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Msci Theoretical Physics Project

A Short Survey of Concepts inString Theory

Candidate:Hassan Mirza 090270709

Supervisor:Prof. David Berman

April 2, 2015

ABSTRACT

The classical non-relativistic string is analysed by picturing it as (a) a col-lection of infinite beads and (b) a collection of point masses connected bysprings. The first picture results in the Klein-Gordon Lagrangian whereasthe second results in the introduction of creation and annihilation operatorsthrough quantisation of its energy spectrum. The relativistic string actionis then found and its gauge invariance is shown in detail. Analysing thestatic gauge shows that the string end points move transverse to the lengthwith velocity v = c. The light cone gauge allows for the mode expansionof the general solution to the wave equation resulting from specific τ and σparameterisations. Finally, the operator formalism is introduced and usedto construct photon and graviton states.

1

Contents

1 Elementary concepts 51.1 Vector analysis . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 The dot product . . . . . . . . . . . . . . . . . . . . . 51.1.2 The cross product . . . . . . . . . . . . . . . . . . . . 51.1.3 Divergence theorem . . . . . . . . . . . . . . . . . . . 6

1.2 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Integral change of variables . . . . . . . . . . . . . . . 61.2.2 Fourier decomposition . . . . . . . . . . . . . . . . . . 71.2.3 Fourier transform . . . . . . . . . . . . . . . . . . . . . 71.2.4 Fourier series expansion . . . . . . . . . . . . . . . . . 7

1.3 Classical mechanics . . . . . . . . . . . . . . . . . . . . . 81.3.1 Hamilton’s principle of least action . . . . . . . . . . . 81.3.2 Euler-Lagrange equations for classical systems . . . . 8

1.4 Special relativity . . . . . . . . . . . . . . . . . . . . . . . 91.4.1 Geometry and the flat metric . . . . . . . . . . . . . . 91.4.2 Covariant and contravariant vectors . . . . . . . . . . 101.4.3 Spacetime separations . . . . . . . . . . . . . . . . . . 101.4.4 Lorentz transformations . . . . . . . . . . . . . . . . . 11

1.5 Field theory . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5.1 When do we need field theory? . . . . . . . . . . . . . 131.5.2 Deriving the field equations of motion . . . . . . . . . 131.5.3 Example: a necklace of beads . . . . . . . . . . . . . . 151.5.4 Noether’s theorem . . . . . . . . . . . . . . . . . . . . 17

1.6 The non-relativistic string . . . . . . . . . . . . . . . . 171.6.1 Wave equation from the total energy . . . . . . . . . . 181.6.2 Boundary conditions . . . . . . . . . . . . . . . . . . . 18

1.7 Quantising the non-relativistic string . . . . . . . . . 191.7.1 Fourier-analysing the total energy . . . . . . . . . . . 191.7.2 Creation and annihilation operators . . . . . . . . . . 20

2 The classical relativistic string 222.1 The string action . . . . . . . . . . . . . . . . . . . . . . . 22

2.1.1 World-sheets . . . . . . . . . . . . . . . . . . . . . . . 222.1.2 The area functional . . . . . . . . . . . . . . . . . . . 222.1.3 The Nambu-Gotto action . . . . . . . . . . . . . . . . 242.1.4 Reparameterisation invariance of the action . . . . . . 24

2.2 String equations of motion . . . . . . . . . . . . . . . . 262.2.1 Method 1: using the Euler-Lagrange equations . . . . 262.2.2 Method 2: extremising the action . . . . . . . . . . . . 27

2.3 Boundary conditions and branes . . . . . . . . . . . . 282.3.1 Free end points: Neumann boundary . . . . . . . . . . 282.3.2 Fixed end points: Dirichlet boundary . . . . . . . . . 28

2

CONTENTS

3 Gauge fixing the relativistic string 303.1 The gauge conditions . . . . . . . . . . . . . . . . . . . . 30

3.1.1 Parameterising τ . . . . . . . . . . . . . . . . . . . . . 303.2 The static gauge . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.1 Action in terms of string velocity . . . . . . . . . . . . 323.2.2 Motion of the open string end points . . . . . . . . . . 333.2.3 Analysing the string equations of motion . . . . . . . 353.2.4 The wave equation and its constraints . . . . . . . . . 363.2.5 Open string motion . . . . . . . . . . . . . . . . . . . 383.2.6 Closed string motion . . . . . . . . . . . . . . . . . . . 40

3.3 The light cone gauge . . . . . . . . . . . . . . . . . . . . 413.3.1 Parametrising σ . . . . . . . . . . . . . . . . . . . . . 413.3.2 Constraints and the wave equation . . . . . . . . . . . 443.3.3 Applying the constraints . . . . . . . . . . . . . . . . . 453.3.4 Mode expanding the general solution . . . . . . . . . . 463.3.5 Light cone solution . . . . . . . . . . . . . . . . . . . . 48

4 Scalar fields and particle states 524.1 Classical scalar fields . . . . . . . . . . . . . . . . . . . . 52

4.1.1 The Klein-Gordon equations of motion . . . . . . . . . 524.1.2 Plane wave solutions . . . . . . . . . . . . . . . . . . . 53

4.2 Quantum scalar fields . . . . . . . . . . . . . . . . . . . . 544.2.1 Quantising the solution . . . . . . . . . . . . . . . . . 544.2.2 The Hamiltonian and spacetime momentum . . . . . . 574.2.3 Quantising the solution . . . . . . . . . . . . . . . . . 59

4.3 Building particle states . . . . . . . . . . . . . . . . . . 614.3.1 Example 1: photon states . . . . . . . . . . . . . . . . 614.3.2 Example 2: graviton states . . . . . . . . . . . . . . . 64

4.4 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.4.1 The non-relativistic string . . . . . . . . . . . . . . . . 674.4.2 The relativistic string . . . . . . . . . . . . . . . . . . 684.4.3 The static gauge . . . . . . . . . . . . . . . . . . . . . 684.4.4 The light cone gauge . . . . . . . . . . . . . . . . . . . 694.4.5 Particle states . . . . . . . . . . . . . . . . . . . . . . 69

References 70

3

INTRODUCTION

Following Barton Zwiebach’s book A First Course in String Theory (2ndEdition), this text examines basic concepts in non-relativistic and relativisticstring theory, focusing more on determining the properties of strings math-ematically using the power of gauge invariance. The difference between thisand A First Course in String Theory is the choice of signature for the flatspacetime metric. The text is structured as follows:

• Chapter 1 introduces some basic mathematical background betweenSection 1.1 through to Section 1.4. Section 1.5 introduces scalar fieldtheory by bridging the gap between classical and field mechanics usingthe example of a necklace of beads. This leads onto the theory of non-relativistic strings, which is examined by modelling such a string as acollection of point masses connected by springs.

• Chapter 2 begins the analysis of the relativistic string by developingits action through the area functional of the world sheet. It followsSections 6.1 up to 6.6 of A First Course in String Theory in order todemonstrate the gauge invariance of the string action and how bound-ary conditions give rise to branes.

• Chapter 3 focuses on the static and light cone gauges. It begins byfollowing Section 7.1 of A First Course in String Theory to look atparameterising the time on the world sheet τ . It then examines thestatic gauge first, following Section 6.6 through to 6.9. The light conegauge is then analysed following Section 7.2 through to the end ofChapter 9.

• Chapter 4 follows Chapter 10 of A First Course in String Theory,developing the operator formalism by mode expanding the wave equa-tion arising from the light cone gauge in the previous chapter. Thephoton and graviton particle states are constructed in the languageof creation and annihilation operators, followed by a conclusion andsummary of the findings in the text.

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CHAPTER

1ELEMENTARY CONCEPTS

In this chapter we take a detailed look at some of themathematical tools and concepts in physics that will beutilised and built upon throughout the remainder of thetext.

1.1 Vector analysis

1.1.1 The dot product

The scalar product of two vector quantities is a commutative operationwhich gives the projection of one vector along the other:

A ·B = |A||B| cos θ = |B||A| cos θ. (1.1)

In component form (within flat Euclidean geometry), this is equivalent to:

A ·B = A1B1 +A2B2 +A3B3. (1.2)

Figure 1.1: The projection of a vector A along another vector B.

1.1.2 The cross product

The cross product of two 3-dimensional vectors A and B is given by

A×B =

∣∣∣∣∣∣i j kA1 A2 A3

B1 B2 B3

∣∣∣∣∣∣= i(A2B3 −A3B2)− j(A1B3 −A3B1) + k(A1B2 −B1A2). (1.3)

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Chapter 1. Elementary concepts

When only a single component of a vector product is of interest - say, theith component - one can employ the Levi-Civita tensor εijk as follows:

(A×B)i = εijkAjBk. (1.4)

Some useful properties of the Levi-Civita tensor (in addition to the fact thatε123 = 1):

• it is symmetric (unchanged) under cyclic permutations of its indices:εijk = εkij = εjki

• it is antisymmetric under swapping of any two of its indices: εijk =−εjik = −εikj

• it vanishes if any two of its indices are equal

• the product of two Levi-Civita tensors, εijkεlmn, is given by the deter-minant ∣∣∣∣∣∣

δil δim δinδjl δjm δjnδkl δkm δkn

∣∣∣∣∣∣ . (1.5)

1.1.3 Divergence theorem

This equates a closed-surface integral to one over the volume that the surfaceencloses. This holds as long as the vector field A is continuous and first orderdifferentiable. In other words, the flux of a vector field A through a closedsurface S can be converted to a volume integral by integrating the divergenceof A over the volume V enclosed by S:∮

SA · da =

∫V

(~∇ ·A) dV. (1.6)

n-dimensional generalisation

In most cases to follow, we will need to use the divergence theorem in spacesof higher than three-dimensions. In such cases, the d-dimensional divergenceof a vector field X in n-dimensional space, integrated over a certain volumeV in that space, may be converted to an integral of the d-dimensional fluxof X through the surface ∂V = S (with outbound normal da) which is aboundary to V : ∫

V

∂Xi

∂xidV =

∮∂V

X · da =

∮∂VXia

idS, (1.7)

where the index i ∈ {1, 2 . . . d}.

1.2 Calculus

1.2.1 Integral change of variables

Suppose we wish to change from one coordinate system (x0, x1), to another(ξ0, ξ1), such that the latter is a mapping of the former. If the startingintegral in x-space has the form

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I =

∮dx dy f(x0, x1), (1.8)

then provided that the mapping ξi = ξi(x0, x1) where i ∈ {0, 1} exists andis well-behaved1, the integral in the ξ-space becomes

I =

∮g(ξ0, ξ1) J dξ0dξ1, (1.9)

where g(ξ0, ξ1) is found by eliminating the xi in favour of the ξi. In theabove equation, J is the Jacobian determinant [6, p. 204] and is defined as

J = det(Jij) = det

(∂xi

∂ξj

)=

∣∣∣∣∣∣∣∣∂x0

∂ξ0

∂x0

∂ξ1

∂x1

∂ξ0

∂x1

∂ξ1

∣∣∣∣∣∣∣∣ . (1.10)

1.2.2 Fourier decomposition

Dirchlet boundary

A function X(σ) which satisfies the Dirchlet boundary condition (i.e. itvanishes at specified end points) can be written as a sum of sines:

X(σ) =∞∑n=1

Xn sinnσ. (1.11)

Neumann boundary

Similarly, a function X(σ) which satisfies the Neumann boundary condition(i.e. it’s derivative with respect to the argument vanishes at specified endpoints) can be written as a sum of cosines:

X(σ) =

∞∑n=0

Xn cosnσ. (1.12)

1.2.3 Fourier transform

A function f(t) can be Fourier transformed by means of the following equa-tion [1, p. 965]:

g(ω) =1√2π

∫ +∞

−∞dt f(t)eiωt. (1.13)

1.2.4 Fourier series expansion

For a periodic function fµ(x), its Fourier series expansion in terms of oscil-lation modes is given by

fµ(x) = fµ0 +

∞∑n=1

(aµn cosnu+ bµn sinnu) . (1.14)

1Well-behaved means that the function is single-valued and first order differentiable.

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1.3 Classical mechanics

Studying how objects move in space (and in spacetime) is central to ourgoal. Here we will go through a simple derivation of the Euler-Lagrangeequations for classical systems having finite degrees of freedom. This willthen be generalised to cases where the number of degrees of freedom diverges,which give way to the introduction of fields.

1.3.1 Hamilton’s principle of least action

Hamilton’s principle states that the laws which govern the motion of a sys-tem are defined by its Lagrangian, L = T − V ; the difference between itskinetic and potential energies. The motion of a given system between twopoints in time t1 < t2 is then given by an equation which makes the lineintegral of L between the times t1 and t2 stationary.

Mathematically, Hamilton’s principle states that, upon extremising theaction

S =

∫ t2

t1

L(qi, qi, t) dt, (1.15)

one obtains the equations of motion for the system described by the La-grangian. Thus, the equations of motion are given by a functional satisfyingδS = 0.

1.3.2 Euler-Lagrange equations for classical systems

In this derivation we will consider one degree of freedom, parameterised bythe generalised coordinate q = x. Since we require δS = 0, we wish to find

δS =

∫ t2

t1

δL(x, x, t) dt.

The infinitesimal change δL due to small variations of its input variablesδx, δx, δt is given by

δL =∂L

∂xδx+

∂L

∂xδx+

∂L

∂tδt. (1.16)

We only consider variations of position and velocity within the time limitst1 < t2, so we apply the constraints

δx(t1) = δx(t2) = 0;

δx(t1) = δx(t2) = 0.

We can now write δS = 0 as

δS =

∫ t2

t1

δL dt

=

∫ t2

t1

(∂L

∂xδx+

∂L

∂xδx

)dt

=

∫ t2

t1

∂L

∂xδx dt+

∫ t2

t1

∂L

∂xδx dt

= 0

8


after which it becomes possible to integrate the second term in the thirdline by parts: ∫ t2

t1

∂L

∂xδx dt =

∂L

∂xδx

∣∣∣∣t2t1

−∫ t2

t1

d

dt

∂L

∂xδx dt

= 0−∫ t2

t1

d

dt

∂L

∂xδx dt, (1.17)

where the first term on the right-hand side of the first line vanishes as itrequires the evaluation of δx at t1 and t2 - which is precisely where we haveset it to be 0 in our second constraint. Substituting (1.17) into δS gives

δS =

∫ t2

t1

∂L

∂xδx dt−

∫ t2

t1

d

dt

∂L

∂xδx dt

=

∫ t2

t1

(∂L

∂x− d

dt

∂L

∂x

)δx dt = 0. (1.18)

As t is a positive quantity, and since the integrand does not contain anoscillating function, the vanishing integral implies that the integrand itselfshould also vanish. Hence, we obtain exactly what were looking for:

d

dt

∂L

∂x=∂L

∂x, (1.19)

which is the Euler-Lagrange equation for one generalised coordinate x. Thequantity ∂L/∂x is known as the momentum conjugate px [3, p. 55], suchthat

px =∂L

∂x. (1.20)

1.4 Special relativity

Before moving onto fields, we will spend some time establishing a firmground in special relativity. We will discuss some concepts and mathematicsbehind spacetime - the four-dimensional continuum which unites the threespatial dimensions of Euclidean space with the single dimension of time.

1.4.1 Geometry and the flat metric

In relativity we study the geometry of spacetime and the behaviour of objectsdue to its structure. Geometry here refers to the surface upon which thetrajectories of objects are traced out. Clearly, it can be either curved (asin general relativity (GR)) or flat (as in special relativity (SR)). Just likewe study distances between points along an object’s trajectory in classicalmechanics, we are required to do the same in relativity. A key player inthis is a certain matrix known as the metric. The metric associated with aspacetime geometry describes its curvature locally.

Since our analyses will mostly involve flat spacetime, the metric we areinterested in for this geometry (known as the Minkowski metric), with ourchoice of signature (+,−,−,−), is

ηµν = diag(1,−1,−1,−1). (1.21)

Note that ηµν = ηµν – i.e. the flat metric is its own inverse, such that

9


ηµνηνλ = δµλ. (1.22)

The other choice of metric signature diag(−1, 1, 1, 1) can also be used and issometimes more convenient. The main function of the metric is to enable theraising and lowering of indices in tensor equations, as will be done numeroustimes throughout the text – so examples will be plentiful.

