see later that it can get modified in quantum theory due to a "conformal anomaly."

l_{0, \pm 1}

and

{\bar{l}}_{0, \pm 1}

generate the finite dimensional subgroup

S O (1, 3) =

S L (2, C) / Z_{2}

, as you will show in the HW. This is the

2 d

case of

S O (D + 1, 1)

which is the conformal group for

D > 2

Euclidean dimensions. In

2 d

, this is konwn as the "restricted conformal group".

For open strings with Neumann boundary conditions, the residual symmetries in conformal gauge are generated by

V_{n} = e^{i n σ^{+}} \frac{\partial}{\partial σ^{t}} + e^{i n σ^{-}} \frac{\partial}{\partial σ^{-}}

as you will show on HW. Hence, there is just one copy of Virasoro algebra. We can Wick-rotate to a Euclidean worldsheet and map to the upper half of the complex plane via

z = e^{i σ^{-}}, \bar{z} = e^{i σ^{+}}

Note that the boundary of the string gets mapped to real axis. The Virasoro generators then become

l_{n} = - (z^{n + 1} \partial_{z} + {\bar{z}}^{n + 1} \partial_{\bar{z}})

8 Light cone gauge

This is based on BBS pg 48-50, Tong pg 35,36, JLBS pg 8,9. Recall the mode expansion for a closed string in conformal gauge:

\begin{aligned} X_{L}^{μ} = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{+} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} {\tilde{α}}_{n}^{μ} e^{- i n σ^{+}} \\ X_{R}^{μ} = \frac{1}{2} x_{0}^{μ} + \frac{1}{2} α^{'} p^{μ} σ^{-} + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} e^{- i n σ^{-}} \end{aligned}

Also recall that in conformal gauge, we have residual symmetry

δ σ^{\pm} = f^{\pm} (σ^{\pm})

where

f^{\pm}

is periodic in

σ

. Hence, we may choose

f^{\pm}

such that:

X_{L}^{+} = \frac{1}{2} α^{'} p^{+} σ^{+}, X_{R}^{+} = \frac{1}{2} α^{'} p^{+} σ^{-}

X^{+} (τ, σ) = α^{'} p^{+} τ

where

X^{\pm} = \frac{1}{\sqrt{2}} (X^{0} \pm X^{D - 1})

and

I \in {1, \dots, D - 2}

labels transverse directions. Equivalently, we set

x_{0}^{+} = α_{n}^{+} = {\tilde{α}}_{n}^{+} = 0, n \neq 0

. This is known as "lightcone gauge." Note that we can't set terms linear in

σ^{\pm}

to zero because they are not periodic in

σ

In lightcone gauge, the mass formula and level-matching condition reduce to

\begin{aligned} M^{2} = \frac{2}{α^{'}} \sum_{n > 0} (α_{- n} \cdot α_{n} + {\tilde{α}}_{- n} \cdot {\tilde{α}}_{n}) \to M^{2} = \frac{2}{α^{'}} \sum_{n > 0} (α_{- n}^{I} α_{n}^{I} + {\tilde{α}}_{- n}^{I} α_{n}^{I}) \\ \sum_{n > 0} α_{- n} \cdot α_{n} = \sum_{n > 0} {\tilde{α}}_{- n} \cdot {\tilde{α}}_{n} ⟶ \sum_{n > 0} α_{- n}^{I} α_{n}^{I} = \sum_{n > 0} {\tilde{α}}_{- n}^{I} α_{n}^{I} \end{aligned}

where we noted that

α_{- n} \cdot α_{n} = - α_{- n}^{+} α_{n}^{-} - α_{- n}^{-} α_{n}^{+} + α_{- n}^{I} α_{n}^{I} = α_{- n}^{I} α_{n}^{I}

in lightcone gauge. Hence in lightcone gauge, only transverse modes contribute to mass. Moreover, the modes of

X^{-}

can be fixed in terms of transverse modes. This follows from the Virasoro constraints:

\begin{aligned} 0 = & {(\partial_{+} X)}^{2} = - 2 \partial_{+} X^{-} \partial_{+} X^{+} + {(\partial_{+} X^{I})}^{2} \\ = & - 2 \partial_{+} X^{-} \frac{1}{2} (\partial_{τ} + \partial_{σ}) (α^{'} p^{+} τ) + {(\partial_{+} X^{I})}^{2} \\ = & - 2 \partial_{+} X^{-} \frac{1}{2} (\partial_{τ} + \partial_{σ}) (α^{'} p^{+} τ) + {(\partial_{+} X^{I})}^{2} \\ = & - α^{'} p^{+} \partial_{+} X^{-} + {(\partial_{+} X^{I})}^{2} \\ \to \partial_{+} X^{-} = \frac{1}{α^{'} p^{+}} {(\partial_{+} X^{I})}^{2} \end{aligned}

Plugging in the mode expansions then gives

\begin{aligned} \sqrt{\frac{α^{'}}{2}} \sum_{n} {\tilde{α}}_{n}^{-} e^{- i n σ^{+}} = & \frac{1}{α^{'} p^{+}} \frac{α^{'}}{2} \sum_{n, m} {\tilde{α}}_{n}^{I} {\tilde{α}}_{m}^{I} e^{- i (n + m) σ^{+}} \\ = \frac{1}{2 p^{+}} \sum_{n, m} {\tilde{α}}_{n - m}^{I} {\tilde{α}}_{m}^{I} e^{- i n σ^{+}} \\ \to {\tilde{α}}_{n}^{-} = \frac{1}{\sqrt{2 α^{'}}} \frac{1}{p^{+}} \sum_{m} {\tilde{α}}_{n - m}^{I} {\tilde{α}}_{m}^{I} \end{aligned}

Similarly,

{(\partial_{-} X)}^{2} = 0 \to

α_{n}^{-} = \frac{1}{\sqrt{2 α^{'}}} \frac{1}{p^{+}} \sum_{m} α_{n - m}^{I} α_{m}^{I}

Recalling that

α_{0}^{-} = {\tilde{α}}_{0}^{-} = \sqrt{\frac{α^{'}}{2}} p^{-}

, for

n = 0

these relations imply the mass
formula:

\begin{aligned} α^{'} p^{-} & = \frac{1}{p^{+}} (\frac{α^{'}}{2} {(p^{I})}^{2} + \sum_{n \neq 0} {\tilde{α}}_{- n}^{I} {\tilde{α}}_{n}^{I}) \\ = \frac{1}{p^{+}} (\frac{α^{'}}{2} {(p^{I})}^{2} + \sum_{n \neq 0} α_{- n}^{I} α_{n}^{I}) \\ \to M^{2} = 2 p^{+} p^{-} - {(p^{I})}^{2} = \frac{4}{α^{'}} \sum_{n > 0} {\tilde{α}}_{- n}^{I} {\tilde{α}}_{n}^{I} = \frac{4}{α^{'}} \sum_{n > 0} α_{- n}^{I} α_{n}^{I} \end{aligned}

For an open string with Neumann boundary conditions, the story is similar except that there is only one set of oscillators and

p^{μ} \to 2 p^{μ}

. Here is a summary for open strings:

Lightcone gauge: $X^{+} = 2 α^{'} p^{+} τ \to x_{0}^{+} = α_{n}^{+} = 0, n \neq 0$
$α_{n}^{-}$ constraint: $α_{n}^{-} = \frac{1}{\sqrt{2 α^{'}}} \frac{1}{2 p^{+}} \sum_{m} α_{n - m}^{I} α_{m}^{I}$
Mass formula: $M^{2} = \frac{1}{α^{'}} \sum_{n > 0} α_{- n}^{I} α_{n}^{I}$

Hence, in lightcone gauge we express the theory in terms of transverse modes, which encode the independent physical degrees of freedom. To quantize the theory in light cone gauge, we must therefore quantize the transverse degrees of freedom.

9 Canonical Quantisation

Based on BBS pg 35-37, Tong pg 28-30. To quantize in a covariant way, we impose the canonical commutation relations:

\begin{aligned} [X^{μ} (τ, σ), Π^{ν} (τ, σ^{'})] = i δ (σ - σ^{'}) η^{μ ν} \\ [X^{μ} (τ, σ), X^{ν} (τ, σ^{'})] = [Π_{μ} (τ, σ), Π_{ν} (τ, σ^{'})] = 0 \end{aligned}

where we evaluate commutators at fixed worldsheet time and

Π^{μ}

is the canonical momentum:

Π_{μ} = \frac{\partial L}{\partial {\dot{X}}^{μ}} = T {\dot{X}}_{μ}, L = \frac{1}{2} ({\dot{X}}^{2} - X^{' 2})

Inserting mode expansions then implies commutation relations for the modes: For closed strings, one gets

[x^{μ}, p^{ν}] = i η^{μ ν}, [α_{m}^{μ}, α_{n}^{v}] = [{\tilde{α}}_{m}^{μ} {\tilde{α}}_{n}^{ν}] = m η^{μ ν} δ_{m + n, 0}, [α_{m}^{μ} {\tilde{α}}_{m}^{ν}] = 0

Lot

a_{m}^{μ} = \frac{1}{\sqrt{m}} α_{m}^{μ}, a_{m}^{μ †} = \frac{1}{\sqrt{m}} α_{- m}^{μ}, m > 0

. Then

[a_{m}^{μ}, a_{n}^{ν †}] = [{\tilde{a}}_{m}^{μ}, {\tilde{a}}_{n}^{ν +}] = η^{μ ν} δ_{m, n}

which is essentially the algebra of raising and lowering operators for harmonic oscillators. For open strings, there is only one set of oscillators.

