String Theory I GEORGE SIOPSIS AND STUDENTS

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1 String Theory I GEORGE SIOPSIS AND STUDENTS Department of Physics and Astronomy The University of Tennessee Knoxvie, TN U.S.A. e-mai: siopsis@tennessee.edu Last update: 26

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3 Contents 1 A first ook at strings Units Why Strings? Point partice Strings The Mode Expansion Cosed strings Gauge invariance iii

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5 UNIT 1 A first ook at strings Foowing String Theory by J. Pochinski, Vo.I. Notes written by students (work sti in progress). For more information contact George Siopsis siopsis@tennessee.edu 1.1 Units First, we must expain the unit convention we are going to use. Take the foowing two resuts from Quantum Mechanics and Specia Reativity: E = ω (1.1.1) E = mc 2 (1.1.2) These two equations ink energy to frequency and mass through some constant of proportionaity. The question is, are these constants fundamenta in nature or created by man? The answer is that they are artificia creations, existing purey because of the units we have chosen to work in. We coud easiy choose units such that = c = 1. By doing this, the number of fundamenta units in the universe is reduced to 1, e.g., energy, a others being reated to it: [energy] = [1/time] = [mass] = [1/ength] (1.1.3) 1.2 Why Strings? Our motivation behind the deveopement of String Theory is our desire to find a unified theory of everything. One of the major obstaces that previous theories have been unabe to overcome is the formation of a quantum theory of gravity, and it is in this respect that String Theory has had notabe success

6 2 UNIT 1. A FIRST LOOK AT STRINGS (in fact, at present, String theory is the ony theory which incudes gravitationa interactions). This eads us to beieve that, whie String Theory may not be the fina answer, it is certainy a step in the right direction. Let us first discuss the probems one runs into when trying to create a quantum theory of gravity using Quantum Fied Theory as our guide. Take the Hydrogen atom, whose energy eves are given by: E n = E 1 n 2, E 1 = 2 2m e a, a = 2 m e e 2 (1.2.1) where E 1 is the ground state energy and n abes the energy eves. Suppose we had ony the most basic knowedge of physics: what woud we guess the energy of the Hydrogen atom to be? The parameters of the system are the mass of the eectron m e, the mass of the proton m p, and the eectron charge e. We may negect m p as we are interested in the energy eves of the eectron. We might guess the energy to be E = m e c 2 (1.2.2) Equations (1.2.1) and (1.2.2) are ceary not the same, but if we take the ratio we obtain ( ) 2 E 1 = e4 1 E 2 c 2 = = α 2 (1.2.3) 137 This is a ratio, so is independent of our choice of units, so α is a fundamenta constant that exists in nature independent of our attempts to decribe the word, and indicates some fundamenta physics underying the situation. In fact α is the fine structure constant and describes the probabiity for an eectromagnetic interaction, e.g. proton - eectron scattering (of which hydrogen is a specia case in which the scattering resuts in a bound state). From the diagram of an e-p interaction, INSERT FIGURE HERE each vertex contributes a factor e to the ampitude for the interaction, so that A 1 Ampitude e 2 (1.2.4) Probabiity Ampitude 2 e 4 α 2 (1.2.5) Now according to cassica anaysis, this is the ony ampitude we woud get for the interaction, but in quantum mechanics there can be intermediate scattering events that cannot be observed: e.g., INSERT FIGURE HERE contributes o(α 2 ) to the overa ampitude for the interaction. Inserting a compete set of states, we obtain the ampitude of this second-order process in terms of A 1, de de A 2 E A 1 2 α 2 E (1.2.6)

7 1.2 Why Strings? 3 This is ogarithmicay divergent. However, a higher-order ampitudes have the same divergence and when we sum the series in α: Ampitude = () + α() + α 2 () +... (1.2.7) it yieds finite expressions for physica quantities. Let us try it for the gravitationa interaction between two point masses, each of mass M, separated by a distance r. The potentia energy is: V = GM 2 r (1.2.8) which shows upon comparison with eectromagnetism that the charge of gravity is e g GM. A gravitationa Hydrogen atom wi have energy eves E n E 1 n 2, E 1 Me4 g 2 2 G2 M 5 (1.2.9) Comparing with E = Mc 2, we obtain the ratio E 1 E G 2 M 4 e 4 g (1.2.1) Immediatey we see probems with using this charge to describe the gravitationa interaction, because e g is energy (mass) dependent. The cassica scattering ampitude is A 1 e 2 g GE2 (1.2.11) where E = Mc 2 and the second-order contribution (exchange of two gravitons) is de A 2 E A 1 2 G 2 de (E ) 3 (1.2.12) which has a quartic divergence. Worse yet, higher-order ampitudes have worse divergences, making it impossibe to make any sense of the perturbative expansion (1.2.7) (non-renormaizabiity of gravity). We may see our way to a possibe soution by considering the probem of beta decay: Initiay it was treated as a three body probem with the proton - neutron - eectron interaction occuring at one vertex. When the energies of the resutant eectrons did not match experiment, the theory was modified to incude a fourth partice, the neutrino, and the interaction was smeared out: the proton and neutron interacted at one vertex, where a W boson was created, which traveed a short distance before reaching the eectron - neutrino vertex.