1.4.2 Covariant and contravariant vectors

Vectors in four-dimensional spacetime are called four-vectors, with compo-nents labelled by Greek indices. Covariant vectors are those with downstairsindices whereas contravariant vectors have upstairs indices.1 The differencebetween covariant and contravariant vectors is a sign reversal of spatial com-ponents. Taking the four-position vector as an example, we have:

xµ = (ct,−~x) ; xµ = (ct, ~x), (1.23)

which are covariant and contravariant vectors respectively. These are relatedthrough the flat metric ηµν as

xµ = ηµνxν . (1.24)

1.4.3 Spacetime separations

Intervals and causality

Suppose two observers measure the spacetime positions of a pair of events.In general, they will not agree on each other’s measurements since they areboth in different frames of reference. However, the quantity they will agreeon is known as the interval, which is the spacetime separation between theevents. The interval ∆S is defined as

∆S = (c∆t)2 − (∆~x)2 = ηµν∆xµ∆xν . (1.25)

∆S can be either 0, positive or negative. This reflects the type of separationbetween the events, as summarised below:

• ∆S > 0 : interval is timelike. Events that are timelike separatedhave a causal connection. In other words, there exists a frame ofreference in which both events happen at the same point in space butat different times. This is because, within the time interval betweenthe occurrences of both events, light will travel a greater distance thanthat which separates the events.

• ∆S = 0 : interval is lightlike. Lightlike separated events also have acausal connection. However they can only be linked by a light ray andnot any other object traveling at speed v < c.

• ∆S < 0 : interval is spacelike. Spacelike intervals are not causallyconnected - i.e. there exists a frame wherein the two events happen

1Latin indices are reserved for three-dimensional Euclidean space and are thus ∈{1, 2, 3}. Greek indices are reserved for four-dimensional spacetime, and as such theyare ∈ {0, 1, 2, 3} where the 0th index represents time.

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at the same time but at different positions in space. The time inter-val between both events is not large enough for light to traverse thedistance between them.

Figure 1.2: Identifying causal relationships using a light cone. As A lieswithin the O’s light cone, O and A are causally connected. This is not truefor O and B as B lies outside the light cone.

Light cones

Following the discussion above, we can picture the different signs of ∆Sthrough the aid of a spacetime diagram [4, p. 58]. These are constructedby designating one axis as the coordinate c×time. As an example, we maychoose one spatial dimension x and then designate the line x = 0 as ctinstead of y. In such a coordinate system, the line x = ct is the path a rayof light would take. This is a trivial consequence of x = ct as it implies thatthe velocity x = v = c.

The diagram we are interested in is formed by revolving the x = ct lineby 2π about the ct axis. The resulting shape - a light cone - helps identifythe causal connection between events by allowing us to visualise the differenttypes of spacetime separations, as shown in Figure 1.2.

1.4.4 Lorentz transformations

Lorentz transformations describe the relationship between the coordinatesof an event as measured in two frames which are in relative motion withrespect to one another. Suppose a frame S′, with coordinates (x′, y′, z′, t′),is moving in the positive x-direction with respect to another frame S, withcoordinates (x, y, z, t). The relationship between the four-position vector ofan event measured in S′ and that in S is then expressed as

xµ → x′µ = Λµνxν , (1.26)

where Λµν is a Lorentz transformation matrix whose components are deter-mined by the type of transformation in question.

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Lorentz invariance

We are interested only in the Lorentz transformations which leave the inter-val (1.25) invariant. The set of matrices Λµν obeying this rule is known asthe Lorentz group [5, p. 16]. The invariance of the interval, ∆S′ = ∆S,under such a Lorentz transformation, can be written as

ηµν∆x′µ∆x′ν = ηµν∆xµ∆xν . (1.27)

Clearly, (1.27) must impose a constraint on the matrix elements of Λµν inorder for it to be true. We can find the constraint as follows. As a start,let us insert (1.26) into the left-hand side of (1.27) to remove the primedcoordinates. Then, with a bit of index-massaging, one obtains

ηµν∆x′µ∆x′ν = ηρσ(Λρµ∆xµ)(Λσν∆xν).

However, the right-hand side of the above must equal that of (1.27):

ηρσ(Λρµ∆xµ)(Λσν∆xν) = ηµν∆xµ∆xν .

Finally, cancelling the coordinates through gives

ηµν = ηρσΛρµΛσν , (1.28)

which can be written in a more friendly manner using matrix notation.As the transpose of a product of matrices (AB)T = BTAT , we can write(gρσΛρµ)T = (ΛT ) ρ

µ gσρ. However, since gρσ = gσρ, we obtain

ηµν = (ΛT ) ρµ ηρσΛσν ,

which we can write in matrix form as

(ΛT )gΛ = g. (1.29)

Equations (1.28) and (1.29) show us the condition that a Lorentz transfor-mation must obey in order for it to leave the interval (1.25) invariant.

1.5 Field theory

Following from our discussion of systems with finite degrees of freedom inSection 1.3, the next step is to develop a similar analysis for systems withinwhich the number of degrees of freedom is infinite. Such systems are knownas fields.

Our treatment of field mechanics will mirror the previous analysis inSection 1.3 for classical systems - we write down the Lagrangian (density),formulate a generic action, and then extremise the action to determine thefield equations of motion. We will then go a step further and write downNoether’s theorem; which will be used later on to derive conserved currentsand charges.

12


1.5.1 When do we need field theory?

Before delving into the mathematics, let us take a moment to understandwhy exactly we need fields.

The need for another framework of analytical mechanics arises when wefind ourselves in either of the two following cases:

1. when our system has many particles (many degrees of freedom), and/or

2. when it has a varying number of particles.

In both of these, we see that the classical Lagrangian treatment is simply in-adequate. Not only must we specify each degree of freedom individually, wemust also have the number of particles in the theory fixed. The Lagrangianmust change if we add or remove particles. In a situation where particlescan be created and annihilated – a more than recurring theme in QED – theformalism breaks down.

Field theory provides a framework to deal with situations where thenumber of degrees of freedom diverges. By doing so, it automatically solvesthe two issues above. Indeed, if a Lagrangian is describing an infinite numberof particles, whether we choose to keep that number fixed, add 10 or subtract1000 from it, the theory will be unaffected.

1.5.2 Deriving the field equations of motion

Now that we know why we need field theory, we can begin understandingits formalism by analogy with classical mechanics.

The Lagrangian density L

In field theory, we construct actions using a Lagrangian density as opposedto a Lagrangian. The relationship between the Lagrangian and its densityin a three-dimensional theory is given by

L =

∫d3x L. (1.30)

However, conforming to the standard literary convention adopted by physi-cists, the term ‘density’ is implied through context, so it needn’t be men-tioned henceforth.

The next step is to establish the functional dependence of L. Takingthe general case where L may contain interaction terms, we can write thegeneric form of the Lagrangian as

L = L(ϕ, ∂µϕ), (1.31)

where ϕ is the field. The reason why (1.31) depends on ∂µϕ and not simplythe time-derivative of the field is to guarantee Lorentz invariance.

The action

Having found the functional dependence of L, writing the action is trivial.We simply integrate L over space to obtain L and then integrate the resultover time to obtain S [5, p. 43]. Thus we find that

13


S =

∫dt L =

∫dt

∫d3x L

=

∫d4x L. (1.32)

The field Euler-Lagrange equations

The Euler-Lagrange equations for fields are found in a way quite similar tothe classical equations of motion – we find a functional which extremises theaction (1.32):

δS =

∫d4x δL =

∫d4x

(∂L∂ϕ

δϕ+∂L

∂(∂µϕ)δ(∂µϕ)

)= 0.

Note that δ(∂µϕ) = ∂µ(δϕ) – varying the derivative is the same as differen-tiating the variation.

There is a neat trick we can use to put the second term in the parenthesesof the above equation in a nicer form. Employing differentiation by parts:(∂u)v = ∂(uv)− u(∂v), we can set

u =∂L

∂(∂µϕ), v = ∂ϕ,

so that

∂L∂(∂µϕ)

∂µ(δϕ) = ∂µ

(∂L

∂(∂µϕ)δϕ

)− ∂µ

∂L∂(∂µϕ)

∂ϕ.

Inserting this into δS gives

δS =

∫d4x

(∂L∂ϕ− ∂µ

∂L∂(∂µϕ)

)δϕ+

∫d4x ∂µ

(∂L

∂(∂µϕ)δϕ

). (1.33)

To make our lives simpler, let us say that the second integral on the right-hand side of (1.33) is

∫d4x ∂µX

µ. Then, using (1.7), we see that∫d4x ∂µX

µ =

∫∂µn

µ dS. (1.34)

Just as in the case of classical equations of motion (where the variation isset to vanish at the path end points), the analogous constraint for fields isthat they vanish at ∞. Since the right-hand side of (1.34) is the flux ofthe field through an arbitrary surface S of our choice, we will convenientlychoose for its boundary to be infinitely far away from the fields; so that theflux through it will vanish. Thus the entire equation (1.34) vanishes, andwe are left with

δS =

∫d4x

(∂L∂ϕ− ∂µ

∂L∂(∂µϕ)

)δϕ = 0,

from which the equations of motion are simply read off as being [5, p. 45]

∂µ∂L

∂(∂µϕ)=∂L∂ϕ

. (1.35)

14


1.5.3 Example: a necklace of beads

Let us now analyse an example to make sense of field mechanics. We willsee that, through our efforts, an absolutely crucial type of Lagrangian - theKlein-Gordon Lagrangian - will emerge nicely.

Consider a closed circular necklace of beads, as shown in figure 1.3.Initially the system has a finite number of beads, each of which has itsposition parameterised by one degree of freedom qi(t) and is separated fromits nearest neighbours by a distance a. (Note that, since the necklace iscircular, qN+1 = q1.) The total number of beads, N , is then increased toreach ∞ where the necklace effectively becomes a continuum of material.At this stage the system can be thought of as a string. Assuming each beadhas the same mass m, the classical Lagrangian will take the form

L =1

2

N∑i=1

mq2i − V (q1, q2, . . . qN ). (1.36)

The next step is to establish the potential V in (1.36). We will approachthis by considering a potential made of two separate sources: one due tothe particles interacting with only their nearest neighbours, and the otherdue to an external potential (a gravitational one, for example). These twopotentials, Vint and Vext, can be written as

Vint =1

2

N∑i=1

p(qi+1 − qi)2 , Vext =1

2

N∑i=1

hq2i . (1.37)

At this point one can certainly ask why the interaction potential Vint hasthe form shown in (1.37). In a conservative system (where the potentialdepends only on the generalised coordinates), provided that the coordinatesdo not involve explicit time dependence, the potential V due to interactingparticles generally has the form

∑a,b kabqaqb +

∑a,b,c labcqaqbqc + . . . where

kab =(

∂2V∂qa∂qb

) ∣∣∣0, as discussed in detail by Goldstein [3, p. 240]. However in

most cases of interest, only the harmonic approximation is needed; meaningthat terms of O(q3) and beyond may be neglected.

Setting V = Vint + Vext and substituting (1.37) into (1.36), we obtain

L =1

2

N∑i=1

mq2i −

1

2

N∑i=1

[hq2i − p(qi+1 − qi)2], (1.38)

where p and q are arbitrary parameters that will change depending on thephysical properties of the string.

At this point we are ready to transition from the classical Lagrangianin (1.38) to one containing fields! Although the procedure is quite ad-hoc,knowing that the Klein-Gordon Lagrangian is waiting for us at the endmakes it all worthwhile. However, before moving forward, some definitionsneed to be made which will be explained in due course.

As it stands, (1.38) has no explicit dependence on a. A good start wouldbe to introduce it into the game, which can be done easily by defining

h = σa, (1.39)

where σ is an arbitrary proportionality constant. Let us also set the quantityp equal to tension per unit length:

15


Figure 1.3: The necklace for increasing values of N from left to right. Wecan approximate a continuous string in the limit N →∞.

p =T

a. (1.40)

Finally, we define a mass per unit length: ρ = ma , where the unit of length

has been chosen to equal a – again for reasons that will follow. Thesedefinitions introduce a into each term within the Lagrangian, which meanswe can factor it out as an overall constant. Rewriting (1.38) then results in

L =a

2

[N∑i=1

ρq2i −

N∑i=1

(σq2

i − T(qi+1 − qi)2

a2

)]. (1.41)

Now we can justify the definitions made above. As we wish to go fromclassical mechanics to fields, we want our number of beads N to be verylarge. The condition that N → ∞ also brings with it other changes thatmust be made in (1.41). Essentially, if N →∞, then:

• the distance between neighbouring beads a tends to 0. This must betrue since, by definition, there is no longer any gap between the beadsonce we set N to diverge

• the mass is no longer localised in the form of beads. It becomes dis-tributed over the length of the string - so it becomes a mass density.Thus we must also set m→ ρa, as defined above

• the index i which labels the particle number in qi is no longer relevant;we do not have a discrete number of particles now. This means that thediscrete index i must be transformed to a continuous variable x uponwhich the position q will now depend along with time: qi(t)→ q(t, x).

To justify (1.40), let us pay attention to the term in (1.41) for which T is aconstant coefficient. This is just the square of the definition of the derivative:

(qi+1 − qi)2

a2=

(∂q(t, x)

∂x

)2

, (1.42)

which is identical to what we normally find in the one-dimensional waveequation for a string, wherein the prefactor is also set to be tension up to aconstant. This justifies the placement of T in (1.41).

Now we must transform the sum over i into an integral over x, so we seta∑

i →∫dx. Using this, (1.41) becomes

16


L =1

2

∫dx

[ρ

(∂q(t, x)

∂t

)2

− σq2(t, x)− T(∂q(t, x)

∂x

)2]. (1.43)

Further analogy with the wave equation – where the double spatial derivativeappears with the prefactor ρc2 = T – motivates us to set ρ = T in naturalunits of c = 1:

L =T

2

∫dx

[{(∂q(t, x)

∂t

)2

−(∂q(t, x)

∂x

)2}− σ

Tq2(t, x)

]. (1.44)

Now the final step arrives! Define the field φ(t, x) ≡ q(t, x) and noting thatL =

∫dnxL, we see that (1.44) becomes the Klein-Gordon Lagrangian

for a (1+1)-dimensional theory:

L =1

2

{(∂φ(x0, x1)

∂x0

)2

−(∂φ(x0, x1)

∂x1

)2

−m2φ2(x0, x1)

}(1.45)

up to overall constants which we can engineer to our choosing. This caneasily be generalised to a (1+3)-dimensional theory:

L =1

2

[(∂µφ)(∂νφ)ηµν −m2φ2

]; (1.46)

which is a central equation in non-interacting scalar field theory and will beof great use to us in the coming chapters.

1.5.4 Noether’s theorem

For a continuous global transformation, we can write down the condition forit to be a symmetry of the Lagrangian:

δJµ =∂L

∂(∂µϕ)δϕ+ Lδxµ. (1.47)

Equation (1.47) is known as the Noether current. The correspondingNoether charge is given by

Q =

∫d3x δJ0(x, t). (1.48)

If for a given transformation the time-derivative of (1.48) vanishes, we saythat it is a conserved charge. That is to say that

Q = 0 (1.49)

is the condition for charge conservation.

1.6 The non-relativistic string

Before moving onto relativistic strings, let us take a quick tour of nonrel-ativistic string dynamics, focusing on the wave equation which we will bemeeting in different forms in later chapters. We will then quantise this stringin order to introduce the creation-annihilation operator formalism.

17


1.6.1 Wave equation from the total energy

In order to derive the wave equation, we imagine the string to be a collectionof N point masses connected by N − 1 springs, each of stiffness κ, as shownin figure 1.4.

In string theory, the standard convention is to set the parameter alongthe length of the string equal to σ, which ranges from 0 to π. This is apurely arbitrary choice, but a clever one nevertheless – it simplifies somecalculations we will be doing later. If the “wiggling” of the string is con-strained to be in the plane of the paper (the x−y plane, say), then the totalenergy of the system can be written as

E =N∑i=1

m

2

(x2i + y2

i

)+κ

2

(∆x2

i + ∆y2i

), (1.50)

assuming each point has a mass m. We can now simplify the above bymaking some considerations based on elementary calculus. Firstly, we knowthat, for a function x(σ) (the same being true for y(σ)):

∆x =∂x

∂σ∆σ, (1.51)

which can be simplified further by noting that ∆σ = π/N and setting m =1/N up to a constant:

E =1

2π

N∑i=1

∆σ(x2i + y2

i

)+

1

2π

N∑i=1

1

∆σ

(∆x2

i + ∆y2i

). (1.52)

We now introduce another coordinate τ which parameterises time as mea-sured on the string. If we allow the time derivatives in (1.52) to be withrespect to τ , then, by absorbing the x and y coordinates into a vector ~x,(1.52) becomes

E =1

2π

∫ π

0dσ

[(∂~x

∂τ

)2

+

(∂~x

∂σ

)2], (1.53)

where one can see that the integrand encloses the 2-D wave equation up toconstants [1, p. 435].

