Light Cone Gauge Quantization of
Strings, Dynamics of D-brane and
String dualities.
Muhammad Ilyas1
Department of Physics
Government College University
Lahore, Pakistan
Abstract
This review aims to show the Light cone gauge quantization of strings. It is divided
up into three parts. The first consists of an introduction to bosonic and superstring
theories and a brief discussion of Type II superstring theories. The second part deals
with different configurations of D-branes, their charges and tachyon condensation. The
third part contains the compactification of an extra dimension, the dual picture of
D-branes having electric as well as magnetic field and the different dualities in string
theories. In ten dimensions, there exist five consistent string theories and in eleven
dimensions there is a unique M-Theory under these dualities, the different superstring
theoies are the same underlying M-Theory.
1
ilyas mia@yahoo.com
M.Phill thesis in Mathematics and Physics Dec 27, 2013
Author: Muhammad Ilyas
Supervisor: Dr. Asim Ali Malik
i
Contents
Abstract
i
INTRODUCTION
1
BOSONIC STRING THEORY
6
2.1
The relativistic string action . . . . . . . . . . . . . . . . . . . . . . . .
6
2.1.1
The Nambu-Goto string action . . . . . . . . . . . . . . . . . .
7
2.1.2
Equation of motions, boundary conditions and D-branes . . . .
7
2.2
Constraints and wave equations . . . . . . . . . . . . . . . . . . . . . .
9
2.3
Open string Mode expansions . . . . . . . . . . . . . . . . . . . . . . .
10
2.4
Closed string Mode expansions
. . . . . . . . . . . . . . . . . . . . . .
12
2.5
Light cone solution and Transverse Virasoro Modes . . . . . . . . . . .
14
2.6
Quantization and Commutations relations . . . . . . . . . . . . . . . .
18
2.7
Transverse Verasoro operators . . . . . . . . . . . . . . . . . . . . . . .
22
2.8
Lorentz generators and critical dimensions . . . . . . . . . . . . . . . .
27
2.9
State space and mass spectrum . . . . . . . . . . . . . . . . . . . . . .
29
SUPERSTRINGS
3.1
32
The super string Action . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.1.1
Equations of motion and Boundary conditions . . . . . . . . . .
33
3.2
Neveu Schwarz sector . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
3.3
Ramond sector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
3.4
Super transverse Virasoro operators . . . . . . . . . . . . . . . . . . . .
40
3.5
Counting states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
3.6
Open superstrings and the GSO projection . . . . . . . . . . . . . . . .
43
3.7
Closed string theories . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
ii
3.7.1
Type IIA Superstring Theory . . . . . . . . . . . . . . . . . . .
45
3.7.2
Type IIB Superstring Theory . . . . . . . . . . . . . . . . . . .
46
3.7.3
Heterotic superstring theories . . . . . . . . . . . . . . . . . . .
47
3.8
Type I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.9
Critical dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
D-BRANES
50
4.1
Tachyons and D-brane decay . . . . . . . . . . . . . . . . . . . . . . . .
4.2
Quantization of open strings in the presence of various kinds of D-branes 52
4.3
4.4
4.2.1
Dp-branes and boundary conditions . . . . . . . . . . . . . . . .
52
4.2.2
Quantizing open strings on Dp-branes
. . . . . . . . . . . . . .
53
4.2.3
Open string between parallel Dp-branes
. . . . . . . . . . . . .
55
4.2.4
Strings between parallel Dp and Dq-branes . . . . . . . . . . . .
58
String charge and electric charge . . . . . . . . . . . . . . . . . . . . . .
60
4.3.1
Fundamental string charge . . . . . . . . . . . . . . . . . . . . .
60
4.3.2
Visualizing string charge . . . . . . . . . . . . . . . . . . . . . .
62
4.3.3
Strings ending on D-branes
. . . . . . . . . . . . . . . . . . . .
63
D-brane charges and stable D-branes in Type II . . . . . . . . . . . . .
65
STRING DUALITIES
5.1
5.2
5.3
5.4
50
69
T-duality and closed strings . . . . . . . . . . . . . . . . . . . . . . . .
69
5.1.1
5.1.1 Mode expansions for compact dimension . . . . . . . . . .
69
5.1.2
Quantization and commutation relations . . . . . . . . . . . . .