Note that

[a_{m}^{0}, a_{m}^{0 †}] = - 1

. This gives negative norm states: Let

| ψ ⟩ =

a_{m}^{0 †} | 0 ⟩, m > 0

. Then

⟨ ψ ∣ ψ ⟩ = ⟨ 0 | a_{m}^{0} a_{m}^{0 †} | 0 ⟩ = ⟨ 0 | a_{m}^{0 †} a_{m}^{0} | 0 ⟩ + ⟨ 0 | [a_{m}^{0}, a_{m}^{0 †}] | ∣ 0 ⟩ = - 1

where the first term after the second equality vanishes and we used the commutation relations in the second term. Negative norm states can be removed by imposing additional constraints on the Hilbert space encoding which encode the Virasoro constraints (see chapter 24 of Zwiebach for more detials).

In lightcone gauge negative norm states do not appear because we eliminate oscillators of

X^{\pm}

and the Hilbert space is then constructed using only transverse oscillators and is manifestly positive definite. We will therefore proceed with quantization in light cone gauge. Our main task will be to show that

[x^{I}, p^{J}] = i δ^{I J}, [α_{m}^{I}, α_{n}^{J}] = [{\tilde{α}}_{m}^{I}, {\tilde{α}}_{n}^{J}] = m δ^{I J} δ_{m + n, 0}, [α_{n}^{I}, {\tilde{α}}_{m}^{J}] = 0

implies the canonical commutation relations

[X^{I} (τ, 6), Π^{J} (τ, σ^{'})] = i δ^{I J} δ (σ - σ^{'})

where

I \in 1, \dots, D - 2

and all other commutators vanish. In light cone gauge

x_{0}^{-}

and

p^{+}

are not eliminated, so we must also impose the commutation relation

[x_{0}^{-}, p^{+}] = - i

Let's prove the commutation relations. Recall that for closed string

\begin{aligned} X^{I} (τ, σ) = x_{0}^{I} + α^{'} p^{I} τ + i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} ({\tilde{α}}_{n}^{I} e^{- i n σ^{+}} + α_{n}^{I} e^{- i n σ^{-}}) \\ \to Π^{I} & = T {\dot{X}}^{I} \\ = T (α^{'} p^{I} + \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} ({\tilde{α}}_{n}^{I} e^{- i n σ^{+}} + α_{n}^{I} e^{- i n σ^{-}})) \\ = T \sqrt{\frac{α^{'}}{2}} \sum_{n} ({\tilde{α}}_{n}^{I} e^{- i n σ^{+}} + α_{n}^{I} e^{- i n σ^{-}}), α_{0}^{I} = {\tilde{α}}_{0}^{I} = \sqrt{\frac{α^{'}}{2}} p^{I} \\ \to & [X^{I} (τ, σ), Π^{J} (τ, σ^{'})] = T (α^{'} [x_{0, p^{I}}^{J}] + \frac{i α^{'}}{2} \sum_{n \neq 0, m \neq 0} \frac{1}{n} A_{n m}^{I J}) \end{aligned}

where

\begin{aligned} A_{n \cdot m}^{I J} & = [{\tilde{α}}_{n}^{I} e^{- i n (τ + σ)} + α_{n}^{I} e^{- i n (τ - σ)}, {\tilde{α}}_{m}^{J} e^{- i m (τ + σ^{'})} + α_{m}^{J} e^{- i m (τ - σ^{'})}] \\ = [{\tilde{α}}_{n}^{I}, {\tilde{α}}_{m}^{J}] e^{- i (n + m) τ} e^{- i (n σ + m σ^{'})} + [α_{n}^{I}, α_{m}^{J}] e^{- i (n + m) τ} e^{i (n σ + m σ^{'})} \\ = n δ^{I J} δ_{m + n, 0} (e^{- i n (σ - σ^{'})} + e^{i n (σ - σ^{'})}) \end{aligned}

Hence

\begin{aligned} [X^{I} (τ, σ), Π^{J} (τ, σ^{'})] & = T (i α^{'} δ^{I J} + \frac{i α^{'}}{2} \sum_{n \neq 0} (e^{- i n (σ - σ^{'})} + e^{i n (σ - σ^{'})})) \\ = i \frac{1}{2 π} \sum_{n} e^{i n (σ - σ^{'})} δ^{I J} \\ = i δ (σ - σ^{'}) δ^{I J}, as expected \end{aligned}

Let us prove that

δ (σ - σ^{'}) = \frac{1}{2 π} \sum_{n} e^{i n (σ - σ^{'})}

Proof:

Any function

f (σ)

defined over

σ \in [0, 2 π]

can be Fourier expanded as

\begin{aligned} f (σ) = \frac{1}{2 π} \sum_{m} C_{m} e^{i m σ} \\ \to \int_{0}^{2 π} d σ f (σ) e^{- i n σ} = \frac{1}{2 π} \sum_{m} C_{m} \underset{2 π δ_{m, n}}{\underset{⏟}{\int_{0}^{2 π} d σ e^{i (m - n) σ}}} = C_{n} \end{aligned}

For

f (σ) = δ (σ - σ^{'})

, we find

\begin{aligned} C_{n} = \int_{0}^{2 π} d σ δ (σ - σ^{'}) e^{- i n σ} = e^{- i n σ^{'}} \\ \to δ (σ - σ^{'}) = \frac{1}{2 π} \sum_{n} e^{i n (σ - σ^{'})} \end{aligned}

as expected.

Now let's look at the other commutation relations:

\begin{aligned} [X^{I} (τ, σ), X^{J} (τ, σ^{'})] & = α^{'} τ ([x_{0}^{I}, p^{J}] + [p^{I}, x_{0}^{J}]) - \frac{α^{'}}{2} \sum_{n, m \neq 0} \frac{1}{n m} A_{n m}^{I J} \\ = \frac{α^{'}}{2} δ^{I J} \sum_{n \neq 0} \frac{1}{n} (e^{- i δ^{I J} (σ - σ^{'})} + e^{i n (σ - σ^{'})}) \end{aligned}

which vanishes after taking

n \to - n

in the first term in the second line. Finally, we have

\begin{aligned} [Π^{I} (τ, σ), Π^{J} (τ, σ^{'})] = \frac{α^{'}}{2} T^{2} \sum_{n, m} A_{n m}^{I J} \\ = \frac{α^{'}}{2} T^{2} δ^{I J} \sum_{n} n (e^{- i n (σ - σ^{'})} + e^{i n (σ - σ^{'})}) \end{aligned}

which vanishes after taking

n \to - n

in the first term of the second line.

This concludes the derivation of the canonic commutation relations for the closed string. For the open string there is only one set of oscillators, so we have

[x^{I}, p^{J}] = i δ^{I J}, [α_{m}^{I}, α_{n}^{J}] = m δ^{I J} δ_{m + n, 0}

The prof of these commutation relations is similar to the closed string case except now we use the identity

δ (σ - σ^{'}) = \frac{1}{π} \sum_{n} \cos (n σ) \cos (n σ^{'})

which you will show on the HW.

10 Spectrum and Critical Dimension

This will be based on BBS pg 50,51,53, Tong sections 2.3,2.4. Recall the in lightcone gauge, states are constructed from transverse oscillators:

[α_{m}^{I}, α_{n}^{J}] = [{\tilde{α}}_{m}^{I}, {\tilde{α}}_{n}^{J}] = m δ^{I J} δ_{m + n, 0}

where

I = 1, \dots, D - 2

label transverse directions. For open strings, there are only one set of oscillators

α_{m}^{I}

. Let

a_{m}^{I} = \frac{1}{\sqrt{m}} α_{m}^{I}, a_{m}^{†} = \frac{1}{\sqrt{m}} α_{- m}^{I} = \frac{1}{\sqrt{m}} {(α_{m}^{I})}^{+}, m > 0

and similar for

{\tilde{α}}_{m}^{I}

. Then

[a_{m}^{I} a_{m}^{J †}] = [{\tilde{a}}_{m}^{I} {\tilde{a}}_{m}^{J †}] = δ^{I J} δ_{m, n}

which we recognize as raising and lowering operators of harmonic oscillators. In particular,

α_{m}^{I} | 0; k ⟩, α_{m}^{I} | 0; k ⟩ = 0, m > 0

where the ground state has momentum

k^{μ}

p^{μ} | 0; k^{μ} ⟩ = k^{μ} | 0; k^{μ} ⟩

We encountered this state when we quantised the relativistic point particle using the world line formalism, which only gives single-particle states. In string theory, we can act on this state with

α_{- m}^{I}

and

{\tilde{α}}_{- m}^{I}, m > 0

to give an infinite tower of massive higher-spin states. Note that these are still single-particle states.

Since all the oscillators are transverse, the Hilbert space is manifestly positive definite. Hence lightcone gauge makes all the physical degrees of freedom manifest. Let's compute the spectrum, starting from the open string with Neumann boundary conditions.

