String Theory IIaesop.phys.utk.edu/strings/part2.pdfString Theory II GEORGE SIOPSIS AND STUDENTS...

String Theory II

GEORGE SIOPSIS AND STUDENTS

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

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

Last update: 2006

ii

Contents

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

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

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

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

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

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

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

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

iii

iv CONTENTS

UNIT 6

Compactification and Duality

6.1 The Kaluza-Klein Mechanism

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

Fµν = ∂µAν − ∂νAµ,

and the action

S =

∫

d4x

(

−1

4FµνF

µν +AµJµ

)

.

leads to the Maxwell equations

∂µFµν = Jν .

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

µ = 0). The conserved charge is given by

Q =

∫

d3xJ0,dQ

dt= 0.

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

pµpµ = −m2

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

This is obtained from the action

S =1

2

∫

d4x ∂µΦ∗∂µΦ +m2|Φ|2

2 UNIT 6. COMPACTIFICATION AND DUALITY

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

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

S =1

2

∫

d2x(

(∂µ − iqAµ)Φ∗(∂µ + iqAµ)Φ +m2|Φ|2)

comparing with∫

d4x AµJµ, we obtain

Jµ = − iq2

(Φ∗∂µΦ− Φ∂µΦ∗).

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

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

xµ = (t, x, y, z, u), u ≡ u+ 2πR.

The line element is

ds2 = GMNdxMdxN = Gµνdx

µdxν + 2Gµudxµdu+Guudu

2.

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

ds2 = gµνdxµdxν +Guu(du+Aµdx

µ)2

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

f(u) =

∞∑

n=−∞ane

inu/R.

For a field φ(xµ) = φ(xµ, u), we have

φ(xµ) =

∞∑

n=−∞an(x

µ)einu/R.

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

∂µ∂µΦn −

n2

R2Φn = 0

6.2 Strings 3

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

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

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

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

∑

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

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

Dµ∂µ(aneinu/R) = 0

where

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

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

ΓLMN =1

2GLP (∂MGPN + ∂NGMP − ∂PGMN ) .

A short calculation reveals

(∂µ +Aµ∂u)2(ane

inu/R) = 0

∂µ + in

RAµ)an =

n2

R2an

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

S ∼∫

d5x√−GR(5),

where

G = detGMN = −1 R(5) = −1

4FµνF

µν .

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

4FµνFµν which is four-dimensional gravity and electromagnetism!

6.2 Strings

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

S =1

2πα′

∫

d2z ∂U∂U


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

Xµ(σ, τ) = xµ+2πα′pµτ

`+i

√

α′

2

∑

m6=0

1

m

(

αµme−2πim(σ+τ)/` + αµme

2πim(σ−τ)/R)

,

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

Xµ(z, z) = xµ + 2πα′pµτ

`+ i

√

α′

2

∑

m6=0

1

m

(

αµmz−m + αµmz

−m) ,

with derivatives

∂Xµ(z, z) = −iα′

2pµz−1 − i

√

α′

2

∑

m6=0

αµmz−m−1,

∂Xµ(z, z) = −iα′

2pµz−1 − i

√

α′

2

∑

m6=0

αµmz−m−1.

Notice that the momentum is p = 12πα′

∮

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

∮

(dz∂X + dz∂X) = 0. (6.2.1)

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

U(σ + `) = U(σ) + 2πRw, w ∈ Z.

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

2πσ/` = −i/2 ln(z/z), so

U(σ, τ) = − i2wR ln

(z

z

)

has to be added. We obtain

U(z, z) = u− iα′

2

n

Rln |z|2 − i

2wR ln

(z

z

)

+ i

√

α′

2

∑

m6=0

1

m(αmz

−m + αmz−m),

= u− iα′

2

(

n

R+wR

α′

)

ln z − iα′

2

(

n

R− wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

m(αmz

−m + αmz−m),

with the derivatives

∂U(z, z) = −iα′

2pLz

−1 − i√

α′

2

∑

m6=0

αmz−m−1,

∂U(z, z) = −iα′

2pRz

−1 − i√

α′

2

∑

m6=0

αmz−m−1,

6.3 Partition Function 5

where PLR

= nR ± wR

α′ . Notice that

1

2πα′

∮

(dz∂X − dz∂X) =1

2(pL + pR) = p =

n

R,

and1

2πα′

∮

(dz∂X + dz∂X) =1

2(pL − pR) =

wR

α′ 6= 0.

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

L0 =α′p2

L

4+

∞∑

n=1

α−nαn, L0 =α′p2

R

4+

∞∑

n=1

α−nαn

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

6.3 Partition Function

First, let us recall the uncompactified case

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

= Tr (qq)α′p2/4−1/24

∏

n

( ∞∑

Nn=0

qnNn

)

∞∑

Nn=0

qnNn

,

=(

Tr e−πτ2α′p2)

∣

∣

∣

∣

∣

∏

n

(1− qn)−1

∣

∣

∣

∣

∣

2

(qq)−1/24,

= |η(τ)|−2V

∫

dp

2πe−πτ2α

′p2 ,

= 6|η(τ)|−2 V

2π

1√α′τ2

,

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

Notice that for Xµ = xµ − iα′

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

tion,we have ∂Xµ = −iα′

2 pµz−1, ∂Xµ = −iα′

2 pµz−1 and the action is

S =1

2πα′

∫

d2 z∂Xµ∂Xµ =

α′

8πp2

∫

d2z

|z|2 , z = ei(σ1+iσ2),

=α′

8πp22VTorus, (VTorus = 2π(2πτ2))

= α′p2τ2π.

Thereforee−S = e−πτ2α

′p2 ,

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


Partition Function for Compactified Space

Z = Tr (qL0−1/24qL0−1/24),

= |η(τ)|−2∑

n,w

qα′p2L/4qα

′p2R/4, pLR

=n

R± wR

α′

= |η(τ)|−2∑

n,w

exp

[

−πτ2α′(

n2

R2+w2R2

α′2

)

+ 2πiτ1nw

]

.

Use the Poisson resummation formula:

∞∑

n=−∞e−πan

2+2πibn =1√a

∞∑

m=−∞e−π(m−b)2/a.

Therefore the partition function for the compactified case is

Z = |η(τ)|−2 R√τ2α′

∑

m,w

exp

[

−πR2

α′τ2

(

(m− τ1w)2 + τ22w

2)

]

,

= |η(τ)|−2 V

2π√α′τ2

∑

m,w

exp

(

−πR2

α′τ2|m− τw|2

)

,

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

τ → τ + 1 ⇔ m→ m+ w,

τ → −1

τ⇔ m→ −w, w → m.

Notice that the solution with the correct boundary conditions satisfying

U(σ1 + 2π, σ2) = U(σ1, σ2) + 2πwR,

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

can be written as

U(σ1, σ2) =wR

τ2(τ2σ

1 − τ1σ2) +mR

τ2σ2

where

∂1U = wR, ∂2U =R

τ2(m− wτ1).

The action is given by

S =1

4πα′

∫

dσ1dσ2

[

(∂1U)2 + (∂2U)2]

=1

4πα′ 2π2πτ2R2

τ22

|m− wτ |2 + ...


Thermodynamics

From thermodynamics we know the partition function is

Z =∑

E

e−βE, β =1

T

and T is the temperature.

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

Z =∑

E

〈E|e−βH |E〉 = Tr e−βH .

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

Z =∑

E,x,y

〈E|x〉〈x|e−βH |y〉〈y|E〉

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

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

〈x(t)|y(0)〉 =∑

p

〈x|p〉e−iHt/~〈p|y〉,

=

∫

dp e−ipxe−iHt/~eipy.

For x− y ' −qt, so

〈x(t)|y(0)〉 =

∫

dp ei(qp−H)t =

∫

dp eiS , S =

∫ t

0

dt′ (q −H)

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

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

〈x(t)|y(0)〉 = eiS .

This integrates for finite t. Then

Z =∑

e,x,y

eiS(x,y)ψ∗E(x)ψE(y) =

∑

x

eiS(x,x)

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


6.4 Vertex Operators

Recall

U(z, z) = u−iα′

2

(

n

r+wR

α′

)

ln z−iα′

2

(

n

r− wR

α′

)

ln z+i

√

α′

2

∑

m6=0

1

m

(

αmz−m + αmz

−1)

Momenta pL, pR are different. Their eigenvalues are

pL =n

r+wR

α′ , pR =n

r− wR

α′ .

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

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

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

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

UL(z) = uL − iα′

2pL ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m,

UR(z) = uR − iα′

2pR ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m.

The operator product expansions are

UL(z)UL(0) ∼ −α′

2ln z, UR(z)UR(0) ∼ −α

′

2ln z, UL(z)UR(0) ∼ 0.

The vertex operator also splits into a holomorphic and antiholomorphic pieces

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

nr−wRα′ )UR(z) :

c.f., the uncompactified case,

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

The OPE for these vertex operators is

V kL (z)V k′

L (0) = : eik·XL(z) :: eik′·XL(0) :,

∼ eik·k′ α′

2ln z : ei(k+k

′)XL(0) :,

= zα′2k·k′ : V k+k

′

L (0) : .

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

factor eπiα′k·k′ (z

α′2k·k′ = e

α′2k·k′ ln z → e

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

6.5 Amplitudes 9

z−α′2k·k′ picks up a factor e−πiα

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

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

′(0, 0)

In the compactified case, let kL = nr + wR

α′ , kR = nr − wR

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

V kL(z)V k′L(0) ∼ |z|α′kL·k′L/2V

kL+k′LL (0) V kR (z)V k

′R(0) ∼ |z|α′kR·k′R/2V kR+k′R(0)

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

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

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

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

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

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

eiπα′2kL·k′L in the left part and eiπ

α′2kR·k′R in the right part. Overall,

eiπα′(kL·k′L−kR·k′R) = e2πi(nw

′+n′w) = ±1

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

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

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

CkL,kr (p) = eiπα′2

(kL−kR)r, p = /2(pL + pR)

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

When we commute two vertices, we pick up the factor

eiπα′2

(kL−kR) k′2 e−iπ

α′2

(k′L−k′R)k2 = eπi(nw

′−n′w)

Thus the overall factor is now

eπi(nw′−n′w)eπi(nw

′+n′w) = e2πinw′= 1.

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

6.5 Amplitudes

Recall in the uncompactified case,

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

i<j

|zi − zj |α′ki·kj

(6.5.1)


where the vertex operators are given by

Vi(zi) =: eiki·X(zi) : .

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

plicitly

〈eik1·(x−iα′

2p ln |z|2)...eikN ·(x−iα′

2p ln |z|2)〉 = 〈eik1·(x−iα

′2p ln |z|2)...eikN−1·(x−iα

′2p ln |z|2)|kN 〉

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

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

∏

i<j

|zi − zj |α′ki·kj →

∏

i<j

(zi − zj)α′kLikLj/2(zi − zj)α

′kRikRj/2.

The zero modes contribute

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

Cocycles give additional± signs.

6.6 Spectrum

Recall in the uncompactified case:

L0 =α′p2

4+N, N =

∞∑

n=1

α−nαn.

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

L0 =α′(p2 + p2

L)

4+N, p2 = pµp

µ

and

N =

∞∑

n=1

αµ−nαnµ +

∞∑

n=1

α−nαn,

and similarly for L0. The Mass-shell condition states

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

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

is the eigenvalue of pµ. Then

L0 − 1 = 0⇒ α′

4(−m2 + k2

L) +N − 1 = 0⇒ m2 = k2L +

4

α′ (N − 1)

6.6 Spectrum 11

From the antiholomorphic piece we get

m2 = k2R +

4

α′ (N − 1)

Subtract the two and we find

N −N =α′

4(k2L = k2

R) = nw

If we add the two conditions we find

m2 =2

α′ (N + N − 2) +k2L + k2

R

2=n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2)

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

Massless States

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

αµ−1αν−1|0; k〉, αµ−1α−1|0; k〉, α−1α

µ−1|0; k〉, α−1α−1|0; k〉.

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

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

The two vectors are

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

µ−1α−1|0; k〉

The amplitude for |B〉 is

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

The relevent parts of U, Xµ are:

UL = uL − iα′

2pL ln z + ...

UR = rR − iα′

2pR ln z + i

√

α′

2α1z

−1 + ...

Xµ = xµ − iα′

2pµ ln |z|2 + i

√

α′

2αµ1 z

−1 + ... where z = 1


So

A ∼ Bµα′

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

ν1k2να

µ−1α−1|0; k〉

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

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

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

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

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

A+A′ ∼ α′(k2 − k3) ·Bn

Rδ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2)

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

Aµ(k)(αµ−1α−1 + αµ−1α−1)|0; k〉.

The other vectorA′µ(k)(α

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

A−A′ ∼ (k2 − k3) ·BwR

α′ δ(k1 + k2 + k)δ(n1 + n2)δ(w1 + w2)

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

6.7 R =

√α′

When R =√α′, m2 = 0 implies n2

α′ + w2

α′ + 2a′ (N + N − 2) = 0, N −N = nw.

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

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

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

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

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

6.7 R =√α′ 13

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

S =

∫

d4x

[

−1

4FµνF

µν + |DµΦ|2 +m2|Φ|2]

,

where

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

µ, A2µ, A

3µ,

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

−1

4

3∑

i=1

F iµνFiµν

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

d4x∑

i AiµJ

iµ, where J iµ is

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

F iµν = ∂µAiν − ∂νAiµ − εijkAjµAkν

and the action is

S = −1

4

∫

d4x∑

i

F iµνFiµν

and leads to nonlinear Maxwell equations given by

∂µFiµν − εijkAjµF kµν = 0.

Gauge transformations: Aiµ → Aiµ − ∂µλi − εijkAjµλ

k (infinitesimal). Finite

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

Aµ =1

2Aiµσ

i

where σi represents the Pauli spins matrices which obey the algebra

[σi, σj ] = 2iεijkσk


The field strength is

Fµν =1

2F iµνσ

i

S = ei2λiσi ∈ SU(2), SS† = I

For λ 1 we may expand S

S ' 1 +i

2λjσj

Let us see how Aµ = 12A

iµσ

i transforms under a finite transformation

Aµ → S (Aµ − i∂µ)S†

→ (1 +i

2λiσi)(Aµ − i∂µ)(1−

i

2λjσj)

→ Aµ −1

2σj∂µλ

j +i

4λi[σi, σj ]Ajµ +O(λ2)

→ Aµ −1

2σj∂µλ

j − 1

2εijkσiλjAkµ

Aiµ → Aiµ − ∂µλi − εijkλjAkµ (6.7.1)

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

Dµ = ∂µ + iAµ.