71
5.1.3
Constraint and mass spectrum . . . . . . . . . . . . . . . . . . .
72
5.1.4
State Space of compactified closed strings . . . . . . . . . . . .
73
T-Duality for Closed Strings . . . . . . . . . . . . . . . . . . . . . . . .
74
5.2.1
Type II superstrings and T-duality . . . . . . . . . . . . . . . .
76
T-duality of open strings . . . . . . . . . . . . . . . . . . . . . . . . . .
76
5.3.1
T-duality and open strings . . . . . . . . . . . . . . . . . . . . .
76
5.3.2
Open strings and Wilson lines . . . . . . . . . . . . . . . . . . .
78
Electromagnetic fields on D-branes and T-duality . . . . . . . . . . . .
80
5.4.1
Maxwell fields coupling to open strings . . . . . . . . . . . . . .
80
5.4.2
D-branes with Electric fields and T-dualities . . . . . . . . . . .
80
5.4.3
D-branes with Magnetic fields and T-dualities . . . . . . . . . .
82
5.5
String coupling and the dilaton . . . . . . . . . . . . . . . . . . . . . .
83
5.6
S-duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
A Appendix A
The Massless States Of Closed String
86
B Appendix B
The Spinors Algebra In 2D
88
INTRODUCTION
General relativity and quantum mechanics were the two major breakthroughs that
revolutionized theoretical physics in the twentieth century. General relativity gives
the idea to understand of the large-scale expansion of the Universe and gives a small
correction to the predictions of Newtonian gravity for the motion of planets and the
deflection of light rays, and it predicts the existence of gravitational radiation and black
holes. It describes the gravitational force in terms of the curvature of spacetime which
has fundamentally changed our view of space and time i.e. they are now viewed as
dynamical [1].
Quantum mechanics, on the other hand, is the essential tool for understanding the sub
atomic particles and microscopic physics. The evidence continues to build that it is
an exact property of Nature [2]. The fundamental law of Nature is surely incomplete
until general relativity and quantum mechanics are successfully reconciled and unified.
String theory is a candidate which resolves this problem and a straightforward attempt
to combine the General relativity and Quantum mechanics. String theory is based on
the idea that particles are not point-like, but rather tiny loops (i.e. closed strings)
or (open) pieces of string i.e. the 0-dimensional point particle is replaced by a 1dimensional string [3]. This assumption leads to some features which are
1
2
General features
Even though string theory is not yet fully formulated, and we cannot yet understand the
detailed description of how the standard model of elementary particles should emerge
at low energies, or how the Universe originated, there are some general features of the
theory that have been well understood.
Vibrating string
String theory predicts that all objects in our universe are composed of Vibrating strings
and different vibrational modes of the strings represent different kinds of particles. Since
there is just one type of string, and all particles arise from string vibrations, all particles
are naturally incorporated into a single theory [3].
Gravity
String theory attempts to reconcile the General relativity and Quantum mechanics.
One of the vibrational modes of strings is the graviton particle, the quantum version of
the gravity, so string theory has the remarkable property of predicting gravity [3].
Unification of forces
There are four fundamental forces that had been recognized to exist in nature.
1. Electromagnetic force
2. Weak force
3. Strong force
4. Gravitational force
As the quantum version of electromagnetism describes the photon (a massless particle)
and its interactions with charged particles, while the Yang-Mills theory describes W
3
and Z bosons and gluons (the mediators of the weak and the strong nuclear forces)
and their interactions. All of these theories make a single theory named the Standard
Model of particle interactions, which is a gauge theory. The Standard Model of particle
physics does not include the graviton particle and its interactions. The graviton which
has spin 2 can not be described by the gauge theory. Since the standard model is
believed to incomplete due to it does not incorporate the gravity forces. String theory
is currently the most promising candidate to unify all the fundamental forces. This is
a general feature of the string theory [3].
Yang-Mills gauge theory
Standard model of particle physics describe the elementary particles in nature. It
reconciles the special relativity and Quantum mechanics [4]. And is based on YangMills theory having the gauge group
SU (3) × SU (2) × U (1)
However it has some shortcomings. It does not include Gravity and it has about 20
parameters that cannot be calculated and we use them as an input. While string
theory predicts the gravity as well as describe all the elementary particles and has one
parameter, the string length. Its value is roughly equal to the typical size of strings.