Open string spectrum

Recall that in light cone gauge, the Virasoro constraints fix

p^{-}

in terms of transverse oscillators implying the following mass formula:

\begin{aligned} M^{2} & = \frac{1}{2 α^{'}} \sum_{n \neq 0} α_{- n}^{I} α_{n}^{I} = \frac{1}{2 α^{'}} \sum_{n > 0} (α_{- n}^{I} α_{n}^{I} + α_{n}^{I} α_{- n}^{I}) \\ = \frac{1}{2 α^{'}} \sum_{n > 0} (2 α_{- n}^{I} α_{n}^{I} + [α_{n}^{I}, α_{- n}^{I}]) \\ = \frac{1}{2 α^{'}} \sum_{n > 0} (2 α_{- n}^{I} α_{n}^{I} + n δ^{I I}) \\ = \underset{classical}{\underset{⏟}{\frac{1}{α^{'}} \sum_{n > 0} α_{- n}^{I} α_{n}^{I}}} + \underset{\begin{array}{c} quantum \\ cosult \end{array}}{\underset{⏟}{\frac{D - 2}{2 α^{'}} \sum_{n > 0} n}} \\ = \frac{1}{α^{'}} (N - a) \end{aligned}

where

N = \sum_{n > 0} α_{- n}^{I} α_{n}^{I} = \sum_{n > 0} n a_{n}^{I †} a_{n}^{I}

is the "number operator" which counts the number of oscillators weighted by their mode number and

a = - \frac{D - 2}{2} \sum_{n > 0} n

is a "normal ordering constant." We need regulate the infinite sum

\sum_{n > 0} n

. Consider the general sum

ζ (s) = \sum_{n = 1}^{\infty} n^{- s} "Riemann zeta function"

where

s \in C

. This sum converges for

Re (s) > 1

, and has a pole at

s = 1

. It can be analytically continued around the pole to give

ζ (- 1) = - \frac{1}{12}

as you will show in the HW. Hence, we find that

a = \frac{D - 2}{24}

Let us consider the first excited state:

α_{- 1}^{I} ∣ 0; k >

. In this case we have

N = 1

M^{2} = \frac{1}{α^{'}} (1 - a)

. Note that the state is a

(D - 2)

-component vector and recall that Lorentz invariance implies that massive states must form represntations of

S O (D - 1)

and massless states must form representations of

S O (D - 2)

. Since we cannot package the

(D - 2)

states into a representation of

S O (D - 1)

, it must be a massless state. Indeed this is the structure of single
particle states that we found when quantizing Maxwell theory. Hence, we find that open strings contain massless spin- 1 particles, i.e. gauge fields, in their spectrum. It follows that

M^{2} = \frac{1}{α^{'}} (1 - a) = 0 \to a = 1 \to D = 26 "Critical dimension"

Hence, string theory predicts the dimension of spacetime! If

D \neq 26

, the theory would not be Lorentz-invariant.

A more rigorous (but much more tedious) way to derive

a = 1, D = 26

is to compute the commutator

[J^{I -}, J^{K -}]

where

J^{μ ν}

are Lorentz generators computed from Noether's theorem and expressed in terms of transverse oscillators after plugging in mode expansion and eliminating

α_{n}^{-}

oscillators in lightcone gauge. Note that generators of Lorentz group must satisfy the following algebra:

[J^{μ ν}, J^{ρ σ}] = - η^{μ ρ} J^{ν σ} - η^{ν σ} J^{μ ρ} + η^{μ σ} J^{ν ρ} + η^{ν ρ} J^{μ σ}

Hence

[J^{I -}, J^{K -}] = 0

. After a tedious calculation one finds that this is only possible if

a = 1, D = 26

. For more details, see section 2.4 of Tong and section 2.3.1 of GSW.

Let us return to the open string spectrum. The ground state

| 0; k ⟩

is a scalar with momentum

k^{μ}

and mass

N = 0 \to M^{2} = - \frac{a}{α^{'}} = - \frac{1}{α^{'}} < 0 "tachyon"

The tachyon indicates an instability of the vacuum of the bosonic string. Recall that the mass squared of a field is the quadratic term of the potential:

M^{2} = {\partial_{ϕ}^{2} V (ϕ) |}_{ϕ = 0}

so if

M^{2} < 0

, we are expanding around a maximum of the potential. The tachyon can be removed by introducing supersymmetry, as you will see next term.

Finally, let us consider the first states with positive mass squared, correpsonding to

N = 2

α_{- 2}^{I} | 0; k ⟩, α_{- 1}^{I} α_{- 1}^{J} | 0; k ⟩ \to M^{2} = (2 - a) / α^{'} = 1 / α^{'}

There are 24 states with a single index while the states with two indices correspond to a symmetric

24 \times 24

matrix giving

(\binom{24}{2}) + 24 = 300

states. In total, there are 324 massive states at this level. Note that a symmetric traceless

25 \times 25

matrix has

(\binom{25}{2}) + 25 - 1 = 324

independent components, so the states indeed form a representation of

S O (D - 1) = S O (25)

, as expected.

Closed string spectrum

For the closed string in lightcone gauge, we construct states using two sets of transverse oscillators. Now the mass formula is

M^{2} = \frac{4}{α^{'}} (N - 1) = \frac{4}{α^{'}} (\tilde{N} - 1) = \frac{2}{α^{'}} (N + \tilde{N} - 2)

where

N

and

\tilde{N}

are defined in terms of

α_{- n}^{I}

and

{\tilde{α}}_{- n}^{I}

, repectively, and we the normal ordering constant to

a = 1

. We also have the level-matching constraint

N = \tilde{N}

Let's look at the first two mass levels:

$N = 0$ : Ground state $| 0; k ⟩$ has $M^{2} = - \frac{4}{α^{'}}$ , tachyon
$N = 1$ : A general state takes the following form:

| Ω^{I J} ⟩ = α_{- 1}^{I} {\tilde{α}}_{- 1}^{J} | 0; k ⟩, M^{2} = 0

This can be understood as the tensor product of two massless vectors, one left-moving and one night-moving. We can decompose this into a symmetric traceless part (graviton), an antisymmetric part (Kalb-Ramond), and the trace part (dilaton) as shown below:

| Ω^{I J} ⟩ = (| Ω^{(I J)} ⟩ - \frac{1}{D} | Ω^{I I} ⟩ δ^{I J}) + | Ω^{[I J]} ⟩ + \frac{1}{D} | Ω^{I I} ⟩ δ^{I J}

We see that massless states indeed form reps of

S O (D - 2)

. These massless fields will also play a rate in superstring.

Summary: String theory predicts gravity! Also unifies gravity with gauge theory since open strings give rise to gauge fields (like photons) while closed strings give rise to gravitons. String theory also predicts infinite tower of higher spin states with mass

O (1 / α^{'})

11 CFT: Virasoro from Stress Tensor

This is based on Zwiebach pg

254, 258, 259

, and BBS pg 63,64 . We will now develop ad CFT techniques for doing worldsheet calculations. This will be much more efficient than working with mode expansions and the oscillator algebra. In particular, we will use it to give a more elegant and rigorous proof of the critical dimension and to compute string amplitudes. We will not impose lightcone gauge.

First let us demonstrate that the modes of the stress tensor

L_{m}, {\tilde{L}}_{m}

are Virasoro generators. To do that, we need the following identities (for the closed string):

\begin{aligned} [L_{m}, α_{n}^{μ}] = - n α_{m + n}^{μ}, [{\tilde{L}}_{m}, {\tilde{α}}_{n}^{μ}] = - n {\tilde{α}}_{m + n}^{μ} \\ [L_{m}, x_{0}^{μ}] = - i \sqrt{\frac{α^{'}}{2}} α_{m}^{μ}, [{\tilde{L}}_{m}, x_{0}^{μ}] = - i \sqrt{\frac{α}{2}} {\tilde{α}}_{m}^{μ} \end{aligned}

Let's check the first one and leave the rest for HW. First compute

\begin{aligned} [L_{m}, α_{n}^{μ}] & = [\frac{1}{2} \sum_{p} α_{m - p} \cdot α_{p}, α_{n}^{μ}] \\ = \frac{1}{2} \sum_{p} (α_{m - p} \cdot α_{p} α_{n}^{μ} - α_{n}^{μ} α_{m - p} \cdot α_{p}) \\ = \frac{1}{2} \sum_{p} (α_{m - p ν} [α_{p}^{ν}, α_{n}^{μ}] - [α_{m - p}^{ν}, α_{n}^{μ}] α_{p ν}) \\ = \frac{1}{2} (- n α_{m + n}^{μ} + (m - (m + n)) α_{m + n}^{μ}) \\ = - n α_{m + n}^{μ} \end{aligned}

Now consider commutator of

L_{m}

with

X^{μ}

\begin{aligned} [L_{m}, X^{μ} (τ, σ)] & = [L_{m}, x_{0}^{μ}] + [L_{m}, α^{'} p^{μ} τ] + [L_{m}, i \sqrt{\frac{α^{'}}{2}} \sum_{n \neq 0} \frac{1}{n} α_{n}^{μ} e^{- i n σ^{-}}] \\ = - i \sqrt{\frac{α^{'}}{2}} \sum_{n} α_{m + n}^{μ} e^{- i n σ^{-}} \\ = - i e^{i m σ^{-}} \sqrt{\frac{α}{2}} \sum_{n} α_{n} e^{- i n σ^{-}} \\ = - i e^{i m σ^{-}} \partial_{-} X^{μ} \\ \to [L_{m}, X^{μ} (τ, σ)] & = - i V_{m}^{-} (X^{μ} (τ, σ)), V_{n}^{\pm} = e^{i n σ^{\pm}} \partial_{\pm} \end{aligned}

Similarly, we find

[{\tilde{L}}_{m}, X^{μ} (τ, σ)] = - i V_{m}^{+} (X^{μ} (τ, σ))

. Hence

L_{m}, {\tilde{L}}_{m}

generate Virasoro symmetry.