8 4 UNIT 1. A FIRST LOOK AT STRINGS p e p e g g n n v So maybe we can sove our probems with quantum gravity by smearing out the interactions, so that the objects mediating the force are no onger point partices but extended one dimensiona objects - strings. This is the genera concept from which we wi proceed. It is a difficut task - the gravitationa interaction must obey a much arger symmetry than Lorentz invariance, it must be invariant under competey genera co-ordinate transformations, and we must of course sti be abe to describe the weak, strong and eectromagnetic interactions. In this chapter we wi take a first ook at strings. Initiay we examine the competey genera equations of motion for a point partice using the method of east action, and then appy that method to the case of a genera string moving in D dimensions. We wi obtain the equations of motion for the string, and then attempt to quantize it and obtain its energy spectrum. This wi highight some basic resuts of string theory, as we as some fundamenta difficuties. 1.3 Point partice We begin by examining the case of a point partice, iustrating the method we wi use for strings. The trajectory of a point partice in D-dimensiona space is decribed by coordinates X µ (τ), where τ is a parameter of the partice s trajectory. For a massive partice, τ is its proper time. X wi be a timeike cordinate, the remaining X spanning space. Infinitesima distances in spacetime: dt 2 = ds 2 = (dx ) 2 + (d X) 2 (1.3.1)

9 1.3 Point partice 5 a n n We wish to derive the equation of motion from an action principe. This is simiar to Fermat s principe of minimizing time aong a ight ray. For the ight ray joining points A and B, this yieds Sne s Law, n 1 sin θ 1 = n 2 sin θ 2 (1.3.2) Aong the trajectory of a free reativistic partice, proper time is maximized. Therefore, trajectories are obtained as extrema of the action S = m b a b dt (1.3.3) where we mutipied by the mass to obtain a dimensioness quantity (in units where = 1). x t t x a and b are the fixed start and end points on the trajectory. There is a probem that when the mass m =, since the action is zero. This probem wi be ceared up a itte ater.

10 6 UNIT 1. A FIRST LOOK AT STRINGS We can write the action as S = m b a dτ dt dτ = m b a dτ ( ) ( dx 2 d ) 2 X (1.3.4) dτ dτ (invariant under reparametrizations τ τ (τ)) from which we can define the Lagrangian for the system: L = m ( ) ( dx 2 d ) 2 X (1.3.5) dτ dτ Using the convention Ẋ = dx/dτ, we write Lagrange s equations: ( ) d L dτ X = L µ X µ (1.3.6) For this Lagrangian, L X µ = (1.3.7) and we obtain the equation of motion for the point partice: ( ) ( ) d L dτ X = d X µ = (1.3.8) µ dτ L To see the physica meaning of this equation, switch from τ to X (or set τ = X to fix the gauge ). Then the 3-veocity is and the equation of motion (1.3.8) reads v i = dxi dx (1.3.9) ü µ =, u µ = γ(1, v), γ = 1 L = 1 1 v 2 (1.3.1) i.e., that the acceeration is constant, as expected. We wi now ook at a better expression for the action: defining an extra fied η(τ), which at the moment is arbitrary, we write a new Lagrangian: L = 1 2η Ẋµ Ẋ µ 1 2 ηm2 (1.3.11) This has the nice feature that it is sti vaid for m =. Lagrange s equation for η is ( ) d L = = L (1.3.12) dτ η η