Yang-Mills gauge theories arise very naturally in string theory. However, it is not yet
fully understood why the gauge group
SU (3) × SU (2) × U (1)
Of the standard model with three generations of quarks and leptons should be singled
out in nature [5, 6].
Supersymmetry
A supersymmetry is a symmetry which relates bosons and fermions. There exists a non
supersymmetric bosonic string theory which is an unrealistic theory due to the lack
4
of fermions. In the order to get a realistic string theory which explains the beauty of
nature, we need a supersymmetry. Hence supersymmetry is a general feature of string
theory [3].
Extra dimensions of space
In quantum field theory, for point particle, we let the dimensions of the spacetime to
be four while superstring theories predict some additional dimensions of the spacetime.
The superstring theories are only able to work in a ten dimensions or eleventh (in some
cases) dimensions of spacetime. To make an ordinary four dimensional space time,
there is a straight forward possibility that is, the additional six or seven dimensions
can be curled up and compactified on an internal manifold having the sufficient small
size, which can not be detectable at the low energies. The idea of an extra dimension
was first introduce by Kaluza and Klein in 1920s. Their aim was to unify the electromagnetic force and the gravitational force. The compactification of an extra dimension
can be imagined as, let us consider we have a cylinder having the radius R. when the
cylinder is viewed from a very large distance or equivalently, when the radius of the
cylinder R becomes too small then the two dimensional cylinder will look like a one
dimensional line.
Generalizing this idea by letting the cylinder as a four dimensional spacetime and replacing the short circle of radius R (compact space) by a six or seven dimensional manifold,
hence at large distance or at the low energies, the additional dimensions (compact manifold) can not be visible. These additional dimensions or compact manifold are called
Calabi-Yau manifolds [3].
The size of the strings
Quantum field theory deals the particles as a mathematical zero dimensional point while
in string theory the ordinary point particles are replaced by a one dimensional string.
These one dimensional strings will have a characteristic length scale which is denoted
5
be and can estimated by the dimensionality analysis. As string theory is a relativistic
quantum theory which also includes the force of gravity, since it must involves the fundamental constants speed of light c Planck’s constant h̄, and the gravitational constant
G From these, we can form a length, called the Planck length
ls =
h̄G
c3
1/2
= 1.6 × 10−33 cm ,
1/2
= 2.176 × 10−5 g ,
1/2
= 5.391 × 10−44 s .
The Planck mass becomes
mp =
h̄c
G
And similarly the Planck time
tp =
h̄G
c5
The Planck length scale is a natural guess for a fundamental string length scale as well
as the characteristic size of compact extra spatial dimensions. At low energies string
can be approximated by the point particle that explains why the quantum field theory
has been successfully in the describing our world. The relativistic quantum gravity
effects can be important on the above three scales of planks length, planks mass and
planks time [3].
BOSONIC STRING THEORY
The bosonic string theory is the simplest string theory that predicts and describes
only a certain set of boson. As the theory does not describe any fermions, so it is an
unrealistic theory but this theory is a natural place to start, because the same techniques
and structures, together with some additional terms are required for analysis of more
realistic theory (super string theories).
String can be regarded as 1-brane moving through a space-time which is a special case
of a p-brane, p-dimensional extended object. Point particle corresponds to 0-brane.
Similarly the two dimensional extended object or 2-brane are called membranes.
2.1
The relativistic string action
In quantum field theory, the action for point particle (0-brane) is proportional to the
invariant length of the word-line or particle trajectory. Similarly when we replace the
point particle by 1 dimensional string (1-brane) then the action is proportional to the
proper area of the word-sheet or the area swept out by the string in D dimensional
space-time.
SN G = − T
Z
dA
(2.1)
The word-sheet is parameterized by the two coordinates ξ 0 = τ which is time-like and
ξ 1 = σ which is space-like. For close string the σ will be periodic while for open string
σ will be have some finite value [3].