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

Φ→ SΦ, Aµ → S(Aµ − i∂µ)S†,

so

DµΦ → ∂µ(SΦ) + iA′µSΦ,

= S(∂µΦ + S†∂µSΦ + iS†A′µSΦ),

= SDµΦ. (6.7.2)

This would have worked nicely had we chosen

Φ→ S′Φ, S′ = ei2qλiσi ,

because iS′†A′µS

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

6.7 R =√α′ 15

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

∂U(z)V (0, 0) ∼ ∂UL(z) : eikLUL :

∼ ikL∂

(

−α′

2ln z

)

: eikLUL(0) :

∼ −ikLα′

2· 1zV (0, 0)

∂U(z)V (0, 0) ∼ −ikRα′

2· 1zV (0, 0)

The charges are

QL =1

2πi

∮

C

dz∂U(z) QR = − 1

2πi

∮

C

dz∂U(z).

so

[QL, V ] = −iα′

2kLV [QR, V ] = −iα

′

2kRV

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

δV = −λ[QL, V ] = −iα′

2kLλV

so V → (1 + iα′

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

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

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

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

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

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

:(since n = w, kR = 0)

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

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

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

where

kL =n

R+wR

α′ = ± 2√α′, kR = ± 2√

α′.

The OPEs are given by


j+(z)j+(0) ∼ : ei 2√

α′ UL(z) :: ei 2√

α′ UL(0) :

∼ e−4

α′ ln(z)−α′2 : e

i 4√α′ UL(0)

∼ z2 : ei 4√

α′ UL(0) : ∼ 0

j−(z)j−(0) ∼ 0

j+(z)j−(0) ∼ z−2 : ei 2√

α′ UL(z)e−i 2√

α′ UL(0) :

= z−2

[

1 + z i2√α′ ∂UL + z2( ) + ...

]

∼ 1

z2+ i

2√α′∂UL

1

z

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

photon.

j+(z)j−(0) ∼ 1

z2+

2

zj3(z) + ...

j3(z)j+(0) ∼ − 2

α′

(

−α′

2∂ ln z

)

j+(0) ∼ 1

zj+(0)

j3(z)j−(0) ∼ −1

zj−(0)

Be defining j1 = 12 (j+ + j−), j2 = 1

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

ja(z)jb(0) ∼ 1

2z2δab + iεabcjc(0).

The corresponding charges

Qa =

∮

dz

2πija(z)

satisfy an SU(2) algebra

[Qa, Qb] = iεabcQc

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

ja(z) =∑

jamz−m−1

6.8 Away fromR =√α′ 17

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

[jaµ, jbν ] =

m

2δm+n,0δ

ab + iεabcjcm+n.

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

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

6.8 Away from R =

√α′

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

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

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

µ−1α

ν−1+

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

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

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

S =

∫

d4x(

|DµΦ|2 + V (Φ))

.

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

µ−1α

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

changes R and corresponds to the vertex

: ∂U(z)∂U(z)eik·X :,

or: j3(z)j3(z)eik·X : .

AtR =√α′, we have additional scalars

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

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

3© : n = ±2, w = 0⇒ kL = kR =n

R= ± 2√

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

4© : n = 0, w = ±2⇒ kL = −kR =wR

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


Putting everything together, we have

: ja(z)jb(z)eik·X :

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

m2 =

(

R2 − α′

Rα′

)2

,

using

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2)

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

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

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

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

Conclusion

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

6.9 T-duality

Recall

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2).

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

R→ R′ =α′

R.

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

and R′ = α′

R are identical.

6.9 T-duality 19

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

UL = uL − iα′

2

(

n

R+wR

α′

)

ln z...

UR = uR − iα′

2

(

n

R− wR

α′

)

ln z...

so

Z = uL − uR − iα′

2

[(

n

R+wR

α′

)

ln z −(

n

R− wR

α′

)

ln z

]

+ ...

(6.9.1)

and theR′ theory has

Z = uL + uR − iα′

2

[(

n

R+wR

α′

)

ln z +

(

n

R− wR

α′

)

ln z

]

+ ...

so

U = uL + uR − iα′

2

[(

w

R+nR

α′

)

ln z −(

w

R− nR

α′

)

ln z

]

+ ...

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

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

√α′ from the string point of view.

T-duality, R ↔ α′

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

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

R −√α′ =√

α′

R (√α′ −R) ' −δR since (R '

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

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

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

√α′ + δR and√

α′ − δR are gauge equivalent.

R =

√α′

k∼= R = k

√α′

Recall

m2 =n2

R2+

(

wR

α′

)2

+2

α′ (N + N − 2).

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

m2 =n2

k2α′ +

(

wk2

α′

)2

− 4

α′ = 0⇒ n = ±2k, w = 0.


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

Therefore, kL = wRα′ = ± 2

α′ = −kR, and the vertex operators are

: eikLULeikRUReik·X :=: e±i2

α′ ULe∓i2

α′ UReik·X :=: j±j∓eik·X :

which is part of the setjajbeik·X :

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

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

√α′ andR =

√α′/2

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

UL = uL − iα′

2

(

n

R+wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m,

UR = uR − iα′

2

(

n

R− wR

α′

)

ln z + i

√

α′

2

∑

m6=0

1

mαmz

−m.

For

R =√α′ : UL = uL − i

√

α′

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

R =

√α′

2: UL = uL − i

√

α′

2(n+

w

2) ln z + ...

To mimic R =√α′/2 at R =

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

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

ψeven =1

2(1 + P )ψ(x) =

1

2(ψ(x) + ψ(−x)),

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

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

′RUR :

The OPE gives

V (z, z)V (z′, z′) ∼ z α′

2kLk

′Le

α′2kRk

′RVkL+kL′ ,kR+k′

R.

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

6.9 T-duality 21

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

√α′/2 circle.

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

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

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

∀k ∈ N.

Open Strings

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

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

m2 =n2

R2+

1

α′ (N − 1).

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

U(z, z) = u− iα′p ln |z|2 + i

√

α′

2

∑

m6=0

1

mαm(z−m + z−m).

For compact U , p = n/R.

z = eπi(σ+τ)/`, z = e−πi(σ−τ)/`

We will set ` = π for simplicity. So

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


so

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

so

Z(σ = π)− Z(σ = 0) =

∫ π

0

dσ∂σZ =

∫ π

0

dσ∂τU =

∫ π

0

dσ∂τ (2α′pτ)

= 2α′pπ = 2α′πn

R= 2πnR′

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

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

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

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

: ∂τXµeik·XeinU/R : (σ = 0) : ∂τUe

ik·XeinU/R :

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

: ∂σZeik·X : .

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

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

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

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

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

Scattering

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

X(z, z)X(0, 0) ∼ −α′

2ln |z|2 + ...

6.9 T-duality 23

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

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

G(z, z; z′, z′) = −α′

2ln |z − z′|2 − α′

2ln |z − z′|2 (6.9.2)

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

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

This shows that the OPE for open strings ought to be

X(z)X(0) ∼ −2α′ ln |z|.

Recall the amplitudes:Closed strings:

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

ikn·X(zn, zn) :〉.

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

∂z1An = 〈: ik1∂Xeik1·X(z1, z1)e

ik2·X(z2, z2)...eikn·X(zn, zn) :〉.

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

21z eik·X (0) we obtain

∂z1An =α′

2An∑

i

k1kiz1 − zi

.

Integrating ...

An ∝∏

i<j

|zi − zj |α′kikj

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

∏

i<j |zi − zj |2α′kikj . For

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


UNIT 7

Superstrings

7.1 Bosons and fermions

Bosonic strings have the action

S =1

2πα′

∫

d2z∂Xµ∂Xµ.

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

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

Sbc =1

2π

∫

d2zb∂c.

and their OPEs are

b(z)c(0) ∼ 1

z.

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

T =: (∂b)c : −λ∂(: bc :)

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

T (z)T (0) ∼ c

2z4+

2

z2T (0) +

1

z∂T (0)

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

26 UNIT 7: Superstrings

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

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

Define

b =1√2(ψ1 + iψ2), c =

1√2(ψ1 − iψ2).

Then the action is

S =1

2π

∫

d2zb∂c =1

4π

∫

d2z(ψ1∂ψ1 + ψ2∂ψ2).

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

T (z) = −1

2ψ1∂ψ1 −

1

2ψ2∂ψ2.

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

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

T = − 1

α′ ∂Xµ∂Xµ −

1

2ψµ∂ψµ.

The TT OPE becomes

T (z)T (0) ∼ (3D/2)

2z4+

2

z2T (0) +

1

z∂T (0)

where we used

Xµ(z, z)Xν(0, 0) ∼ −α′

2ηµν ln |z|2, ψµ(z)ψν(0) ∼ 1

zηµν .

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

TF = i

√

α′

2ψµ(z)∂Xµ(z)

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

TF (z)Xµ(0, 0) ∼ −i√

α′

2

α′

2

1

zψµ(0) = −i

√

α′

2

1

zψµ(0)

TF (z)ψµ(0) ∼ i

√

2

α′1

z∂Xµ(0, 0)

7.1 Bosons and fermions 27

So

δXµ = −iε∮

dz

2πiη(z)TF (z)Xµ(0, 0) = −

√

α′

2εηψµ(0)

δψµ = −iε∮

dz

2πiη(z)TF (z)ψµ(0) = −

√

2

α′ εη∂Xµ(0, 0)

The other OPEs are given by

T (z)TF (0) ∼ 2

(

− 1

a′

)

(

i

√

α′

2

)

(

α′

2∂2 ln |z|

)

∂Xµ(z)ψµ(0) +

(

−1

2

)

(

i

√

α′

2

)

(

∂1

z

)

ψµ(z)∂Xµ(0)

+

(

−1

2

)

(

i

√

α′

2

)

1

z∂ψµ(z)∂X

µ(0)

T (z)TF (0) ∼ 3

2z2TF (0) +

1

z∂TF (0)

TF (z)TF (0) ∼(

i

√

α′

2

)21

z

(

α′

2∂2 ln |z|

)

D +

(

i

√

α′

2

)21

z∂Xµ∂Xµ

+

(

i

√

α′

2

)2(

α′

2∂2 ln |z|

)

ψµ(z)ψµ(0)

∼ D

z3+

2

zT (0).

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

S =1

4π

∫

d2zψµ∂ψµ.

The wave equation is given by

∂ψµ = 0,

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

ψµ(z)ψν(0) ∼ 1

zηµν .

The stress-energy tensors are

T = −1

2ψµ∂ψµ, TF = i

√

α′

2ψµ∂Xµ.

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


7.2 The ghosts

Recall,

Sbc =1

2π

∫

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

z.

The weights and central charge for the bc system are

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

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

Sβγ =1

2π

∫

d2zβ∂γ,

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

Tβγ = (∂β)γ − λ′∂(βγ).

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

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

The central charge for the combination of the two systems becomes

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

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

Linear Dilaton

Recall

T (z) = − 1

α′ ∂Xµ∂Xµ + Vµ∂

2Xµ,

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

c = D + 6α′V µVµ.

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

µ∂ψµ and c = D/2, we obtain

c =3D

2+ 6α′V µVµ,

and

TF = i

√

2

α′ψµ∂Xµ − i

√2α′Vµ∂ψ

µ.

7.3 Mode Expansions 29

7.3 Mode Expansions

Let us do closed strings first. Recall the expansion

∂Xµ(z) = −i√

α′

2

∑

m

αµmz−m−1,

where αµ0 =√

α′

2 pµ and [αµm, α

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

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

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

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

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

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

ψµ(σ + τ) =√i∑

r∈Z+ 12

ψµr e−ir(σ+τ)

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

z-picture, z = ei(σ+τ), we have

ψµ(z) =

(

∂w

∂z

)h

ψµ(σ + τ)

=1√izψµ(σ + τ)

=∑

r∈Z+ 12

ψµr z−r− 1

2

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

ψµ(z)ψν(0) ∼ 1

zηµν .


The anti-commutation relations are

ψµr , ψνs = ηµνδr+s,0.

We find similar results for the right-moving sector

ψµ(z) =∑

r∈Z+ 12

ψµr z−r−1

2 , ∂Xµ(z) = −i√

α′

2

∑

m∈Z

αµmz−m−1,

and the anti-commutation relations are

ψµr , ψνs = ηµνδr+s,0.

The stress-energy tensor is

T (z) =∑

m∈Z

Lmz−m−2, h = 2.

The OPE gives the Virasoro algebra with central extension

[Lm, Ln] = (m− n)Lm+n +c

12m(m− 1)(m+ 1)δm+n,0.

In terms of the OPEs we find

TF (z)TF (0) ∼ 3

2z2TF (0) +

1

z∂TF (0).

We may expand TF (z) in terms of modes

TF (z) =∑

r∈Z+ 12

Grz−r− 3

2 .

Recall

[Lm, Gr] = ((h− 1)m− r)Gr+m = (1

2m− r)Gr+m.

Finally

TF (z)TF (z′) ∼ D

(z − z′)3 +2

z − z′T (z′),3D

2= c, so D =

2c

3.

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

Gr =

∮

dz

2πirr+

12TF (z),

and∮

dz

2πizr+

12 TF (z)TF (z′) =

∮

dz

2πizr+

12

D

(z − z′)3 + 2z′r+12 T (z′)

7.3 Mode Expansions 31

f(z) = zr+12 , f ′(z) =

(

r +1

2

)

rr−12 , f ′′(z) =

(

r2 − 1

4

)

zr−32 ,

so∮

dz

2πizr+

12

D

(z − z′)3 =D

2

(

r2 − 1

4

)

z′r−32

Second step: apply∮

dz′

2πiz′s+ i

2 to isolateGs:

Gr, Gs =D

2

(

r2 − 1

4

)∮

dz′

2πiz′r+s−1 + 2

∮

dz′

2πiz′r+s+1T (z′)

= 2Lr+s +D

2

(

r2 − 1

4

)

δr+s,0

= 2Lr+s +c

12

(

4r2 − 1)

δr+s,0

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

∂Xµ = −i√

α′

2

∑

m∈Z

αµmz−m−1, ψµ =

∑

r∈Z+ 12

ψµr z−r−1

2 ,

and

T (z) = − 1


1

2ψµ∂ψµ = − 1


1

4(ψµ∂ψµ − (∂ψµ)ψµ)

we have

Lm =

∮

dz

2πizm+1T (z) =

1

2

∑

n,n′

∮

dz

2πiαµnαn′µz

−n−n′−m−1 +1

4

∑

r,r′

∮

dz

2πiψµr ψr′µ(r − r′)z−r−r

′+m−1

=1

2

∑

n∈Z

αµm−nαnµ +1

4

∑

r∈Z+ 12

(2r −m)ψµm−rψrµ.