11 1.3 Point partice 7 which impies 1 2η 2 Ẋµ Ẋ µ 1 2 m2 = η = ẊµẊ µ m 2 (1.3.13) Using this equation to eiminate η from the Lagrangian (1.3.11), we get back the origina Lagrangian (1.3.5). Varying X µ,we obtain from (1.3.11) ( ) d X µ = (1.3.14) dτ η which agrees with the previous eq. (1.3.8). We wi now examine the meaning of the fied η(τ). The trajectory is parameterized by some co-ordinate of the system in terms of which an infinitesima distance aong the trajectory, ds, can be expressed. Let us suppose our trajectory is aong the y axis. Then it woud be easiest to parameterize the system with the co-ordinate y, and then ds = dy. We coud, however, choose the parameter to be θ, the the ange between a ine drawn from a fixed point on the x axis at a distance to a position on the y axis. Then our new distance woud be ( ) 1 ds = cos 2 dθ (1.3.15) θ The factor /cos 2 θ is our η 2. It represents the geometry of the system due to our choice of co-ordinates. In Minkowski space, ds 2 = γ ττ dτ 2, γ ττ = η 2 (1.3.16) where γ ττ is the (singe component) metric tensor. Under a reparametrization, ( ) 2 dτ τ τ (τ), γ ττ dτ γ ττ (1.3.17) i.e. γ ττ transforms as a tensor (this foows from the invariance of ds 2 ). We can see that the action is invariant under this transformation: we have and X µ transforms as X µ = dxµ dτ Thus the Lagrangian transforms as and the action transforms as S = m b a η = dτ dτ η (1.3.18) = dτ dx µ dτ dτ (1.3.19) L = dτ dτ L (1.3.2) dτl = m b a dτ L (1.3.21)

12 8 UNIT 1. A FIRST LOOK AT STRINGS thus proving its invariance. Now et us form the Hamitonian for the system: the conjugate momenta to co-ordinate X µ and the parameter η are P µ = Then the Hamitonian is: L X = X µ µ η, P η = L η = (1.3.22) H = P µ X µ + P η η L = 1 2 η(p µ P µ + m 2 ) (1.3.23) Here the roe of η is that of a Lagrange mutipier; it is not a dynamica variabe. From Hamiton s equation: H η = P η = = P µ P µ + m 2 (1.3.24) which is Einstein s equation for the reativistic energy of a parice of mass m. Define χ as: χ = 1 2m P µ P µ m = H (1.3.25) ηm Then χ = is a constraint which generates reparametrizations through Poisson brackets: δx µ {X µ, χ} = P µ m, δp µ {P µ, χ} = (1.3.26) We may identify X with time and sove for its conjugate momentum P = P 2 + m 2 (1.3.27) This is the true Hamitonian of the system. Equations of motion: Ẋ i = P P i = P i P, P i = (1.3.28) same equation as before, if we note v i = Ẋi, 1 v 2 = m 2 /P 2 and therefore, v i = P i 1 v 2 m (1.3.29) This system may be quantized by Eigenstates of the Hamitonian: [P i, X j ] = iδ j i (1.3.3) H k = ω k, ω = k2 + m 2 (1.3.31)

13 1.3 Point partice 9 Aternativey, we may define ight-cone coordinates in spacetime: X ± = 1 2 (X ± X 1 ) XT = (X 2,..., X D 1 ) (1.3.32) The Lagrangian reads L = 1 2η Ẋµ Ẋ µ 1 2 ηm2 = 1 η Ẋ+ Ẋ + 1 2η X 2 T 1 2 ηm2 (1.3.33) Let X + = τ pay the roe of time; then P + is the Hamitonian. X and X T are the coordinates and P and P T are their conjugate momenta. The Lagrangian becomes: L = 1 η Ẋ + 1 2η Ẋ2 i 1 2 ηm2 (1.3.34) yieding P = L = 1 Ẋ η (1.3.35) P i = L Ẋi = 1 η Ẋi (1.3.36) The Hamitonian is P + = Ẋ P + Ẋi P i L = P 2 T + m2 2P (1.3.37) Note that there is no term P + Ẋ + because X + is not a dynamica variabe in the gauge-fixed theory. Quantization: Eigenstates of the Hamitonian: [P i, X j ] = iδ j i, [P, X ] = i (1.3.38) P + k, k T = ω + k, k T, ω + = k 2 T + m2 2k (1.3.39) To compare with our earier resut, define ω and k 1 by ω + = 1 2 (ω + k 1 ), k = 1 2 (ω k 1 ) (1.3.4) Then ω 2 k 2 = ω 2 k 2 1 k 2 T = 2ω +k k 2 T = m2.