6
7
2.1.1
The Nambu-Goto string action
As one-dimensional string having1 Xµ (τ, σ) space-time coordinates tracing out a two
dimensional word-sheet. In the order to calculate the area of word-sheet, we will need
a metric. Let ξ i (i = 0, 1) denote the word-sheet coordinates and take the metric
gµν having the signature ( − , +, +,......, +) with µ, ν = 0, 1,...., D − 1 , describe the
background geometry in which the string propagates then
− ds2 = gµν dX µ dX ν = gµν
∂X µ ∂X ν i j
dξ dξ ≡ gij dξ i dξ j
∂ξ i ∂ξ j
(2.2)
Here ξ 0 = τ ,ξ 1 = σ and the minus sign is used with ds2 is for having real time-like
trajectory. In Minkowski flat space-time gµν = η µν . The Nambu-Goto action then takes
the form
SN G
Here Ẋ µ =
Z p
Z q
2
− det gij d ξ = −T◦
(Ẋ.X ′ )2 − (Ẋ 2 )(X ′2 )dτ dσ
= −T
∂Xµ
,
∂τ
X ′µ =
∂Xµ
∂σ
(2.3)
and by the dimensional analysis T = Tco , To is the string
tension and we will use natural units in which speed of light is 1.
2.1.2
Equation of motions, boundary conditions and D-branes
Let us start from the Nambu-Goto string action which is in the form of lagrangian
density
Zτf
Zτf
dτ
L(Ẋ , X ) = − T◦
q
S=
τi
dτ L=
Zσ1
dσL(Ẋ µ , X ′µ )
(2.4)
(Ẋ.X ′ ) − (Ẋ)2 (X ′ )2
(2.5)
0
τi
Where L is given as
µ
′µ
The equations for the motion of string can be obtained by the action principle.
δS =
Zτf
τi
1
dτ
Zσ1
0
τ
∂Pµ
∂Pµσ
∂
∂
µ
µ τ
µ σ
dσ
δX Pµ +
δX Pµ − δX
+
∂τ
∂σ
∂τ
∂σ
As we will use the X µ for string coordinate while for general space-time we will called xµ
(2.6)
8
Where
Pµτ ≡
And
Pµσ
(Ẋ.X ′ )Xµ′ − (X ′ )2 Ẋµ
δL
= −T◦ q
δ Ẋ µ
(Ẋ.X ′ )2 − (Ẋ)2 (X ′ )2
(Ẋ.X ′ )Ẋµ − (X ′ )2 Xµ′
δL
= −T◦ q
≡
δX ′µ
(Ẋ.X ′ )2 − (Ẋ)2 (X ′ )2
(2.7)
(2.8)
For δS = 0 we get some conditions (boundary conditions) and equations of motion,
which are, the equations of motion for relativistic string2 . (closed or open) are.
∂Pµτ
∂Pµσ
+
=0
∂τ
∂σ
And the boundary conditions are
(2.9)
3
A) The boundary condition for closed string is, as for the closed string the word-sheet
is like a tube (cylinder type), so their boundary conditions are
X µ (τ, σ + 2π) = X µ (τ, σ)
(2.10)
B) While the word-sheet of open string is like a sheet, there are two types of boundary
conditions for open string
i) Neumann conditions
Pµσ (τ, σ∗ ) = 0
(2.11)
∂X µ
(τ, σ∗ ) = 0, µ 6= 0
∂τ
(2.12)
ii) Dirichlet conditions
Here σ∗ are the end points of the open string. No momentum flows off the ends of the
String by implying the Neumann conditions while the endpoints are fixed in spacetime
by implying the Dirichlet conditions. When the end points are fixed in space-time this
means that the string is attached with some physical object, which is called D-brane.
While when the Dirichlet conditions are applied to a subset of the d indices (spatial
2
3
For closed string σ ∼ [0, 2π]. While for open string σ ∼ [0, π]. The length of string is σ1
These are the momentum densities, Pµτ , Pµσ . While Pµτ is called momentum conjugate.
9
dimensions) of X µ , i.e. to the indices p+1, . . . ,d, this mean that the endpoints of the
string are restricted to move on a p-dimensional hyper-plane. This higher dimensional
object is called a Dp-brane, where p indicates its dimension and D stands for Dirichlet.
It turns out that these objects should be considered to be dynamical as well [3, 7, 8, 9,
10].