Let us briefly comment on the factor of

- i

on appearing in these relations. Consider a scalar theory in

D

spacetime dimensions. For symmetry

δ ϕ

with Noeither charge

Q

, in the quantum theory, we have

[Q, ϕ] = - i δ ϕ

Proof:

From Noether's theorem we have

Q = \int d^{D - 1} \cdot j^{0} = \int d^{D - 1} x Π (t, x) δ ϕ (t, x), Π = \frac{\partial L}{\partial \dot{ϕ}}

Hence,

\begin{aligned} [Q, ϕ (t, y)] & = \int d^{D - 1} x [Π (t, x) δ ϕ (t, x), ϕ (t, y)]] \\ = \int d^{D - 1} x δ ϕ (t, x) \underset{- i δ^{D - 1} (x - y)}{\underset{⏟}{[Π (t, x), ϕ (t, y)]}} = - i δ ϕ (t, y) \end{aligned}

Now let's map the worldsheet to the complex plane:

τ \to - i τ

so that

z =

e^{i σ^{-}} = e^{τ - i σ}

and

\bar{z} = e^{i σ^{+}} = e^{τ + i σ}

. In terms of

z, \bar{z}

we then have

\partial_{z} X^{μ} \equiv \partial X^{μ} = - i \sqrt{\frac{α}{2}} \sum_{n} α_{n}^{μ} z^{- n - 1}, \partial_{\bar{z}} X^{μ} \equiv \bar{\partial} X^{μ} = - i \sqrt{\frac{α}{2}} \sum_{n} {\tilde{α}}_{n}^{μ} {\bar{z}}^{- n - 1}

and

T \equiv - \frac{1}{α} (\partial X)^{z} = \sum_{n} L_{n} z^{- n - 2}, \bar{T} \equiv - \frac{1}{α} (\bar{\partial} X)^{2} = \sum_{n} {\bar{z}}^{- n - 2}

The commutators above are then written as

[L_{n}, X^{μ}] = - i e^{i n σ^{-}} \partial_{-} X^{μ} = - i z^{n} \underset{i z}{\underset{⏟}{\partial_{-} z}} \partial X^{μ} = z^{n + 1} \partial X^{μ}, [{\tilde{L}}_{n}, X^{μ}] = {\bar{z}}^{n + 1} \bar{\partial} X^{μ}

Hence for a general conformal transformation

δ z = ϵ (z) = \sum_{n} ϵ_{n} z^{n + 1}, δ \bar{z} = \tilde{ϵ} (\bar{z}) = \sum_{n} {\tilde{ϵ}}_{n} {\bar{z}}^{n + 1}

we have

\begin{aligned} δ X^{μ} (w, \bar{w}) = ϵ (w) \partial X (w, \bar{w})^{μ} + \tilde{ϵ} (\bar{w}) \bar{\partial} X^{μ} (ω, \bar{w}) \\ = \sum_{n} ϵ_{n} w^{n + 1} \partial X (ω, \bar{w})^{μ} + \sum_{n} {\tilde{ϵ}}_{n} {\bar{w}}^{n + 1} \bar{\partial} X_{(w, \bar{w})}^{μ} \\ = \sum_{n} ϵ_{n} [L_{n}, X (ω, \bar{w})^{μ}] + \sum_{n} {\tilde{ϵ}}_{n} [{\tilde{L}}_{n}, X^{μ} (w, \bar{w})] \end{aligned}

Noting that

\frac{1}{2 π i} ∮ d z z^{n + 1} T (z) = \frac{1}{2 π i} \sum_{m} ∮ d z z^{n - m - 1} L_{m} = \frac{1}{2 π i} \sum_{m} 2 π i δ_{m, n} L_{m} = L_{n}

where the contour encloses

z = 0

and we used Cauchy's theorem:

∮ \frac{d z}{2 π i} z^{k} =

δ_{k, - 1}

. Similarly, we find

{\tilde{L}}_{n} = \frac{1}{2 π i} ∮ d \bar{z} {\bar{z}}^{n + 1} \tilde{T} (\bar{z})

. More generally Cauchy's theorem can be stated as follows:

Cauchy's Theorem

\frac{1}{2 π i} ∮_{Γ} d z f (z) = \sum_{i} Res [f (z_{i})]

where the sum runs over the simple poles

z_{i}

f

inside the contour

Γ

, and near each pole

lim_{z \to z_{i}} f (z) = \frac{Res [f (z_{i})]}{z - z_{i}}

Putting everything together,

\begin{aligned} δ X^{μ} (w, \bar{w}) & = \frac{1}{2 π i} ∮ d z \sum_{n} ϵ_{n} z^{n + 1} [T (z), X^{μ} (w, \bar{w})] \\ + \frac{1}{2 π i} ∮ d \bar{z} \sum_{n} {\tilde{ϵ}}_{n} {\bar{z}}^{n + 1} [\tilde{T} (\bar{z}), X^{μ} (w, \bar{w})] \\ = \frac{1}{2 π i} ∮ d z ϵ (z) [T (z), X^{μ} (w, \bar{w})] \\ + \frac{1}{2 π i} ∮ d z \tilde{ϵ} (\bar{z}) [\tilde{T} (\bar{z}), X^{μ} (w, \bar{w})] \\ = [Q_{ϵ}, X^{μ}] + [Q_{\tilde{ϵ}}, X^{μ}] \end{aligned}

where

Q_{ϵ} = \frac{1}{2 π i} ∮ d z T (z) ϵ (z), Q_{\tilde{ϵ}} = \frac{1}{2 π i} ∮ d \bar{z} \tilde{T} (\bar{z}) \tilde{ϵ} (\bar{z})

are the generators of conformal transformations. More generally, for

δ z = ϵ (z)

and

δ \bar{z} = \tilde{ϵ} (\bar{z})

an operator

Φ (w, \bar{w})

transforms as

δ_{ϵ} Φ = [Q_{ϵ}, Φ], δ_{\tilde{ϵ}} Φ = [Q_{\tilde{ϵ}}, Φ]

12 Operator Product Expansion

This is based on BBS pg 64-66 and Tong sections 4.3.1,4.3.3. Under a conformal transformation

δ z = ϵ (z)

, an operator

Φ

transforms as

δ_{ϵ} Φ (w, \bar{w}) = \frac{1}{2 π i} ∮ d z ϵ (z) [T (z), Φ (w, \bar{w})]

The contour encluses the origin and is specified by "radial ordering": operators to the left should be located further from from the origin (recall that langer radius in the complex plane corresponds to larger

τ

on the cylinder so this is essentially time ordering). Hence, the commutator gives a contour which encircles the point

w

In general, we can express the product of local operators as a series of local operators located at one of their positions. This is known as an "Operator

Product Expansion" (OPE). Hence, the contour integral for

δ_{ϵ} Φ

is just the residue of the pole in the OPE of

T

and

Φ

ϵ (z) T (z) Φ (w, \bar{w}) = \dots + \frac{Res [ϵ (z) T (z) Φ (w, \bar{w})]}{z - w} + \dots

where the ... on the left denote higher order poles in

z - w

while the ... on the right are finite as

z \to w

\to δ_{ϵ} Φ = \frac{1}{2 π i} ∮_{Γ_{w}} d z \frac{Res [ϵ (z) T (z) Φ (w, \bar{w})]}{z - w} = Res [ϵ (z) T (z) Φ (w, \bar{w})]

Φ

is a "conformal primary with weight

(h, \tilde{h})

" if

\begin{aligned} T (z) Φ (w, \bar{w}) = \frac{h}{(z - w)^{2}} Φ (w, \bar{w}) + \frac{1}{z - w} \partial Φ (w, \bar{w}) + \dots \\ \tilde{T} (\bar{z}) Φ (w, \bar{w}) = \frac{\tilde{h}}{(\bar{z} - \bar{w})^{2}} Φ (w, \bar{w}) + \frac{1}{\bar{z} - \bar{w}} \bar{\partial} Φ (w, \bar{w}) + \dots \end{aligned}

where.. are non-singular as

z \to w

. From this OPE, we find

\begin{aligned} δ_{ϵ} Φ (w, \bar{w}) & = \frac{1}{2 π i} ∮_{Γ_{w}} d z [\frac{h ϵ (z)}{(z - w)^{2}} Φ (w, \bar{w}) + \frac{ϵ (z)}{z - w} \partial Φ (w, \bar{w}) + \dots] \\ = \frac{1}{2 π i} ∮_{Γ_{w}} d z [\frac{h (ϵ (w) + (z - w) \partial ϵ (w) + \dots)}{(z - w)^{2}} Φ (w, \bar{w}) + \frac{(ϵ (w) + \dots)}{z - w} \partial Φ (w, \bar{w}) + \dots] \\ = \frac{1}{2 π} ∮_{Γ_{w}} d z \frac{1}{z - w} (h \partial ϵ (w) Φ (w, \bar{w}) + ϵ (w) \partial Φ (w, \bar{w})) \\ = h \partial ϵ (w) Φ (w, \bar{w}) + ϵ (w) \partial Φ (w, \bar{w}) \end{aligned}

Similarly, we find that

δ_{\tilde{ϵ}} Φ (w, \bar{w}) = \tilde{h} \bar{\partial} \tilde{ϵ} (\bar{w}) Φ (w, \bar{w}) + \tilde{ϵ} (\bar{w}) \bar{\partial} Φ (w, \bar{w})

Note that these are just infinitesimal versions of

Φ (z, \bar{z}) \to {(\frac{\partial z^{'}}{\partial z})}^{h} {(\frac{\partial {\bar{z}}^{'}}{\partial \bar{z}})}^{\tilde{h}} Φ (z^{'}, {\bar{z}}^{'})

under

(z, \bar{z}) \to (z^{'} (z), {\bar{z}}^{'} (\bar{z}))

. To see this, let

z^{'} = z + ϵ (z), {\bar{z}}^{'} = \bar{z} + \tilde{ϵ} (\bar{z})

, where

ϵ, \tilde{ϵ}

infinitesimal:

\begin{aligned} Φ (z, \bar{z}) \to (1 + \partial ϵ)^{h} (1 + \bar{\partial} \tilde{ϵ})^{\tilde{h}} Φ (z + ϵ, \bar{z} + \tilde{ϵ}) \\ = (1 + h \partial ϵ + \dots) (1 + \tilde{h} \bar{\partial} \tilde{ϵ} + \dots) (Φ (z, \bar{z}) + ϵ \partial Φ + \tilde{ϵ} \bar{\partial} Φ + \dots) \\ = Φ (z, \bar{z}) + \underset{δ_{ϵ} Φ}{\underset{⏟}{h \partial ϵ \bar{Φ} + ϵ \bar{Φ}}} + \underset{δ_{\tilde{ϵ}} \bar{Φ}}{\underset{⏟}{\tilde{h} \bar{\partial} \tilde{ϵ} Φ + \hat{ϵ} \bar{\partial} Φ}} + \dots \end{aligned}

where... are higher order terms in

ϵ

and

\tilde{ϵ}

. Hence,

δ Φ = δ_{ϵ} Φ + δ_{ϵ} Φ

found via the OPE is indeed an infinitesimal conformal transformation.