1.6.2 Boundary conditions

Boundary conditions are important to discuss as they will be needed whenwe look at relativistic strings in the next chapter. The concept behindboundary conditions is simple. It arises from the fact that the end points ofa wave can either be fixed or free to move.

Neumann (free ends)

Let us take a quick look at how Neumann boundary conditions are useful.Referring to figure 1.4, the force that the Nth point particle experiences isequal to κ∆x. With our choice of constants, this becomes

F = mx =N

π2

∂x

∂σ∆σ =

1

π

∂x

∂σ. (1.54)

18


Figure 1.4: The string is considered to be an assembly of N point massesconnected by N − 1 springs of stiffness κ. Positions along the string rangefrom σ = 0 to σ = π.

Noting that m = 1/N and rearranging for F gives

x =N

π

∂x

∂σ. (1.55)

Here we see something interesting. The string is an infinite number of pointmasses; meaning that N →∞. However, looking at (1.55), we can immedi-ately tell that such a condition will mean that the acceleration of the Nthpoint mass is infinite! To prevent this we impose the Neumann boundarycondition:

∂x

∂σ= 0, (1.56)

which corresponds to a free end point of the string. The derivative vanishingsimply means that the slope of the wave at the end point is zero.

Dirichlet (fixed ends)

Although not entirely relevant in this context, the other type of boundarycondition is, of course, one in which the end points are fixed. This is theDirichlet boundary condition, given by

x(σ) = 0. (1.57)

1.7 Quantising the non-relativistic string

Quantising the string means that we transform its continuous energy spec-trum into one which is discrete. This is done by means of the Fourierdecomposition formulae we met in Section 1.2.

1.7.1 Fourier-analysing the total energy

Our aim here is to transform (1.53) using (1.12). The choice of using (1.12)is due to the fact that we have imposed the Neumann boundary conditionin (1.55). However upon applying (1.12) to the first term in the integrandof (1.53), we observe that

~x(σ) =∂

∂τ

( ∞∑n=0

~xn cosnσ

)= 0, (1.58)

19


since there is no explicit time dependence on τ in (1.12). The only way toovercome this is to set the Fourier coefficients xn = xn(τ). Then we obtain

~x(σ) =

∞∑n=0

~xn cosnσ. (1.59)

Squaring this gives

~x2(σ) =∞∑

n,m=0

~xn · ~xm cosnσ cosmσ. (1.60)

The next term in (1.53) is a σ-derivative, which we can write as

∂~x

∂σ= −

∞∑n=0

n~xn sinnσ −→(∂~x

∂σ

)2

=1

2π

∞∑n,m=0

nm~xn · ~xm sinnσ sinmσ.

(1.61)Putting all this together gives us the total energy

E =1

2π

∞∑n,m=0

~xn · ~xm∫ π

0dσ [cosnσ cosmσ + nm sinnσ sinmσ] . (1.62)

Upon analysing (1.62), it is clear that the integral is only non-trivial forwhen n = m. However, the integral over the cosines does have a finite valuewhen n = m = 0, and that value is precisely π. Thus, setting n = m, wecan evaluate the integrals in (1.62) with ease, obtaining

E =1

2~x2

0 +1

4

∞∑n=1

(~x2n + n2~x2

n

). (1.63)

The first term in the above is the velocity of the centre of mass of thesystem. Provided we leave this constant, it can be dropped without harmingthe physics of our non-relativistic string. We can also drop the summationnotation as the subscripts n already tell us that the system has differentmodes of oscillation. With these two facts in mind, (1.63) can be written as

E =1

4

(~x2n + n2~x2

n

). (1.64)

1.7.2 Creation and annihilation operators

Canonical quantisation is when we use the total energy of the system to gen-erate two operators – the creation operator a†n and the annihilation operatoran – which satisfy the commutation relation

[an, a†n] = 1. (1.65)

For the purposes of this part of the chapter, we will consider motion in thex-direction only and hence drop the vectorial notation in (1.64). The termswhich correspond to motion in other directions can be easily recovered bysimple relabelling as they are identical to what happens in the x-direction.

Before deriving the creation and annihilation operators, one must firstfind the Hamiltonian of the system. Given that we already have the totalenergy in (1.64), all we have to do is construct the Lagrangian

20


L =1

4

(x2n − n2x2

n

)(1.66)

and use it to change variables in (1.64) from velocity xn to momentum pn.This can be done by means of (1.20):

pxn = 2xn. (1.67)

substituting into (1.64) gives the Hamiltionian

H = (pxn)2 +n2x2

4=(nxn

2+ ipxn

)(nxn2− ipxn

), (1.68)

where the sum of two squares is used to change the form of the Hamiltonian.Each of the two parentheses in (1.68) enclose either the creation or the

annihilation operator – but which is which? In order to find the answer tothis, let us take a guess and impose the commutation relation in (1.65):

[(nxn2

+ ipxn

),(nxn

2− ipxn

)]=[(nxn

2+ ipxn

),nxn

2

]−[(nxn

2+ ipxn

), ipxn

]= 2

[ipxn,

nxn2

]= n

(1.69)using the fact that [x, p] = i in natural units. Comparing the above with(1.65), we notice that we are off by a factor of n. This can easily be corrected

by introducing a factor of (√n)−1 into the definitions of an and a†n. Thus

we find the creation and annihilation operators respectively as

a†n =

√nx

2− i√

npxn,

an =

√nx

2+

i√npxn.

(1.70)

The system that we have quantised in this section is simple; however thetechniques used here will be required later when we come to quantising thephoton and graviton fields.

21

CHAPTER

2THE CLASSICAL

RELATIVISTIC STRING

Here we will develop the foundations of classical relativis-tic string theory. The chapter begins with a derivation ofthe Nambu-Gotto string action, which is then used exten-sively to understand the string’s properties and conservedquantities after showing the action’s gauge invariance.

2.1 The string action

2.1.1 World-sheets

Just as particles trace out world-lines in classical mechanics, the stringswe study here trace out world-sheets, as shown in figure 2.1. A world-sheet differs from an ordinary surface because it lives in four-dimensionalspacetime as opposed to simply existing in three-dimensional space. Weparameterise the world-sheet using two coordinates: one to measure thestring’s motion through time: ξ1 = τ , and the other to measure lengthsalong the string: ξ2 = σ. Thus, a point on the world-sheet described bythe coordinates (ξ1, ξ2) = (τ, σ) is described in spacetime by the vectorXµ(τ, σ). This mapping relationship between the world sheet and spacetimecoordinates (henceforth known as string coordinates) can be expressed as

f : (world-sheet space)→ (string space) = Xµ(τ, σ). (2.1)

2.1.2 The area functional

Our first mission is to find the area of the parallelogram spanned by thevectors d~v1 and d~v2 in figure 2.1. We do this by first finding the infinitesi-mal area element and then integrating over the ranges of both world-sheetcoordinates. The area element is given by

22

Chapter 2. The classical relativistic string

Figure 2.1: On the left is the two-dimensional world-sheet space, param-eterised by world-sheet coordinates (τ, σ). On the right, the sheet is em-bedded in a four-dimensional spacetime, with string coordinates Xµ. Theparallelogram of sides d~v1, d~v2 is a projection of the rectangle of sides dτ, dσon the spacetime.

dA = |d~v1||d~v2|| sin θ|

= |d~v1||d~v2|√

1− cos2 θ

=(|d~v1|2|d~v2|2 − |d~v1|2|d~v2|2 cos2 θ

) 12

=((d~v1 · d~v1)(d~v2 · d~v2)− (d~v1 · d~v2)2

) 12 , (2.2)

where the last line is obtained using equation (1.1).We can now eliminate the explicit d~vi dependence in (2.2) using the

mapping functions Xµ(τ, σ). Provided that the vector d~vi in string spacedepends only on the corresponding dξi in the world-sheet space (so thatd~vi = f(dξi) only), we may write

d~vi =∂xµ

∂ξidξi. (2.3)

Upon inserting (2.3) into (2.2) and integrating over τ and σ, we obtain thearea functional

A =

∫dτdσ

((∂Xµ

∂τ· ∂X

µ

∂τ

)(∂Xµ

∂σ· ∂X

µ

∂σ

)−(∂Xµ

∂τ· ∂X

µ

∂σ

)2) 1

2

=

∫dτdσ

√(∂Xµ

∂τ

)2(∂Xµ

∂σ

)2

−(∂Xµ

∂τ· ∂X

µ

∂σ

)2

=

∫dτdσ

√(X)2(X ′)2 − (X ·X ′)2,

(2.4)

where we have dropped the relativistic indices as they are being summedover through the dot product. Unfortunately, though, our job isn’t finishedyet! As it stands, the quantity under the square root in (2.4) is less thanzero, which must be corrected. This can be understood by noting thatdifferent points along the length of the string must be spacelike, while twostrings separated by a time ∆τ must be timelike (or perhaps lightlike insome limit). Therefore the world-sheet must have spacelike and timelike

23


directions which are mutually perpendicular – they form a basis set on theworld-sheet. This idea is highlighted in figure 2.1.

The question now is: how do we fix the area functional in (2.4)? It turnsout that we just have to swap the order of the terms appearing in the squareroot. This can be proven by considering the solutions to the square of anarbitrary vector on the world-sheet originating from a point P . The tangentvector to P in the σ direction is ∂Xµ

∂σ ≡ vσ, and ∂Xµ

∂τ ≡ vτ in the τ direction.Thus any arbitrary vector v on the world-sheet can be found by taking alinear combination of these: v = avσ + bvτ . However, we only want to scalethe vector in the σ direction, so we may set b = 1. Then, taking the squareof v and solving for a gives

a =−vτvσ ±

√(vτ · vσ)2 − (vτ )2(vσ)2

vτ. (2.5)

Referring to figure 1.2, we know that if the interval ∆S is less than zero,it is spacelike, and timelike if it’s greater than zero. This means that oursolutions for a must be both positive and negative. In order for this to betrue, the quantity under the square root of (2.5) must be positive. Thus,the correct area functional is

A =

∫dτdσ

√(X ·X ′)2 − (X)2(X ′)2. (2.6)

2.1.3 The Nambu-Gotto action

Now we can write down the Nambu-Gotto action for the relativistic string,which is simply the integral over the ranges of τ and σ of the area functional,multiplied by a constant for dimensional consistency:

S = − 1

2πα′~c2

∫dτdσ

√−γ, (2.7)

where α′ = (2πT0~c)−1 and T0 is the string tension (more on this in Chapter3). Using natural units by setting ~ = c = 1, (2.7) becomes

S = − 1

2πα′

∫dτdσ

√−γ, (2.8)

where γ = det(γαβ), such that

γαβ =

(X2 X ·X ′

X ·X ′ X ′2

). (2.9)

The content of (2.9) describes what is known as an induced metric asit determines distances on the world-sheet. However, since the world-sheetlives within our flat spacetime described by the metric ηµν , γαβ is said to beinduced on the world-sheet.

2.1.4 Reparameterisation invariance of the action

The reparameterisation (or gauge) invariance of the action (2.8) is an im-portant fact to demonstrate. It can be done by first setting (τ, σ) = (ξ0, ξ1),and then introducing a new parameterisation (ξ0, ξ1), such that ξ0 = (ξ0, ξ1)and ξ1 = ξ1(ξ0, ξ1). The functions ξi(ξ0, ξ1) are therefore a mapping which

24


transforms the original ξ-space coordinates to their projections in the newξ-space. Using our ξ-notation, (2.8) becomes

S = − 1

2πα′

∫dξ0dξ1√−γ. (2.10)

In order to establish the gauge invariance of (2.10), we have to transformeach object under the new parameterisation and then construct a new ac-tion, S′, by putting the transformed objects together. If the form of S′

is identical to that of S in (2.10), we will have proven that the action isreparameterisation invariant.

Beginning with the integration measure dξ0dξ1, we transform this tothe measure in the new parameterisation using the Jacobian determinant in(1.10):

dξ0dξ1 = det

(∂ξi

∂ξj

)dξ0dξ1 = Jdξ0dξ1. (2.11)

Notice that, if we had a mapping from the ξ-space to the ξ-space (i.e, ifwe wanted to perform the reverse transformation to the one in (2.11)), wewould need the determinant J = |Jij |, defined as

dξ0dξ1 = det

(∂ξi

∂ξj

)dξ0dξ1 = Jdξ0dξ1, (2.12)

which, together with (2.11), tells us that JJ = 1.Having dealt with the integration measure, our second and last target is

the quantity appearing in the square-root of (2.8). This must also transformto its equivalent in the new ξ-space. To do so, let us consider a vector dXµ

which exists on the world-sheet. The interval (1.25) of this vector should beindependent of parameterisation – the lengths on the world-sheet should bethe same regardless of which coordinate system we choose to measure them.Mathematically, this means that

γµν(ξ)dξµdξν = γρσ(ξ)dξρdξσ. (2.13)

We can safely multiply and divide the right-hand side by ∂ξµ∂ξν , giving

γµν(ξ)dξµdξν = γρσ(ξ)∂ξρ∂ξσ

∂ξµ∂ξνdξµdξν , (2.14)

from which we can deduce the following equation, within which we haveinserted the Jacobian Jij twice:

γµν(ξ) = γρσ(ξ)∂ξρ∂ξσ

∂ξµ∂ξν= γρσJρµJσν = (JT )µργρσJσν . (2.15)

Taking the determinant of the above, we obtain

γ = γJ2 −→√−γ =

√−γJ . (2.16)

Now we can construct the new action, S′, using the transformed integrationmeasure in (2.11) and the equation in (2.16):

S = − 1

2πα′

∫dξ0dξ1√−γ −→ S′ = − 1

2πα′

∫Jdξ0dξ1

√−γJ ,

25


where S′ can be simplified using the fact that JJ = 1 to give

S′ = − 1

2πα′

∫dξ0dξ1

√−γ, (2.17)

which demonstrates the gauge invariance of the action in (2.8).

2.2 String equations of motion

Now that we are fully confident in the gauge invariance of the string action,our next mission is to study the behaviour of the strings. This is done byfinding the equations of motion using the action in (2.8).

It turns out that the equations of motion can be obtained in two ways.The first method is by treating the string coordinates Xµ as fields and thendirectly applying the field Euler-Lagrange equations in (1.35). The secondoption is to start from scratch and derive the Euler-Lagrange equations fora system whose Lagrangian has the form L(Xµ, Xµ′); essentially repeatingthe exercises in Section 1.4 and in Section 1.7.

Although each method will give us the same equations of motion, it willbecome apparent that the second method gives rise to boundary condi-tions, which are not explicitly demanded by the result of the first method.We will now study each method in turn.

2.2.1 Method 1: using the Euler-Lagrange equations

The application of this method is straightforward. Beginning with the La-grangian

L(Xµ, Xµ′) = − 1

2πα′√−γ = − 1

2πα′

√(Xµ ·Xµ′)2 − (Xµ)2(Xµ′)2, (2.18)

we first notice that the right-hand side of (1.35) vanishes since L dependsonly on the derivatives of the string coordinates Xµ and not on the actualcoordinates themselves. Setting the dummy index to α so that it representsthe world-sheet coordinates (i.e. α ∈ {0, 1} = (τ, σ)), the left-hand side of(1.35) becomes

∂α∂L

∂(∂αXµ)= ∂0

∂L∂(∂0Xµ)

+ ∂1∂L

∂(∂1Xµ)= 0. (2.19)

The above equation can be simplified by noting that ∂0Xµ = Xµ and

∂1Xµ = Xµ′:

∂ρ∂L

∂(∂ρXµ)= ∂0

∂L∂Xµ

+ ∂1∂L∂Xµ′ . (2.20)

If we now define two new quantities; Pτµ and Pσµ , such that

Pτµ :=∂L∂Xµ

and

Pσµ :=∂L∂Xµ′ ,

(2.21)

our equations of motion become

26


∂Pτµ∂τ

+∂Pσµ∂σ

= 0, (2.22)

which can also be written as

∂αPαµ = 0. (2.23)

As noted earlier, this method provides us with no information about thestring end points. Common sense suggests that the end points can eithermove or remain fixed. But which is true? The short answer is: both arepossible. However, there is simply no way of explaning why this is trueusing (2.23) alone.

What we are about to see in the second method is that, not only will itgive us the same equations of motion, but it will also allow the emergenceof another term. In order for this term to drop out we will need to thinkabout the string endpoints, which will be the focus of Section 2.3.

2.2.2 Method 2: extremising the action

In order to regain the lost boundary information, we must re-derive theequations of motion from first principles. This is a matter of repeating theexercise of deriving the Euler-Lagrange equations by extremising the action.