14 1 UNIT 1. A FIRST LOOK AT STRINGS 1.4 Strings x t=2 t=1 s= s= x Parametrize the string with σ [, ]. In anaogy to Fermat s minimization of time, we wi minimize the area of the word-sheet mapped out by the string. To find an expression for the area, pick a point on the wordsheet and draw the tangent vectors t τ = X τ = X, t σ = X σ = X (1.4.1) The infinitesima paraeepiped with sides t τ dτ and t σ dσ has area da = t τ t σ dτdσ (1.4.2) We obtain the tota area by integrating over the wordsheet coordinates (τ, σ). The action to be minimized is the Nambu-Goto action S NG = T da (1.4.3) where T is a constant that makes the action dimensioness (tension of the string). Using ( a b) 2 = a 2 b 2 ( a b) 2 a = 2 a b a b b 2 (1.4.4) we deduce S NG = T dτdσ det h ab, h ab = a X µ b X µ (1.4.5) where the minus sign in the square root is because we are working in Minkowski space. h ab is the two-dimensiona metric induced on the wordsheet, ( ) X h ab = 2 X X X X (1.4.6) X 2

15 1.4 Strings 11 and the Lagrangian density is L = det h ab = where we ignored T (or set T = 1). ( X X ) 2 X 2 X 2 (1.4.7) σ a = (τ, σ) a =, 1 (1.4.8) Lagrange Equation: a L ( a X µ ) = ( L Ẋµ ) ( ) L + X µ = (1.4.9) We have L = Ẋ µ X 2 X µ X X, Ẋµ L L X µ = X µ X 2 Ẋµ X X L (1.4.1) Let us choose the wordsheet coordinates (τ, σ) so that the metric h ab becomes proportiona to the two-dimensiona Mikowski metric, ( ) 1 h ab (1.4.11) 1 Upon comparison with (1.4.6), we deduce the constraints X 2 + X 2 =, X X = (1.4.12) which may aso be cast into the form ( X ± X ) 2 = (1.4.13) Using the constraints, the Lagrange eq. (1.4.9) reduces to which is the wave equation. Introducing coordinates 2 X µ τ 2 = 2 X µ σ 2 (1.4.14) σ ± = 1 2 (τ ± σ) (1.4.15) so that ± = 1 2 ( τ ± σ ) (1.4.16) the constraints and the wave equation become, respectivey, ( ± X) 2 =, + X µ = (1.4.17)

16 12 UNIT 1. A FIRST LOOK AT STRINGS The genera soution to the wave equation is X µ = f(σ + ) + g(σ ) (1.4.18) where f and g are arbitrary functions. Let us write out the action expicity with the simpifications made above. The parameter τ takes vaues from - to + and σ takes vaues between and, the ength of the string. Anticipating future resuts, we write the constant T T = 1 2πα (1.4.19) where α is caed the Regge sope. Then, from equation (81), S NG = 1 4πα = 1 4πα + + dτdσ( X 2 X 2 ) (1.4.2) dτdσ( a X µ a X µ ) (1.4.21) We now examine the effects of boundary conditions which have yet to be taken into account in the equations of motion. We start off by varying the co-ordinates: X µ X µ + δx µ (1.4.22) Starting from equation (93), the variation in the action is δs NG = 1 4πα + = 1 2πα And noting the tota derivative dτdσ ( a δx µ a X µ + a X µ a δx µ ) (1.4.23) + dτdσ ( a δx µ a X µ ) (1.4.24) a (δx µ a X µ ) = a δx µ a X µ + δx µ a a X µ (1.4.25) The ast term is just the wave equation, which equas zero, so we are eft with: δs NG = 1 2πα = 1 2πα + + dτ dτdσ a (δx µ a X µ ) (1.4.26) [ δx µ Xµ ] (1.4.27) We wi now introduce two sets of boundary conditions that wi get rid of this term and eave the equations of motion unchanged at the boundary: Open string (Neumann) boundary conditions, which correspond to there being no forces at the boundary:

17 1.4 Strings 13 X µ (σ = ) = X µ (σ = ) = (1.4.28) Cosed string boundary conditions, which means there is no boundary and the string co-ordinates are periodic: X µ (σ = ) = X µ (σ = ) (1.4.29) X µ (σ = ) = X µ (σ = ) (1.4.3) We sha now ook at deriving invariant quantities in the theory from symmetries using Noether s theorem. We wi start with Poincare invariance, which is invariant under the transformation X µ Λ µ νx ν + Y µ (1.4.31) where Λ and Y are constant quantities. We construct the Noether current by appying this symmetry to the action. Taking the second term, we write the change in X µ as δx µ = Y µ (1.4.32) The change in the action is found by inserting this into equation (96): δs NG = 1 2πα + dτdσ a Y µ a X µ (1.4.33) The Noether current Pµ a is defined by δs = dτdσ a Y µ Pµ a (1.4.34) So in this case, and P a µ = T a X µ (1.4.35) a P a µ = T a a X µ = (1.4.36) i.e. the Poincare symmetry has ed to a conserved quantity in the Noether current. Now et s do the same for the variation δx µ = ɛλ µ νx ν (1.4.37) where ɛ is a sma quantity. The variation in the action is now δs NG = 1 2πα + From this we can define the current as dτdσ (X µ a X ν X ν a X µ ) Λ µν a ɛ (1.4.38) J µν a = T (X µ a X ν X ν a X µ ) (1.4.39)

18 14 UNIT 1. A FIRST LOOK AT STRINGS which is conserved: a J µν a = T ( a X µ a X ν a X ν a X µ +X µ a a X ν X ν a a X µ ) = (1.4.4) since the first two terms cance and the ast two terms are the wave equation, which equas zero. Anaogous to eectromagnetism, we can define a charge. In EM, the charge is the integra over a voume of the zeroth (time) component of the current 4-vector j µ, so here we define the charge for the current P a µ = (P τ µ, P σ µ ) as P µ = dσp τ µ (1.4.41) P τ µ can be seen to be the momentum of the string at a certain point, so P µ, µ > is the tota momentum of the string, and P is the tota energy of the string. Differentiating P µ with respect to time: dp µ dτ = dσ P τ µ = dσ P µ σ = Pµ σ = (1.4.42) where the second equaity foows from the wave equation and the ast equaity form the boundary conditions at and. This is simpy conservation of momentum. Simiary, for the current Jµν a, which we interpret as the anguar momentum of the string, we define the charge J µν = and we find dσj a µν (1.4.43) dj µν dτ = (1.4.44) So from Poincare invariance we obtain conservation of momentum and anguar momentum. We now give two exampes to demonstrate the above concepts: Exampe 1

19 1.4 Strings 15 t R y t= x We take a cosed string whose initia configuration is a circe centered on the x-y origin with radius R, and whose initia veocity is v =. Then X µ = (t, x, y). The soution to the wave equation satisfying these boundary conditions is: x = R cos 2πτ cos 2πσ (1.4.45) y = R cos 2πτ sin 2πσ (1.4.46) Let s check the constraints Ẋ2 + X 2 = : t = 2πR τ (1.4.47) ṫ 2 + ẋ 2 + ẏ 2 t 2 + x 2 + ý 2 (1.4.48) ( ) 2 ( ) 2 2πR 2πR = + sin 2 2πτ ( 2πR + ) 2 cos 2 2πτ = (1.4.49) The tota energy of the string is given by P = dσp τ = T dσ τ X = T dσ τ t = 2πRT dσ = 2πRT = E (1.4.5) The ength of the string is 2πR, so T can be identified as energy per unit ength of the string, i.e. the tension. Exampe 2

20 16 UNIT 1. A FIRST LOOK AT STRINGS y x Now we consider an open string rotating in the x-y pane. The soution to the wave equation is x = R cos πτ cos πσ (1.4.51) y = R cos πτ sin πσ (1.4.52) t = πr τ (1.4.53) The speed of each point on the string is given by v = ( dx dt, dy ) = ( dx dt πr dτ, dy ) = cos πσ ( dτ From which we see that v 2 = cos 2 πσ sin πτ πτ ), cos (1.4.54) (1.4.55) Thus at the ends of the string σ =, we see that v 2 = 1, i.e. the ends of the string trave at the speed of ight. This is to be expected, since the string is massess and there are no forces on the ends of the string (Neumann boundary conditions). The intermediate points on the string don t trave at the speed of ight because they experience the tension of the string. The energy of the string is worked out exacty as before, and is found to be: P = T Rπ (1.4.56) Let us now work out the z component of the anguar momentum of the string:

21 1.5 The Mode Expansion 17 = T dσ πr2 J x y = T dσ(x a y y a x) (1.4.57) ( cos 2 πτ cos 2 πσ + sin 2 πτ = T πr2 dσ cos 2 πσ cos 2 πσ ) (1.4.58) = 1 2 T πr2 (1.4.59) From the tota energy and anguar momentum we can form the quantity J J xy E 2 = 1 2πT = α (1.4.6) α E 2 Regge Sope Now if we want our strings to correspond to fundamenta partices, their anguar momenta correspond to the spins of the partices, which are either integer or haf integer. Thus the above reation suggests that if we pot the spins of partices against their energies squared, we woud observe a straight ine. This was indeed observed for strongy interacting partices. In fact, string theory started out being a theory of the strong interaction, but then QCD came aong. Now string theory become a theory of everything! 1.5 The Mode Expansion We must first introduce the Hamitonian formaism for future reference: we have the Lagrangian The conjugate momentum to the variabe X is L = 1 2 T (Ẋ2 X 2 ) (1.5.1)

22 18 UNIT 1. A FIRST LOOK AT STRINGS Π µ = L = T Ẋµ (1.5.2) Ẋµ Then the Hamitonian is given by H = = 1 2 T dσ(π µ Ẋ µ L) (1.5.3) ( ) Π 2 dσ T 2 + X 2 (1.5.4) = 1 2 T dσ(ẋ2 + X 2 ) = (1.5.5) So this is not a good Hamitonian. We wi find a good one ater. Now et us examine the mode expansion for open strings, which obey the Neumann boundary conditions. The genera soution to the wave equation is a fourier expansion. Here we write such an expansion as foows: X µ = x µ + 2α π pµ τ + i 2α x 1 ( n αµ n e πinτ nπσ ) cos (1.5.6) where x µ is a constant and the first mode p µ has been written out expicity. The constants have been chosen on dimensiona grounds The fact that X µ must be a rea number yieds the condition: The conjugate momentum to X µ is Π µ = T Ẋµ = 2α T π pµ + α T α n = (α n ) (1.5.7) ( α µ n e πinτ nπσ ) cos x The centre of mass position of the string is given by X µ = 1 (1.5.8) dσx µ = x µ + 2α π pµ τ (1.5.9) so the centre of mass moves in a straight ine. The tota momentum is P µ = dσπ µ = 2α π pµ T = p µ (1.5.1) In both cases a the harmonic terms vanish on integration. We now move to the ight cone gauge mentioned before: we defined the transverse co-ordinates X + and X and fix the gauge by imposing the condition: X + = x + + 2α π p+ τ α + n = n (1.5.11)

23 1.5 The Mode Expansion 19 and set so that 2α π p+ = 1 (1.5.12) X + = x + + τ (1.5.13) which is the ight cone gauge condition from before with an arbitrary constant x +. The centre of mass position can now be written The constraint from before: X µ = x µ + pµ p + τ (1.5.14) (Ẋ ± X) 2 = (1.5.15) 2(Ẋ+ ± X + )(Ẋ ± X ) + (Ẋi ± X i ) 2 = ((Ẋ + Ẋ1 ) ± ( X + X 1 ))(( X Ẋ1 ) ± ( X X 1 )) + (Ẋi ± X i ) 2 = ((Ẋ ± X ) + (Ẋ1 ± X 1 ))(( X ± X 1 ) (Ẋ1 ± X 1 )) + (Ẋi ± X i ) 2 = (Ẋ ± X ) 2 + (Ẋ1 ± X 1 ) 2 + (Ẋi ± X i ) 2 = (Ẋµ ± X µ ) 2 = (1.5.16) 2(Ẋ+ ± X + )(Ẋ ± X ) = (Ẋi ± X i ) 2 (1.5.17) The mode expansion of the word-sheet fieds: 2(Ẋ ± X ) = (Ẋi ± X i ) 2 (1.5.18) X µ = x µ + 2α π pµ τ + i 2α x ẋ ± x == 2α π p + 2α π (ẋ i ± x i ) 2 = 2α π pi + 2α π 1 n αµ n e πinτ cos x x ( nπσ ) α n e πin (τ±σ) α i n e πin (τ±σ) 2 (1.5.19) (1.5.2) (1.5.21)