All of the OPE's we need in practice can be derived from the following OPE:

X^{μ} (z) X^{ν} (w) = - \frac{α^{'}}{2} n^{μ ν} \ln (z - w) + \dots

where we write

X (z, \bar{z}) = X (z) + X (\bar{z})

and

\dots

are non-singular as

z \to w

. The singular part of the OPE can be read of from the correlation function

⟨ X^{μ} (\bar{z}, \bar{z}) X^{ν} (z^{'}, {\bar{z}}^{'}) ⟩

, which can in turn be deduced from a path integral:

0 = \int D X \frac{δ}{δ X_{μ} (z, \bar{z})} [e^{- S} X^{ν} (z^{'}, {\bar{z}}^{'})]

where the integral is over all field configurations and the integral vanishes because the integrand contains a total functional derviative. The action is given by

\begin{aligned} S & = \frac{T}{2} \int d^{2} σ ({\dot{X}}^{2} + X^{' 2}), τ \to - i τ \\ = \frac{1}{2 π α^{'}} \int d^{2} z \partial X \cdot \bar{\partial} X, d^{2} z = d z d \bar{z} \\ = - \frac{1}{2 π α^{'}} \int d^{2} z X \cdot \partial \bar{\partial} X \end{aligned}

We then find

\begin{aligned} 0 & = \int D X [- \frac{δ S}{δ X_{μ} (z, \bar{z})} X^{ν} (z^{'}, {\bar{z}}^{'}) + \frac{δ X^{ν} (z^{'}, {\bar{z}}^{'})}{δ X_{μ} (z, \bar{z})}] e^{- S} \\ = \int D X [- \frac{1}{π α^{'}} \partial \bar{\partial} X^{μ} (z, \bar{z}) X^{ν} (z^{'}, {\bar{z}}^{'}) + δ (z - z^{'}, \bar{z} - {\bar{z}}^{'}) η^{μ ν}] e^{- S} \\ \to \partial \bar{\partial} ⟨ X^{μ} (z, \bar{z}) X^{ν} (z^{'}, {\bar{z}}^{'}) ⟩ = - π α^{'} δ (z - z^{'}, \bar{z} - {\bar{z}}^{'}) η^{μ ν} \end{aligned}

In the HW you will show that

\partial \bar{\partial} \ln {| z - z^{'} |}^{2} = 2 π δ (z - z^{'}, \bar{z} - {\bar{z}}^{'})

Hence, the 2-point function (or "propagator") is given by

⟨ X^{μ} (z, \bar{z}) X^{ν} (z^{'}, {\bar{z}}^{'}) ⟩ = - \frac{α^{'}}{2} \ln {| z - z^{'} |}^{2} η^{μ ν}

Writing this in terms of holomorphic and antiholomorphic parts,

⟨ X^{μ} (z) X^{ν} (z^{'}) ⟩ = - \frac{α^{'}}{2} \ln (z - z^{'}) η^{μ ν}, ⟨ X^{μ} (\bar{z}) X^{ν} ({\bar{z}}^{'}) ⟩ = - \frac{α^{'}}{2} \ln (\bar{z} - {\bar{z}}^{'}) η^{μ ν}

from which we see that

X (z)^{μ} X (w)^{ν} = - \frac{α^{'}}{2} \ln (z - w) η^{μ ν} + \dots

as claimed. From this, we see that

\begin{aligned} \partial X^{μ} (z) \partial X^{ν} (w) & = \partial_{z} \partial_{w} (X^{μ} (z) X^{ν} (w)) \\ = \partial_{z} \partial_{w} (- \frac{α^{'}}{2} \ln (z - w) η^{μ ν} + \dots) \\ = - \frac{α^{'}}{2} \frac{η^{μ ν}}{(z - w)^{2}} + \dots \end{aligned}

Let us denote

⟨ X^{μ} (z) X^{ν} (w) ⟩ = \overset{⏜}{X^{μ} (z) X^{ν}} (w)

, which is known as a "Wick contraction." We can then compute OPE of more general operators by summing over all possible contractions. We also define "normal ordering":

: O (z_{1}) \dots O (z_{n}) := O (z_{1}) \dots O (z_{n}) - all contractions

Hence, these operator have no singular terms in their OPE. We an then define composite operators like

T

and

\tilde{T}

more precisely as

T \equiv - \frac{1}{α^{'}} : \partial X^{μ} (z) \partial X^{ν} (z) : η_{μ ν}

where

\begin{aligned} : \partial X^{μ} (z) \partial X^{ν} (z) & := lim_{z \to w} (\partial X^{μ} (z) \partial X^{ν} (w) - \partial X^{μ} (z) \partial X^{ν} (w)) \\ = lim_{z \to w} (\partial X^{μ} (z) \partial X^{ν} (w) + \frac{α^{'}}{2} \frac{n^{μ ν}}{(z - w)^{2}}) \end{aligned}

In summary, when computing OPEs we never contract two operators inside of the same normal ordering symbol : :, but we sum over all other contractions.

Examples

$\partial X^{μ} (w)$ is a primary operator with $(h, \tilde{h}) = (1, 0)$ .

\begin{aligned} T (z) \partial X^{μ} (w) = - \frac{1}{α^{'}} : \partial X (z) \cdot \partial X (z) : \partial X^{μ} (w) - \frac{1}{α^{'}} : \partial X (z) \cdot \partial X (z) : \partial X^{μ} (w) + \dots \\ = - \frac{2}{α^{'}} : \partial X (z) \cdot \partial X (z) : \partial X^{μ} (w) + \dots \\ = - \frac{2}{α^{'}} \partial X^{μ} (z) \partial_{z} \partial_{w} (- \frac{α^{'}}{2} \ln (z - w)) + \dots \\ = \frac{\partial X^{μ} (z)}{(z - w)^{2}} + \dots = \frac{(\partial X^{μ} (w) + (z - w) \partial^{2} X^{μ} (w) + \dots)}{(z - w)^{2}} + \dots \\ = \frac{\partial X^{μ} (w)}{(z - w)^{2}} + \frac{\partial^{2} X^{μ} (w)}{z - w} + \dots \end{aligned}

Note that

\tilde{T} (\bar{z}) \partial X^{μ} (w) = \dots

since

⟨ X^{μ} (z) X^{ν} (w) ⟩ = 0

: $e^{i k \cdot X (ω, \bar{w})}$ is a primary operator with $h = \tilde{h} = α^{'} k^{2} / 4$ .

First compute the following OPE (suppressing Lorentz indices for simplicity)

\begin{aligned} \partial X (z) : e^{i k X (w)} : & = \sum_{n = 0}^{\infty} \frac{(i k)^{n}}{n!} \partial X (z) : X (w)^{n} : \\ = \sum_{n = 0}^{\infty} \frac{(i k)^{n}}{n!} n \partial X (z) : X (w) X (w)^{n - 1} : + \dots \\ = \sum_{n = 0}^{\infty} \frac{(i k)^{n}}{(n - 1)!} \partial_{z} (- \frac{α^{'}}{2} \ln (z - w)) : X (w)^{n - 1} : + \dots \\ = - \frac{α^{'}}{2} \frac{1}{z - w} \sum_{n = 1}^{\infty} \frac{(i k)^{n}}{(n - 1)!} : X (w)^{n - 1} : + \dots \\ = - \frac{i α^{'} k}{2 (z - w)} : e^{i k X (w)} : + \dots \end{aligned}

From this, we find

\begin{aligned} T (z) : e^{i k X (w)} : & = - \frac{1}{α^{'}} : \partial X (z) \partial X (z) :: e^{i k X (w)} : \\ = - \frac{1}{α^{'}} [{(- \frac{i α^{'} k}{2 (z - w)})}^{2} : e^{i k X (w)} : + 2 (\frac{- i α^{'} k}{2 (z - w)}) : \partial X (z) e^{i k X (w)} :] \end{aligned}

where the first term came from contracting both

\partial X^{'}

's in the stress tensor with the exponential and the second term came from contracting only one

\partial X

. This gives a factor of two since there are two possible ways to perform this contraction. We then obtain

\begin{aligned} T (z) : e^{i k X (w)} : & = \frac{α^{'} k^{2} / 4}{(z - w)^{2}} : e^{i k X (w)} : + \frac{i k : \partial X (z) e^{i k X (w)} :}{z - w} + \dots \\ = \frac{α^{'} k^{2} / 4}{(z - w)^{2}} : e^{i k X (w)} : + \frac{i k : (\partial X (w) + \dots) e^{i k x (w)} :}{z - w} + \dots \\ = \frac{α^{'} k^{2} / 4}{(z - w)^{2}} : e^{i k X (w)} : + \frac{\partial_{w} : e^{i k X (w)} :}{z - w} + \dots \end{aligned}

Similarly

\tilde{T} (\bar{z}) : e^{i k X (\bar{w})} := \frac{α^{'} k^{2} / 4}{(\bar{z} - \bar{w})^{2}} : e^{i k X (\bar{w})} : + \frac{\partial_{\bar{w}} : e^{i k X (\bar{w})} :}{\bar{z} - \bar{w}} + \dots

13 Critical Dimension from CFT

This is based on Tong pg 82, sections 5.1, 5.2, BBS pg 73-76, GSW pg 121-127. We will revisit the calculation of critical dimension using 2d CFT. In lightcone
gauge, we found that Lorentz invariance was amalous unless

D = 26

. Now we will work covariantly and show that conformal symmetry is anomalous unless

D = 26

. The conformal anomaly shows up in central extension of the Virasoro algebra which can be derived from the OPE of the stress tensor with itself:

\begin{aligned} T (z) T (w) & = \frac{1}{{(α^{'})}^{2}} : \partial X (z) \cdot \partial X (z) :: \partial X (w) \cdot \partial x (w) : \\ = \frac{1}{{(α^{'})}^{2}} (2 : \partial X \cdot \partial X : \partial X \cdot \partial X : + 4 : \partial X \cdot \partial X :: \partial X \cdot \partial X : + \dots) \end{aligned}

where the numerical coefficients in front of each term encode the number of ways of performing the contracitons. Recalling that