2.2
Constraints and wave equations
As the word-sheet have the two parameters, (τ, σ) by reparameterization (gauge) of τ
and σ of the word-sheet, simply using a class of gauges 4 which fix the parameterization
(τ and σ) of the word-sheet [11] and give the some constraints equations which are
Ẋ.X ′ = 0, Ẋ 2 + X ′2 = 0
(2.13)
Both these constraints equations combine to give
(Ẋ ± X ′ )2 = 0
(2.14)
Using these two constraints equation to simplify the momentum densities P τ µ and P σµ
which give5
P τµ =
1
1
X ′µ
Ẋ µ , P σµ = −
′
2πα
2πα′
(2.15)
Putting these momentum densities in the equation of motion which is ∂τ P τ µ + ∂σ P σµ =
0, we get
Ẍ µ − X ′′µ = 0
(2.16)
So by reparameterization we got the equation of motion just as a wave equation.
4
Reparameterization of word-sheet In static gauge (t = τ ), our solution is not fully explicit so we
use the more general gauges. A class of choices for τ is n.X(τ, σ) = βα′ (n.p)τ , Where β is equal to
two for open string and one for closed string and ”n” is chosen in such a way that (n.p) is conserved.
And the conservation of (n.p) means n.P σ = 0 for both closed as well as open strings And similarly
τ
the associated σ parameterization gives n.p = 2π
β n.P
1
5
The string tension T◦ is equal to T◦ = 2πα′ , and α′ is the Regge-slope parameter.
10
2.3
Open string Mode expansions
Let we have a space-filling D-brane for this all the string coordinates X µ satisfy the
boundary condition (Neumann) at the end points. And as the solution of a wave
equation can be written in the form of superposition of waves moving to the left and
right on the string.
X µ (τ, σ) =
1 µ
(f (τ + σ) + g µ (τ − σ))
2
(2.17)
Where f µ and g µ are the arbitrary functions. As the Neumann boundary conditions
P σµ = 0 at the endpoints
∂X µ
∂σ
= 0 , at σ = 0, π
The Neumann boundary conditions at σ = 0 give
1
∂X µ
(τ, 0) = (f ′µ (τ ) − g ′µ (τ )) = 0
∂σ
2
(2.18)
This equation means that the derivative of f µ and g µ are same then these two functions
are differ by some constant i.e. g µ = f µ +cµ , so replacing this and absorbing the constant
into the definition of f µ . This give
X µ (τ, σ) =
1 µ
(f (τ + σ) + f µ (τ − σ))
2
(2.19)
Now consider boundary conditions at σ = π
∂X µ
1
(τ, π) = (f ′µ (τ + π) − f ′µ (τ − π)) = 0
∂σ
2
(2.20)
Since from this equation we see that f ′µ is a periodic with period of 2π. so we can write
it in term of Fourier series
′µ
f (u) =
f1µ
+
∞
X
n=1
(aµn cos nu + bµn sin nu)
(2.21)
11
Now integrating this equation and then putting the result f µ (u) back into equation
(2.17) and simplifying, we get
X µ (τ, σ) = f0µ + f1µ τ +
∞
X
(Aµn cos nτ + Bnµ sin nτ ) cos nσ
(2.22)
n=1
Now replace the coefficients in above equation (2.22) by new coefficients which have
some simple physical interpretation.
√
2α′ µ∗ inτ
an e − aµn e−inτ
cos nτ +
sin nτ = −i √
(2.23)
n
√
Here * denote the complex conjugate and 2α′ factor is to make the aµn dimensionless
Bnµ
Aµn
and the physical interpretation of these new constant and their conjugate is, they
become annihilation and creation operators when considering Quantum field theory
Similarly f1µ in equation (2.22) has also a simple physical interpretation which is as
from the momentum conjugate equation
Z π
µ
P τ µ dσ =
P =
0
1
2πα′
Z
π
Ẋ µ dσ =
0
1 µ
f
2α′ 1
(2.24)
This equation tells us that the quantity f1µ is proportional to the space-time momentum
which is carried by the string. From above equation (2.24), putting the value of f1µ with
the new coefficients equation (2.23) into equation (2.22), then we gets6
µ
X (τ, σ) =
xµ0
′ µ
+ 2α p τ − i
√
2α′
∞
X
n=1
inτ
aµ∗
− aµn e−inτ
n e
cos nσ
√
n
(2.25)
In the order to make the simple expression of above equation (2.25), we replace the
constant foµ = xµo and defining7
α0µ =
√
√
√
µ
= aµ∗
2α′ pµ , αnµ = aµn n and α−n
n,n≥1
n
µ
= (αnµ )∗ , using these new coefficients (modes) and simplifying by which
And also α−n
we get
X µ (τ, σ) = xµ0 +
6
7
√
X1
√
2α′ α0µ τ + i 2α′
αnµ e−inτ cos nσ
n
n6=0
(2.26)
If all the coefficients (oscillators) aµn vanishes, then the equation represents the point particle.