Let us start with finding the variation of the action with respect to avariation of the Lagrangian:

δS =

∫dτ

∫dσ δL(Xµ, Xµ′)

=

∫dτ

∫dσ

(∂L∂Xµ

δXµ +∂L∂Xµ′ δX

µ′)

=

∫dτ

∫dσ

(∂L∂Xµ

∂

∂τ(δXµ) +

∂L∂Xµ′

∂

∂σ(δXµ)

)=

∫dτ

∫dσ(Pτσ∂tδXµ + Pσµ∂σδXµ

). (2.24)

Now we can use differentiation by parts u∂αv = ∂α(uv) − (∂αu)v on eachterm appearing in the parentheses of (2.24) by setting α = τ for the firstterm and α = σ for the second:

Pτσ∂τδXµ = ∂τ (PτµδXµ)− (∂tPτµ)δXµ, and

Pσµ∂σδXµ = ∂σ(PσµδXµ)− (∂σPσµ )δXµ,(2.25)

which we can substitute into (2.24) to obtain

δS =

∫dτ

∫dσ(∂τ (PτµδXµ)− (∂tPτµ)δXµ + ∂σ(PσµδXµ)− (∂σPσµ )δXµ

)=

∫dτdσ

(∂τ (PτµδXµ) + ∂σ(PσµδXµ)

)−∫dτdσ

(∂τPτµ + ∂σPσµ

)δXµ.

The first term in the last line of the above contains a total derivative in τ ,which will vanish if we assume that the variation vanishes at the end points.The second term can be integrated over σ, giving us

δS =

∫dτ[δXµPσµ

]σ10−∫dτdσ

(∂αPαµ

)δXµ = 0, (2.26)

27


where the limits (0, σ1) are the end points of the string.In the above equation, we find the information missing from (2.23). Be-

fore we get to that, however, let us first state two obvious things about(2.26). The first is that, since it must be equal to zero, each term has tovanish – and by setting the second term equal to zero we immediately obtainthe equations of motion as found previously in (2.23). The second thing isthat, for a closed string, the limits (0, σ1) represent the same point along itslength, so the first term vanishes automatically.

The not-so-obvious part of (2.26) is the emergence of the first term,and even more so, how to deal with it. It is meaningless to simply imposea vanishing condition without considering the physics behind why we can.So how does one make it vanish? The answer is: by imposing boundaryconditions!

2.3 Boundary conditions and branes

The purpose of imposing boundary conditions is to ensure that the first termin (2.26) vanishes. This only makes sense for an open string; a closed onerequires no such constraint.

There are two types of conditions we can impose in order to achieve thiseffect while also gaining an understanding about the behaviour of the stringend points and we will now detail each one below.

2.3.1 Free end points: Neumann boundary

This boundary condition allows the strings to move freely in spacetime. Weknow that in classical mechanics, if the momentum conjugate πx = ∂L

∂x = 0,then the momentum in the x−direction is conserved; the motion in the xdirection is free and unconstrained. Generalising this to all four spacetimedimensions, the Neumann boundary can be written as

∂L∂Xµ′ = Pσµ (τ, σe) = 0. (2.27)

2.3.2 Fixed end points: Dirichlet boundary

Here the end points of the strings are fixed, or constrained to move in afixed number of dimensions which we can choose. If we let σe represent theend point of the string – so that σe = 0 or σe = σ1 – then we can write theDirichlet boundary condition as

∂τXi(τ, σe) = 0. (2.28)

Notice that the index on the string coordinates is i and not µ – meaningthat only the spatial components of Xµ are of interest. Since τ , the measureof time along the string, is related to t (the time in spacetime), we knowthat at the end points:

∂X0

∂τ6= 0, (2.29)

so (2.28) must exclude the µ = 0 term. Thus, (2.28) simply tells us that thevelocity of the string end points is zero, which means that their positionsare constant in time.

28


Figure 2.2: A D2-brane over the (x1, x2) plane. The end points of the stringare free to move about over the plane while any translation along x3 isforbidden.

An alternative Dirichlet boundary condition

Since our target is to make the first term in (2.26) disappear, we can do it aslightly different way. Suppose that we fix the coordinates of the string endpoints, such that

Xµ(τ, σe) = kµ, (2.30)

where kµ is a constant four-vector. Due to this constancy, the variationδXµ(τ, σe) = δkµ vanishes and, in doing so, it also solves our problem.This is a slightly different condition to that in (2.28) as it also fixes theµ = 0 component on the string; which in turn fixes time in spacetime. Thiscondition describes a string whose four-position is constant – not just itsspatial configuration.

Dp-branes

Dirichlet boundary conditions require that the string end points remainfixed. In order for this to happen, they must be attached to some objectswhich constrain their motion. These objects are called Dirichlet branes,or D-branes. The dimensionality of a given D-brane determines how manydegrees of freedom the end point has. For example, a D0-brane allows forno motion at all (or, motion in 0-dimensional space) whereas a D2-braneallows the string end points to move freely to over a 2-dimensional surface,as shown in figure 2.2.

29

CHAPTER

3GAUGE FIXING THE

RELATIVISTIC STRING

Gauge fixing involves choosing different sets of param-eterisations for τ and σ, which is only possible due tothe reparameterisation invariance of the action. We willclosely examine the static and light cone gauges and seehow strings behave in each.

3.1 The gauge conditions

3.1.1 Parameterising τ

The gauges we will consider here can be neatly packaged into one equation,which is found by setting the string time τ equal to a linear combination ofthe string coordinates Xµ(τ, σ):

nµXµ(τ, σ) = λτ. (3.1)

Note that the right-hand side is independent of σ. In spacetime, if twopoints xµ1 and xµ2 satisfy (3.1), then we see that nµ(xµ2 − x

µ1 ) = 0, meaning

that the set of all such vectors forms a hyperplane in spacetime which isperpendicular to the plane for which nµ is a normal. Since the Xµ lie onthe world-sheet, the points which satisfy (3.1) are the intersection betweenthe hyperplane and the world-sheet.

String momentum

In order to proceed further we first set the parameters τ, σ to be dimension-less. This is necessary in order for us to mode expand the solutions to thelight cone equations of motion, which we will be finding later in this chap-ter. Next, we want to write the gauge condition (3.1) in a more meaningfulway. This can be done by writing it in terms of the string momentum pµ byreplacing the constant λ as follows

n ·X(τ, σ) = λτ −→ βα′(n · p)τ, (3.2)

30

Chapter 3. Gauge fixing the relativistic string

where

β =

{2 open strings

1 closed strings.

We now note two very important assumptions:

• the vector nµ is chosen such that nµpµ = n·p is a world-sheet constant,

and

• the dot product nµPσµ = n · Pσ = 0 at the open string end points.

The first assumption can actually be proven quite easily. Starting with(2.22), note that the object Pτµ is given by ∂L

∂Xµ. Comparing this with the

classical momentum analogue in (1.20), it is clear that Pτµ is a momentumdensity (as it is derived from the Lagrangian density). Therefore, the mo-mentum carried by the string is given by

pµ(τ) =

∫ σ1

0Pτµ(τ, σ)dσ. (3.3)

In order for n · p to be a constant, (3.3) must be conserved in time. Thiscan be checked by finding its time-derivative:

dpµ(τ)

dτ=

∫ σ1

0

∂

∂τPτµ(τ, σ) dσ. (3.4)

However, in the above, we can substitute for∂Pτµ∂τ , using the equations of

motion in (2.22):

dpµ(τ)

dτ= −

∫ σ1

0

∂

∂σPσµ (τ, σ) dσ = −

[Pσµ]σ10

= 0, (3.5)

for both closed and open strings. For closed strings, this holds because thepoints {0, σ1} represent the same point along the length of the string, andfor open strings, it holds due to the Neumann boundary condition in (2.27).This shows that the object nµp

µ is indeed constant – its time-derivativevanishes.

3.2 The static gauge

For the remainder of this chapter we will look at the results of fixing gaugesin our string theory. The first example we will consider is the static gauge.In this, the time on the world-sheet τ is made to coincide with the time inspacetime t, such that τ = t. This is equivalent to setting the vector nµ

in (3.77) equal to (1, 0 . . . 0) and all constants on the right-hand side to c.Note that in this section, it is more convenient to work in non-natural units,so we will include factors of c and ~. The defining condition for the staticgauge is then given by

X0(τ, σ) = ct. (3.6)

31


Figure 3.1: The component ~v⊥ perpendicular to ∂ ~X∂t can be written in terms

of the velocity ∂ ~X∂t and unit vector tangent to the string ∂ ~X

∂s .

3.2.1 Action in terms of string velocity

We will proceed by defining a string velocity ~v⊥ which is transverse to thelength of the string. The differential of ~X = ~X(t, σ) is given by

d ~X =∂ ~X

∂σdσ, (3.7)

since dt = 0. If we define a new parameter s which measures lengths alongthe string, we can set ds = |d ~X|. By dividing the above equation by ds, weobtain

∂ ~X

∂σ

dσ

ds=∂ ~X

∂s= 1, (3.8)

where ∂ ~X∂s is a unit vector tangent to the string, as shown in figure 3.1.

Using the fact that, given a vector ~a, its component perpendicular to a

normal vector ~n can be written as ~a− (~a · ~n)~n, setting ~a = ~v⊥, ~u = ∂ ~X∂t and

~n = ∂ ~X∂s , we obtain

~v⊥ =∂ ~X

∂t−

(∂ ~X

∂t· ∂

~X

∂s

)∂ ~X

∂s. (3.9)

Squaring the above gives

~v2⊥ =

(∂ ~X

∂t

)2

− 2

(∂ ~X

∂t· ∂

~X

∂s

)2

+

(∂ ~X

∂t· ∂

~X

∂s

)2(∂ ~X

∂s

)2

=

(∂ ~X

∂t

)2

+

(∂ ~X

∂t· ∂

~X

∂s

)2−2 +

(∂ ~X

∂s

)2

=

(∂ ~X

∂t

)2

−

(∂ ~X

∂t· ∂

~X

∂s

)2

, (3.10)

since ∂ ~X∂s = 1. We can now calculate the derivatives Xµ and Xµ′ in terms

of ~v⊥ and their squares:

32


Xµ =∂Xµ

∂t=

(∂X0

∂t,∂ ~X

∂t

)=

(c,∂ ~X

∂t

),

−→ X2 =

c2 −

(∂ ~X

∂t

)2 ,

(3.11)

and

Xµ′ =∂Xµ′

∂t=

(0,∂ ~X

∂σ

)

−→(Xµ′)2 = −

(∂ ~X

∂σ

)2

,

(3.12)

so the mixed term is then

(X ·X ′)2 =

(∂ ~X

∂t· ∂

~X

∂σ

)2

. (3.13)

The quantity under the square root in the Nambu-Gotto action (2.7) canthen be written as

(X ·X ′)2−(X)2(X ′)2 =

(∂ ~X

∂t· ∂

~X

∂σ

)2

+

(∂ ~X

∂σ

)2c2 −

(∂ ~X

∂t

)2 . (3.14)

Using ∂ ~X∂σ = ∂ ~X

∂sdsdσ = ds

dσ , the above can be written as

(X ·X ′)2 − (X)2(X ′)2 =

(ds

dσ

)2(∂ ~X

∂t· ∂

~X

∂σ

)2

+ c2 −

(∂ ~X

∂t

)2

=

(ds

dσ

)2 (c2 − v2

⊥)

= c2

(ds

dσ

)2(1−

v2⊥c2

). (3.15)

Inserting the above into (2.7) gives us the Nambu-Gotto action in terms ofthe velocity ~v⊥:

S = − 1

2πα′~c

∫dt

∫ σ1

0dσ

(ds

dσ

)√1−

v2⊥c2. (3.16)

3.2.2 Motion of the open string end points

The static gauge, combined with the Neumann boundary condition in (2.27),provides us with interesting information regarding the motion of the endpoints of an open string. To begin this analysis, let us first find the momen-tum density Pσµ in the static gauge, using the transverse velocity ~v⊥:

33


Pσµ = − 1

2πα′~c2

(X ·X ′)Xµ − (X)2Xµ′√(X ·X ′)2 − (X)2(X ′)2

= − 1

2πα′~c3

(dσ

ds

)(∂ ~X

∂t· ∂

~X

∂σ

)∂Xµ

∂t−(dσ

ds

)c2 −

(∂ ~X

∂t

)2 ∂Xµ

∂σ√1−

v2⊥c2

= − 1

2πα′~c3

(∂ ~X

∂t· ∂

~X

∂s

)∂Xµ

∂t−

c2 −

(∂ ~X

∂t

)2 ∂Xµ

∂s√1−

v2⊥c2

.

(3.17)The boundary condition in (2.27) tells us that at the string end points, theabove must vanish ∀µ ∈ {0, 3}. Finding the µ = 0 component and settingit equal to zero using (3.17) gives

Pσ0 = − 1

2πα′~c2

(∂ ~X

∂t· ∂

~X

∂s

)√

1−v2⊥c2

−→ ∂ ~X

∂t· ∂

~X

∂s= 0, (3.18)

which holds for the end points of the open string. This tells us that the end

points move transverse to the string’s length, with velocity ~v⊥ = ∂ ~X∂t .

Inserting the final equation in (3.18) back into (3.17) gives

Pσµ =1

2πα′~c3

c2 −

(∂ ~X

∂t

)2 ∂Xµ

∂s√1−

v2⊥c2

=1

2πα′~c3

c2

(1−

v2⊥c2

)∂Xµ

∂s√1−

v2⊥c2

=1

2πα′~c

√1−

v2⊥c2

∂Xµ

∂s. (3.19)

Although the above does vanish at the end points of the string, it isn’t

very useful for us since ∂Xµ

∂s = 0 and

√1− v2⊥

c2= 0 are both valid solutions.

However, we can force

√1− v2⊥

c2= 0 to be the only solution if we look at the

spatial components of the momentum density Pσi instead of all spacetime

components. This will set ∂Xµ

∂s →∂ ~X∂s = 1. So we then obtain

~Pσ =1

2πα′~c

√1−

v2⊥c2−→ v⊥ = c, (3.20)

which shows that the string end points move at the speed of light!

34


3.2.3 Analysing the string equations of motion

The µ = 0 component

Here we will analyse the µ = 0 and µ = i components of the equations ofmotion in (2.22) separately. To proceed further, we first choose a parame-terisation for σ. As the string evolves in time, we want a particular pointon the string, say at σ = σ0, to trace a line which is perpendicular to thetangent to the string at that point. So the parameterisation is

∂ ~X

∂σ· ∂

~X

∂t= 0. (3.21)

This allows us to write the momentum density Pσµ as (3.19), but this timeit is valid for the entire string and not just the end points. Pτµ can bewritten as

Pτµ = − 1

2πα′~c2

(X ·X ′)Xµ′ − (X ′)2Xµ√(X ·X ′)2 − (X)2(X ′)2

= − 1

2πα′~c3

dσ

ds

(∂ ~X

∂t· ∂

~X

∂σ

)∂Xµ

∂σ+dσ

ds

(∂ ~X

∂σ

)2∂Xµ

∂t√1−

v2⊥c2

= − 1

2πα′~c3

ds

dσXµ√

1−v2⊥c2

. (3.22)

The µ = 0 component of (3.22) is then given by

Pτ0 = − 1

2πα′~c2

ds

dσ√1−

v2⊥c2

, (3.23)

Noting that the µ = 0 component of Pσµ vanishes as can be seen by makingthis substitution in (3.19), the equations of motion for the µ = 0 componentof (2.22) becomes

∂

∂t

1

2πα′~c2

ds

dσ√1−

v2⊥c2

= 0. (3.24)

The above equation provides us with very useful information regarding theenergy carried by the string. Since dσ is a constant in (3.24), by multiplyingthrough by cdσ, we see that

∂

∂t

T0ds√1−

v2⊥c2

= 0, (3.25)

35


where we used the fact that T0 = 12π~α′c . Equation (3.25) tells us that

the quantity within the square brackets – which has units of energy – isconserved in time. We deduce that this is the relativistic energy of a pieceof string of length ds. The Hamiltonian is then given by the integral overall ds:

H = E =

∫T0ds√1−

v2⊥c2

. (3.26)

The µ = i components

We start by obtaining the µ = i components for both densities Pσµ and Pτµfrom (3.19) and (3.22) respectively:

~Pσ =1

2πα′~c

√1−

v2⊥c2

∂ ~X

∂s, (3.27)

and

~Pτ = − 1

2πα′~c3

ds

dσ√1−

v2⊥c2

~v⊥ (3.28)

Inserting (3.27) and (3.28) into the µ = i components of (2.22) gives theequation

∂

∂σ

1

2πα′~c

√1−

v2⊥c2

∂ ~X

∂s

=∂

∂t

1

2πα′~c3

ds

dσ√1−

v2⊥c2

~v⊥

. (3.29)

Multiplying through both sides by dσds and putting the constants in terms of

T0 = 12πα′~c gives

T0

√1−

v2⊥c2

∂2 ~X

∂s2=T0

c2

1√1−

v2⊥c2

∂~v⊥∂t

, (3.30)

which we can cast into a more familiar form:

Teff∂2 ~X

∂s2− µeff

∂2 ~X

∂t2= 0, (3.31)

The above equation has the form of an effective wave equation, the generalsolution to which is very well known and fairly simple to analyse.