\partial \overset{⏜}{X^{μ} (z) \partial X^{ν}} (w) = η^{μ ν} \partial_{z} \partial_{w} (- \frac{α^{'}}{2} \ln (z - w)) = - \frac{α^{'}}{2} \frac{η^{μ ν}}{(z - w)^{2}}

we obtain

\begin{aligned} T (z) T (z) & = \frac{1}{{(α^{'})}^{2}} (2 D {(- \frac{α^{'}}{2} \frac{1}{(z - w)^{2}})}^{2} + 4 (- \frac{α^{'}}{2} \frac{1}{(z - w)^{2}}) : \partial X (z) \cdot \partial X (w) : + \dots) \\ = \frac{D / 2}{(z - w)^{4}} - \frac{2}{α^{'}} \frac{: \partial X (w) \cdot \partial X (w) :}{(z - w)^{2}} - \frac{2}{α^{'}} \frac{\partial^{2} X (w) \cdot \partial X (w) :}{(z - w)} + \dots \\ = \frac{D / 2}{(z - w)^{4}} + \frac{2 T (w)}{(z - w)^{2}} + \frac{\partial T (w)}{z - w} + \dots \end{aligned}

To obtain the second line, we wrote

\partial X (z) = \partial X (w) + (z - w) \partial^{2} X (w) + \dots

in the first line. Hence, we see that

T

is not a primary operator because of the

1 / (z - w)^{4}

term.

More generally,

T (z) T (w) = \frac{c / 2}{(z - w)^{4}} + \frac{2}{(z - w)^{2}} T (w) + \frac{1}{z - w} \partial T (w) + \dots

where

c

is the "conformal anomaly" or "central change". For the bosonic string

c = D

, ie. each

X^{μ} (z)

contributes 1 to the central charge. More generally, c counts the number of riigt-moving degrees of freddom.

T

is only a primary it

c = 0

, in which case it has weight

(2, 0)

. Similarly, if the left-moving central charge

\tilde{c} = 0

, then

\tilde{T} (\bar{z})

is a primary with weight

(0, 2)

Let's verify that

c

corresponds to central extension of Virasoro algebra. First recall that

L_{m} = ∮ \frac{d z}{2 π i} z^{m + 1} T (z) \leftrightarrow T (z) = \sum_{n} L_{n} z^{- n - 2}

Hence,

[L_{m}, L_{n}] = ∮ \frac{d z}{2 π i} z^{m + 1} ∮ \frac{d w}{2 π i} w^{n + 1} [T (z), T (w)]

Let us do the

z

integral first, holding

w

fixed. Using radial ordering, the commutator is computed by considering the

z

integral along a small path encircling

w

(which we will refer to as

Γ_{w}

) so we can compute it from the OPE of

T (z) T (w)

[L_{m}, L_{n}] = ∮ \frac{d w}{2 π i} w^{n + 1} ∮_{Γ_{w}} \frac{d z}{2 π i} z^{m + 1} [\frac{c / 2}{(z - w)^{4}} + \frac{2}{(z - w)^{2}} T (w) + \frac{1}{z - w} \partial T (w) + \dots]

Compute the residues by Taylor expanding

z^{m + 1}

around

z = w

\begin{aligned} [L_{m}, L_{n}] & = ∮ \frac{d w}{2 π i} w^{n + 1} ∮_{Γ_{w}} \frac{d z}{2 π i} [\frac{c / 2}{(z - w)^{4}} (\dots + {\frac{1}{3!} \partial_{z} (z^{m + 1}) |}_{z = w} (z - w)^{3} + \dots) \\ + \frac{2}{(z - w)^{2}} (\dots + {(z - w) \partial_{z} (z^{m + 1}) |}_{z = w} + \dots) T (w) + \frac{1}{z - w} (w^{m + 1} + \dots) \partial T (w)] \\ = ∮ \frac{d w}{2 π i} [\frac{c}{12} (m^{3} - m) w^{n + m - 1} + 2 (m + 1) w^{n + m + 1} T (w) + w^{n + m + 2} \partial T (w)] \end{aligned}

Recalling that the contour encircles

w = 0

and using (13) we obtain

\begin{aligned} [L_{m}, L_{n}] & = ∮ \frac{d w}{2 π i} [\frac{c}{12} (m^{3} - m) w^{n + m + 1} + 2 (m + 1) \sum_{l} w^{n + m - l - 1} L_{l} - \sum_{l} (l + 2) w^{n + m - l - 1} L_{l}] \\ = \frac{c}{12} (m^{3} - m) δ_{m + n, 0} + 2 (m + 1) \sum_{l} δ_{m + n - l, 0} L_{l} - \sum_{l} (l + 2) δ_{m + n - l, 0} L_{l} \\ = \frac{c}{12} (m^{3} - m) δ_{m + n, 0} + 2 (m + 1) L_{n + m} - (n + m + 2) L_{n + m} \\ = (m - n) L_{m + n} + \frac{c}{12} (m^{3} - m) δ_{m + n, 0} \end{aligned}

Hence, we the Virasaro algebra recieves quantum correction proportional to the central charge

c

Ghost fields

In order to have zero central charge, we need to add worldsheet fields with negative central charge. These violate spin statistics theorem and are known as "ghosts." In fact ghosts arise when gauge-fixing path integrals using the "Fadeev-Popov procedure." Let's sketch how this works for bosonic string. For details see GSW pg 121-127 and Tong sections 5.1 and 5.2. Recall that in light cone gauge the worldsheet metric is

h_{α β} = η_{α β} \to h_{+ +} = h_{- -} = 0

We also have a residual gauge symmetry:

δ h_{α β} = \partial_{α} ξ_{β} + \partial_{β} ξ_{α} \to δ h_{\pm \pm} = 2 \partial_{\pm} ξ_{\pm}

Due to this gauge symmetry, path integral contains and infinite redundancy coming over the sum over physically equivalent field configurations related by a
gauge transformations. We can remove this redundancy by inserting

1 = \int D ξ_{\pm} δ (h_{+ +}^{'}) δ (h_{- -}^{'}) \det (\frac{δ h_{+ +}^{'}}{δ ξ_{+}}) \det (\frac{δ h_{- -}^{'}}{δ ξ_{-}})

where

h_{+ +}^{'} = h_{+ +} + 2 \partial_{+} ξ_{+}, h_{- -}^{'} = h_{- -} + 2 \partial_{-} ξ_{-}

This is the infinite-dimensional analogue of

1 = \int d x δ (f (x)) | f^{'} (x) |, since \int d x δ (f (x)) = \frac{1}{| f^{'} (x_{i}) |}, where f (x_{i}) = 0

The other key fact we need is that determinants can be represented by path integrals over anticommuting scalar fields (ghosts), as you will show in the HW:

\begin{aligned} \det (δ h_{+ +}^{'} / δ ξ_{+}) = \int D c D b \exp (- \frac{1}{π} \int d^{2} σ b \partial_{+} c) \\ \det (δ h_{- -}^{'} / δ ξ_{-}) = \int D \tilde{c} D \tilde{b} \exp (- \frac{1}{π} \int d^{2} σ \tilde{b} \partial_{-} \tilde{c}) \end{aligned}

After converting to

z, \bar{z}

coordinates the get

S_{ghost} = \frac{1}{2 π} \int d^{2} z (b \bar{\partial} c + \tilde{b} \partial \tilde{c})

The equations of motion imply that the fields

(b, c)

are holomorphic while

(\tilde{b}, \tilde{c})

are antiholomprhic. Moreover, one can show that they have the following stress tensor:

T = - 2 : b \partial c : + : c \partial b :, \tilde{T} = - 2 : \tilde{b} \bar{\partial} \tilde{c} : + : \tilde{c} \bar{\partial} \tilde{b} :

The ghost fields have the following OPE:

b (z) c (w) = - c (w) b (z) = \frac{1}{z - w} + \dots

As you will show in the HW,

\begin{aligned} T (z) c (z) & = - \frac{c (w)}{(z - w)^{2}} + \frac{\partial c (w)}{z - w} + \dots, T (z) b (w) = \frac{2 b (w)}{(z - w)^{2}} + \frac{\partial b (w)}{z - w} + \dots \\ T (z) T (w) & = \frac{- 13}{(z - w)^{4}} + \frac{2 T (w)}{(z - w)^{2}} + \frac{\partial T (w)}{z - w} + \dots \end{aligned}

Hence

c_{ghost} = - 26

. It follows that

c_{ghost} + c_{matter} = 0

D = 26

since

c_{matter} = D

14 Vertex operators

This will be based on Tong pg 121-123 and BBS pg 85-88. Recall that a primary operator of weight

(h, \tilde{h})

transforms as

Φ (z, \bar{z}) \to {(\frac{\partial w}{\partial z})}^{h} {(\frac{\partial \bar{w}}{\partial \bar{z}})}^{\hat{h}} Φ (w, \bar{w})

under

(z, w) \to (w (z), \bar{w} (\bar{z}))

. Note that

d z d \bar{z} \to \frac{\partial z}{\partial w} \frac{\partial \bar{z}}{\partial \bar{w}} d w d \bar{w}

d z d \bar{z}

has weight

(- 1, - 1)

under a conformal rescaling. Hence

V = \int d^{2} z \hat{V}

will be conformal invariant if

\hat{V}

is a primary of weight

(+ 1, + 1)

. In that case,

V

is "vertex operator." Vertex ops correspond to the physical states.

Examples

Tachyon

V_{tachyon} = \int d^{2} z : e^{i p \cdot X (\bar{z}, \bar{z})} :

First note that the operator has no Lorentz indices, so it describes a scalar particle. In chapter 12 we showed that

= e^{i p - X}

: is a primary with weight

h = \tilde{h} = α^{'} p^{2} / 4

. If

h = \tilde{h} = + 1

, then

M^{2} = - p^{2} = - \frac{4}{α^{'}}

, which is indeed the tachyon mass found in chapter 10.