TF (z) = i

√

2

α′ψµ∂Xµ ⇒ Gr =

∮

dz

2πizr+

12TF (z) =

∑

n,r′

∮

dz

2πiαµnαr′µz

−n+r+r′−1 =∑

n∈Z

αµnψr−n µ.

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

−1 = L1.Now L−1|0〉 = 1

2

∑

αµ−1−nαnµ|0〉 + 14

∑

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

L−1|0〉 = 0, ||L−1|0〉||2 = 0,

so〈0|2L0|0〉 = 2a = 0 ⇒ a = 0.

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

† = αµn, (ψµ−r)† = ψµr .


The ghosts

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

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

2.

We are interested in the λ = 2 case, in order to couple this system to the(Xµ, ψµ) system. Then

hb = 2, hc = 1− 1, hβ =3

2, hγ = −1

2,

and the expansions are

b =∑

m∈Z

bmz−m−2, c =

∑

m∈Z

cmz−m+1, β =

∑

r∈Z+ 12

βrz−r− 3

2 , γ =∑

r∈Z+ 12

γrz−r+ 1

2 .

From the operator product expansions, we get standard (anti) commutators

bm, cn = δm+n,0, [γr, βs] = δr+s,0.

bm, cm, βr, γr are all annihilation operators for r,m > 0. Recall the subtletywith the zero modes b0, c0, satisfying b0, c0 = 1. We have two choices for thevacuum. Choose c0|0〉 = 0. The conformal generators are

Lm =

∮

dz

2πizm+1T (z),

T (z) = (∂b)c− λ∂(bc) + (∂β)γ − λ′∂(βγ)

= (∂b)c− 2∂(bc) + (∂β)γ − 3

2∂(βγ)

=∑

n,n′

(−n′ − 2)bn′z−n′−3cnz

−n+1 − 2(−n− n′ − 1)bn′cnz−n−n′−2

+∑

r,r′

(

−r′ − 3

2

)

βr′z−r′− 5

2 γrz−r+1

2 − 3

2(−r − r′ − 1)βr′γrz

−r−r′−2

=∑

n,n′

(n′ + 2n)bn′ccz−n−n′−2 +

1

2

∑

r,r′

(3r + r′)βr′γrz−r−r′−2

So

Lm =

∮

dz

2πizm+1T (z) =

∑

n

(m+ n)bm−ncn +1

2

∑

r

(m+ 2r)βm−rγr

7.4 Open Strings 33

The SUSY generators are

Gr =

∮

dz

2πizr+

12TF (z), TF (z) = −1

2(∂β)c+ λ′∂(βc)− 2bγ

where

TF (z) =∑

s,n

−1

2

(

−s− 3

2

)

βsz−s− 5

2 cnz−n+1 +

3

2

(

−s− n− 1

2

)

βscnz−s−n− 3

2 − 2bnγsz−n−s− 3

2

=∑

s,n

−1

2(2s+ 3n)βscnz

−n−s− 32 − 3bnγsz

−n−s− 32 .

So

Gr =

∮

dz

2πizr+

12TF (z) = −

∑

n

1

2(2r + n)βr−ncn + 2bnγr−n.

Normal ordering: again, only L0 has a problem; should be L0 + a. To find a,consider [L1, L−1] = 2L0.

L−1 =∑

n

(n− 1)b−1−ncn +1

2

∑

r

(2r − 1)β−1−rγr.

When applied to the ground state, |0〉, only the terms n = −1, r = −1/2contribute (recall c0|0〉 = 0), soL−1|0〉 = −2b0c−1|0〉−β−1/2γ−1/2|0〉. Similarly,we obtain 〈0|c−1 = 〈0|(2b1c0 + β1/2γ1/2). So

〈0|[L1, L−1]|0〉 = 〈0|L1L−1|0〉 = −2〈0|(b1c0b0c−1−β 12γ 1

2β− 1

2γ− 1

2|0〉 = −2+1 = −1.

So 2a = 〈0|2L0|0〉 = −1, so a = −1/2 (−1 from the bc and 1/2 from the βγ).

7.4 Open Strings

Open strings do not have independent ocsillators αµn, αµn. Instead, αµn = αµn.

Thus,

∂Xµ(z) = −i√

α′

2

∑

m

αµmz−m−1, ∂Xµ(z) = −i

√

α′

2

∑

m

αµmz−m−1.

where αµ0 =√

2α′pµ (c.f. αµ0 =√

α′

2 pµ for closed strings). Similarly for the

ψµ’s:

ψµ(z) =∑

r

ψµr z−r−1

2 , ψµ(z) =∑

r

ψµr z−r− 1

2 .


The spectrum

For a physical state, |ψ〉, we demand

Ln|ψ〉 = Gr|ψ〉 = 0, for r, n > 0.

Also, L−n|ψ〉, Gr|ψ〉 are orthogonal to all physical states |ψ′〉 : 〈ψ′|L−n|ψ〉 =〈ψ|Ln|ψ′〉 = 0, and similarly for Gr|ψ〉. They are in the equivalence class ofzero. Check also L−n|ψ〉 is null: ||L−n|ψ〉||2 = 0. Physical states also obey theconstraint

(

L0 −1

2

)

|ψ〉 = 0

i.e., the Hamiltonian H = L0 − 1/2 = 0 (vanishes).We build the Hilbert space by applying αµ−n, ψ

µ−r oscillators only (no ghost

modes- the lead to states in the same equivalence classes as above) to theground state.

H =1

2

∑

n∈Z

: αµ−nαnµ : +1

2

∑

r

r : ψµ−rψrµ : −1

2

plus the ghost oscillators, but they do not contribute. Since α0 =√

2α′pµ foropen strings, we have

H = α′p2 +N − 1

2, N =

∞∑

n=1

αµ−nαnµ +

∞∑

r= 12

rψµ−rψrµ.

The lowest state: |0; k〉 for which N = 0, so α′k2 − 1/2 = 0, so m2 = −k2 =−1/2α′, a tachyon!So we still have a tachyon. This was to be expected, because we took thebosonic theory and enlarged it therefore we should expect the new SUSY the-ory to contain all the states of the bosonic theory and more.The next state: |1〉

Aµ(k)ψµ

− 12

|0; k〉

has N = 12 . We see that this is a massless state since

α′k2 +1

2− 1

2= 0, ⇒ m2 = −k2 = 0.

Also, Gr =∑

n αµnψr−nµ, so when G1/2 acts on our state, only the n = 0 term

contributes. So

G 12|1〉 = αµ0Aµ(k)|0; k〉 =

√2α′k ·A|0; k〉 = 0, ⇒ k · A = 0,

i.e., transverse polarization. Also note that this is a null state:

G− 12|0; k〉 = αµ0ψ− 1

2µ|0; k〉 =

√2α′k · ψ− 1

2|0; k〉

7.4 Open Strings 35

i.e., the state with longitudinal polarization is null (and orthogonal to all phys-ical states).Thus the massless state is aD−2 = 8 dimensional vector. It transforms underthe group SO(8). For closed strings, the situation is similar.TheH = 0 constraint translates into L0 = L0 = 0, i.e.,

α′

4p2 +N − 1

2=α′

4p2 + N − 1

2= 0

Notice the difference in α′p2 → α′

4 p2, which is due to the different definitions

of αµ0 between closed and open string.

The lowest state: |0; k〉with α′

4 k2 = 1

2 , som2 = −k2 = − 2α′ which is a tachyon!

The next level: Aµνψµ−1/2ψ

ν−1/2|0; k〉, with α′

4 k2 = 0, i.e., m2 = 0. This de-

omposes into a scalar, an antisymmetric tensor, and a traceless symmetrictensor:

Aµν =1

D − 2Aρρηµν +

1

2(Aµν −Aνµ) +

1

2(Aµν + Aνµ +

2

D − 2Aρρηµν .

SectionGetting rid of the tachyon Comparing the tachyon with the masslessstates, there is a clear difference: the tachyon has one less fermionic excita-tion then the massless states. If we select the states with as odd number offermionic excitations, that will get rid of the tachyon. This is similar to theharmonic oscillator, where we could select, e.g., all the odd states and stillhave a perfectly well defined physical system.The operator that did the trick there was P (parity) which commuted withthe Hamiltonian and could therefore be simultaneously diagonalized with it.Here we need to find an operator that has two eigenvalues and commuteswith all generators of space-time symmetries (not just the Hamiltonian). Thespace-time symmetries from the Lorentz group (Poincare group rather, but

Lorentz suffices). Let us review briefly. The angular momentum ~L = ~r × ~p. Interms of components we have

Lx = ypz − zpy, Ly = zpx − xpz , Lz = xpy − ypx.

where x and p obey the commutation relations [xi, pj ] = δij .

Define the antisymmetric tensorLij = xipj−xjpi, thenLi = 12εijkLjk . An an-

tisymmetric tensor is a vector in three-dimensions. Not so in four-dimensions.So generalize Lij → Lµν = xµpν − xνpµ, [xµ, pν ] = iηµν which includes time.~L generates rotations:

δxi = − i2ωkl[Lkl, xi] = ωijxj

where ωij is an anti-symmetric tensor. In terms of the vector ~ω we have δ~x =~ω × ~x. This generalizes to Lµν : δxµ = ωµνx

ν . For e.g., L01, we have δt =ω01x, δx = −ω01t, a boost! L0i is a boost in the xi-direction. The algebra of


these Lorentz generators is

[Lµν , Lρσ] = [xµpν − xνpµ, xρpσ − xσpρ]= i(ηνρLµσ − ηµρLνσ − ηνσLµρ + ηµσLνρ)

Lie algebra of SO(3, 1), or in D-dimensions, SO(D − 1, 1). Introduce spinors:we need to add a piece to Lµν that will rotate the spinor (or boost it). Callthis piece Σµν . It needs to satisfy the same SO(D − 1, 1) algebra and willcommute with Lµν by constuction (since Lµν involves space-time and Σµνinvolves fermionic operators).Guess:

Σµν = −i∑

r

ψµr ψν−r = − i

2

∑

r

[ψµr , ψν−r].

Then the algebra is

[Σµν ,Σρσ ] = −1

4

(

∑

r

[ψµr , ψν−r],

∑

s

[ψρs , ψσ−s]

)

= −(

∑

r

ψµr ψν−r,∑

s

ψρs , ψσ−s

)

= i(ηνρΣµσ − ηµρΣνσ − ηνσΣµρ + ηµσΣνρ)

where we used ψµr , ψνs = ηµνδr+s,0.Σµν generates Lorentz transformations on the fermionic fields ψµ(z). No-tice that in D = 10, there are five operators that commute with each other:Σ01, Σ23, Σ45, Σ67, Σ89 (trivial - they contain different ψµr modes). They canbe simultaneously diagonalized. How do they act? Let us be specific and con-sider Σ23. It acts on ψ2

r , ψ3r as follows:

[

Σ23, ψ2r

]

= −i∑

s

[ψ2sψ

3−s, ψ

2r ] = −

∑

s

ψ2s , ψ

2rψ3

−s = iψ3r

[

Σ23, ψ3r

]

= −[Σ32, ψ3r ] = −iψ2

r

Eigenstates: ψ2r + iψ3

r , ψ2r − iψ3

r .

[

Σ23, ψ2r + iψ3

r

]

= ψ2r + iψ3

r eigenvalue : +1[

Σ23, ψ3r − iψ3

r

]

= −ψ2r + iψ3

r eigenvalue : −1

Consider a finite transformation (rotation) U(θ) = eiθΣ23

. Then U(θ)(ψ2r +

iψ3r)U

†(θ) = eiθ(ψ2r + iψ3

r).Proof:

Σ23(ψ2r + iψ3

r) = (ψ2r + iψ3

r)(1+Σ23)⇒ (Σ23)n(ψ2r + iψ3

r = (ψ2r + iψ3

r)(1+Σ23)n

⇒ U(θ)(ψ2r + iψ3

r)U†(θ) = (ψ2

r + iψ3r)e

iθ(1+Σ23) = eiθ(ψ2r + iψ3

r)U(θ).

7.4 Open Strings 37

Similarly, U(θ)(ψ2r + iψ3

r)U†(θ) = e−iθ(ψ2

r + iψ3r). In particular, for θ = π,

the action of U(π) on both ψ2r ± iψ3

r is the same. Therefore U(π)ψ2,3r U †(π) =

eiπψ2,3r = −ψ2,3

r , i.e., U(π) and ψ2,3r anti-commute!

On the other handU(π) commutes with all otherψµr , µ 6= 2, 3. ThusU(π) onlyhas two eigenvalues, ±1, like parity! If a state has an even number of ψ2

−r, 3’s

(r > 0), then it belongs to eigenvalue +1 - with an odd number of ψ2,3−r ’s, it has

U(π) = −1. E.g.:

ψ2−r|0〉 : U(π)ψ2

−r |0〉 = −ψ2−rU(π)|0〉 = −ψ2

−r|0〉 : (−1)

U(π)ψ2−r1ψ

3−r2 |0〉 = −ψ

2−r1U(π)ψ3

−r2 |0〉 = ψ2−r1ψ

3−r3 |0〉 (+1)

etc.We can do the same with all other Σ’s. Thus we have

U1(π) = eπΣ12

, U2(π) = eiπΣ23

, U3(π) = eiπΣ45

, U4(π) = eiπΣ67

, U5(π) = eiπΣ89

.

Notice that U1(π) has no i in the exponential. This is because ψ0r , ψ

0s =

−δr+s,0. The product

U1(π)U2(π)...U5(π) = eiπ(−iΣ01+Σ23+Σ45+Σ67+Σ89) = eiπF .

This anti-commutes with allψµr . F is a fermion number operator. eiF will playthe role of parity in the harmonic oscillator case. Correction: eiπFVgh will. Vgh

is the ghost contribution. Since there are no ghost oscillators, all it does is acton the vacuum: Vgh|0〉 = −|0〉. Thus restrict Hilbert space to eigenstates ofeiπFVgh of eigenvalue +1 (invariant states). This gets rid of the tachyon, foreiπFVgh|0; k〉 = −|0; k〉 but keeps all massless states ψµ−1/2|0; k〉.