3.2.4 The wave equation and its constraints

Before moving onto solving (3.30), let us first put it in a more convenientform using the energy E as a parameterisation of the string, such that eachsegment σ of it has equal energy. We start with

36


1

c2

∂2 ~X

∂t2=

(1−

v2⊥c2

) 12 ∂

∂s

((1−

v2⊥c2

) 12 ∂ ~X

∂s

)

=

(1−

v2⊥c2

) 12

ds

dσ

∂

∂σ

(

1−v2⊥c2

) 12

ds

dσ

∂ ~X

∂σ

=

1

A2(σ)

∂2 ~X

∂σ2, (3.32)

where A(σ) ≡ dsdσ/

(1− v2⊥

c2

)and must be set equal to 1 in order to simplify

(3.32). The value of σ assigned to each portion of the string must be changedin order to accomplish this. Using the energy in (3.26), we can write

dσ =ds(

1−v2⊥c2

) 12

=1

T0dE −→ dE

dσ= T0 (3.33)

which actually constrains the dsdσ derivative:(ds

dσ

)2

+v2⊥c2

= 1 (3.34)

which, using the fact that ∂ ~X∂s is a unit vector and rewriting v2

⊥ in derivativeform, can be written as(

∂ ~X

∂σ

)2

+1

c2

(∂ ~X

∂t

)2

= 1. (3.35)

Thus, up to a constant, we can set A(σ) = 1 to obtain a simplified versionof (3.30):

1

c2

∂2 ~X

∂t2=∂2 ~X

∂σ2. (3.36)

Note that Pσµ in (3.19) and Pτµ in (3.22) can be simplified using A(σ) = 1:

Pσµ = T0∂Xµ

∂σ, (3.37)

and

Pτµ = −T0

c2

∂Xµ

∂t. (3.38)

We will also consider the free end point condition which corresponds to theNeumann boundary in (2.27). This means that we have to set (3.27) equalto zero, implying that the Neumann boundary condition, in this context, isequivalent to the constraint

∂ ~X

∂σ

∣∣∣∣∣σ=0

=∂ ~X

∂σ

∣∣∣∣∣σ=σ1

= 0. (3.39)

37


3.2.5 Open string motion

Having established the simplified wave equation in (3.36), the goal now is tosolve it while also catering for the σ parameterisation in (3.21), the Neumann

boundary condition in (3.35), and the ∂ ~X∂σ derivative condition in (3.35).

To start, let us recall the solution for (3.36):

~X(t, σ) =1

2

(~f(ct+ σ) + ~g(ct− σ)

). (3.40)

The above solution is very general; it satisfies (3.36) for any arbitrary func-tions ~f and ~g. We can, however, make it more specific to what we need.This is done by first examining how it behaves as we apply the Neumannboundary condition in (3.39) and then the condition in (3.35).

Applying the Neumann boundary condition

For the string end point at σ = 0, (3.40) becomes

1

2

(~f ′(ct)− ~g′(ct)

)= 0 −→ ∂σ ~f(ct) = ∂σ~g(ct), (3.41)

which shows that the functions ~f and ~g are equivalent up to a constantof integration which can be absorbed into either of their definitions. It isimportant to note that the above is valid for all arguments of the functions.This means we can rewrite (3.40) as

~X(t, σ) =1

2

(~f(ct+ σ) + ~f(ct− σ)

). (3.42)

A similar analysis can be made for the σ = σ1 end point of the string.Setting the σ-derivative equal to zero implies that the two functions areagain equal up to a constant of integration (which we will set to 2σ1

~v0c for

convenience):

~f ′(ct+ σ1) = ~f ′(ct− σ1) −→ ~f(u+ 2σ1) = ~f(u) + 2σ1~v0

c, (3.43)

where u ≡ ct − σ1. We learn that, up to a constant, ~f is periodic in 2σ1.This means that if we want to know ~f(u) for every possible value of u, wejust need to know ~f for u ∈ {0, 2σ1}. Once ~f(u) is known, the solution forthe wave equation (3.42) is fully defined as we know that ~v0 is the averagevelocity of the string in the time interval t ∈ {0, 2σ1

c }.

Applying the restrictions

Now we must analyse the effect of the other restrictions on the vector func-tion ~f . To begin, let us first note that (3.21) can be multiplied by 2

c likeso:

∂ ~X

∂σ· ∂

~X

∂t= 0 −→ 2

∂ ~X

∂σ· 1

c

∂ ~X

∂t= 0, (3.44)

without changing the fact that it vanishes. This is now equal to the crossterm in the expansion of (A+B)2, where

A =∂ ~X

∂σ, B =

1

c

∂ ~X

∂t. (3.45)

38


The point of doing this is so that we can combine the constraints (3.21) and(3.35) into one. Noting that (3.35) is A2 + B2 by comparing with (3.45),we can add and subtract (3.44) to (3.35) and factorise to obtain a singleconstraint (

∂ ~X

∂σ± 1

c

∂ ~X

∂t

)2

= 1, (3.46)

which encapsulates both (3.21) and (3.35). The next step is to plug thewave equation solution (3.42) into (3.46):

∂ ~X

∂σ=

1

2

(~f ′(ct+ σ)− ~f ′(ct− σ)

),

1

c

∂ ~X

∂t=

1

2

(~f ′(ct+ σ) + ~f ′(ct− σ)

)−→ ∂ ~X

∂σ± 1

c

∂ ~X

∂t= ±~f ′(ct± σ).

(3.47)Letting the argument of ~f ′ equal u once again, the combined constraint(3.46) shows us that ∣∣∣~f ′(u)

∣∣∣2 = 1, (3.48)

which means that ~f ′(u) is a unit vector. In a (D−1)-dimensional spacetime,the length of the curve of the function ~f is parameterised by u.

Rigid rotating string

We will now analyse the motion of an open string rotating about its midpointin order to find a function ~f(u) which satisfies (3.48) and (3.43). If werestrict the motion to the x − y plane (such that the angular momentumis in the z-direction) and let the length of the string equal l and angularfrequency equal ω, then we can immediately write down the solution of thewave equation for the σ = 0 end point as

~X(t, σ = 0) = ~f(ct) =l

2(cosωt, sinωt) , (3.49)

or, in terms of u = ct:

~f(u) =l

2

(cosω

u

c, sinω

u

c

). (3.50)

Since the above solution is periodic, the average velocity ~v0 must vanish.(Consider the average value of, for example, sinx.) So we now have a simplerperiodicity condition than the one in (3.43) due to the vanishing of ~v0, and

is given by ~f(u+ 2σ1) = ~f(u). This means that ωuc + ω(2σ1)

c = ωuc , which in

turn implies that ω(2σ1)c = 0. However this is the case at time t = 0. After

one cycle of rotation, ω(2σ1)c = 2π. Thus, in general, we have ω(2σ1)

c = 2πn.For simplicity we will choose n = 1, which gives

ω

c=

π

σ1=T0π

E, (3.51)

where the last equation uses the final equation in (3.33). We can now findthe energy of string in terms of its length, using (3.48) and (3.51):

39


~f ′(u) =ωl

2c

(− sinω

u

c, cosω

u

c

)→∣∣∣~f ′(u)

∣∣∣2 = 1 −→ l =2c

ω=

2σ1

π=

2E

T0π,

(3.52)

which means that

E =T0πl

2. (3.53)

To conclude the analysis of the open string’s motion, let us find the solution~X(t, σ) for any σ ∈ {0, σ1}. Replacing l with 2σ1

π and ωc with π

σ1in (3.50)

gives

~f(u) =σ1

π

(cos

πu

σ1, sin

πu

σ1

). (3.54)

The above can be inserted into (3.42) using u = ct± σ:

~X(t, σ) =σ1

2π

[cos

π(ct+ σ)

σ1+ cos

π(ct− σ)

σ1, sin

π(ct+ σ)

σ1+ sin

π(ct− σ)

σ1

],

(3.55)which can be simplified by a great deal using elementary trigonometric iden-tities for sin(A±B) and cos(A±B), giving

~X(t, σ) =σ1

πcos

πσ

σ!

(cos

cπt

σ1, sin

cπt

σ1

). (3.56)

3.2.6 Closed string motion

Having dealt with open strings, the next task is to look at the motion ofa free closed string. We begin in a similar fashion to the case of the openstring, which is by stating the general solution to the wave equation in (3.36):

~X(t, σ) =1

2

(~f(ct+ σ) + ~g(ct− σ)

). (3.57)

Letting u ≡ ct + σ and v ≡ ct− σ, we can take the derivatives required by(3.36):

∂ ~X

∂σ=

1

2

(~f ′(u)− ~g′(v)

),

1

c

∂ ~X

∂t=

1

2

(~f ′(u)− ~g′(v)

)−→ ∂ ~X

∂σ+

1

c

∂ ~X

∂t= ~f ′(u),

∂ ~X

∂σ− 1

c

∂ ~X

∂t= ~g′(v). (3.58)

This time, the constraint in (3.48) holds for both ~f ′ and ~g′. We now imposethe periodicity condition on ~X:

~X(t, σ) = ~X(t, σ + σ1) (3.59)

since, for a given point σ on the string, going around the length by a furtherσ1 will result in reaching σ again. Replacing σ with σ + σ1 in (3.57), weobtain

~f(u+ σ1) + ~g(v − σ1) = ~f(u) + ~g(v). (3.60)

40


Differentiating the above by u and v separately (since these are independentvariables) gives

~f ′(u+ σ1) = ~f ′(u), ~g′(v + σ1) = ~g′(v), (3.61)

which means that ~f ′ and ~g′ are both periodic vectors which exist on thesurface of a unit 2-sphere.

Cusps

Cusps occur when the two vectors ~f ′ and ~g′ intersect on the surface of theunit 2-sphere. If we define u0 ≡ ct0 + σ0 and v0 ≡ ct0 − σ0 to be the pointat which ~f ′ and ~g′ coincide, then ~f ′(u0) = ~g′(v0), and

∂ ~X(t0, σ0)

∂σ=

1

2

(~f ′(u0)− ~g′(v0)

)= 0,

1

c

∂ ~X(t0, σ0)

∂t=

1

2

(~f ′(u) + ~g′(v)

)= ~f ′(u0).

(3.62)What do the above equations mean? Recall that, on account of (3.48), thevector ~f ′(u0) squares to unity. Using the second equation above, this meansthat

1

c2

(∂ ~X

∂t

)2

= 1 −→ ∂ ~X

∂t= ~X = c, (3.63)

which shows that, at the cusps, the string reaches the speed of light.

3.3 The light cone gauge

Reverting back to natural units of c = ~ = 1, we now look at using thelight cone gauge to parameterise the string. The analysis will, to an extent,resemble that of the static gauge. As far as τ is concerned, given that thestatic gauge is a special choice for the vector nµ = (1, 0, . . . , 0) in (3.2),the light cone gauge uses the same τ parameterisation but the choice nµ isdifferent. For now we will treat nµ generally and it’s specific value will bedeclared in subsection 3.3.5. The parameterisation for σ is a little bit moreinvolved and is examined in detail below.

3.3.1 Parametrising σ

The parameterisation for σ is simply that the vector nµ dotted with Pτµis constant, which for the time being, we will denote as k. In order todetermine k, we integrate nµPτµ with respect to σ:

nµPτµ = k −→∫ π

0nµPτµ dσ =

∫ π

0k dσ

= πk.

(3.64)

Now, using the fact that∫dσPτµ = pµ, we also have∫ π

0nµPτµ dσ = nµp

µ = n · p, (3.65)

41


from which we observe that k = n·pπ . We now have our desired σ parame-

terisation:

n · Pτ =n · pπ

. (3.66)

We now take a look at what effect the sigma parameterisation in (3.66) hason the equations of motion in (2.22). Noting that n · Pτ is constant, we findthat the first term in (2.22) vanishes if we dot the equations of motion withnµ. The result shows us that n · Pσ independent of σ:

∂

∂σ(n · Pσ) = 0, for open and closed strings. (3.67)

n · Pσ for open strings

The next task is to show that n · Pσ vanishes for both open and closedstrings. For open strings, this is easy. Since it is assumed that n · Pσ = 0at the string end points, the fact n · Pσ is independent of σ as shown by(3.67) means that n · Pσ will remain at zero for every value that σ can takefrom the σ = 0 end point all the way through to the σ = π point. In otherwords, changing the value of σ has no effect on n · Pσ being equal to zero.This gives us the condition

n · Pσ = 0 for open strings. (3.68)

n · Pσ for closed strings

In order to show that n · Pσ vanishes for closed strings also, we first needthe explicit form of Pσµ. This is found by inserting the final right-hand sideof (2.18) into the expression for Pσµ given by (2.21):

Pσµ = − 1

2πα′(X ·X ′)Xµ − (X)2Xµ′[(X ·X ′)2 − (X)2(X ′)2

] 12

= − 1

2πα′(X ·X ′)∂τXµ − (X)2∂σX

µ[(X ·X ′)2 − (X)2(X ′)2

] 12

.

(3.69)

Although we will not need it yet, for the record, we can also write themomentum density Pτµ explicitly as

Pτµ = − 1

2πα′(X ·X ′)Xµ′ − (X ′)2Xµ[(X ·X ′)2 − (X)2(X ′)2.

] 12

. (3.70)

Dotting (3.69) with nµ gives

n · Pσ = − 1

2πα′(X ·X ′)∂τ (n ·X)− (X)2∂σ(n ·X)[

(X ·X ′)2 − (X)2(X ′)2] 1

2

, (3.71)

which can be simplified by using the fact that ∂σ(n ·X) = 0, since (3.2) isindependent of σ:

42


n · Pσ = −β(n · p)2π

(X ·X ′)[(X ·X ′)2 − (X)2(X ′)2

] 12

, (3.72)

Figure 3.2: The point σ = 0 is chosen to be at P . The tangent space tothe world-sheet at P is constructed by declaring a spacelike vector Xµ′ tobe tangent to the string at P and a timelike vector Xµ to be tangent to theworld sheet at P . We require these vectors to be perpendicular.

where the expression for ∂τ (n ·X) has been inserted using (3.2). The aboveequation now makes our task clear: we have to show that X · X ′ = 0 atsome general point on every string. Then, since n · Pσ is independent of σ,the same argument as for the open string will hold.

The task is approached as follows. Referring to figure 3.2, one chooses apoint P on a string at time τ0 to be where σ = 0. At P , a spacelike vectortangent to the string is labelled Xµ′ and a timelike vector tangent to P islabelled Xµ. Since these two vectors are perpendicular, their dot productXµ · Xµ′ = 0, as required. This process is repeated for a string at a latertime τ1. However, for strings which exist subsequent to τ0, the σ = 0 pointis given by Xµ(P ) + εXµ. As ε scales, the points σ = 0 forms a line, and ateach point on this line, the spacelike tangent vector Xµ′ is always normal tothe timelike tangent vector Xµ, thus guaranteeing that Xµ ·Xµ′ = 0 at anytime.It can be shown that if Xµ is perpendicular to a spacelike tangent vectorXµ′, then Xµ is timelike. Since both of these vectors form the tangent spaceto the world sheet at P , we can write, in terms of an arbitrary vector tµ:

Xµ = tµ + bXµ′. (3.73)

In order to solve for b, we simply impose the condition Xµ ·Xµ′ = 0:

Xµ ·Xµ′ = t ·X ′ + b(X ′ ·X ′) −→ b = − t ·X ′

X ′ ·X ′. (3.74)

Plugging the above into (3.73) gives us

Xµ = tµ − t ·X ′

X ′ ·X ′Xµ′, (3.75)

43


which is timelike if its interval is greater than zero:

XµXµ = tµ

(tµ − t ·X ′

X ′ ·X ′Xµ′

)− t ·X ′

X ′ ·X ′Xµ′

(tµ − t ·X ′

X ′ ·X ′Xµ′

)= t2 − (t ·X ′)2

X ′ ·X ′− (t ·X ′)2

X ′ ·X ′+

(t ·X ′)2

X ′ ·X ′

−→ t2 − (t ·X ′)2

X ′ ·X ′> 0, (3.76)

where the final inequality is drawn from the fact that, by definition, Xµ′

is a spacelike vector so its interval X ′ · X ′ < 0, thereby making the entireleft-hand side positive for any value of tµ and Xµ′.