Massless states

Now consider

V_{massless} = \int d^{2} z : e^{i p \cdot X} \partial X^{μ} \bar{\partial} X^{ν} : ϵ_{μ ν}

Let's compute the OPE of

\hat{V} =: e^{i p \cdot X} \partial X^{μ} \bar{\partial} X^{ν} : ϵ_{μ ν}

with

T

- \frac{1}{α^{'}} : \underset{A}{\underset{⏟}{\partial X (z) \cdot \partial X (z)}} : \underset{B}{\underset{⏟}{e^{i p \cdot X (w, \bar{w})}}} \underset{C}{\underset{⏟}{\partial X^{μ} (w)}} \bar{\partial} X^{ν} (\bar{w}) : ϵ_{μ ν}

We have previously computed contractions of

A

with

B

and

A

with

C

so we can recycle those results. Contracting of

A

with

B

gives

\frac{α^{'} p^{2} / 4}{(z - w)^{2}} e^{i p \cdot X} + \dots

while contracting

A

with

C

gives

\frac{+ 1}{(z - w)^{2}} \partial X + \dots

Hence, the operator has conformal weight

h = 1 + α^{'} p^{2} / 4

. Similarly,

\tilde{h} =

1 + α^{'} p^{2} / 4

. Hence

h = \tilde{h} = 1 \to p^{2} = 0

We still need to check that

\hat{V}

is primary. In fact, there is a contribution to OPE which we haven't previously computed which can spoil this property. It comes from contracting one

\partial X

A

with

B

and the other

\partial X

A

with

C

\begin{aligned} - \frac{2}{α^{'}} η_{ρ λ} \partial \overset{⏜}{X^{ρ} (z) \partial X^{λ} (z) \exp [i p \cdot X (w, \bar{w})] \partial X^{μ}} (w) ϵ_{μ ν} \\ = - \frac{2}{α} n_{ρ λ} (- \frac{α^{'}}{2} \frac{1}{(z - w)^{2}} η^{ρ μ}) (- \frac{i α^{'} p^{λ}}{2 (z - w)} e^{i p \cdot X}) ϵ_{μ ν} \\ = - \frac{i α^{'}}{2} \frac{p^{μ} ϵ_{μ ν}}{(z - ω)^{3}} e^{i p \cdot X} \end{aligned}

where the factor of 2 in the first line comes from performing two equivalent contractions. Similarly, when we compute OPE

\tilde{T}

we find a cubic pole

- \frac{i α^{'}}{2} \frac{p^{ν} ϵ_{μ ν}}{(\bar{z} - \bar{w})^{3}} e^{i p \cdot x}

In order tor

\hat{V}

to be primary, the polarisation tensor must be transverse to momentum:

p^{μ} ϵ_{μ ν} = p^{μ} ϵ_{ν μ} = 0

as expected for massless spinning fields. Finally, we can decompose

ϵ_{μ ν}

into symmetric traceless, antisymmetric, and a trace piece, corresponding to tje graviton, Kalb-Ramond, and dilation fields respectively. Hence, we recover the massless states of the closed bosonic string.

General states

In general, a state of the form

| ϕ ⟩ = \prod_{i} α_{- m_{i}}^{μ_{i}} \prod_{j} {\tilde{α}}_{- m_{j}}^{ν_{j}} | 0; k ⟩

corresponds to a vertex operator of the form

{\hat{V}}_{ϕ} = \int d^{2} z : \prod_{i} \partial^{m_{i}} X^{μ_{i}} (z) \prod_{j} {\bar{\partial}}^{n_{j}} X^{v_{j}} (\bar{z}) e^{i k \cdot X (z, \bar{z})}

15 String interactions

This will be based on Tong sections 6.1, 6.2, and GSW section 1.1. Scattering amplitudes describe the probability for states to interact in a certain way. In QFT, they can be computed perturbatively by summing over Feynman diagrams. In string theory, one sums over all worldsheet topologies. For closed strings (which we will focus on) the number of loops corresponds to the number of handles or genus of the worlsheet. Note that many different Feynman diagrams can arise from a single string diagram at low energies because they correspond to different degenerations of the worldsheet. To compute string amplitudes in practice, we use conformal symmetry of the worldsheet to map infinite tubes corresponding to external legs to punctures:

At each puncture, we then place a vertex operator corresponding to an external state.

Let's consider the tree-level

n

-point scattering of tachyons. This is given by the correlation function of

n

tachyon vertex operators on sphere:

A_{n} (p_{1}, \dots, p_{n}) = \int \frac{Π_{i = 1}^{n} d^{2} z_{i}}{Vol (S L (2, C))} ⟨ \hat{V} (z_{1}, p_{1}) \dots \hat{V} (z_{n}, p_{n}) ⟩

where

p_{i}^{μ}

is the momentum of particle

i

and we can use the global conformal group

S L (2, C)

to fix the location of three punctures. More explicitly, the correlator is given by

\begin{aligned} ⟨ \hat{V} (z_{1}, p_{1}) \dots \hat{V} (z_{n}, p_{n}) ⟩ & = \int D X \exp (- \frac{1}{2 π α^{'}} \int d^{2} z \partial X \cdot \bar{\partial} X) \exp (i \sum_{i = 1}^{n} p_{i} \cdot X (z_{i}, {\bar{z}}_{i})) \\ = \int D X \exp [\int d^{2} z (X \cdot \hat{D} X + i J \cdot X)] \end{aligned}

where

J^{μ} (z, \bar{z}) = \sum_{i = 1}^{n} p_{i}^{μ} δ^{2} (z - z_{i})

and

\hat{D} = \frac{1}{2 π α^{'}} \partial \bar{\partial}

. This is a Gaussian integral. You worked out a finite-dimensional version of this in HW:

\int_{- \infty}^{\infty} d x e^{a x^{2} + i J x} \propto e^{J^{2} / 4 a}

Similar manipulations of the functional integral give (see chapter 8 of Srednicki)

⟨ \hat{V} (z, {\bar{z}}^{\bar{z}}, p_{1}) \dots \hat{V} (z_{n_{n}}, {\bar{z}}_{n}, p_{n}) ⟩ \propto \exp [\frac{1}{4} \int d^{2} z d^{2} z^{'} J (z, \bar{z})^{μ} \underset{G (z, \bar{z}; z^{'}, {\bar{z}}^{'})}{\underset{⏟}{{\hat{D}}^{- 1} (z, \bar{z}; z^{'}, {\bar{z}}^{'})}} J_{μ} (z^{'}, {\bar{z}}^{'})]

where

G

satisfies

\begin{aligned} \hat{D} G (z, \bar{z}; z^{'}, {\bar{z}}^{'}) = δ^{2} (z - z^{'}) \\ \to \partial \bar{\partial} G (z, \bar{z}; z^{'}, {\bar{z}}^{'}) = 2 π α^{'} δ^{2} (z - z^{'}) \\ \to G (z, \bar{z}; z^{'}, {\bar{z}}^{'}) = α^{'} \ln {| z - z^{'} |}^{2} \end{aligned}

It follows that

\begin{aligned} ⟨ \hat{V} (z_{1}, {\bar{z}}_{1}, p_{1}) \dots \hat{V} (z_{n}, {\bar{z}}_{n}, p_{n}) ⟩ & \propto \exp [\frac{α^{'}}{4} \int d^{2} z^{2} d^{2} z^{'} J^{μ} (z, \bar{z}) \ln {| z - z^{'} |}^{2} J_{μ} (z^{'}, {\bar{z}}^{'})] \\ = & \exp [\frac{α^{'}}{4} \sum_{i, j} p_{i} \cdot p_{j} \ln {| z_{i} - z_{j} |}^{2}] \\ \to \exp [α^{'} \sum_{i < j} p_{i} \cdot p_{j} \ln | z_{i} - z_{j} |] \\ = & \prod_{i < j} {| z_{i} - z_{j} |}^{α^{'} p_{i} \cdot p_{j}} \end{aligned}

where we dropped the

i = j

terms in the third line since they are divergent and can be removed by normal ordering. Hence, we obtain

A_{n} \propto \int \frac{\prod_{i = 1}^{n} d^{2} z_{i}}{Vol (S L (2, C))} \prod_{i < j} {| z_{i} - z_{j} |}^{α^{'} p_{i} \cdot p_{j}}

Now let's specialise to

n = 4

and use

S L (2, C)

to set

z_{1} = \infty, z_{2} = 0, z_{4} = 1

Calling

z_{3} = z

, we get

\begin{matrix} (5) & A_{4} \propto \int d^{2} z | z |^{α^{'} p_{2} \cdot p_{3}} | 1 - z |^{α^{'} p_{3} \cdot p_{4}} \end{matrix}

where we have dropped

| z_{1} - z_{j} | = \infty

terms. In general, we have

\begin{matrix} (6) & \int d^{2} z | z |^{2 a - 2} | 1 - z |^{2 b - 2} = \frac{2 π Γ (a) Γ (b) Γ (c)}{Γ (1 - a) Γ (1 - b) Γ (1 - c)} \end{matrix}

where

c = 1 - a - b

(see see Tong section 6.5 for a derivation). Hence,

A_{4} \propto \frac{Γ (- 1 - α^{'} s / 4) Γ (- 1 - α^{'} t / 4) Γ (- 1 - α^{'} u / 4)}{Γ (2 + α^{'} s / 4) Γ (2 + α^{'} t / 4) Γ (2 + α^{'} u / 4)}

This is known as the "Virasoro-Shapiro" amplitude. It is expressed in terms of the Mandelstam variables

s = - {(p_{1} + p_{2})}^{2}, t = - {(p_{1} + p_{4})}^{2}, u = - {(p_{1} + p_{3})}^{2}