24 2 UNIT 1. A FIRST LOOK AT STRINGS You can write α n in terms of αi n : (Ẋi ± X i ) 2 = pi p i p + p + + π2 2 2α n Concentrating on the zeroth mode: From Einstein s equation: 2 p p + = pi p i (p + ) α (p + ) 2 p = 1 2p + p i p i + 1 α α i nα i n + 2 π2 2 (2α ) 3/2 p i f(σ, τ) (1.5.22) n n α i n αi n α i nα i n (1.5.23) = H (1.5.24) p µ p µ + m 2 = (1.5.25) m 2 = p µ p µ = 2p + p p i p i = 1 α n α i n αi n (1.5.26) Now quantizing the system: [ p i, x j] = iδ ij [ Π i (σ), X j (σ ) ] = iδ ij δ(σ σ ) (1.5.27) Π µ = T Ẋµ = pµ + 2α π T X µ = x µ + 2α π pµ τ + i 2α [ Π i (σ), X j (σ ) ] = 1 [ p i, x j] + i m,n x x 1 n α µ n e πinτ 1 n αµ ne πinτ cos ( nπσ ) cos ( nπσ ) [ α i m, α j πiτ n] e (n+m) cos (1.5.28) (1.5.29) ( nπσ ) ( ) mπσ cos [ (1.5.3) α i m, α j ] n = mδ ij δ m+n, (1.5.31) [ Π i (σ), X j (σ ) ] = iδ ij δ(σ σ ) (1.5.32) This is just the commutation reation for the harmonic osciator operators with nonstandard normaization a = 1 m α i m a = 1 m α m i (1.5.33) [ a, a ] = 1 (1.5.34)

25 1.5 The Mode Expansion 21 The state, k is defined to be annihiated by the owering operators and to be an eigenstate of the center-of-mass momenta M 2 = 1 α M 2, k = 1 α n n a = a, k = (1.5.35) Π m,i, k =, k (1.5.36) α i nα i n = 1 α i= i We have to perform the sum: n ma a = 1 α N (1.5.37) i= i (D 2)n (D 2), k = 2 2α n n n, k (1.5.38) n (1.5.39) To perform this sum, we wi mutipy by the sum by e 2πnɛ imit of ɛ n ne 2πnɛ = C M 2 = 1 α n n ne nc = ( ) 1 C 1 e C = where C = 2πɛ ne nc = 1 c n α i nα i n = i= i (D 2) 2α The ast equaity was obtained using(64) M 2 = (D 2) 2α ( 1 c 2 1 ) = 12 n and then take the e C (1 e C ) 2 = 1 C (1.5.4) (1.5.41) (1.5.42) n = 2p + p p i p i (1.5.43) (D 2) 2 2) (D 2) 2α (D (2πɛ) 2 24α = (D 2) (2πɛ) 2 πp+ 24α = 2p + p p i p i (1.5.44) The mass of each state is thus determined in terms of the eve of excitation. M 2 = 1 ( α N D 2 ) N = (1.5.45) 24

26 22 UNIT 1. A FIRST LOOK AT STRINGS This operator acting on the ket yieds: M 2 (D 2) = 24α (1.5.46) The mass-squared is negative for D > 2. The state is a tachyon The owest excited states of the string are obtained by exciting one of the n = 1 modes once: M 2 (α i 1 ) = 1 ( α 1 D 2 ) (α i 1 24 ) = 1 ( ) 26 D α (1.5.47) 24 Lorentz invariance now requires that this state be massess, so the number of spacetime dimensions is D = Cosed strings We must now ook at the mode expansion and quantization os cosed strings, which are required when ooking at string interactions. The procedure is very simiar to that of open strings, except now we have Dirichet boundary conditions, i.e. X µ is periodic. The mode expansion is now made up of eft and right moving parts: such that X µ = X µ R (τ σ) + X µ L (τ + σ) (1.6.1) X µ R = 1 2 xµ + α p µ 1 (τ σ) + i 2 α X µ L = 1 2 xµ + α p µ 1 (τ + σ) + i 2 α x x 1 n αµ n e 2πin(τ σ) (1.6.2) 1 n αµ n e 2πin(τ+σ) (1.6.3) The sum of these is periodic. In the sum the integer n is in effect 2n so there are now twice as many modes as for the open string. Once again, the fact that X µ is rea means that Quantizing as before: α µ n = (αµ n ) α µ n = ( αµ n ) (1.6.4) [ α i m, α j ] n = i [ α m, α j ] n = mδ ij δ m+n, (1.6.5) [ α i m, α j ] n = (1.6.6)