Consistent truncation

Since eiπF is made of Lorentz generators it is guaranteed to be conserved bythe OPEs of vertex operators. So even states will produce even states whenthey interact with other even states.Thus, we now have a consistent string theory without a tachyon! Or do we?We still need to check modular invariance. TheXµ part of the partition func-tion is modular invariant by itself,

ZX(τ) =

(

1

2π√α′τ2|η(q)|−2

)D

, η(q) = q1/24∞∏

n=1

(1− qn), q = e2πiτ .

the fermionic part of the partition function is similarly calculated. The resultis a Jacobi-theta function. But, alas, it is not modular invariant. This can beseen without doing any calculation as follows.Before we demanded ψµ(σ + 2π) = −ψµ(σ) (anti-periodic boundary condi-tions). On a torus, we demand ψµ(z + 2π) = −ψµ(z) and also ψµ(z + 2πτ) =−ψµ(z). But then,

ψµ(z + 2π(τ + 1)) = −ψµ(z + 2πτ) = +ψµ(z).


Therefore the transformation τ → τ + 1 changes the boundary conditionsto periodic! Therefore τ → τ + 1 is not a symmetry of the theory. Our the-ory is not modular invariant. The above argument also shows how to fix thetheory. We need to include (somehow) the sector in which ψµ obeys periodicboundary conditions. That is the Ramond sector and we study it next.

7.5 The Ramond (R) sector

The R-sector can only exist in two-dimensions, because there is no spin-statisticstheorem there. The mode expansion is

ψµ(z) =∑

n∈Z

ψµnz−n− 1

2

indices are integers, since ψµ(z) is periodic. The expansion has a factor ofz−1/2, because the weight of ψµ is h = 1/2. Therefore this is not a Laurentexpansion and has a branch cut. We still have

ψµm, ψνn = ηµνδm+n,0

as in the NS-sector. We also have the same algebra for Lm, Gr (note it is nowGm, m ∈ Z).

Normal ordering

We only have a problem with L0. Using [L1, L−1] = 2L0, we have

2〈0|L0|0〉 = 〈0|L+1L−1|0〉,

L−1|0〉 =(

1

2

∑

n

αµ−1−nαnµ +1

4

∑

n

(2n+ 1)ψµ−1−nψnµ

)

|0〉

For the α’s we need −1 − n, n < 0, so −1 < n < 0, which is impossible. Forthe ψ’s, we need−1− n, n ≤ 0, so−1 ≤ n ≤ 0, so n = 0, or n = −1. Therefore

L−1|0〉 =1

4

(

−ψµ0ψ−1µ + ψµ−1ψ0µ

)

|0〉 = 1

2ψµ−1ψ0µ|0〉.

Therefore

〈0|L1L−1|0〉 =1

4〈0|ψ0νψ

ν1ψ

µ−1ψ0µ|0〉,

=1

4〈0|ψµ0ψ0µ|0〉

=1

8〈0|ψµ0 , ψ0µ|0〉

=D

8

Therefore 〈0|L0|0〉 = D/16 = a (i.e., L0 =: L0 : −D/16).

7.5 The Ramond (R) sector 39

The ghosts

b =∑

m∈Z

bmz−m−2, c =

∑

m∈Z

cmz−m+1, β =

∑

r∈Z+ 12

βrz−r−3

2 , γ =∑

r∈Z+ 12

γrz−r+ 1

2 .

where β, γ are not Laurent expansions. The algebras are

bm, cn = δm+n,0, [γm, βn] = δm+n,0,

which are the same as before, but in addition, the zero modes: [γ0, β0] = 1,i.e., γ0, β0 are creation and annihilation operators respectively. This define|0〉 by bm|0〉 = 0, m > 0, βm|0〉 = 0, m ≥ 0 and cm|0〉 == gm|0〉 = 0 form > 0.

Normal ordering

L0 again has a problem. We can solve as we did before.

L−1|0〉 =

(

∑

n

(n− 1)b−1n−1cn +1

2

∑

n

(2n− 1)β−n−1γn

)

|0〉

=

(

−b−1c0 −1

2β−1γ0

)

|0〉

There is only one possibility since −1 < n ≤ 0, so

〈0|L1L−1|0〉 = −〈0|b0c1b−1c0|0〉 −1

4〈0|β0γ1β−1γ0|0〉

= −1− 1

4= −5

4

and

〈0|L0|0〉 =1

2〈0|L1L−1|0〉 = −

5

8= a.

The spectrum

First observe that the defintion |0〉 is ambiguous. Indeed |0〉 is defined byψµm|0〉, m > 0. But then ψµ0 |0〉 is as good as |0〉, for ψνmψ

ν0 |0〉 = −ψν0ψµm|0〉 =

0, m > 0. the ground state is then a representation of the algebra of the zeromodes, ψµ0 , ψν0 = ηµν (Clifford - Dirac algebra). |0〉 therefore is a spinor.Instead of one spin, here we have five, because we are in ten-dimensions.The spin operators are Σ01,Σ23,Σ45,Σ67,Σ89. They commute with each otherso they can be simultaneously diagonalized. We can then define a basis ofground states |s1, s2, s3, s4, s5〉 where si = ±1/2(i = 1, 2, 3, 4, 5). We will usethe notation ~s = (s1, s2, s3, s4, s5). There are 25 = 32 such states (c.f. 22 = 4states in the four-dimensional Dirac spinor). All states built from |~s〉 have


integer +1/2 spin, because ψµ−m has spin one (eigenstate of SXXXX+1 witheigenvalue +1). to be contrasted with NS-sector where all states have integerspin. Thus, the inclusion of the R-sector is important, because we need allspins to describe Nature.

The Hamiltonian (L0) has const. D/16 − 5/8 = 10/16 − 5/8 = 0, so H =α′p2 +N (c.f. H = α′p2 +N − 1/2 in the NS-sector)

The lowest level: N = 0, soH = 0 andm2 = −p2 = 0. There is no tachyon! Thelowest states, |0; k〉 are massless! Non-trivial constraint: G0|~s; k〉 = 0. Relevantpiece: G0 =

√2α′pµψ

µ0 , so kµψ

µ0 |~s; k〉 = 0 which is the Dirac equation (γµ =

1√2ψµ0 , then kµψ

µ|~s; k〉 = 0). Notice also that the algebra G0, G0 = 2L0, i.e.,

G20 = L0. G0 is the square root of the Hamiltonian!

This is just like in the Dirac case. It is also a generic feature of a SUSY theory:the Hamiltonian can be written as the square of a SUSY charge.

Notice that this also implies that the ground state has zero eigenvalue, be-cause G0|0〉 = 0, which makes it very hard to have a finite cosmological con-stant in a SUSY theory. In terms of the fields, the contribution of the bosonalways exactly cancels the contribution of the fermions (due to SUSY boson↔ fermion) and we get zero vacuum expectation energy (cosmological con-stant). The R-sector can also be split into two eigenspaces of eiπF with eigen-values±1. The ground states belongs to +1.

7.6 Superstring Theories

We may now combine the NS and R-sectors to form a consistent superstringtheory. We need to have analycity in the OPEs (which is not guaranteed inthe R-sector, due to branch cuts in the expansions of the fields). This severelyconstrains the possibilities (we also do not want a tachyon) to ...

IIA : (NS+, NS+) (R+, NS+) (NS+, R−) (R+, R−)IIB : (NS+, NS+) (R+, NS+) (NS+, R+) (R+, R+)IIA′ : (NS+, NS+) (R−, NS+) (NS+, R+) (R−, R+)IIB′ : (NS+, NS+) (R−, NS+) (NS+, R−) (R−, R−)

It can be shown that IIA′ is the same as IIA (also, similarly, IIB′ is the sameas IIB)

Proof: Transform X9 → −X9, ψ9 → −ψ9, ψ9 → −ψ9. Then eiπS89 → e−iπS

89

(same eigenvalue), but S89|0〉 → −S89|0〉, so the sign is reversed in the R-sector (S89 annihilates the NS vacuum (no zero modes), so no change there).Therefore this transformation maps R+→ R− and vice versa. QED

Open Strings

Only one possibilty: type I: NS+,R+. The projection of eigenspaces of eiπF and

eiπF is known as the Gliozzi-Scherk-Olive (GSO) projection.

7.6 Superstring Theories 41

The resulting theories turn out to have space-time SUSY and obey the spin-statisics theorem (which has to be obeyed for D > 2). The fact that space-time SUSY and the spin-statistics theorem emerge is rather unexpected. Onewould expect that these two should be evident from the start - built in formal-ism. This fact remains elusive.

Modular Invariance

We have already seen that modular invariance for the NS-NS sector alonecannot possibly work. Now we have a multitude of sectors and a hope thatmodular transformations will map one onto others and somehow the combi-nation will be invariant. Let us start with the NS-sector. Only the NS+ subsec-tor appears. The partition function for the Xµ’s is the same as before and wehave already established it is modular invariance, so we will concentrate onthe ψµ’s.

The partition function is as always

ZNS+ = Tr (qH ), q = e2πiτ .

If |ψ〉 is in NS+, then eiπF |ψ〉 = |ψ〉. To find such a |ψ〉, we can start with anarbitrary state |ψ′〉 and project onto the eigenspace of eπiF of eigenvalue +1.The projection operator is

P =1

2(1 + eiπF ), P 2 = P.

Also, eiπFP |ψ′〉 = P |ψ′〉, so eigenvalue +1. Thus, to compute the Tr NS(PA) =12Tr NSA+ 1

2Tr NS(eiπFA). First trace: for each µ, we have the creation oper-ators ψµ−r, r > 0 where of course r ∈ Z + 1

2 . A state can have 0 or 1 ψµ−r, since(ψµ−r)

2 = 0 (fermionic mode). So for fixed r, µ we get a factor q0 + qr = 1 + qr

(since N = 0, r) the rest of H has already been considered in theXµ part).

Varying r, we get a product

∏

r>0

(1 + qr) =

∞∏

m=1

(1 + qm−1/2).

Varying µ, we get eight copies of this product (because only the transverse µ’scontribute and there are 10− 2 = 8 of them). Thus

Tr NSqH =

(

q−1/48∞∏

m=1

(1− qm− 12 )

)8

.

NB the factor of q−1/48 which comes from the new tensor transformation ofT (stress-energy “tensor”) as we go from z to σ + τ (z = ei(σ+τ)) c.f. in thebosonic case we got q−1/24, double becuase for a boson c = 1 whereas for a


fermion c = 1/2. We can write this partiion function in terms of the Jacobiϑ-function. Recall ... (z = e2πiν , q = e2πiτ )

ϑ00(ν, τ) =

∞∏

m=1

(1− qm)(1 + zqm−1/2)(1 + z−1qm−1/2)

ϑ01(ν, τ) =∞∏

m=1

(1− qm)(1− zqm−1/2)(1− z−1qm−1/2)

ϑ10(ν, τ) = 2eπiτ/4 cosπν∏

m=1

(1− qm)(1 + zqm)(1 + z−1qm)

ϑ11(ν, τ) = −2eπiτ/4 sinπν∏

m=1

(1− qm)(1− zqm)(1− z−1qm)

For ν = 0, z = 1, so

ϑ00(ν, τ) =

∞∏

m=1

(1− qm)(1 + qm−1/2)(1 + qm−1/2)

ϑ01(ν, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)(1− qm−1/2)

ϑ10(ν, τ) = 2q1/8∏

m=1

(1− qm)(1 + qm)(1 + qm)

ϑ11(ν, τ) = −2q1/8 sinπ0∏

m=1

(1− qm)(1− qm)(1− qm) = 0!

Also η(τ) = q1/24∏∞m=1(1− qm). Thus,

ϑ00(0, τ) = q−1/24η(τ)

( ∞∏

m=1

(1 + qm−1/2)

)2

.

So

Tr NSqH =

(

ϑ00(0, τ)

η(τ)

)

.

Next, let us do Tr NSeπiF qH . Let us fix µ and r again. If the state has zero

ψµ−r’s then F = 0 and N = 0. So we get q0 = 1. If the state has one ψµ−r, theneiπF = −1 (recall eiπFψµ−re

−πiF = −ψµ−r) and N = r, so we get−qr.So this case differs from the previous one by a mere sign change which impliesthatϑ00 → ϑ01. Moreover, the ground state |0〉has eigenvalue−1 (eiπF |0〉NS =−|0〉NS), so we get an overall “-” sign. Thus

Tr NSeiπF qH = −

(

ϑ01(0, τ)

η(τ)

)4

.

Similarly, in the Ramond sector,

ZR+ = Tr R+PqH =

1

2

(

Tr RqH + Tr Re

πiF qH)

.


The creation modes are ψµ−m ,m > 0 and we need to take special care of thezero modes ψµ0 .Fix µ andm > 0. Then, for zeroψµ−m’s, we obtain q0 = 1 and for one ψm−mu, weobtain qm, so overall, 1 + qm. The ground state has energy H = 1/16 (normalordering constant we obtained earlier, so 1 + qm → q1/16(1 + qm). Varyingµ, m we obtain the product

Tr RqH =

(

q116 q−

148

∞∏

m=1

(1 + qm)

)8

× “0”

where “0” is the contribution of the zero-modes. Recall the ground state |~s〉.Consider the Σ23 spin, for example. There are two states | ↑〉, | ↓〉, with spin± 1

2 respectively. Each contributes 1, so overall 1 + 1 = 2. we have four inde-pendent such states (since we have eight transverse dimensions Σ01 does notproduce independent physical states). Therefore the overall factor “0” = 24

c.f. with

ϑ01(0, τ) = 2q1/8∞∏

m=1

(1− qm)(1 + qm)2 = 2q1/8q−1/24η(τ)

( ∞∏

m=1

(1 + qm)

)2

Tr RqH =

2q1/8q−1/24

( ∞∏

m=1

(1 + qm)

)2

4

= −(

ϑ10(0, τ)

η(τ)

)4

where the minus sign comes from space-time spin-statistics (ghosts). Finally,Tr Re

iπF qH gives a similar product, but with two changes

• 1 + qm → 1− qm (- form eiπF , as in NS-sector

• | ↑〉 and | ↓〉 have opposite eigenvalues, contributing 1− 1 = 0!

Therefore

Tr R = qH = −(

ϑ10(0, τ)

η(τ)

)4

= 0.

Putting everything together, the partition function for ψµ in NS+ and R+ sec-tor is

Zψ(τ) = Tr NS+qH + Tr R+q

H

=1

2Tr NSq

H + Tr NS1

2eπiF qH +

1

2Tr Rq

H +1

2Tr Re

πiF qH

=1

2(η(τ)4(

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 + ϑ11(0, τ)

4)

.