To see how the arguments of the Gamma functions arise note that

\begin{aligned} s = - p_{1}^{2} - p_{2}^{2} - 2 p_{1} \cdot p_{2} = + 2 M^{2} - 2 p_{1} \cdot p_{2} = - 8 / α^{'} - 2 p_{1} \cdot p_{2} \\ \to \frac{α^{'}}{2} p_{1} \cdot p_{2} = - 2 - α^{'} s / 4 \end{aligned}

Similarly,

\frac{α^{'}}{2} p_{1} \cdot p_{4} = - 2 - α^{'} t / 4

and

\frac{α^{'}}{2} p_{1} \cdot p_{3} = - 2 - α^{'} u / 4

Another useful identity is

\begin{aligned} s + t + u & = - {(p_{1} + p_{2})}^{2} - {(p_{1} + p_{4})}^{2} - {(p_{1} + p_{3})}^{2} \\ = 6 M^{2} - 2 p_{1} \cdot (p_{2} + p_{2} + p_{3}) \\ = 4 M^{2} = - 16 / α^{'} \\ \to s + t + u = - 16 / α^{'} \end{aligned}

where we used

p_{i}^{2} = - M^{2}

and momentum conservation

\sum_{i = 1}^{n} p_{i}^{μ} = 0

. Also note that

{(p_{2} + p_{3})}^{2} = {(p_{1} + p_{4})}^{2} \to \underset{- 2 M^{2}}{\underset{⏟}{p_{2}^{2} + p_{3}^{2}}} + 2 p_{2} \cdot p_{3} = \underset{- 2 M^{2}}{\underset{⏟}{p_{1}^{2} + p_{4}^{2}}} + 2 p_{1} \cdot p_{4} \to p_{2} \cdot p_{3} = p_{1} \cdot p_{4}

and similarly,

p_{3} \cdot p_{4} = p_{1} \cdot p_{2}

. Applying (6) to (5) and using the above identities, we see that the arguments of the Gamma functions are

\begin{aligned} a = \frac{α^{'}}{2} p_{2} \cdot p_{3} + 1 = \frac{α^{'}}{2} p_{1} \cdot p_{4} + 1 = (- 2 - α^{'} t / 4) + 1 = - 1 - α^{'} t / 4 \\ b = \frac{α^{'}}{2} p_{3} \cdot p_{4} + 1 = - 1 - α^{'} s / 4 \\ c = 1 - a - b = 1 + (1 + α^{'} t / 4) + (1 + α^{'} s / 4) \\ = 3 + \frac{α^{'}}{4} (s + t) = - 1 - α^{'} u / 4 \end{aligned}

Hence we obtain VS amplitude which describes the scattering of four tachyons in bosonic string theory. Let's analyse some of it's properties.

Poles and Residues

Recall that

Γ (z)

has poles at

z = 0, - 1, - 2, \dots

If we fix

t

and vary

s

, we see that

Γ (- 1 - α^{'} s / 4)

has poles when

s = \frac{4}{α^{'}} (n - 1) n = 0, 1, 2, \dots

Recall that the spectrum of the closed string is given by

M^{2} = \frac{4}{α^{'}} (N - 1) = \frac{4}{α^{'}} (\tilde{N} - 1)

Hence, poles correspond to masses af exchanged particles. We can read off the spin of the exchanged particles from the residue of the pole. Near

s = \frac{4}{α} (n - 1)

u = - \frac{16}{α^{'}} - s - t = - \frac{16}{α^{'}} - \frac{4}{α^{'}} (n - 1) - t = - \frac{4}{α^{'}} (n + 3) - t

so the amplitude is approximately

\begin{aligned} A_{4} & \propto \frac{1}{s - M_{n}^{2}} \frac{Γ (- 1 - α^{'} t / 4) Γ (- 1 - α^{'} u / 4)}{Γ (2 + α^{'} t / 4) Γ (2 + α^{'} u / 4)} \\ = \frac{1}{s - M_{n}^{2}} \frac{Γ (- 1 - α^{'} t / 4) Γ (n + 2 + α^{'} t / 4)}{Γ (- n - 1 - α^{'} t / 4) Γ (2 + α^{'} t / 4)} \end{aligned}

where

M_{n}^{2} = \frac{4}{α^{'}} (n - 1)

. Noting that

\begin{aligned} \frac{Γ (n + 2 + α^{'} t / 4)}{Γ (2 + α^{'} t / 4)} & = (n + 2 + α^{'} t / 4) (n + 1 + α^{'} t / 4) \dots (3 + α^{'} t / 4) \\ = {(α^{'} t / 4)}^{n} + \dots \\ \frac{Γ (- 1 - α^{'} t / 4)}{Γ (- n - 1 - α^{'} t / 4)} & = (- 1 - α^{'} t / 4) (- 2 - α^{'} t / 4) \dots (- n - α^{'} t / 4) \\ = {(- α^{'} t / 4)}^{n} + \dots \end{aligned}

where ... indicates lower order terms in

t

, we see that near

s = \frac{4}{α} (n - 1)

the amplitude is

\begin{matrix} (7) & A_{4} \sim \frac{t^{2 n}}{s - M_{n}^{2}} \end{matrix}

This indicates that the highest spin of a particle with mass

M_{n}

J = 2 n

. To see this, note that coupling of a scalar field

ϕ

to a spin-

J

field

χ^{μ_{1} \dots μ_{J}}

is of the form

ϕ \partial_{μ_{1}} \dots \partial_{μ_{J}} ϕ χ^{μ_{1} \dots μ_{J}}

From this, we see that an s-channel Feynman diagram with four external scalars exchanging spin-

J

particle scales like

t^{J}

. It follows from (7) that the maximum spin being exchanged is

2 n

, which is precisely what we expect from the closed string spectrum. Indeed, recall that the spectrum is given by

M^{2} = \frac{4}{α^{'}} (N - 1) = \frac{4}{α^{'}} (\tilde{N} - 1)

and

J_{max} = N + \tilde{N} = 2 N

, so for example

\begin{array}{ll} N = 0 : & M^{2} = - 4 / α^{'}, J_{max} = 0 \\ N = 1 : & M^{2} = 0, J_{max} = 2 \end{array}

Hence, all states in the spectrum get exchanged and the amplitude has the schematic form

\begin{matrix} (8) & A_{4} \sim \sum_{n = 0}^{\infty} \frac{t^{2 n}}{s - M_{n}^{2}} \end{matrix}

High energy limit

Let's take

α^{'} s, α^{'} t \to \infty

holding

s / t

fixed. In this limit, each term in (8) separately diverges for

n > 0

. Hence exchange of spin-2 particle (gravity) diverges
like

E^{2}

as large energy

E

and the exchange of higher-spin particles gives rise to even worse divergences. Hence, if we truncate the sum in (8) the amplitude is ill-behaved at high energies. Remarkably, including all states in the sum yields a result that is well-behaved at high energies.

In the high energy limit, we can neglect the masses of the external particles and take the external momenta to be

\begin{aligned} p_{1} = \frac{\sqrt{s}}{2} (1, 1, 0, \dots) \\ p_{2} = \frac{\sqrt{s}}{2} (1, - 1, 0, \dots) \\ p_{3} = \frac{\sqrt{s}}{2} (- 1, \cos θ, \sin θ_{1} \dots) \\ p_{4} = \frac{\sqrt{s}}{2} (- 1, - \cos θ, - \sin θ, \dots) \end{aligned}

where

θ

is the scattering angle and we take all the momenta to be outgoing so

\sum_{i = 1}^{4} p_{i}^{μ} = 0

. Then

\begin{aligned} {(p_{1} + p_{2})}^{2} & = - \frac{s}{4} (2, 0, 0)^{2} = - s \\ t & = - {(p_{1} + p_{4})}^{2} = - \frac{s}{4} (0, 1 - \cos θ, - \sin θ)^{2} \\ = - \frac{s}{4} ((1 - \cos θ)^{2} + \sin^{2} θ) \\ = - \frac{s}{2} (1 - \cos θ) \\ u & = - \frac{16}{α^{'}} - s - t = - \frac{16}{α^{'}} - \frac{s}{2} (1 + \cos θ) \sim - \frac{s}{2} (1 + \cos θ) \end{aligned}

Plugging this into the VS amplitude then gives a function of

s

and

θ

. Taking

s \to \infty

holding

θ

fixed and using Stirling's approximation

Γ (x) \sim \sqrt{2 π x} (x / e)^{x}

, I find that

lim_{s \to \infty} \ln A_{4} = - α^{'} s F (θ)

where

F (θ) = \frac{1}{8} [\ln 16 - (1 - \cos θ) \ln (1 - \cos θ)^{2} - (1 + \cos θ) \ln (1 + \cos θ)^{2}]

see Mathematica for more details. Since

F (θ) ⩾ 0

we see that the amplitude is exponentially suppressed at high energies:

lim_{s \to \infty} A_{4} = e^{- α^{'} s F (θ)}

Hence, string theory cures the UV divergences of Einstein gravity!

Superstrings Term 1

Contents

1 Introduction

2 Basics

2.1 Special Relativity

2.2 Field theory

2.2.1 Klein-Gordon

2.2.2 Maxwell

2.3 General Relativity

3 Relativistic Point Particle

4 Nambu - Goto Action

5 Polyakov Action

5.1 Symmetries and Gauge-fixing

5.2 Equations of motion and boundary conditions

5.3 Conserved Charges

Noether's theorem

Corollary

Proof

6 Mode expansion

7 Virasoro algebra

8 Light cone gauge

9 Canonical Quantisation

Proof:

10 Spectrum and Critical Dimension

Open string spectrum

Closed string spectrum

11 CFT: Virasoro from Stress Tensor

Proof:

Cauchy's Theorem

12 Operator Product Expansion

Examples

13 Critical Dimension from CFT

Ghost fields

14 Vertex operators

Examples

15 String interactions

Poles and Residues

High energy limit