The above analysis shows us that n · Pσ = 0 at the σ = 0 point on thestring. In doing so, it shows that this condition holds for both open andclosed strings. We can now write the gauge conditions for both types ofstring in full generality:

n ·X(τ, σ) = βα′(n · p)τ, (3.77)

and

n · p =2π

βn · Pτ , (3.78)

where

β =

{2 open strings

1 closed strings.

3.3.2 Constraints and the wave equation

As discussed subsection 3.3.1, the factor of ∂τ (n ·X) is a non-zero constant,and for both open and closed strings, we have

X ·X ′ = 0 (3.79)

which can now be used to simplify Pτµ in (3.70):

Pτµ =1

2πα′X ′2Xµ√−X2X ′2

. (3.80)

Dotting the above with nµ gives

n · Pτ =1

2πα′X ′2(n · X)√−X2X ′2

, (3.81)

which allows us to find n · p using (3.78). Doing so, while also inserting theexpression for n · X in (3.77) gives

n · p =1

βα′X ′2(n · X)√−X2X ′2

−→ 1 =X ′2√−X2X ′2

, (3.82)

from which it is clear that

X2 +X ′2 = 0 (3.83)

44


Note that (3.83) and (3.79) together form the expansion of a quadraticexpression, which allows us to combine both constraints into one:

(X ±X ′)2 = 0. (3.84)

The identification that X2 = −X ′2 in (3.83) can be used to simplify (3.80)by replacing −X2 with X ′2 in the denominator:

Pτµ =Xµ

2πα′. (3.85)

In a similar fashion, we now simplify Pσµ in order to find the equations ofmotion explicitly. Beginning with the first line of (3.69), we can first insert(3.79) and then (3.83) to give

Pσµ = − 1

2πα′(X ·X ′)Xµ − (X)2Xµ′[(X ·X ′)2 − (X)2(X ′)2

] 12

=1

2πα′(X2)Xµ′√−(X2)(X ′)2

=1

2πα′(X2)Xµ′

X ′2

= −Xµ′

2πα

(3.86)

The equations of motion in (2.22) now become the wave equation:

1

2πα′

(∂

∂τXµ − ∂

∂σXµ′

)= 0 −→ Xµ −Xµ′′ = 0, (3.87)

which we will now analyse by mode expanding its general solution.

3.3.3 Applying the constraints

Just as in the case of the static gauge, here we consider strings with free endpoints, which corresponds to the Neumann boundary condition in (2.27). Inthe present case, this translates to

Xµ′ = 0, (3.88)

since Pσµ depends on Xµ′. The most general solution for the wave equationin (3.87) is given by

Xµ(τ, σ) =1

2(fµ(τ + σ) + gµ(τ − σ)) . (3.89)

Note that this time, the solution involves Lorentz vectors.We can now follow the same procedure we did as in the static gauge –

find how the boundary condition applied at σ = 0 and σ = π constrains thesolution (3.89). Starting with the σ = 0 end point, imposing ∂σX

µ = 0 on(3.89) gives

1

2

(fµ′(τ)− gµ′(τ)

)= 0 −→ fµ(τ) + cµ = gµ(τ). (3.90)

Absorbing the constant of integration into the definition of fµ, we can write

45


Xµ(τ, σ) =1

2(fµ(τ + σ) + fµ(τ − σ)) . (3.91)

Imposing the ∂σXµ = 0 at the σ = π end point gives

1

2

(fµ′(τ + π) + fµ′(τ − π)

)= 0. (3.92)

The above shows us that fµ′ is periodic in 2π. This can be seen easily if weshift the argument of both vectors by +σ:

fµ′(τ + 2π) = fµ′(τ). (3.93)

Imposing the boundary condition at both end points of the string tells usthat the vectors fµ and gµ are equal up to a constant, and that the functionfµ′ is periodic in 2π.

3.3.4 Mode expanding the general solution

To find the function fµ, the strategy is to first mode expand fµ′ using theFourier series expansion in (1.14) and then integrating it to obtain fµ:

fµ′(x) = fµ1 (x) +∞∑n=1

(aµn cosnx+ bµn sinnx) , (3.94)

which we can now integrate term by term with respect to x, obtaining

fµ(x) = fµ0 + fµ1 x+

∞∑n=1

(aµnn

sinnx− bµnn

cosnx

), (3.95)

where fµ0 is a constant of integration. If we now define a new set of Fouriercoefficients B = a

n and A = − bn , then we can write

fµ(x) = fµ0 + fµ1 x+

∞∑n=1

(Aµn cosnx+Bµn sinnx) . (3.96)

We now need to replace the variable x with our usual τ and σ. This is doneby first replacing x with (τ + σ) in fµ(x):

fµ(τ + σ)

= fµ0 + fµ1 (τ + σ) +

∞∑n=1

(Aµn cosn(τ + σ) +Bµn sinn(τ + σ))

= fµ0 + fµ1 (τ + σ) +

∞∑n=1

(Aµn[cosnτ cosnσ −(((((

((sinnτ sinnσ]

+Bµn [sinnτ cosnσ +((((

(((cosnτ sinnσ]

).

(3.97)

Similarly for fµ(τ − σ):

fµ(τ − σ)

= fµ0 + fµ1 (τ − σ) +

∞∑n=1

(Aµn[cosnτ cosnσ +((((

(((sinnτ sinnσ]

+Bµn [sinnτ cosnσ −(((((

((cosnτ sinnσ]

).

(3.98)

46


In both of the above equations, the crossed out terms cancel when we takethe sum fµ(τ + σ) + fµ(τ − σ) as required by (3.91). In doing this sumwe also pick up factors of 2 which cancel with the 1

2 outside the brackets in(3.91), so we obtain

Xµ(τ, σ) = fµ0 + fµ1 τ +∞∑n=1

(Aµn cosnτ +Bµn sinnτ) cosnσ. (3.99)

Using Euler’s identity for eix, we will now rewrite (3.99) in terms of complexexponentials. We will define

Aµn cosnτ +Bµn sinnτ = −i

√2α′

n

(aµ∗n e

inτ − aµne−inτ), (3.100)

where aµn ≡ (Bµn − iAµn) and the (∗) denotes complex conjugation. We will

also let the zero mode fµ0 = xµ0 , so that the only unfamiliar object in (3.99)is fµ1 , which we can find by integrating the momentum density Pτµ:

Pτµ =1

2πα′

(fµ1 + ∂τ

[−i√

2α′

n

∞∑n=1

(aµ∗n e

inτ − aµne−inτ)

cosnσ

])

→ pµ =1

2πα′

∫ π

0

(fµ1 + ∂τ

[−i√

2α′

n

∞∑n=1

(aµ∗n e

inτ − aµne−inτ)

cosnσ

])dσ

pµ =1

2�πα′�πf

µ1

−→ fµ1 = 2α′pµ,(3.101)

where between the second and third lines we used the fact that∫ π

0 cosnσ dσvanishes. With these new identifications, the solution in (3.99) can be writ-ten as

Xµ(τ, σ) = xµ0 + 2α′pµ − i√

2α∞∑n=1

(aµ∗n e

inτ − aµne−inτ) cosnσ√

n

= xµ0 +√

2α′αµ0τ + i√

2α′∞∑n 6=0

1

nαµne

inτ cosnσ,

(3.102)

where in the last line, along with combining the sum and excluding the zeromode, the following definitions have been used:

αµ0 ≡√

2α′pµ

αµn ≡ aµn√n

αµ−n ≡ (αµn)∗ = aµ∗n√n.

(3.103)

Using (3.102), we can find the τ and σ derivatives:

Xµ =√

2α′∑n∈Z

αµn cosnσe−inτ , (3.104)

and

47


Xµ′ = −i√

2α′∑n∈Z

αµn sinnσe−inτ , (3.105)

where the zero modes of both oscillators have been absorbed into the sum-mand so the summation now spans the entire domain of integers.

3.3.5 Light cone solution

The solution in (3.102), as of yet, does not satisfy the constraints in (3.84).In order that it does, we must introduce the light cone coordinates, whichcan be done by choosing explicitly the vector nµ:

nµ =

(1√2,

1√2, 0, . . . , 0

). (3.106)

Before moving further, however, a brief recap of the light cone coordinatesystem is due. For a D-dimensional spacetime, an arbitrary vector xµ exist-ing within it can have its components split into the coordinates {x0, x1, xI}where the xI are (D−2)-dimensional transverse coordinates. The light conegauge designates two new light cone coordinates x+ and x− in place ofthe original x0 and x1, which are specific linear combinations of x0 and x1,such that

x± =x0 ± x1

√2

. (3.107)

In this coordinate system, the metric which is responsible for the scalarproduct of two vectors is (not surprisingly) the light cone metric η, and in4-dimensions, is defined as

η =

0 1 0 01 0 0 00 0 −1 00 0 0 −1

. (3.108)

The light cone metric in a D-dimensional space is defined the same way butwith (D − 2) elements equalling unity in the lower diagonal. The scalarproduct of two vectors aµ and bµ in D-dimensional space is then given by

a · b = ηµνaµbν = a−b+ + a+b− − aIbI , (3.109)

where the repeated indicies imply summation.Having reviewed the basics, we can now start taking our solution (3.102)

into the light cone coordinates. The first step is to note that, with nµ definedas (3.106), we can rewrite (3.77) and (3.78):

n ·X =X0 +X1

√2

= X+, n · p =p0 + p1

√2

= p+,

X+ = βα′p+τ =√

2α′α+0 τ −→ X+ = βα′p+ =

√2α′α+

0 , (3.110)

where the definitions for the zero modes in (3.103) has been used. Thetwo equations (3.104) and (3.105) can be used to rewrite the square root ofcondition in (3.84) as

48


Xµ ±Xµ′ =√

2α′∑n∈Z

αµne−in(τ±σ). (3.111)

However, as is evident from (3.84), we need the square of (3.111). Thiscan be found using the light cone coordinates and metric. If we defineaµ ≡ Xµ ±Xµ′, then

a2 = 2a+a− − (aI)2, (3.112)

and given that a± = X± ±X±′, this becomes

(XI ±XI′)2 = 2(X+ ±X+′)(X− ±X−′), (3.113)

which can be simplified by noting that X+′ = 0 and X+ = βα′p+:

X− ±X−′ = (XI ±XI′)2

2βα′p+, p+ > 0. (3.114)

We will now focus on open strings and set β = 2. The solution Xµ in (3.102)will now be defined piecewise in terms of 3 sets of coordinates: one for X+,one for XI and the third for X−. For X+, the solution is already given in(3.110):

X+(τ, σ) =√

2α′α+0 τ, (3.115)

which shows that, for the X+ coordinate, there is no zero mode and nooscillations (comparing with (3.102) makes this evident). The solution forthe transverse coordinates is simply given by replacing µ −→ I in (3.102):

XI(τ, σ) = xI0 +√

2α′αI0τ + i√

2α′∞∑n6=0

1

nαIne

inτ cosnσ. (3.116)

The only solution which remains to be found is for the X− coordinates. Itturns out that the (−) oscillators can be found as functions of the transverse(I) oscillators. This follows quite naturally by letting µ −→ − and µ −→ Iin (3.111) and inserting the results into (3.114). Focusing on the right-handside first (since it’s harder!), we get

1

2βα′p+

(√

2α′∑n∈Z

αIne−in(τ±σ)

)2

. (3.117)

In order to deal with the above, we will first let n = p and introduce anothersummation variable q, since we have a sum being squared. These are dummyindices; we can call them whatever we want. Then, setting β = 2 as we aredealing with open strings now, we get

1

2p+

∑p∈Z

∑q∈Z

αIpe−ip(τ±σ)αIqe

−iq(τ±σ)

=

1

2p+

∑p∈Z

∑q∈Z

αIpαIqe−i(p+q)(τ±σ)

(3.118)

The left-hand side of (3.114) is given by

49


√2α′∑n∈Z

α−n e−in(τ±σ). (3.119)

Equating (3.119) and (3.118) gives

√2α′∑n∈Z

α−n e−in(τ±σ) =

1

2p+

∑p∈Z

∑q∈Z

αIpαIqe−i(p+q)(τ±σ)

. (3.120)

Comparing the exponentials in both sides of the above equation, it can beseen that n must equal p + q. We can therefore eliminate q in favour of nusing q = n− p:

√2α′∑n∈Z

α−n e−in(τ±σ) =

1

2p+

∑n∈Z

∑p∈Z

αIpαIn−pe

−in(τ±σ)

, (3.121)

from which we can now cancel factors of∑

n∈Z e−in(τ±σ), giving us our

desired result in terms of the transverse Virasoro mode Ln

√2α′α−n =

1

2p+

∑p∈Z

αIpαIn−p =

1

p+Ln, (3.122)

where

Ln ≡1

2

∑p∈Z

αIpαIn−p. (3.123)

The solution for the X− coordinates is also given by (3.102), but with µ = −:

X−(τ, σ) = x−0 +√

2α′α−0 τ + i√

2α′∞∑n6=0

1

nα−n e

inτ cosnσ. (3.124)

Clearly there is room to simplify this using the transverse Virasoro modesin (3.123). We start by noting that

+i√

2α′∞∑n6=0

1

nα−n e

inτ cosnσ = i∑n6=0

1

p+Lne

−inτ cosnσ, (3.125)

and that

√2α′α−0 =

1

p+L0. (3.126)

Back into (3.124), we get

X−(τ, σ) = x−0 +1

p+L0τ +

i

p+

∑n 6=0

1

nLne

−inτ cosnσ, (3.127)

which, combined with the solutions for X+ and XI coordinates

X+(τ, σ) =√

2α′α+0 τ (3.128)

50


and

XI(τ, σ) = xI0 +√

2α′αI0τ + i√

2α′∞∑n6=0

1

nαIne

inτ cosnσ, (3.129)

provides a complete solution for the wave equation (3.87) in the light conegauge.

51

CHAPTER

4SCALAR FIELDS ANDPARTICLE STATES

Here we will further develop the powerful language ofquantum field theory. The primary focus will be on es-tablishing the operator formalism and seeing how this isused to describe particle states. By the end we will havedeveloped all the tools necessary to study the quantisa-tion of strings.

4.1 Classical scalar fields

4.1.1 The Klein-Gordon equations of motion

From here on our work will involve the Klein-Gordon Lagrangian (1.46) (andin particular, its solutions) quite extensively. We will begin by writing itout once more:

L =1

2

[(∂µφ)(∂νφ)ηµν −m2φ2

]. (4.1)

Our initial targets are to solve the equations of motion arising from (4.1),translate the solution to phase space and then proceed with quantisation byintroducing operators. Once the quantised solution is found, we will lookat the relativistic point particle, after which we will be ready to study openand closed strings.

The equations of motion that are due to (4.1) can be found using theEuler-Lagrange equations in (1.35):

∂L∂(∂µφ)

= (∂νφ)ηµν −→ ∂µ∂L

∂(∂µφ)= ∂µ∂νφη

µν = �φ (4.2)

where � ≡ ηµν∂µ∂ν = ∂20 − ~∇2 is the flat space d’Alambertian operator

(this can be thought of as the generalisation of the Laplace operator ∂2i to

the four spacetime dimensions of special relativity). Note that the factor of12 drops out due to the symmetry of the Lagrangian’s kinetic term underswapping of the µ and ν indices.

52

Chapter 4. Scalar fields and particle states

The right-hand side of (1.35) is given by

∂L∂φ

= −m2φ, (4.3)

so the equations of motion are

(� +m2)φ(x) = 0. (4.4)

4.1.2 Plane wave solutions

The equations of motion in (4.4) have the general solution φ = a(t)eip·x,where p · x = pµx

µ. This, however, is complex; we want our solution to bereal. Since the reality of an object X in position space requires that it besymmetric under complex conjugation (i.e. X∗ = X), we are able to writea real solution as

φ(x) = a(t)eip·x + a∗(t)e−ip·x. (4.5)

The scalar product p · x can be split into time and space coordinates innatural units: pµ = (p0, ~p) = (Ep, ~p). Using this, we may write the dotproduct as p · x = Ept− ~p · ~x, so that (4.5) becomes

φ(x) = a(t)eiEpt−i~p·~x + a∗(t)e−iEpt+i~p·~x, (4.6)

however this form will only be invoked when convenient. Since we wishto analyse these solutions in the light cone gauge, we must now let xµ =(x+, x−, xI). Inserting this into (4.4) and using the light cone metric ηµν inplace of the flat space metric ηµν in the definition of the d’Alambertian, oneobtains

(� +m2)φ(x+, x−, xI) = (ηµν∂µ∂ν +m2)φ(x+, x−, xI)

=

(2∂

∂x−∂

∂x+− ∂

∂xI∂

∂xI+m2

)φ(x+, x−, xI).