This is complicated combination of Jacobi-theta functions, yet not only is itmodular invariant, but it vanishes identically! This should not be too surpris-ing, since we have space-time SUSY and so the cosmological constant should


vanish. This fact was known to Jacobi himself, for he proved the “abstruseidentity”

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 = 0,

and we have already seen ϑ11(0, τ)4 = 0.

Of course the total Z is a product of Zψ and Z→ = Z∗ψ in the case of IIB and

(Z ′ψ)∗ =

1

2η(τ)4(

ϑ00(0, τ)4 − ϑ01(0, τ)

4 − ϑ10(0, τ)4 − ϑ11(0, τ)

4)

which of course Z ′ψ = Zψ.

Modular invariance of type-I

Type-I is an open string theory. Instead of a torus, we have a cylinder.The cylinder can easily be deduced from the torus. Recall for the torus

Z =

∫

F0

dτdτ

4τ2Z(τ), Z(τ) = Tr qH , q = e2πiτ

where F0 is the fundamental region and dτdτ4τ2

is a modular invariant mea-

sure on the torus (τ2(2π)2 is the volume of the torus = volume of the groupof translations). The cylinder defines a more honest partition function, be-cause τ → t ∈ R and Z(t) = Tr qH , q = e−2πt, i.e., τ = iτ2, τ2 = t, and

Z =

∫ ∞

0

dt

2tTr e−2πtL0 .

Notice that thee is no fundamental region, so we have potential divergencesfrom both limits t → ∞ and t → 0. t → ∞ is usually associated with theIR region (long-distance, low energy). t → 0 is associated with UV diver-gences (short-distances - high energies). In closed strings, there is no t → 0limit, for it is cut by the restriction to the fundamental region F0. In the openstring case, it is there. But does open string theory have UV divergences? Thatwould make it as bad as field (particle) theory. To answer this, concentrate onXµ, µ = 0, 1, .., D − 1. The partition function (easily deduced from the torus)is

Z(t) = Tr e−2πtL0 = iV(√

8π2α′t)−D

(η(it))−(D−2)

c.f. on torus:

Z(t) = Tr e−2πiτL0e−2πiτL0 = iV(

√

4π2α′τ2)−D

|η(it)|−2(D−2)

where

η(it) = e−πt/12∞∏

m=1

(1− e−2πmt).


LetD = 26. In the t→∞ limit, we may expand

(η(it))−24 = e2πt∞∏

m=1

(1− e−2πmt)−24 = e2πt(1 + 24e−2πt + ...)

= e2πt + 24 + ...

Each term in the expansion comes from a certain mass level. The first termis from the tachyon, and diverges, because m2 < 0. The second term is fromthe massless modes (24 transverse photons). Again, it diverges, but only log-arithmically. This is expected and is similar to field theory. These divergencescancel in physical quantities.Now look at t → 0. This appears to be a high energy effect, but it is not! Thecylinder becomes very thin and it looks like a closed string is being created,propogating and disappearing again (NB: t does not represent a physical dis-tance). So t → 0 is still an IR effect (long-distance). To show this, use themodular property, η(−1/τ) =

√iτη(τ). For τ = it, we get η(i/t) =

√tη(it), so

η(it) =1√tη

(

i

t

)

.

Change variables to s = πt . Then, apart from constants

Z ∼∫ ∞

0

dt

tt−13η(it)−24 =

∫ ∞

0

dt

t2η

(

i

t

)−24

∼∫ ∞

0

dsη

(

is

π

)−24

.

t→ 0 is obtained by expanding in large s,

η

(

is

π

)−24

= e2s + 24 + ...

(same expansion as before). The first term is from the tachyon (pathological).The second term is from the massless modes. The propagator for them is 1/k2

and since k2 = −m2 = 0, we have 1/0 =∞. The pole is due to the propagatorfor a long time (on-shell).Let us return to type-I. In this case d = 10, so for theXµ’s, we have

ZX(t) = iV (8π2α′t)−5η(it)−8.

Moreover, there is a subtlety: define the world-sheet parity Ω by

Ω : Xµ(σ)→ Xµ(π − σ).

In terms of modes, Ωαmn uΩ−1 = (−1)nαµn (recall Xµ(σ) ∼ ∑αµne

−inσ). Ob-viously, Ω2 = 1, sp Ω has two eigenvalues, ±1. We need to restrict to the+1 eigenspace for consistency of the theory. This is easily implemented: weneed to keep the states with and even number of αµ−n modes. [NB: In bosonictheory, this would give garbage, for it would exclude the photon! Here, thephoton is ψµ−1/2|0; k〉, so it has 0 (even!) αµ−n’s.


The various partition functions are not affected by the presence of Ω, but weget an extra factor of 1/2 from the projection 1

2 (1 + Ω). Thus

Z =

∫ ∞

0

dt

2t

1

2

1

2ZX(t)Zψ(t)

whereZψ(t) = ϑ00(0, it)

4 − ϑ01(0, it)4 − ϑ10(0, it)

4 − ϑ11(0, it)4

and the two factors of 1/2 come from Ω and the GSO projections respectively.To study the divergences (even though Zψ = 0! - we still need to study the,otherwise Zψ = 0 is a∞−∞ = 0 statement; also these divergences appear(and did not cancel) in other amplitudes) Define s = π/t. Then

ZX(t) = iV

8π(8π2α′)5

∫ ∞

0

ds η

(

is

π

)−8

Zψ

(π

s

)

.

Modular properties

η(it) =1√tη(i/t), ϑ00(0, it) =

1√tϑ00(0, i/t).

Separate NS and R. Then in the NS-sector

ZNS(t) = iV

8π(8π2α′)5

∫ ∞

0

ds η

(

is

π

)−12(

ϑ00(0, is/π)4 − ϑ10(0, is/π)4)

.

To leading order, η(

isπ

)−12= q−1/2 = es and

ϑ10(0, is/π)4 = 24q1/2 = 24es, ϑ00(0, is/π)4 = 1 + ...

So

ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + o(e−2s)).

Notice that the tachyon has disappeared, but of course, we still have the di-vergence from the sixteen massless modes, as expected. What can we do?Well, the cylinder is not the only possibility. We also have the Mobius stripand the Klein bottle.

The Mobius Strip

Same as the cylinder, but we twist before we identify the ends. In other words,Xµ(ω, 2πt) = Xµ(π − σ, 0) = ΩXµ(σ, 0)Ω−1. The partition function is verysimilar to the cylinder. The only difference is the insertion of the parity op-erator, Ω. Thus ZMobius = Tr (qL0Ω). The action of Ω is simple. If a state hasan even (odd) number of αµ−n’s, Ω = +1(−1). Thus, (1− qm)−1 is replaced by(1− (−)mqm)−1, q = e=2πt and so

η(it) = e−πt/12∞∏

m=1

(1− e−2πmt)⇒ e−πt/12∞∏

m=1

(1− (−)me−2πmt)


which can be written in terms of

ϑ00(0, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)2.

As follows: let τ = 2it,

ϑ00(0, 2it) =

∞∏

m=1

(1− qm)(∏

(1 + e−2πt(2m−1)))2

=

∞∏

m=1

(1− qm)−1[

∏

(1 + e−2πt(2m))(1 + e−2πt(2m−1))]2

= e−πt/161

η(2it)

[

∏

(1− (−)me−2πtm)]2

soe−πt/12

∏

(1− (−)me−2πtm) =√

ϑ00(0, 2it)η(2it)

replaces η(it). zero modes are still the same , so... Recall the cylinder

ZX = iV

(

1√8π2α′t

)D

η(it)−(D−2)

The partition function for the Mobius strip is

ZX = iV

(

1√8π2α′t

)D

(ϑ00(0, 2it)η(it))−(D−2)/2

.

Next, do the ψ’s. Easier to work in the R-sector (only one contribution)

Zψ,R = Tr ΩqNψ = −24

[

q1/16q−1/48∞∏

m=1

(1 + (−)mqm)

]8

which can be written in terms of Jacobi-theta functions as follows:

ϑ01(0, τ) =

∞∏

m=1

(1− qm)(1− qm−1/2)

ϑ10(0, τ) = 2q1/8∞∏

m=1

(1− qm)(1 + qm)2

so

ϑ01(0, τ)ϑ10(0, τ) = 2q1/8∞∏

m=1

(1− qm)∏

m

(1 + (−)(√q)m)2,

so let q = e−4πt.

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2= 2eπt/3e−πt/2

∏

m

(1 + (−)(√q)m)2


so

Zψ,R = −(

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2

)4

.

AtD = 10

ZR =

∫

0

dt

2t

1

2

1

2ZXZψ,R = iV

∫

0

dt

8t(8π2α′t)−5

(

ϑ01(0, τ)ϑ10(0, τ)

η(0, 2it)2

)4

.

To study the t→ 0 limit, switch the variable to s = π/t. Then

ZR = iV8

(8π2α′)5

∫ ∞

0

ds

(

ϑ01(0, 2is/π)ϑ10(0, 2is/π)

η3(2is/π)ϑ00(2is/π)

)4

for small s, we have

ϑ01(0, 2is/π) ' 1 + ...,

ϑ01(0, 2is/π) ' 2q1/8 + ...

ϑ00(0, 2is/π) ' 1 + ...,

η(2is/π) ' q1/24 + ...

soϑ01ϑ10

η3ϑ00=

2q1/8

q1/8= 2 + ...⇒

(

ϑ01ϑ10

η3ϑ00

)4

= 24 + ... = 16 + ...

no tachyon, and sixteen massless modes contributing, as expected. c.f. forthe cylinder,

ZR = −ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + ...).

of opposite sign, but they do not cancel!

The Klein Bottle

Even though a bottle looks more appropriate for closed strings, and ampli-tudes are defined in terms of closed string modes, the Klein bottle contributesto open strings.DEFINITION: Consider a torus with τ = it. we identify the sides σ = 0 andσ = 2π and obtain a cylinder, but just like with the Mobius strip, we identifythe sides τ = 0 and τ = 2πτ by twisting them first

Xµ(σ, 0) = Xµ(−σ, 2πt) = ΩXµ(σ, 2πt)Ω−1

The partition function is given by

ZX = Tr Ωe−2πtL0e−2πtL0


In this case, ΩαµnΩ−1 = −αµn (unlike for open strings, whereαµn → −αµn) There-

fore, the diagonal elements if Ω have exactly the same αµn’s as αµn’s.So the trace is effectively over theαµn’s only, which explains why this is an openstring amplitude.For the diagonal elements of Ω we have Ω = +1 (even total # of αµn, α

µn.) and

Ł0 = L0, soZX = Tr e−4πtLoΩ

∣

∣

Ω=1

which is the same as open string partition function, but with q2 instead of q(or 2t instead of t)

ZX = iV (4πα′t)−D/2(η(2it))−(D−2)

Note the first factor has a 4 rather than an 8 due to the closed string.The partition function for theψµ’s is obtained similarly. The result is the sameas the open string (cylinder) again, but with t→ 2t.

ZNSψ =1

(η(2it))4[

ϑ400(2it)− ϑ4

10(2it)]

and ZNSψ = −ZRψ . Overall

ZNS =

∫ ∞

0

dt

2t

1

2

1

2ZXZ

NSψ = iV

∫ ∞

0

dt

8t(4π2α′t)−s(η(2it))−12

[

ϑ400(2it)− ϑ4

10(2it)]

andZR = −ZNS. The study of the t→ − limit can be copied from the cylinderwith an extra 210 factor

ZNS = i210V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

Again there is no tachyon, but alas, Zcylinder +Zmobius +Zklein still has a non-vanishing divergence. What do we do? We need to introduce Chan-Patonfactors!Chan-Paton factors were first introduced in QCD, where the string was madeof glue. They attached quarks at the ends of the string which carried indiceslabeling color.In the present setting, we will introduce them because we can. They do notspoil Lorentz invariance, because they live at the ends of the string. They areuseful because they give us extra degrees of freedom, which are needed todescribe gauge interactions.e.g. E&M: Kaluza-Klein added an extra index throughout the string (didn’tknow about strings, but that is what they effectively did.) (X0, ..., X3, X4): X4

was the extra-dimension. This spoiled Lorentz invariance, but that was ok,because we only care about Lorentz invariance in four dimensions. The ex-tra dimension gave us a gauge group (U(1)) corresponding to a photon. Moredimensions give us more complicated gauge groups and extra degrees of free-dom.


With Chan-Paton factors, the gauge group does not come from extra dimen-sions, but from extra degrees of freedom at the ends of the string (open ofcourse). Yet another innovation of string theory!So all states now carry two more indices |0〉 → |0, ij〉, so, e.g. we now haven2 tachyons or photons, if i, j = 1, 2, ..., n. Thus, the photon can be the weakboson multiplet (W±, Z0), or the gluon.How does this effect the partition function? For the cylinder, all n2 states con-tribute equally, so Z is multiplied by n2. For the Mobius strip ,because of thetwist, i needs to be identified with j, and there are n possibilities, ZMobius →nZMobius.For the Klein bottle, we have no indices, because we have closed strings, soZKlein → ZKlein.Overall, the partition function is now

Z = n2Zcylinder + nZMobius + ZKlein.

Recall for the R-sector

Zcylinder = −i V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

ZMobius = i26V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

ZKlein = −i 210V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

Z = −i(n− 25)2V

8π(8π2α′)5

∫ ∞

0

ds(16 + ...)

We obtain a finite answer if and only if n = 25 = 32. This implies that out ofall possible gauge groups, type I string theory makes a unique choice: SO(32).This was a crucial discovery that led to the explosion of interest in string the-ory.

UNIT 8

Heterotic Strings

8.1 Introduction

The Heterotic string was introduced in 1985 by the “string quartet” (GrossHarvey, Martinec and Rohm).

Basic idea: left and right movers need not be in the same theory. not even inthe same dimension!

Thus, take the holomorphic part as bosonic (∂Xµ(z), µ = 0, 1, ..., 25) and theanti-holomorphic part as a superstring (∂Xµ(z), ψµ(z), µ = 0, 1, ..., 9).

Then the holomorphic piece has c = 26, so we need to throw in the bc ghostswith c = −26 in order to have a vanishing central charge. The anti-holomorphicpiece has a central charge c = 15, so we need couple it to the superconformalghosts (b,c and γ, β) with central charge, c = −15.

It is convenient to split ∂Xµ into ∂Xµ, µ = 0, 1, ..., 9, which will then combinewith the anti-holomorphic piece toXµ(z, z) and the rest ∂Xµ, µ = 10, 11, ..., 25have c = 16 and have no anti-holomorphic partners. We may replace themwith ψA(z), A = 1, 2, ..., 32 which have the same central charge (c=32/2=16).The two theories are the same even though it may not be obvious.