As a is oscillators while α is mode.
12
This is the solution of wave equation and a simple expression for the open string.
Similarly we can get
Ẋ µ ± X ′µ =
2.4
√
2α′
X
αnµ e−in(τ ±σ)
(2.27)
n∈Z
Closed string Mode expansions
Let us consider the general solution for the wave equation, which is
X µ (τ, σ) = XLµ (τ + σ) + XRµ (τ − σ)
(2.28)
Where XLµ is for left-moving wave while XRµ is for right-moving wave of the string.
The word-sheet of the closed string is a cylinder, so to describe the closed string, we
need some conditions, as the closed string have no endpoints but having the periodicity
condition.
(τ, σ) ∼ (τ, σ + 2π)
(2.29)
So by this periodicity, we can write
X µ (τ, σ) = X µ (τ, σ + 2π) for all τ and σ
(2.30)
Now let introduce two variables
u = τ + σ and v = τ − σ
By putting these new variable in equation (2.28), which becomes
X µ (τ, σ) = XLµ (u) + XRµ (v)
(2.31)
Now by periodicity condition (τ, σ) ∼ (τ, σ + 2π), the above equation becomes
XLµ (u + 2π) − XLµ (u) = XRµ (v) − XRµ (v − 2π)
(2.32)
From this equation we can say that both XLµ (u) and XRµ (v) are periodic with the period
of 2π. Therefore we can write the mode expansions
13
′µ
XL (u) =
′µ
XR (v) =
r
α′ X µ −inu
ᾱ e
2 n∈Z n
r
α′ X µ −inv
α e
2 n∈Z n
(2.33)
(2.34)
By integrating these equations, we get
r
r
α′ µ
α′ X 1 µ −inu
1 Lµ
µ
ᾱ0 u + i
ᾱ e
XL (u) = x0 +
2
2
2 n6=0 n n
1
+
XRµ (v) = xRµ
2 0
r
α′ µ
ᾱ v + i
2 0
r
α′ X 1 µ −inv
ᾱ e
2 n6=0 n n
(2.35)
(2.36)
Now putting these two equations (2.35) and (2.36) into equation (2.32) and solving by
which we get
ᾱ0µ = α0µ v
(2.37)
Now by using this equality and equations (2.35) and (2.36), we get the equation (2.31)
as
r
√
α′ X 1 −inτ µ inσ
1
Lµ
Rµ
e
(αn e + ᾱnµ e−inσ )
X µ (τ, σ) = (x0 + x0 ) + 2α′ αnµ + i
2
2 n6=0 n
(2.38)
As from the canonical momentum conjugate equation, we can write
Pµ =
Z
2π
0
1
P τ µ dσ =
2πα′
α0µ
=
r
Z
2π
Ẋ µ dσ =
0
α′ µ
P
2
r
2 µ
α
α′ 0
(2.39)
(2.40)
This equation (2.40) differs by a factor of two from the corresponding open string
result and also tells the same idea i.e. the quantity α0µ is proportional to the space-time
momentum which is carried by the closed string.
Rµ
And let xLµ
0 and x0 be equal i.e. having one conjugate coordinate zero mode, without
any loss of generality,
Rµ
µ
xLµ
0 = x0 ≡ x0
14
Finally using the above equation then equation (2.38) becomes
r
√
α′ X 1 −inτ µ inσ
e
(αn e + ᾱnµ e−inσ )
X µ (τ, σ) = xµ0 + 2α′ α0µ τ + i
2 n6=0 n
(2.41)
Here we required these two relations for the reality of the above equation
µ
µ
= (αnµ )∗ , ᾱ−n
α−n
= (ᾱnµ )∗
As when αnµ = ᾱnµ then the above equation (for closed string) is equal to equation (2.26)
which is for open string. Since the closed string can be viewed as a two copies of the
open strings. This is a complete mode expansion for a closed string [3, 7].