(4.7)Note that the plane wave solution is oscillatory – so in D = (d+ 1) dimen-sions, it can be Fourier transformed into phase space as

φ(x) =

∫dDp

(2π)Deip·x φ(p), (4.8)

where φ(p) is the field in phase space.Just like we require the position space field φ(x) to be real, the reality

condition must also hold for φ(p). Looking at (4.8), this means that thefollowing equation∫

dDp

(2π)Deip·x φ(p) =

∫dDp

(2π)De−ip·x φ∗(p) (4.9)

must be true. Since the integration is over all of p, changing p→ −p has noeffect on the integral. Making this change in the left-hand side of the abovegives: ∫

dDp

(2π)De−ip·x φ(−p) =

∫dDp

(2π)De−ip·x φ∗(p), (4.10)

53


from which one can read off the reality condition in phase space:

φ(−p) = φ∗(p). (4.11)

If we now Fourier transform the x−, xI coordinates into phase space – i.e. weset φ(x+, x−, xI)→ φ(x+, p+, pI), then using the fact that p ·x = ηµνp

µxν =p−x+ + p+x− − pIxI , we can write

φ(x+, x−, xI) =

∫dp+

2πeip

+·x−∫

dD−2pI

(2π)D−2e−ipI ·x

Iφ(x+, p+, pI)

=

∫dp+

2π

∫dD−2pI

(2π)D−2eix−·p+−ipI ·xIφ(x+, p+, pI),

(4.12)

which must then be substituted into (4.7):

(2∂

∂x−∂

∂x+− ∂

∂xI∂

∂xI+m2

)∫dp+ dD−2pI

(2π)D−1eix−·p+−ipI ·xIφ(x+, p+, pI).

(4.13)The above is simplified by evaluating the derivatives, which can be writtenin shorthand as

∂

∂x−:→ ip+,

∂

∂xI:→ −ipI , (4.14)

and results in

(2∂

∂x+ip+ + pIpI +m2

)∫dp+ dD−2pI

(2π)D−1eix−·p+−ipI ·xIφ(x+, p+, pI) = 0.

(4.15)If we now divide the above by 2p+, we get(

i∂

∂x++

1

2p+(pIp

I +m2)

)φ(x+, p+, pI) = 0, (4.16)

allowing us to define

p− =1

2p+

(pIp

I −m2). (4.17)

4.2 Quantum scalar fields

4.2.1 Quantising the solution

The solution (4.5) is quantised by imposing a commutation relation on thecoefficients a(t) and a∗(t). Before we arrive at this, we need to first find theexplicit action, from which the total energy E (and thus the Hamiltonian)can be obtained. Once the Hamiltonian is found, quantising will be a matterof simply imposing the relevant commutation relations.

54


Finding the Hamiltonian

Let us begin by introducing a normalised solution to (4.4) while making thecoefficients a and a∗ depend on time explicitly:

φ(x) =1√V

1√2Ep

(a(t)ei~p·~x + a∗(t)e−i~p·~x

), (4.18)

where V = l1l2 . . . ld is the volume of space in d dimensions (assuming thatspace is a box of sides l1, l2, . . . , ld). The Klein-Gordon action correspondingto (4.18) is

S =

∫dDx

1

2

((∂0φ)2 − (∇φ)2 −m2φ

)(4.19)

which we can now calculate explicitly term-by-term using (4.18). Startingwith the time derivative:

∂0φ =1√V

1√2Ep


), (4.20)

so

(∂0φ)2 =1

2V Ep


)(a(t)ei~p·~x + a∗(t)e−i~p·~x

)=

1

2V Ep

(a2(t)e2i~p·~x + (a∗(t))2e−2i~p·~x + a(t)a∗(t) + a∗(t)a(t)

).

(4.21)Now for the spatial derivative:

∇φ =i~p√

V√

2Ep

(a(t)ei~p·~x − a∗(t)e−i~p·~x

)→ (∇φ)2 =

−(~p)2

2V Ep

(a2(t)e2i~p·~x − a(t)a∗(t)− a∗(t)a(t) + (a∗(t))2e−2i~p·~x

).

(4.22)And finally:

φ2 =1

2V Ep

(a2(t)e2i~p·~x + a(t)a∗(t) + a∗(t)a(t) + (a∗(t))2e−2i~p·~x

). (4.23)

Truncating the time dependence of a and a∗ and inserting (4.21), (4.22) and(4.23) back into (4.19) gives

S =

∫dt

∫ddx

1

4V Ep

[(a2e2i~p·~x + (a∗)2e−2i~p·~x + aa∗ + a∗a

)+ ~p2

(a2e2i~p·~x − aa∗ − a∗a+ (a∗)2e−2i~p·~x

)−m2

(a2e2i~p·~x + aa∗ + a∗a+ (a∗)2e−2i~p·~x

)].

(4.24)This can be simplified a great deal by applying periodic boundary conditionson the momenta pi:

55


pi =2πniLi

. (4.25)

Taking a closer look at the∫ddx e±2i~p·~x integrals, it can be seen that, if we

restrict ~p such that ~p > 0 always, then we have∫ddxe±2i~p·~x ∼ δ(d)(~p) = 0 , ∀~p > 0. (4.26)

Another simplification follows from the fact that, at this point, the objectsa, a and their complex conjugates are not operators, so the terms of the formaa∗ and a∗a double up. With these simplifications, equation (4.24) reducesto

S =

∫dt

∫ddx

1

4V Ep

[2a∗a− 2a∗a~p2 − 2m2a∗a

](4.27)

However since φ is on-shell (i.e. it obeys E2 = p2 +m2), the above becomes

S =

∫dt

∫ddx

1

2V Ep

[a∗a− E2

pa∗a]. (4.28)

The spatial integral over∫ddx is now trivial: each term in the square brack-

ets contributes a factor of V . Restoring the time dependence of a and a∗,we find the action

S =

∫dt( 1

2Epa∗(t)a(t)− 1

2Epa

∗(t)a(t)), (4.29)

so our initial Hamiltonian is given by

H =1

2Epa∗(t)a(t) +

1

2Epa

∗(t)a(t), (4.30)

which will be simplified in due course.

Finding and solving the equations of motion

Our target is to first find the equations of motion of a(t) arising from (4.29)and then to determine their solution. This is done in the usual way byapplying the classical Euler-Lagrange equations given by (1.19) on (4.29).However, the fact that the Lagrangian in question contains complex numbersmakes the direct application of (1.19) problematic. Fortunately, the solutionis simple. We replace a(t) with x+ iy where x, y are real functions of time.Then:

a∗a = x2 + y2, a∗a = x2 + y2, (4.31)

so the action becomes:

S =

∫dt

(1

2Ep(x2 + y2)− 1

2Ep(x

2 + y2)

). (4.32)

Notice that the brackets contain the Lagrangian of a particle with two de-grees of freedom parameterised by the generalised coordinates x and y! Thismeans that we can find their equations of motion independently. Applyingthe Euler-Lagrange equations is now a simple exercise, and results in

56


x− E2px = 0, y − E2

py = 0. (4.33)

Here we can be smart here and argue that, if a(t) is a complex linear com-bination of x and y, it must obey the same equations of motion. So we canessentially write down our target equations of motion for a(t) as

a(t) = −E2pa(t). (4.34)

This is straightforward to show. Using a = x+ iy, differentiating twice withrespect to time gives a = x+ iy. Substituting x and y from (4.33) into thisgives

a = E2px+ iE2

py −→ a = −E2pa, (4.35)

which is the result we guessed. The equations of motion in (4.34) are easilysolved in terms of complex exponentials:

a(t) = ape−iEpt + a∗−pe

iEpt, (4.36)

where the coefficients ap and a−p are constants. Note that since the realitycondition X∗ = X does not hold in (4.36), we say that the function a(t) isa complex dynamical variable.

4.2.2 The Hamiltonian and spacetime momentum

The Hamiltonian

In this subsection we will focus on the tools needed for quantisation: theHamiltonian and momentum. We will start with the Hamiltonian as it canbe found easily by inserting (4.36) into (4.30). In order to do so, we willneed

a(t) = −iEpape−iEpt + iEpa∗−pe

iEpt −→ a∗(t) = iEpa∗peiEpt − iEpa−pe−iEpt,

(4.37)along with

a∗a = −E2p

(a∗−pa

∗pe

2iEpt − a∗−pa−p − apa∗p + apa−pe−2iEpt

)(4.38)

and

a∗a = a∗pap + a∗pa∗−pe

2iEpt + a−pape−2iEpt + a−pa

∗−p. (4.39)

Plugging the above three into (4.30) gives

H =1

2Ep

[−��

��a∗−pa∗pe

2iEpt + a∗−pa−p + apa∗p −��

��

apa−pe−2iEpt

+ a∗pap +��a∗pa

∗−pe

2iEpt + a−pa∗−p +��

��

a−pape−2iEpt

] (4.40)

which finally gives

H = Ep(a∗pap + a∗−pa−p

)=∑p

Ep a∗pap. (4.41)

57


The momentum

Finding the momentum is a much trickier task. Unlike with the Hamiltonian,we have no starting expression for it, which means one has to be found. Theresult we are after is an integral expression for the spacetime momentum,which will be labelled ~P . This expression is given, in terms of the stress-energy tensor Tµν , by

~P =

∫ddx T 0i. (4.42)

Looking at the above, we know we have some work to do since, at thisstage, we have no expression for Tµν . It turns out that imposing translationinvariance on the Klein-Gordon Lagrangian in (4.1) gives rise to Tµν . Letus see how this works. We start by translating the field φ in (4.1) so thatit’s input coordinates go from xµ → xµ′, such that xµ′ = xµ + aµ. Here wenotice that the infinitesimal translation of the coordinates is given by

δxµ = aµ. (4.43)

Since this is a global and continuous transformation, we can invoke Noether’stheorem (1.47) to find the conserved current δJµ if the transformation givenby (4.43) is a symmetry. We know it is a symmetry – we do not expectthe equations of motion to change if we move the field around a little. Inorder to use Noether’s theorem, however; just knowing δxµ is not enough.We also need to know δφ – the transformation of the field due to δxµ – andfor this we need φ′(x). These are found using a slightly different definitionof the derivative:

∂νφ =−(φ(xν − δxν)− φ(xν))

δxν. (4.44)

If we now define φ′(x) = φ(xν − δxν), then the above becomes

−δxν(∂νφ) = φ′(x)− φ(x), (4.45)

and here the right-hand side is the variation of the field δφ:

δφ = −δxν(∂νφ) = −aν∂νφ. (4.46)

We now have all the ingredients to find the conserved current. Inserting(4.43) and (4.46) into (1.47) gives

δJµ =∂L

∂(∂µφ)δφ+ Lδxµ −→ δJµ = − ∂L

∂(∂µφ)(av∂

νφ) + Laµ. (4.47)

This can be simplified by noting that aµ = ηµνaν and factoring out −aν :

δJµ = −[

∂L∂(∂µφ)

∂νφ− Lηµν]aν , (4.48)

where the square brackets enclose the definition of the stress-energy tensorTµν :

Tµν =∂L

∂(∂µφ)∂νφ− Lηµν . (4.49)

58


We can now proceed with finding the spacetime momentum! The aboveequation can be used to find T 0i:

T 0i =∂L

∂(∂0φ)∂iφ =

∂L∂(∂tφ)

~∇φ (4.50)

since η0i = 0. Substituting this into (4.42) gives us our target expression:

~P =

∫ddx

∂L∂(∂tφ)

∇φ. (4.51)

Having found the required expression, the momentum is found by substitut-ing the Lagrangian

L =1

2(∂tφ)2 − 1

2(∇φ)2 − 1

2m2φ2 (4.52)

and the solution

φ(x) =1√V

1√2Ep


)(4.53)

into (4.51):

∂L∂(∂tφ)

= ∂tφ =1√V

1√2Ep


),

∇φ =1√V

1√2Ep

i~p(a(t)ei~p·~x − a∗(t)e−i~p·~x

),

−→ ~P =

∫ddx

i~p

2V Ep

(aae2i~p·~x − aa∗ + a∗a+ aa∗e−2i~p·~x

) (4.54)

which can be simplified (just as in the case of the Hamiltonian) by usingthe fact that the

∫ddx e±2i~p·~x integrals vanish, and that the d-dimensional

integral cancels the factor of V :

~P =i~p

2Ep

(a∗(t)a(t)− a(t)a∗(t)

)(4.55)

This calculation involves quantities that have been found previously in (4.37),together with the solution of a(t) in (4.36). Plugging these in and simplifyingthe result gives

~P = ~p(a∗pap − a∗−pa−p

)=∑p

~p a∗pap (4.56)

which gives us our target expression for the spacetime momentum ~P .

4.2.3 Quantising the solution

Quantisation is achieved by first promoting the object a(t) to an operatorand then by imposing commutation relations between a(t) and its Hermitianconjugate a†(t). If a(t) becomes an operator then so too must ap and a−p.The first commutator we will impose is

[ap, a†k] = δpk, (4.57)

59


and all other commutators of ap and a−p vanish. Next we will impose thecommutation on a(t), a†(t) and their time-derivatives:

[a(t), a†(t)] = [a†(t), a(t)] = 2iEp. (4.58)

We now have to promote the Hamiltonian in (4.41) and the momentum in(4.56) to operators:

H → H =∑p

Ep a†pap, (4.59)

and

~P → ~P =∑p

~p a†pap. (4.60)

Now we have successfully found the creation and annihilation operators: a†pand ap respectively. This corresponds exactly to what we found in Section1.8.2.

The vacuum state

We now introduce a state of space which is completely empty – we haveto put particles into it. This is known as the vacuum and is defined as|0〉. This state is made to vanish once the operator ap is applied to it, andit contains one particle of momentum ~p and energy Ep when the creation

operator a†p is applied. These statements can be written mathematically as

ap |0〉 = 0, (4.61)

and

a†p |0〉 contains one particle of momentum ~p and energy Ep. (4.62)

Similarly, one can construct a state with more particles:

a†p1a†p2 . . . a

†pk|0〉 . (4.63)

which is a state containing k particles with momenta ~p1, ~p2, . . . ~pk and ener-gies Ep1 , Ep2 , . . . Epk .

Particle states in the light cone gauge

The above ideas can easily be transferred to the light cone gauge which ismore relevant and useful for our purposes. The momentum operator in the

light cone gauge ~P = (P+, P I , P−) as

P+ =∑p+,pI

p+ a†p+,pI

ap+,pI ,

P I =∑p+,pI

pI a†p+,pI

ap+,pI ,

P− =∑p+,pI

1

2p+

(pIp

I −m2)a†p+,pI

ap+,pI ,

(4.64)

60


where in the last line we used the fact that p− = 12p+

(pIp

I −m2). This can

be proven by simply expanding p2 in the light cone gauge and rearrangingfor p− while using p2 +m2 = 0:

p2 +m2 = 0 −→ (2p+p− − pIpI) +m2 = 0 −→ p− =1

2p+

(pIp

I −m2).

(4.65)Notice that the equations in (4.64) contain no sums over p−. This is alsoexplained by the above equation. Since p− is determined completely in termsof p+ and pI , a summation over it would offer no uniqueness; it would justcontribute a multiplicity of terms that are already summed over p+ and pI .

4.3 Building particle states

4.3.1 Example 1: photon states

We will now consider two examples of how the idea of particle states canbe applied to different theories. The goal is to start in position space, thenmove over to phase space and then analyse the gauge condition in light conecoordinates.