Now we can write the action

S =

∫

d2z

(

1

2πα′ ∂Xµ∂Xµ + ψµ∂ψµ + ∂A∂ψA

)

, µ = 0, 1, ..., 9, A = 1, 2, ..., 32.

so µ is a space-time index and A is an internal index which represents thegauge degrees of freedom. Think of ψA as a 32 dimensional vector rotatedin an abstract space. The gauge group is then the group of rotations, SO(32),which is large enough (too large!) to accommodate all interactions we see inNature.

Notice that SO(32) is also the gauge group in type-I theory, yet they are differ-ent, for in type-I, SO(32) is at the ends of the strings, whereas in the heterotictheory, SO(32) is along the entire string.

52 UNIT 8: Heterotic Strings

Modern wisdom holds that type-I and heterotic only look different. Deep in-side they are different manifestations of the same theory. The operator prod-uct expansions are as the usual

Xµ(z, z)Xν(0, 0) ∼ −ηµν α′

2ln |z|2

ψµ(z)∂ν(0) ∼ ηµν1

z

ψA(z)ψB(0) ∼ δAB1

z← Euclidean signature!

Energy momentum tensor:

T (z) = − 1

α∂Xµ∂Xµ −

1

2ψA∂ψA , T (z) = − 1

α∂Xµ∂Xµ −

1

2ψA∂ψA.

SUSY currents:

TF = i

√

2

α′ ψµ∂Xµ, , TF = 0

so this theory has N = 0, N = 1 SUSY (world-sheet)To build the Hilbert space, we need to specify the boundary conditions onψA and ψµ. ψµ is as before, leading to NS and R sectors - we may apply GSOprojection to split the sectors in NS± and R±.ψA is tricky. It is not restricted by Lorentz invariance, because A is an in-ternal index. We may only require that T (z) be periodic, so ψA(σ + 2π, τ) =OABψA(σ, τ) where O is a 32×32 orthogonal matrix. This leaves the quadraticform ψA∂ψA and T (z) invariant. A host of possibilities, but only two work!Possibility 1

ψA(σ + 2π, τ) = ±ψA(σ, τ)

same sign from all components. This is easy and is the same as before. Definethe fermion number operator F = Σ12 + Σ34 + ... + Σ31 32 (we now have 16spins). The GSO projection is onto eigenspaces of eiπF .We will select eiπF = +1, thus restricting to NS+, R+ forψA. Partition FunctionRecall for ψµ:

Zψ =1

2

[

(

ϑ00(0, τ)

η(τ)

)4

−(

ϑ10(0, τ)

η(τ)

)4

−(

ϑ01(0, τ)

η(τ)

)4

±(

ϑ11(0, τ)

η(τ)

)4]

= 0

which vanished by the abstruse identity.In our case, instead of 4=8/2, we have 16=32/2 (no time-like coordinate there-fore all components contribute) Answer:

Zψ =1

2

[

(

ϑ00(0, τ)

η(τ)

)16

+

(

ϑ10(0, τ)

η(τ)

)16

+

(

ϑ01(0, τ)

η(τ)

)16

±(

ϑ11(0, τ)

η(τ)

)16]

Where the first “+” is due to the ghosts (β, γ) from which are absent in thiscase. The second “+” sign is due to space-time statistics, ψA is a scalar.

8.2 The spectrum 53

This partition function is multiplied byZψ, which vanishes, but it is still usefulto demonstrate the modular invariance of ZψZ

∗ψ

.

Under τ → − 1τ

ϑ01(0,−1/τ) =√−iτϑ00(0, τ)

ϑ01(0,−1/τ) =√−iτϑ10(0, τ)

ϑ10(0,−1/τ) =√−iτϑ01(0, τ)

η(−1/τ) =√−iτη(τ)

It is obvious that both Zψ and Zψ are invariant since ϑ10 ↔ ϑ01.Under τ → τ + 1

ϑ00(0, τ + 1) = ϑ01(0, τ)

ϑ01(0, τ + 1) = ϑ00(0, τ)

ϑ10(0, τ + 1) = eiπ/4ϑ10(0, τ)

η(τ + 1) = eiπ/12η(τ)

Zψ → e−16πi/12Zψ = e2πi/3Zψ and Zψ → −e−4πi/12Zψ = e2πi/3Zψ. Thereforethe product ZψZ

∗ψ

is invariant under modular transformations.

8.2 The spectrum

Constraints: L0 = L0 = 0. Do L0 first (anti-holomorphic part). This is thesame as before.

NS : L0 =α′

4p2 + N − 1

2= 0.

Lowest state: |0, k〉 has no N = 0, so m2 = −k2 = −2/α′, a tachyon. This has

eiπF |0, k〉 = −|0, k〉, where the minus sign is due to the (β, γ) ghosts. Thereforeit is not a NS+.Next level: ψµ− 1

2

|0, k〉 has N = 12 , so m2 = −k2 = 0. It is a vector transforming

under SO(8). It has eiπF ψµ− 12

|0, k〉 = ψµ− 12

|0, k〉 ∈ NS+.

R-Sector L0 = α′

4 p2 +N + a

a can be deduced as follows: recall in bosonic theory a = −1. This is becausea = −(D − 2)/24. Here D = 10, so we get a = −1/3. In the NS sector, weget a = −1/2, because the fermions contributed−1/6(1/2 = 1/3 + 1/6)-eachfermion contributes −1/48. Now we have 32 fermions, so they contribute−32/48 = −2/3. Overall a = −1/3− 2/3 = −1 in NS.In the R sector, we get a = 0, so fermions contribute 1/24 each. Overall, a =−1/3+ 32/24 = +1 in the R sector which implies all modes are massive in theR sector.In NS, the lowest state has m2 = −k2 = − 4

a′ , which represents a tachyon! Ithas eiπF |0, k〉 = |0, k〉, so we keep it.


The next level has N = 1/2 : ψA−1/2|0, k〉, so m2 = k2 = − 2α′ , which is another

tachyon, but eiπFψA−1/2|0, k〉 = −ψA−1/2|0, k〉, so we must reject it.

The next level hasN = 1 : m2 = k2 = 0, massless! There are two possibilities:

Aµ(k)αµ−1|0; k〉 , BABψ

A1/2ψ

B−1/2|0; k〉

Both possibilities have the correct GSO projection i.e., eiπF = +1, so we keepthem.Aµ(k) represents a photon with 8 transverse polarizations. BAB is an anti-symmetric 32× 32 matrix with 32×31

2 = 496 components.Summary

m2 NS+ R+ NS+ R+−4/α′ |0; k〉 − − −

0 Aµ, BAB − ψµ−1/2|0; k〉 |~s; k〉

Closed strings states must have L0 = L0, so the tachyon in NS+ is rejected,because there is no tachyon in NS+ or R+. At the massless level we have

Aµναµ−1ψ

ν−1/2|0; k〉, Aµ,~sαµ−1|~s; k〉

whereAµν may be decomposed via a scalar, antisymmetric and traceless sym-metric tensors as 8×8 = 64 = 1+28+35. Aµ~s is the supersymmetric partner toAµν and may be decomposed into 8+56 irreducible representations of SO(8).We also have

B(~s)ABψ

A−1/2ψ

B−1/2|~s; k〉, BµABψA−1/2ψ

B−1/2ψµ −1/2|0; k〉

so BµAB represents a gauge boson with 496 components. c.f. gluon has Aµi ,where i = 1, 2, ..., 8. So SO(32) is a gauge symmetry in space-time, just likeSU(3) is a gauge symmetry for QCD.Comparing with SO(32) type-I theory, we see big differences. In type-I theorySO(32) resides at the ends of the strings and the spectrum does not match thatof the heterotic string (except at the massless level). Yet, these two theoriesare one and the same (to be proved)!Possibility #2

Divide ψA(z) into two groups, ψA, A = 1, 2, ..., 16, ψB , B = 17, 18, ..., 32. Thisis possible because

ψA(σ + 2π) = −ψA(σ) (NS1) ψA(σ + 2π) = +ψA(σ) (R1)ψB(σ + 2π) = −ψB(σ) (NS2) ψB(σ + 2π) = +ψB(σ) (R2)

Total of four possibilities: NS1 +NS2, NS1 +R2, R1 +NS2, R1 +R2. Thereare two GSO projections because we have two fermion number operators. Wewill restrict to eiπF1 = eiπF2 = +1.Partition Function

Zψ = ZψAZψB = Z2ψA ,

8.2 The spectrum 55

because ZψA = ZψB . ZψA is the same as Zψ we derived in possibility #1, ex-cept 16→ 8 (we have half as many ψAs now). Therefore

ZψA =1

2

[

(

ϑ00

η

)8

+

(

ϑ10

η

)8

+

(

ϑ01

η

)8

∓(

ϑ11

η

)8]

.

Modular invariance: τ → −1/τ leave the partition function invariant (trivial).τ → τ + 1 takes ϑ00 ↔ ϑ01, ϑ10 → eiπ/4ϑ10, so ϑ8

10 → ϑ810. So only η → eiπ/12η,

i.e., η−8 → e−2πi/3η−8, and the partition transforms as ZψA → e−2πi/3ZψA ⇒Z2ψA → e−4πi/3Z2

ψA = e2πi/3Z2ψA . The additional factor cancels when we mul-

tiple by Z∗ψ→ e−2πi/3Z∗

ψ.

Gauge Group: Obviously, this theory has SO(16) × SO(16) symmetry. How-ever, this is only a subgroup os the full gauge group, which is E8 ×E8 (excep-tional group).To summarize: there exist only two possibilities for heterotic strings, withgauge groups SO(32) and E8 ×E8, respectively.


UNIT 9

Low Energy Physics

9.1 Type IIA Superstring

Sectors: (Ns+,NS+), (R+,NS+), (NS+,R-), (R+,R-).(NS+, NS+): massless states: Aµνψ

µ−1/2ψ

ν−1/2|0; k〉 Aµν is decomposed into a

scalar, antisymmetric field and traceless symmetric field: 8× 8 = 1 + 28 + 35The scalar field represents the dilaton. We do not see it in Nature and it isbelieved to have settled into it’s ground state value and not affect dynamicsany further. We will set it to zero to avoid complications in an already com-plicated discussion (it should be set to a constant, but we can always tweakcouplings, etc., so setting it to zero will be fine). Let Bµν be the antisymmet-ric tensor and gµν the traceless symmetric tensor (graviton). The dynamics ofgµν is described by the Einstein action

Sg =1

4πG10

∫

d10x√−g R,

where G10 is the ten-dimensional Newton’s constant. This can be derivedfrom string theory tree-level amplitudes and loop amplitudes introduce cor-rections. Bµν has field strength

Hµνρ = ∂µBνρ − ∂νBµρ + ∂ρBµν .

Hµνρ is totally antisymmetric in it’s indices. The action is given by

SB = − 1

8πG

∫

d10x√−gHµνρHµνρ,

where indices are raised and lowered by gµν .

(R+,R-): massless states: |~s; k〉 ⊗ |~s′; k〉, 8× 8 of them.Recall ψµ0 |~s; k〉 is also a ground state (annihilated by all ψµr , r > 0).

States decompose into Cµψµ0 |0〉 and Cµνρψ

[µ,0 ψ

[ν,0 ψ

ρ]]0 |0〉 where we antisym-

metrize over all indices. There are 8Cµ (transverse µ) and 56Cµνρ (8+56 = 64)

58 UNIT 9: Low Energy Physics

so they span the ground states. The action is given by

SR = − 1

8πG10

∫

d10x√−g

(

FµνFµν + Fµνρσ F

µνρσ)

where Fµν = ∂µCν − ∂νCµ is the field strength of Cµ.

Fµνρσ = Fµνρσ −1

4(CµHνρσ + CνHρσµ + CρHσµν + CσHµνρ)

where Fµνρσ = ∂µCνρσ + ... (add terms such that Fµνρσ is completely antisym-metric) and is the field strength of Cµνρ.There is one more contribution to the action that does not involve the metric(topological). This is a Chern-Simons term given by

SCS = − 1

8πG10

∫

d10xεµ1µ2...µ10Bµ1µ2Fµ3µ4µ5µ6

Fµ7µ8µ9µ10

The total action is the sum of all the actions and is given by

S = Sg + SB + SR + SCS

There is a fermionic counterpart which we will not discuss.

9.2 Supergravity

Let us compare with supergravity (SUGRA). Unfortunately, SUGRA lives in 11dimensions, yet it looks so much like the type-II superstring, that is hard toignore. It turns out that (modern wisdom holds) we really live in eleven di-mensions and ten dimensional strings are really an eleven dimensional the-ory! What theory? Nobody knows ... M-Theory.We have seen this problem with dimensions before. We compactified one di-mension a la Kaluza-Klein, then let R → 0 and the extra dimension wouldnot go away. Same here. We will compactify the eleventh dimension to get10D superstrings, but the eleventh dimension will remain lurking in the back-ground.Action:

SSUGRA11 =1

4πG11

∫

d11x√−G

(

R(11) − 1

2FMNQRF

MNQR

)

− 1

24πG11

∫

d11x εM1M2...M11AM1M2M3FM4M5M6M7

FM8M9M10M11

where FMNQR is the field strength ofAMNQ (FMNQR = ∂MANQR+...), whereF is completely antisymmetric. The last term is a Chern-Simons term and isgauge invariant

δAMNQ = ∂MλNQ + ...

9.2 Supergravity 59

even though it doesn’t look like it.We need to reduce the dimension from eleven to ten to compare with super-strings. We will do that a la Kaluza-Klein. Assume nothing depends on theeleventh coordinate and call it “u”.The metric: ds2 = GMN (xµ)dxµdxν , M,N = 0, 1, ..., 10, µ = 0, 1, .., 9 We maydecompose the metric as such

ds2 = Gµνdxµdxν + 2Gµudx

µdu+Guudu2

LetGuu = 1 for simplicity (fixes the size of the extra dimension, which can berescaled later).Introduce vector Amu = Gµu and metric gµν = Gµν − AµAν . Then we maywrite ds2 = gµνdx

µdxν + (du + Aµdxµ)2. The potential AMNQ may also be

grouped into Aµνρ and Aµν = Aµνu (no other components exist, becauseAMNQ is antisymmetric, so we can not have two u indices). So now the fieldcontent becomes (from the 10D perspective) gµν , Aµ, Aµν , Aµνρ, very similarto the type-II superstring.Futhermore,

R(11) = R(10) − 1

2FµνF

µν

where Fµν = ∂µAν − ∂νAµ.