27 1.6 Cosed strings 23 The mass operator is now M 2 = 2p + p p i p i = 2 α ( N R + N L + ) 2(D 2) = 2 24 α n α i nα i n + There are two symmetries in these expansions: the transformations n (1.6.7) τ τ + constant (1.6.8) α i n α i n 2(D 2) 24 σ σ + constant (1.6.9) don t change the physics of the string. The first symmetry is shared with the open string, but the spatia transationa symmetry is new. The generator of the time transations is H, which we saw before to be H = dσ(ẋ2 + X 2 ) (1.6.1) and the fact that H = eads to the invariance under time transations. In the case of cosed strings, H = where we have used the fact that we can write dσ[(ẋr + ẊL) 2 + ( X R X L ) 2 ] (1.6.11) dσ(ẋ2 R + Ẋ2 L ) = H R + H L (1.6.12) σ τ (1.6.13) since τ and σ are interchangeabe in the expansions of X R and X L to within a minus sign. This generation of time transations comes from the constraint Ẋ2 + X 2 =, so it is reasonabe to suppose that spatia transations come from the other constraint Ẋ X =. Defining the operator D: D = So dσẋ X dσẋ2 = dσ(ẋ2 R Ẋ2 L ) N R N L = (1.6.14) N R = N L (1.6.15) which common sense tes us must be the case.

28 24 UNIT 1. A FIRST LOOK AT STRINGS Now et us determinant the owest states in the mass spectrum. For the vacuum state M 2 4(D 2) = 24α (1.6.16) so the mass squared is negative for the vacuum state, as for the open string. The first excited state is Ω ij = α i 1 αj 1, where we must remember to keep N R = N L. We obtain M 2 Ω ij = 2 α ( 2 2(D 2) 24 ) Ω ij (1.6.17) As before, we wish M 2 = for the first excited state, so again we get D = 26. The situation is a bit more compicated than for the open string case, so et s ook in a bit more detai. The state Ω ij can be spit into three parts: a symmetric, traceess part; an antisymmetric part and a scaar part: [ Ω ij 1 = 2 ( Ωij + Ω ij ) 2 ] D 2 δij Ω kk ( Ωij Ω ij ) + 1 D 2 δij Ω kk (1.6.18) We ca the three states G ij, B ij, Φ. The symmetric, traceess, spin 2 state G ij can now be identified with the graviton, which didn t exist in the open string theory, so it seems some progress has been made. The spin scaar state is caed the diaton. Now we impose a further symmetry: invariance under the transformation σ σ (1.6.19) which is the condition for unoriented strings. This means X R X L (1.6.2) α n α n (1.6.21) This condition immediatey disaows the antisymmetric state, since under the transformation, B ij B ij (1.6.22) whie G ij and Φ remain unchanged. Next et us turn our attention to the fact that we have been working in ight cone gauge, which is not Lorentz invariant. We woud ike to reassert orentz invariance. We can do this by generaizing the commutation identity: This gives [ α i m, α j ] n = mδ ij δ m+n, [α µ m, αν n ] = mηµν δ m+n, (1.6.23)

29 1.7 Gauge invariance 25 Now et us define a state [ α m, α ] n = mδm+n, (1.6.24) The norm of this state is φ = α 1 (1.6.25) φ φ = α 1α 1 = [α 1, α 1] + α 1α 1 = = 1 (1.6.26) Thus we have a negative norm state, which is physicay meaningess. This tes us there is something wrong with our theory. We wi come back to this in the next coupe of chapters, and sove this probem. 1.7 Gauge invariance Let us examine the open string state under the transformation φ = A µ (k)α µ 1, k (1.7.1) A µ A µ + k µ ω(k) (1.7.2) where ω(k) is an arbitrary function. This is anaogous to a gauge transformation in eectromagnetism A A + ω. The change in φ is and the norm is δφ = k µ ωα µ 1, k (1.7.3) δφ δφ = k µ k ν ω 2, k α µ 1 αν 1, k = k2 ω 2 = (1.7.4) since k = (the mass is zero, and m 2 = k µ k ν ). Thus the theory has produced gauge invariance! The equivaent state for the cosed string is The gauge transformation is This time we find the norm state to be φ = g µν α µ 1 αν 1, k (1.7.5) g µν g µν + k µ ω ν + k ν ω µ (1.7.6) δφ δφ k 2 = (1.7.7) Thus gauge invariance hods for the cosed string state too. What s more, we can identify g µν with the gravitationa potentia, a sign that genera reativity might be incuded in the theory.