Starting with the electromagnetic theory in the absence of sources (i.e.jµ = 0), we write the field strength Fµν in terms of the vector potential Aµ

as

Fµν = ∂µAν − ∂νAµ (4.66)

and consider the gauge transformation in terms of the gauge parameter ε:

δAµ = ∂µε. (4.67)

The field equations of motion are then found by imposing ∂µFµν = 0:

∂νFµν = ∂ν∂

µAν − ∂ν∂νAµ, (4.68)

which gives

�Aµ − ∂µ (∂νAν) = 0. (4.69)

Vector potential and gauge condition in phase space

The Fourier transform of Aµ is given by

Aµ(x) =

∫dDp

(2π)Deip·xAµ(p), (4.70)

where the same reality condition as in (4.11) holds:

Aµ(−p) = (Aµ(p))∗. (4.71)

In order to find the equations of motion in phase space we insert (4.70) into(4.69):

�Aµ = −p2Aµ, ∂νAν = ipνA

ν , ∂µAν = −pµ(p ·A), (4.72)

61


so we get

p2Aµ − pµ(p ·A) = 0. (4.73)

The next step is to also bring the gauge transformation to phase space. Bynow we have enough experience with Fourier transformations to know that aderivative in position space is ∼ ipµ in phase space. So we will immediatelywrite

δAµ = ipµε(p). (4.74)

Light cone gauge condition

Since the vector potential is related to the gauge parameter through (4.74),both of these objects can be analysed in light cone coordinates together. Asusual we will write (4.74) by splitting its coordinates:

δA+ = ip+ε, δA− = ip−ε, δAI = ipIε. (4.75)

We now arrive at the light cone gauge condition. Notice that, if we let

A+ −→ A′+ = A+ + ip+ε, (4.76)

then choosing ε = iA+

p+will make the above vanish. This vanishing is the

definition of the light cone gauge in Maxwell theory:

A+(p) = 0. (4.77)

The effect of this gauge condition on the equations of motion in (4.73) mustnow be analysed. Letting µ = + means that the first term of (4.73) vanishesautomatically, leaving the equations of motion as

p+(p ·A) = 0 −→ p ·A = 0. (4.78)

This in turn implies that

p+A− + p−��>

0A+ − pIAI = 0 −→ A− =

pIAI

p+. (4.79)

Gauge equivalence

Since the dot product p · A is independent of the choice of index for µ, itmust vanish for any choice on account of (4.78). This means that (4.73)reduces to

p2Aµ(p) = 0 ∀µ ∈ {+,−, I}. (4.80)

An interesting situation arises when we consider the transverse fields AI .Due to the above equation, we know that p2AI = 0. This gives rise to twopossible solutions:

• p2 = 0, AI 6= 0. In this case, the AI are completely unconstrained,meaning that there are (D − 2) degrees of freedom for each point onthe mass shell, and

62


• p2 6= 0, AI = 0. In this case the full gauge field (that is, all possiblechoices for µ in the ligh tcone gauge) vanishes.

The latter of the two cases provides an example of what is known as a puregauge. The AI vanishing means that, for each of the (D−2) possible choicesof the index I, the field is gauge equivalent to the zero field. So a puregauge is one which differs from the zero field only by a gauge transformation.To put it more mathematically: if

Aµ = A′µ + ∂µχ, (4.81)

where χ is a scalar function, then Aµ and A′µ are gauge equivalent. If A′µ isnow set to vanish, then Aµ = ∂µχ. However since Aµ is gauge equivalent toA′µ – which is now a zero field – then it only differs from the zero field bya gauge transformation ∂µχ. Therefore it is a pure gauge. In phase space,the pure gauge is written as

Aµ(p) = ipµχ(p). (4.82)

One-photon states

Now we will see how to insert photons into the vacuum. Choosing the casewhere p2 6= 0, let us first show explicitly that Aµ is a pure gauge by rewritingthe equations of motion in (4.73) as

p2Aµ = pµ(p ·A) −→ Aµ = pµ

(p ·Ap2

)= ipµ

(−i(p ·A)

p2

), (4.83)

from which, by comparison with (4.82), we deduce that Aµ is indeed a puregauge:

χ(p) =−i(p ·A)

p2. (4.84)

In order to build photon states, we need creation and annihilation opera-tors, which both run through the set of transverse coordinates I. One cantherefore naively write a one-photon state as

aI†p+,pI

|0〉 . (4.85)

However the above does not take into account the possible polarisationsof the photons. Classically, we know that a photon in four-dimensionalspacetime can be be written as a superposition two plane waves. This isbecause D − 2 = 2 when D = 4. So for arbitrary D, we must have D − 2polarisation states. If we now define a polarisation vector εI , where I is thepolarisation index, then we can correctly write a general one-photon stateas

D−1∑I=2

εIaI†p+,pI

|0〉 . (4.86)

63


4.3.2 Example 2: graviton states

Linearisation of gravity

The analysis of gravitational fields here will source from the approximationthat gravity is weak. This is the large distance limit of string theory. Amore formal statement of this approximation is that the curved spacetimemetric – gµν(x) – is described as a perturbation of the flat metric ηµν , suchthat:

gµν(x) = ηµν + hµν(x), (4.87)

where hµν is a small perturbation. When gravity is present, the line elementds2 is written in terms of the metric gµν :

ds2 = gµν(x)dxµdxν . (4.88)

In order to derive linearised equations of motion for the perturbation hµν ,we first recall Einstein’s equation in a vacuum [4, p. 456]:

Rµν = ∂γΓγµν − ∂νΓγµγ + ΓγµνΓδγδ − ΓγµδΓδνγ = 0, (4.89)

where Γσµν is the Christoffel symbol, defined as [2, p. 203]:

Γσµν =1

2gσρ (∂µgρν + ∂νgµρ − ∂ρgµν) . (4.90)

With the above two equations in mind, the equations of motion for hµν arefound by evaluating the variation of (4.89). Before starting out, we firstnotice that the last two terms of the initial right-hand side of (4.89) willgive us terms O(h2). Since hµν is a small perturbation, we will ignore suchterms. Evaluating the variation then gives

δRαβ = ∂γ(δΓγαβ)− ∂β(δΓγαγ)

= ∂γηγδ (∂βhδα + ∂αhδβ − ∂δhαβ)− 1

2∂βη

γδ (∂γhδα + ∂αhδγ − ∂δhαγ)

= ηγδ (∂γ∂βhδα + ∂γ∂αhδβ − ∂γ∂δhαβ − ∂β∂γhδα − ∂β∂αhδγ + ∂β∂δhαγ)

= ηγδ∂γ∂αhδβ −�hαβ − ∂β∂αh+ ηγδ∂β∂δhαγ

= �hµν + ηµαηνβ∂β∂αh− ηµαηνβηγδ∂γ∂αhδβ − ηµαηνβηγδ∂β∂δhαγ ,(4.91)

where h = ηµνhµν and hµν = ηµαηνβhαβ. After contracting with the flatspacetime metric and relabelling the summed indices, the above can besimplified to

�hµν + ∂µ∂ν − ∂α(∂µhνα + ∂νhµα) = 0. (4.92)

Gauge invariance of the linearised gravity

Since we are interested in phase space, we will immediately write the Fouriertransform of (4.92), which we will define as Sµν(p):

Sµν(p) = p2hµν − pα (pµhνα + pνhµα) + pµpνh = 0. (4.93)

We want to determine whether (4.93) is invariant under the following gaugetransformation:

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∂0hµν(p) = ipµεν(p) + ipνεµ(p). (4.94)

Note that the zero subscript on δ in the above is there to emphasise thatwe are looking at the terms linear in hµν when we consider its variation. Tocheck this gauge invariance, we apply the variation δ0 on Sµν [?]

δ0Sµν =p2(ipµεν + ipνεµ)− pα (pµ(ipνεα + ipαεν) + pν(ipµεα + ipαεµ))

+ 2i(p · ε)pµpν ,(4.95)

which after explicitly expanding out gives a result of zero.As the matrix hνν is symmetric under exchange of its indices, and since

each index is ∈ {+,−, I}, we have a total of six independent sets of indexpairs for µ and ν: (hIJ , h+I , h−I , h+−, h++, h−−). The goal is to make hµν

vanish for any pair of indices involving +. If we apply the variation δ0 ontoh++, h+− and h+I in turn, we get

δ0h++ = 2ip+ε+, (4.96)

δ0h+− = ip+ε− + ip−ε+, (4.97)

andδ0h

+I = ip+εI + ipIε+. (4.98)

We see that choosing ε+, ε− and εI appropriately can make the above threeequations vanish. This leaves us with three fields instead of the six westarted out with:

(hIJ , h+I , h−I , h+−, h++, h−−) −→ (hIJ , h−I , h−−). (4.99)

The fact that h+ν vanishes for any ν ∈ {+,−, I} has a significant effect onthe equations of motion in (4.93). Setting µ = ν = +, (4.93) becomes

(p+)2h = 0 −→ h = 0 (4.100)

since p+ > 0. The vanishing of h can be analysed in greater detail. Sincewe know that h = hµνηµν , we can set h = hµν ηµν and find a light coneexpansion:

h = 2h+− − hII = 0 −→ hII = 0, (4.101)

since both h and h+− vanish independently. The above equation shows thatthe matrix hIJ is traceless.

Let us now take a look at what happens if we set one index µ = + in(4.93). Every term containing hµσ must now vanish for any σ:

p2��*0

hµν − pαpµhνα − pαpν��*0

hµα + pµpν��0h −→ pµ(pαh

να) −→ pαhνα = 0,

(4.102)which must be true for any choice of ν since it is a free index. Considernow the expansion of the final right-hand side of the above in light conecoordinates:

pαhνα = 0 −→ p+hν− + p−hν+ − pIhνI = 0, (4.103)

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which allows us to solve for hν−:

hν− =1

p+pIh

νI . (4.104)

The above equation holds for when ν = − and when ν = I. Therefore wecan write:

hν− =1

p+pJh

νJ , ∀ν ∈ {−, I}. (4.105)

Constructing graviton states

Together with the fact that h = 0, the final equation in (4.102) reduces theequations of motion in (4.93) quite significantly as it holds for any choice ofν. In light of these two conditions, (4.93) reduces to

p2hµν = 0. (4.106)

If we now multiply (4.105) by p2, the result must vanish on account of(4.106). Since we have already ruled out the + indices, the above equationholds also for purely transverse indices, meaning that

p2hIJ = 0. (4.107)

Now we can apply a similar method of analysis to the one we employed whenconstructing photon states. The above equation has two solutions:

• p2 6= 0, hIJ = 0. In this case the full gauge field (that is, all possiblechoices for µ and ν in the light cone gauge) vanishes.

• p2 = 0, hIJ 6= 0. In this case, the only constraint on hIJ is that itmust be traceless.

The states are thus constructed by analogy with the photon states. Thistime we have two transverse indices, so a good starting point is

aIJ†p+,pI

|0〉 . (4.108)

However, the above equation fails to capture the tracelessness conditionof hIJ . In order to introduce this into the above, we have to understandhow many degrees of freedom a square matrix of size (D − 2) has. Notethat degrees of freedom here refers to the number of independent elementswithin the matrix.

Consider a square matrix Xij of dimension d. The maximum number ofindependent elements this matrix can have is d2. If we now require that Xij

is symmetric, the upper diagonal elements must equal the lower diagonalelements. Since the diagonal has d elements, the diagonal line just above ithas (d − 1) elements, and the one just above that has (d − 2) independentelements. This equates to

(d + (d − 1) + (d − 2) + · · · + (d − d + 1) + 1

)degrees of freedom, so we have

Number of degrees of freedom N =d∑i=1

d =1

2d(d+ 1). (4.109)

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Now we must impose the tracelessness condition. To find how this changesthe number of degrees of freedom, we write the trace explicitly:

Tr(X) =d−1∑i=0

Xii = X0,0 +X1,1 + . . . . (4.110)

In order to make the above vanish, the final diagonal element – X(d−1),(d−1)

– must equal the negative sum of all the diagonal elements preceding it:

X(d−1),(d−1) = −d−2∑i=0

Xii. (4.111)

The conclusion is that, before imposing tracelessness, the diagonal had ddegrees of freedom. Once it becomes traceless, the final element is fixed interms of the preceding ones, so it is no longer independent. This means thatwe have lost one degree of freedom, so the total number of them is now givenby

N =1

2d(d+ 1)− 1. (4.112)

Coming back to our case of interest where the dimensionality of hIJ is D,the number of degrees of freedom it has can be rewritten as

N =1

2D(D − 3). (4.113)

We can now write the graviton states. All we have to do is include a gravitonpolarisation vector εIJ which fixes the polarisation for given values of I andJ . Then, in complete analogy with the photon states discussed earlier, weconclude with the statement that the one-graviton states can be written as

D−1∑I,J=2

εIJaIJ†p+,pI

|0〉 . (4.114)

4.4 Conclusion

4.4.1 The non-relativistic string

Through our analysis of the classical non-relativistic string in Chapter 1, wefound that by picturing the string in two different ways, we were able toextract different information. This is summarised as follows:

1. Picturing the string as a collection of beads enabled us to vary thenumber of beads all the way up to infinity. This resulted in us findingthe Klein-Gordon Lagrangian in (1.46).

2. Picturing the string as a collection of point masses connected by springsof stiffness κ, on the other hand, allowed us to find the total energyof the string in a straightforward way. The result was an integral ofthe wave equation up to overall constants, as given by (1.53). Moreimportantly we were able to quantise the string’s energy spectrum andderive the creation and annihilation operators given by (1.70).

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4.4.2 The relativistic string

Relativistic string action

Next, we developed the action of a classical relativistic string. This wasdone by establishing an area functional for the world-sheet traced out bythe string in spacetime, as shown in (2.7).

Gauge invariance of the action

Gauge invariance was central to our analysis in the static and light conegauges. The reparameterisation / gauge invariance of the action was provenby transforming the action in (2.10) from ξ-space to a new arbitrary ξ-space, which is a mapping of the original ξ-coordinates. Transformation ofthe action in such a way showed that the new action has exactly the sameform as (2.10), which is shown in (2.17).

Equations of motion and boundary conditions

Having established the relativistic string action, the next step was to findthe equations of motion using field mechanics. This was done in two ways,the second of which gave rise to boundary conditions as shown by (2.26).The boundary terms are made to vanish using either of the two following:

1. Applying Dirichlet boundary conditions means that the string endpoints are fixed as in (2.28)

2. Applying Neumann boundary conditions requires that the derivativeof Xµ vanishes at the end points, as in (2.27).

4.4.3 The static gauge

In this gauge, the time on the world sheet τ is set to coincide with that ofspacetime, t. The action was then re-written in terms of a transverse velocityof the string, ~v⊥, resulting in (3.16). From this, we were able to deduce that,upon applying Neumann boundary condition in (2.27) and finding the µ = 0component of the momentum density Pσµ:

1. The string end points move transverse to the string’s length, as shownin (3.18)

2. The end points of the string move at the speed of light, as shown in(3.20).

Analysing the equations of motion

By analysing the µ = 0 and µ = i components of the equations of motion in(2.22) separately, we were able to find:

1. The µ = 0 component gives the Hamiltonian of the string as in (3.26)

2. The µ = i components give rise to an effective wave equation given by(3.31).

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4.4.4 The light cone gauge

This gauge requires a general class of parameterisations for τ given by (3.1).By parameterising σ as in equations (3.77) and (3.78), we again found a waveequation and simplified it to yield (3.87). Once the Neumann boundarycondition and the constraints in (3.84) were applied, we found that thefunctions fµ and gµ are equal up to a constant and that fµ′ is periodic in3π.

Light cone solution

The mode expansion of the general solution of the wave equation was anal-ysed in light cone coordinates. This gave rise to the transverse Virasoromodes Ln as shown in (3.123).

4.4.5 Particle states

Using field mechanics, we were able to analyse and construct particle statesfor two different theories:

1. Photons. Photon states were constructed by imposing a light conegauge condition (4.77) on the Maxwell fields Aµ as shown in (4.86).

2. Gravitons. The graviton states were constructed in a similar fashion.First the gauge invariance of the phase space solution to Einstein’sequations for linearised gravity was shown, after which we were ableto construct the states shown in (4.114).

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REFERENCES

[1] Arfken, G. B., Weber, H. J., and Harris, F. E., 2013. MathematicalMethods for Physicists. 7th edition. Oxford: Academic Press.

[2] Collier, P., 2013. A Most Incomprehensible Thing: Notes towards a gentleintroduction to the mathematics of relativity. Incomprehensible Books

[3] Goldstein, H., Poole, C. P., and Safko, J. L., 2002. Classical Mechanics.2nd edition. San Francisco: Addison Wesley.

[4] Hartle, J. B., 2003. Gravity: An Introduction to Einstein’s General Rel-ativity. San Francisco: Addison Wesley.

[5] Maggiore, M., 2005. A Modern Introduction to Quantum Field Theory,Oxford: Oxford University Press.

[6] Riley, K. F., Hobson, M. P., and Bence, S. J., 2006. Mathematical Meth-ods for Physics and Engineering. 3rd edition. New York: CambridgeUniversity Press.

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