FMNQRFMNQR = FµνρF

µνρ + Fµνρσ Fµνρσ

whereFµνρ = ∂µAµρ+... , Fµνρσ = Fµνρσ−AµFνρσ+... andFµνρσ = ∂µAνρσ+...and

1

6εM1M2...M11AM1M2M3

FM4M5M6M7FM8M9M10M11

= εµ1µ2...µ10Aµ1µ2Fµ3µ4µ5µ6

Fµ7µ8µ9µ10

= εµ1µ2...µ10Aµ1µ2µ3Fµ4µ5µ6

Fµ7µ8µ9µ10+ total derivative

The last equality can easily be verified by integrating by parts.Also

√−G =

√−g.If the eleventh dimension has length 2πR, then the action becomes

SSUGRA11 =1

4πG10

∫

d10x√−g

(

R(10) − 1

2FµνF

µν − 1

2FµνρF

µνρ − 1

2Fµνρσ F

µνρσ

−1

2εµ1µ2...µ10Aµ1µ2

Fµ3µ4µ5µ6Fµ7µ8µ9µ10

)

where G10 = 2πRG11 is the 10D Newton’s constant and we have rescaledAµν → 1√

2πRAµν , Aµνρ → 1√

2πRAµνρ.

This action is identical to the one obtained from type-IIA superstring if weidentify

gµν → gµν , Aµ → Cµ, Aµν → Bµν , Aµνρ → Cµνρ.

60 UNIT 9: Low Energy Physics

UNIT 10

D-Branes

10.1 T-duality (again)

Consider type-II theories. We have

IIA : (NS+, NS+) (R+, NS+) (NS+, R−) (R+, R−)IIB : (NS+, NS+) (R+, NS+) (NS+, R+) (R+, R+)

Compactify the 10th dimension on a circle of radius R in IIA, say. As weshowed in the bosonic theory (argument is the same) the theory atR is iden-

tical to the theory at R′ = α′

R (T-duality).

To show this, we started with the coordinate X9 = U = UL(z) + UR(z) andintroduced the coordinate z = UL(z)− UR(z). The resulting theory is at R′ =α′

R . In other words, the parity transformation on the right-moving part (only!).

X9R(z)→ −X9

R(z)

relates the theory atRwith the theory atR′ = α′

R . Because of the superconfor-mal invariance, this parity transformation is also applied tp the superpartner,ψ9(z)

ψ9(z)→ −ψ9(z).

This, in particular, reverses the chirality of the states in the antiholomorphicpart, so R- ↔ R+. Therefore IIA ↔ IIB, because that is the only difference

between the two theories. Therefore IIA at R is equivalent to IIV at R′ = α′

R .

In particular, the IIA R-R fields, Cµ, Cµνλ are mapped onto the IIB R-R fields,C, Cµν , Cµνρσ as follows:

C9 → C, Cµ → Cµ9, Cµν9 → Cµν , Cµνλ → Cµνλ9.

Of course, e.g., Cµνρσ is obtained from Cµνρσ9 in IIA, but Cµνρσ9 is not an in-dependent field (can be expressed in terms of Cµ, Cµνλ) 8 + 56 = 64.

62 UNIT 10: D-Branes

Type-I Strings If we compactify the 10th dimension, X9(z, z), then the theory

in the R → 0 limit is mapped onto a T-dual theory at R = α′

R → ∞ whichcontains a D-brane.Recall the argument, in theR′ theory, the 10th dimension is

Z(z, z) = X9L(z)−X9

R(z), ∂σZ = ∂τX9

so

Z(σ = π)− Z(σ = 0) =

∫ π

0

dσ∂σZ =

∫ π

0

dσ∂τX9 =

∫ π

0

∂τ (2α′pτ)

= 2α′pπ = 2α′ n

Rπ = 2πnR′ = 0.

Translation invariance is broken in the T-dual theory. Massless modes (sameas in uncompactified theory)

NS : Aµψµ−1/2|k〉, Aψ

9−1/2|k〉, R : |~s; k〉

where Aµ represents a photon tangent to the brane. The second state shiftsthe position of the brane making it a dynamical object. (A is a function of k→ its F.T. is a function of Xµ, µ = 0, 1, ..., 8).Even though the translation invariance is broken, the original theory has 32supersymmetries! Of these, only half are broken. Thus the brane is a super-symmetric object with 16 supersymmetries! This large amount of symmetryimplies the existence of conserved charges. What are they?Our brane has 8+1 dimensions, so its volume element couples to the R-R po-tential, Cµ1µ2,...,µ9

(dV ∼ εµ1µ2...µ9dxµ1 ...dxµ9 ).

Recall familiar examples:

• A point charge qmoving along a trajectory xµ(τ) has the action q∫

dτvµAµ =∫

dτjµAµ = q∫

dxµAµ. The charge q is conserved.

• The magnetic flux: Φ =∫

~B · d~s , ~B = ∇ × ~A Define a field strength:Fij = ∂iAj − ∂jAi. Then Bi = 1

2εijkFjk , so Φ =

∫

FjkdSjk where

dSjk = 12εijkdSi is the surface element. This is the magnetic charge,

i.e., 0. Similarly, for the electric charge field, ΦE =∫

F0idΣ0i ∝ q.

For the R-R charge on the D8 brane, we have

Q ∝∫

dxµ1 ...dxµ9Cµ1...µ9

If we dualize two more dimension, the brane becomes a 6+1 dimensional ob-ject, (D6-brane). Two more gives D4, two more gives D2 and two more gives aD0 brane which represents a point particle. The charges are

∫

dxµCµ,∫

dxµ1dxµ2 ...Cµ1µ2...

which are the R-R fields in type IIA theory! On the other hand, the D(2p+1)-branes couple to Cµ1µ2

, Cµ1µ2µ3µ4, etc., which are the potentials in the type-

IIB theory!

10.1 T-duality (again) 63

Not all these potentials are independent. Consider, e.g., D0-brane coupledto Cµ. The D0-brane is a point particle (with strings attached- hairy) withcharge q which is the source of Cµ and field strength Fµν = ∂µCν − ∂νCµ. q isan electric charge. The flux

∫

FµνdΣµν ∝ q (Gauss’ Law).

In four dimensional electromagnetism we may define the dual of Fµν as Fµ =12εµνρσF

ρσ which interchanges ~E ↔ ~B. Then the electric charges become

magnetic charges. One may define a vector potential Aµ corresponding to

Fµν and describe electromagnetics in terms of Aµ instead of Aµ. Aµ can notbe defined globally, since the magnetic flux around a charge is no longer zero,but it can be defined in patches, or almost everywhere apart from the string(Dirac string). If we include both electric and magnetic charges, then noaction can be defined, yet the theory still makes sense. The existence of amonopole leads to quantization of the electric charge (Dirac).Proof: Consider a point particle moving from ~x1 → ~x2. Its wavefunctionchanges ψ(~x1) → ψ(~x2). If I want to compare ψ(~x1) and ψ(~x2), then I willdefine the quantity ψ(~x2) ∗ ψ(~x1). In the limit ~x2 → ~x1 (closed path) weobtain |ψ(~x1)|2. Gauge invariance: ψ(~x) → eiqλ(x)ψ(~x), so ψ∗(~x2)ψ(~x1) →eiq(λ(x2)−λ(x1))ψ∗(~x2)ψ(~x1) This is not a gauge-invariant oblect. To make it

gauge-invariant, multiply by eiq~A·d~, ~A→ ~A−∇λ, so δeiq

~A·d~ = e−iq(λ(~x1)−λ(~x2)),

so ψ∗(~x2)eiq

~A·d~ψ(~x1) is gauge-invariant (physical)!

Go around a loop: we have eiq ~A·d~|ψ(~x1)|2. By Stoke’s theorem,∮

C~A · d~ =

∫

S~B · d~s (flux through S).

If the path shrinks to zero, then∮

C~A · d~=

∫

S~B · d~s = 0.

In the presence of a magnetic monopole,∫

S~B · d~s = m, the magnetic charge,

so eiqS~B·d~s = eiqm. We must have eiqm = 1, therefore qm = 2πn, i.e., q is

quantized even if only one magnetic monopole exists in the entire Universe.Returning to D-branes, the Cµ potential on the D0-brane has field strengthFµν whose dual is εµ1µ2...µ10Fµ9µ10

(8 indices). It corresponds tp a potentialwith seven indices, Cµ1µ2...µ7

which resides on a D6-brane.Thus the D0 electric charge is a source for the same field for which the D6-branes magnetic charge is a source. More generally, the electric Dp-branecharge and the magnetic D(6-p)-brane charge are sources for the same field.Action for D0-branes electromagnetism:

S = −1

2

∫

d10x√−gFµνF µν + q

∫

dxµAµ

The potential between two points (D0-branes) is a Coulomb potential (in10D)

V (y) ∝ q2

y7

In momentum space, this is obtained from the propagator − ik2 where kµ is

the momentum of the exchanged boson (photon). Then

V (y) = −i∫

d10kei~k·~y q

2

k2= −i 15V

32π4

q2

y7


where V =∫

dω.With D-branes, the potential comes from the exchange of closed strings. Thismay also be viewed as an open string with ends at y = 0 and y = y movingaround a loop. We already know how to calculate it.The answer is

Z =

∫ ∞

0

dt

2tZ(t), Z(t) = Tr e−2πtL0 .

Recall our result earlier

ZNS = iV

8π(8π2α′)5

∫ ∞

0

ds(16 + o(e−2s)), s =π

t.

Now we have 9 dimensions (16 compactified), so s = 92 . Also there is no

integral over spatial momenta, only the energy D0-branes have world-lines,

so the contribution from 0-modes(

8π2α′t)−D/2 →

(

8π2α′t)−1/2

, therefore,

there is an additional factor(

8π2α′t)−(1−D)/2 →

(

8π2α′t)9/2

.An extra factor of 4 = 2 × 2 ( 2 from XXXXX and no need to average overorientations). Finally, since Z(σ = π)−Z(σ = 0) = y, the expansion contains

an extra termZ = y σπ + ...which gives an extra contribution toL0 = y2

4π2α′ + ....

Therefore the extra factor is given by e−2πt y2

4π2α′ = e−ty2/2πα′

. The partitionfunction becomes

Z → iV (4× 16)

8π(8π2α′)5

∫ ∞

0

πdt

t2(8π2α′t)9/2e−ty

2/2πα′

= iV (2π)(4π2α′)315

32π4

1

y7

This is compared to the potential V (y) = −i 15V32π4

q2

y7 . In fact it is the R-sector

ZR = V (y), but ZR = −ZNS, so q2 = 2π(4π2α′)3.This generalizes to Dp-branes: (8π2α′t)9/2 → (8π2α′t)(9−p)/2. The potentialgeneralizes to Vp(y) ∼ 1

y7−p and the charges become q2p = 2π(4π2α′)3−p.

For the D6-brane, q26 = 2π(4π2α′)3 , so q6q0 = 2π, the Dirac quantization condi-

tion with n = 1! In general, qpq6−p = 2π, confirming that the Dp-brane andD(6-p)-brane act as electric and magnetic sources for the same field.

10.2 D-branes at angles

So far we have considered similar D-branes separated by a distance y. Theseare parallel D-branes. More generally, we can have a Dp-brane and a Dp’-brane along different subspaces and they may even intersect e.g., a D8-braneobtained by dualizing X9 and a D8-brane from X8 dualization. These twobranes have the space (X1, X2, ..., X7) common. One may be obtained by ro-tating the other by 90 in the (X8, X9) plane. An open string may stretch be-tween these two branes. Then itsX9 coordinate will obey Dirichlet boundary

10.2 D-branes at angles 65

conditions at one end and Neumann boundary conditions at the other. TheX8 coordinate is reversed: Neumann boundary conditions at one end andDirichlet at the other. Thus the modes expansions will be different. Recall forNeumann boundary conditions on both ends (set τ = 0 for simplicity)

XµNN(σ) = xµ + i

√2α′

∑ 1

nαµn cos(nσ).

Check ∂σXNN6µ = 0 at σ = 0, π.For Dirichlet boundary conditions on both ends,

XµDD(σ) =

yσ

π− i√

2α′∑ 1

nαµn sin(nσ)

where y is the separation of the two (parallel in the µ-direction) branes. CheckXµDD(0) = 0, Xµ

DD(π) = y. XµNN is split into holomorphic and antiholomor-

phic pieces as such

XµL =

1

2xµ + i

√

α′

2

∑ 1

nαµne

inσ

XµR =

1

2xµ + i

√

α′

2

∑ 1

nαµne

−inσ

XDD is split as XµDD = Xµ

L − XµR (dual!) For DN-b.c., i.e., Xµ

DN (σ = 0) =0 , ∂σX

µDN (σ = π) = 0, we obtain

XµDN (σ) = −

√2α′

∑

r∈Z+1/2

αµrr

sin(rσ).

For ND-boundary conditions, we haveXµND(σ) = i

√2α′∑

r

Z+1/2αµrr cos(rσ).

The superpartners ψµ and ψµ are similar.Generalize to general angles. Suppose that there is an angle φ between thebranes and consider strings stretched between the two. DefineZ = X8 + iX9

(the brane at X9 = 0 is not rotated-no loss of generality). At σ = 0, X9 = 0and ∂σX

8 = 0, so Im(Z) = 0, Re(Z) = 0. At σ = π, the brane is rotated by φ,so Z → eiφZ, so Im(Ze−iφ = ∂σ Re(Ze−iφ) = 0.We may expand in terms of the modes

Z =

√

2

α′

∑

r∈Z+φπ

αrreirσ +

√

2

α′

∑

r∈Z−φπ

α+r

re−irσ

αr and α+r are independent, because they involve α8

r and α9r (αr = α8

r + α9r).

At σ = 0: Z ∼ O(αr+α+r ), so, Im(Z) = 0, and ∂σZ ∼ i(αr−α+

r ), so, Re(∂σZ) =0.At σ = π: Z ∼ (αr + α+

r )eiφ, so, Im(Ze−iφ) = 0, and ∂σZ ∼ i(αr − α+r )eiφ, so,

Re(∂σZe−iφ) = 0.


10.3 Partition Function

The partition function has contributions from both the αr’s and α+r ’s. It is

easy to see that for q = e−2πt

Z = qa∏

r∈Z+φπ

(1− qr)−1∏

r∈Z−φπ

(1− qr)−1,

= qa∞∏

m=0

(1− qm+ φπ )−1

∞∏

m=1

(1− qm−φπ )−1,

where a is the Casimir energy (normal ordering constant in L0 =: L0 : −a).Recall a = −1/24 for a boson, because a = 1

2

∑∞n=1 n = 1

2ζ(1) = −1/24. Herethe sum becomes

1

2

∑

r∈Z−φπ

r =1

2

∞∑

m=1

(

m− φ

π

)

=1

2

[

1

24− 1

8

(

2φ

π− 1

)2]

.

To prove this, look at the twisted sum problem (Polchinski 2.9.19) done lastsemester.Also,

1

2

∑

r∈Z+φπ

r>0

r =1

2

∞∑

m=0

(

m+φ

π

)

=1

2

[

1

24− 1

8

(

2

(

1− φ

π

)

− 1

)2]

=1

2

[

1

24− 1

8

(

1− 2φ

π

)2]

,

which is the same as before. So, a = 124 − 1

8

(

1− 2φπ

)2

. Therefore,

Z = qa(1− z)−1

[ ∞∏

m=1

(1− zqm)(1− z−1qm)

]−1

, z = qφ/π = e−2φt = e2πiν

This can be expressed in terms of

ϑ11(ν, it) = −2q1/8 sinπν

∞∏

m=1

(1− qm)(1− zqm)(1− z−1qm),

and

η(it) = q1/24∞∏

m=1

(1− qm).

Indeed,

η(it)

ϑ11(ν, it)= −1

2q−1/24−1/8−a 1

sinπν

[

∏

(1− zqm)(1− z−1qm)]−1

= −1

2q1/24−1/8−a 1− z

sinπνZ

= iqφ2/2π2

Z


Therefore

Z = −1eφ

2t/πη(it)

ϑ11(ν, it).

Similarly for the fermion, we obtain

Z =ϑab(ν, it)

eφ2t/πη(it),

for a, b = 0, 1 (NS-NS, NS-R, etc.)Notice that the bosonic Z diverges as ν → 0, i.e., φ → 0. In this limit thetwo branes become parallel to each other, and the string is free to move alongthem, i.e., it has an additional (continuous) momentum, whose trace givesTr qL0 ∼ V√

8πα′t, where V is the volume of the dimension along the brane.

Therefore,

Z = qa1√

8π2α′t

∞∏

m=1

(1− qm)−2, a = 1/24− 1/8 = −1/12

=V√

8π2α′t(η(it))−2.

The fermionic partition functions Zab do not change. Suppose as φ→ 0, bothbranes are in the X8 direction. Now take the dual of X8. Since we have Neu-mann boundary conditions in X8 (Dirichlet in X9), in the dual, we will haveDirichlet in X8. So in the dual picture, the two branes will become distinctpoints separated by a distinct y.If originally we had Dp-branes, we end up with D(p-1) branes in the dualspace. Open strings are stretched between the two branes. Thus, instead of

a continuous momentum, we now have a contribution y2

4π2α′ in L0, therefore

e−ty2/2πα′

, , y2 = y28 + y2

9 , in general. The partition function is

Z = qae−ty2/2πα′

∞∏

m=1

(1− qm)−2 = e−ty2/2πα′

(η(it))−2

Example: Consider two D4-branes at an angle φ1 in the 23-plane, φ2 in the45-plane, φ3 in the 67-plane, φ4 in the 89-plane and separated by a ditancey in the 1-directioin. In each plane, we obtain a partition function for thefermions:

Zab(φi, it) =ϑab(νi, it)

eφ26t/πη(it)

, , ν=iφit/pi, i = 1, 2, 3, 4.

Putting them together, the fermionic partition function is

Zf =1

2

[

4∏

i=1

ϑ00(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ10(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ01(νi, it)

eφ2i t/πη(it)

−4∏

i=1

ϑ11(νi, it)

eφ2i t/πη(it)

]

.


Generalizing our earlier result, when φi = 0 = νi,

Zψ =1

2η4(it)

(

ϑ400(0, iτ)− ϑ4

10(0, iτ)− ϑ401(0, iτ)− ϑ4

11(0, iτ))

.

Earlier we used the abtruse identity to show Zψ = 0. Now, we shall use thegeneralization of the abstruse identity:

∞∏

m=1

ϑ400(0, iτ)−

∞∏

m=1

ϑ410(0, iτ)−

∞∏

m=1

ϑ401(0, iτ)−

∞∏

m=1

ϑ411(0, iτ) = 2

∞∏

m=1

ϑ11(ν′i, it)

ν′i = iφ′it/π, φ′1 = 12 (φ1 + φ2 + φ3 + φ4), φ′2 = 1

2 (φ1 + φ2 − φ3 − φ4)φ′3 = 1

2 (φ1 − φ2 + φ3 − φ4), φ′4 = 12 (φ1 − φ2 − φ3 + φ4)

Notice∑4

i=1 φ′2i =

∑4i=1 φ

2i , so

∏4i=1 e

φ2i t/π =

∏4i=1 e

φ′2i tπ and

Zf =

∏4i=1 ϑ11(ν

′i, it)e

−φ′2i t/π

η4(it).

Bosons: Recall in the 89-plane

Zboson = −ieφ2t/πη(it)

ϑ11(ν, it)

so in the 234...9 direction

Zboson = η4(it)

4∏

i=1

eφ2i t/π

ϑ11(νi, it).

In the 0(time)-direction, we have a continuous distribution, so Z ∼ 1√8π2α′t

.

In the 1-direction, we have branes separated by a distance y, so L0 = y2

4π2α′ +

..., so Z1 ∼ e−ty2/2πα′

.Multiplying everything, the partition function becomes (potential)

V = −∫ ∞

0

dt

t

1√8π2α′t

e−ty2/2πα′

4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)

This is a complicated function of y. We will calculate it for large distances. If yis large, the dominant contribution to the integral comes from small t (due to

the ety2/2πα′

factor). If we set t = 0 in the ϑ-function, we obtain a constant andthe integral diverges. We will calculate the force, which is a physical quantityand define the potential on the integral, V = −

∫

Fdy.

F = −dVdy

= −y∫ ∞

0

dt

πα′e−ty

2/2πα′

√8π2α′t

4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)

10.4 Scattering 69

From

ϑ11(ν, it) = −2q1/8 sinπν

4∏

m=1

(1− qm)(1− zqm)(1− z−1qm),

andϑ11(−iν/t, i/t) = −i

√teπν

2/tϑ11(ν, it),

we obtain4∏

i=1

ϑ11(ν′i, it)

ϑ11(νi, it)=

4∏

i=1

ϑ11(−iν′i/t, i/t)ϑ11(−iνi/t, i/t)

.

As t→ 0, q = e−2π/t → 0, so Π→ ∏4i=1

sin iπν′i/t

sin iπνi/t

νi = iφt/π → iπνi/t = −φi, Π =

4∏

i=1

sinφ′isinφi

.

So

F ∼ Cy∫ ∞

0

dt√te−ty

2/2πα′, y →∞ const. :

1

πα′√

8π2α′

4∏

i=1

sinφ′isinφi

.

and the potential is V ∼ Cy.

10.4 Scattering

How do you make a D-brane move? Simple. Motion in e.g., the 1-directionis motion in Minkowski space (X0, X1) just like a rotation in Euclidean space(X8, X9) we studied above.

(

X0

X1

)

→(

cosh ζ sinh ζsinh ζ cosh ζ

)(

X0

X1

)

,

(

X8

X9

)

→(

cos ζ sin ζ− sin ζ cos ζ

)(

X8

X9

)

.

whereX1 = vX0 and the speed (v) is defined by the rapidity (ζ) as v = tanh ζ.The rapidity is related to the velocity via

cosh ζ =1√

1− v2, sinh ζ =

v√1− v2

.

Consider two parallel Dp-branes moving with relative velocity v in the X1-direction and separated by a distance y in the z-direction (branes are perpen-dicular to bot X1 and X2). In the 01-plane (Minkowski), we may copy ourearlier result with the substitution φ = −iζ: The bosonic part of the partitionfunction is

Zbosonic (01) = −ieφ2t/πη(it)

ϑ11(ν, it), φ = −iζ, ν = iφt/π − ζt/π.


The fermionic part is

Zab =ϑab(ν, it)

eφ2t/πη(it).

In the rest of the direction, the D-branes are parallel, so all other angles arezero. Therefore, the fermionic piece is

Zf =1

2η4(it)e−φ

2t/π[

ϑ00(ν, it)ϑ300(0, it)− ϑ10(ν, it)ϑ

310(0, it)

−ϑ01(ν, it)ϑ301(0, it)− ϑ11(ν, it)ϑ

311(0, it)

]

This may be computed by applying the generalized abstruse identity. We have

φ1 = φ, φ2 = φ3 = φ4 = 0,

so

φ′1 = φ′2 = φ′3 = φ′4 =1

2φ

and therefore

Zf =1

2η4(it)e−φ

2t/πϑ411(

1

2φ, it).

The bosonic piece in the other directions (X2, X3, ..., X9 total of eight ... sixof which are transverse) is

Z bosonic2,3,...,9

= Vp

(

1√8π2α′t

)p

e−ty2/2πα′

(η(it))−6

Therefore the partition function is

Z = −iVp∫ ∞

0

dt

t(8π2α′t)−p/2e−ty

2/2πα′ ϑ411(ν/2, it)

ϑ11(ν, it)(η(it))−9, ν = ζt/π

As the branes move the distance changes to r2 = y2+v2τ2. The potential maybe extracted from

Z = −1

∫ ∞

−∞dτV [r(τ), v].

We easily obtain

V (r, v) = i2Vpv

(√

8π2α′)p+1

∫ ∞

0

dt t(5−p)/2e−tr2/2πα′ ϑ4

11(iζ/2π, i/t)

η9(i/t)ϑ11(iζ/π, i/t),

where we used the modular properties of the ϑ and η functions.Note: as v → 0, u→ 0, so V → 0.Since

ϑ11(ν, it) = −2q1/8 sinπν∏

(1−qm)(1−zqm)(1−z−1qm), η(it) = q1/24∏

(1−qm)

10.4 Scattering 71

we have, as v → 0, ν → 0, Z → 1.

ϑ411(iζ/2π, i/t)

η9(i/t)ϑ11(iζ/π, i/t)=

8i sinh4(ζ/2)

sinh(ζ)+ ... =

1

2v3 + ..., ζ → v

So

V (r, v) = − 2Vpv4

(√

8π2α′)p+1

∫ ∞

0

dt t(5−p)/2e−tr2/2πα′

+ o(v6)

∼ − v4

r7−pVp

α′p−3

Problem: as r → 0, V → ∞! How can string theory claim finiteness at shortdistances (r is real distance - not bogus!)?

Answer: Let r → 0 before expanding in v. r only appears in e−tr2/2πα′

. If werescale t→ t/r2, the r → 0 corresponds to large t. If t is large in ϑ, η, then

ϑ411

η9ϑ11→ sinh4

(

vt4

)

sinh(vt), ζ ∼ v

From the exponential, t ∼ 2πα′/r2 dominates. ut ∼ 2πα′u/r2, so in the limitthat r → 0, ut becomes large and the integral oscillates rapidly. OScillation ona scale ut ∼ 1, i.e., 2πα′u ∼ r2, i.e., r ∼

√α′v. This is the effective scale probed

by the brane: r0 =√α′v. A slow brane (v → 0) probes scales smaller than the

string scale! Moreover, we obtain an uncertainty in the position

δx ≥√α′v.

The time it takes for this scattering process is

δt ∼ δx/v

Therefore,

δxδt ≥ δx

v

√α′v ' α′v

v= α′.

A new uncertainty principle! It implies that coordinates do not commute! Itseems that Nature is described by noncommutative geometry. What can thispossibly mean??Consider two branes separated by a distance y. Strings ending on the samebrane have a massless mode each, so we have two massless modes. A stringstretched between the two branes has

L0 =y2

4π2α′ + ...

This extra term makes L0 > 0 i.e., there are no massless modes. At low ener-gies, we only see two massless particles from the two branes. However, wheny → 0, the stretched string develops a massless mode, and there are two of


them. So when the two branes coincide, there are four massless modes. Thesefour modes can be grouped into a matrix Xij in an obvious notation.Each Xij is the position of the brane! When we develop a particle theory weneed to treat the position of the brane as a 2 × 2 matrix. More generally, ndistinct branes have n massless modes. The particle theory is just n copiesof the same theory. When all n branes coincide, we have n2 massless modes.Each massless mode corresponds to a symmetry of the theory (U(1)). With n2

massless states the symmetry is enhanced to U(n) (n2 generators).Familiar ExamplesPhoton: Fµν = ∂µAν − ∂νAµ, U(1) symmetry (Aµ → Aµ + ∂µλ).3 Photons: F iµν = ∂µA

iν − ∂νAiµ, U(1)3 symmetry.

Weak Bosons: DemandSU(2) symmetry which has three generators, soF iµν 6=∂µA

iν = ∂νA

iµ. There is a correction, to obey the enhanced symmetry Fµν =

∂µAν − ∂νAµ + [Aµ, Aν ] (Aµ is a matrix ... Aµ = Aiµσi)Gluons: Demand SU(3) - eight gluons (32 − 1).

Fµν = ∂µAν − ∂νAµ + [Aµ, Aν ], Aµ = Aiµλi

where λi represent the Gell-Mann matrices. The action is given by

S ∼∫

d4x Tr FµνFµν .

If we only had eight copies of electromagnetism, we would have

S ∼∫

d4x Tr F iµνFµνi .

Now we have interactions between gluons - enhanced symmetry (gluons andweak bosons, unlike photons have charge).Potential: SetAµ =constant, then

Tr FµνFµν ∼ Tr [Aµ, Aν ]

2.

Back to D-branes: Xµ is like aµ (that can be made precise - see Polchinski8.6). So the enhanced symmetry contains a potential

V ∼ Tr [Xµ, Xν ]2

where µ, ν run over that dimension transverse to the branes. Expand aroundXµ = 0 in a Taylor series. There are no linear or quadratic terms in Xµ, sothere is no mass term (which would come from V (φ) = V (0) + V ′(0)φ +12V

′′(0)φ2/m2 + ...)So we have kn2 massless modes, where k is the number of transverse dimen-sions. Also, V = 0 if and only if all [Xm, Xn] = 0, i.e., all Xm commute. Thiscan be accomplished if we make them all diagonal. There are n diagonal el-ements, each corresponding to one of the D-branes. Thus this potential cor-rectly describes n coincident non-interacting free D-branes.

Date post:	22-Jul-2020
Category:	Documents
Upload:	others
View:	2 times
Download:	0 times

String Theory IIaesop.phys.utk.edu/strings/part2.pdfString Theory II GEORGE SIOPSIS AND STUDENTS...

Documents