2.5
Light cone solution and Transverse Virasoro Modes
The light cone solution is, to represents the motion of string by using the light cone
coordinates and this impose a set of conditions which is called light cone gauge. The
light cone gauge is one of the choices among the general gauges. As we know that the
more general gauges
n.X(τ, σ) = βα′ (n.p) τ , n.p =
2π
n.P τ
β
(2.42)
Where β is equal to two for open string and one for closed string. In the order to select
the light cone gauge, we need to impose the above conditions with a vector nµ such
that n.X = X + So
µ
n =
1 1
√ , √ , 0...., 0
2 2
P0 + P1
X0 + X1
√
= X + , n.P = √
= P+
2
2
Now using this pair of equations (2.44) into equation (2.42), we get
n.X =
X + (τ, σ) = βα′ P + τ , P + =
2π +τ
P
β
(2.43)
(2.44)
(2.45)
15
This choice of gauge is called the light cone gauge. Let the transverse coordinates be
denoted as
X l = X 2 , X 3 , ...X d
Here (l = 2, 3, ..., d). Now using the constraints equation (2.14) and expanding it into
light cone coordinates.
(Ẋ ± X ′ )2 = 0
= (Ẋ ± X ′ ).(Ẋ ± X ′ ) = 0
= −2(Ẋ + ± X ′+ ). (Ẋ − ±X ′− ) + (Ẋ l ±X ′l )2 = 0
(2.46)
Here we use the dot products of light cone coordinates.
Now using the equation (2.45) and calculating the equation (2.46), we get8
Ẋ − ± X ′− =
1 1
(Ẋ l ±X ′l )2
′
+
βα 2P
(2.47)
So we have developed the relation between the light cone coordinates and the transverse
coordinates. And the mode expansions for open string (β = 2) in the term of light cone
coordinates are
X l (τ, σ) = xl0 +
√
X1
√
2α′ α0l τ + i 2α′
αnl e−inτ cos nσ
n
n6=0
(2.48)
And as our gauge conditions give
X + (τ, σ) = 2α′ P + τ =
√
2α′ α0+ τ
(2.49)
With position zero mode and oscillators for the X + are
+
+
x+
0 = 0, αn = α−n = 0, n = 1, 2, ..., ∞
Similarly
8
We assume that P
1
+
6= 0, the vanishing of P
negative x direction. While P
+
+
certainly satisfied P
means that a massless particle moving in the
+
≥ 0.
16
X − (τ, σ) = x−
0 +
√
X1
√
2α′ α0− τ + i 2α′
αn− e−inτ cos nσ
n
n6=0
(2.50)
These are the complete set of mode expansions in light cone coordinate [12]. Now using
equation (2.27) with µ = − and µ = l, then we get
X
√
αn− e−in(τ ±σ)
Ẋ − ± X ′− = 2α′
(2.51)
n∈Z
Ẋ l ± X ′l =
√
2α′
X
αnl e−in(τ ±σ)
(2.52)
n∈Z
Now putting these two equations (2.51) and (2.52) into equation (2.47), by simplifying
we get
αn− =
1 X l
1
√
α αl
2P + 2α′ p∈Z n−p p
(2.53)
From above equation (2.53)9 , we have the explicit expression for the minus oscillators
αn− in the term of transverse oscillators αnl . This represents the full solutions [3, 11].
From the above equation (2.53), the right side has given a special type name, which is
the Transverse Virasoro mode L⊥
n,
√
2α′ αn− =
1X l
1 ⊥
⊥
L
,
L
≡
α αl
n
P+ n
2 p∈Z n−p p
(2.54)
Now by using the equation (2.53) and the equation which we defined earlier that is
√
α0µ = 2α′ pµ , then we get
Now using the value of αn−
1 ⊥
L = 2P + P −
(2.55)
α′ 0
from equation (2.54) then equation (2.51) and (2.47) are
written
Ẋ − ± X ′− =
9
1
1 X ⊥ −in(τ ±σ)
L
e
=
(Ẋ l ± X ′l )2
n
+
′
+
P n∈Z
4α P
(2.56)
A critical string only have transverse excitations, just like for a massless particle only has transverse
polarization states.
17
Similarly for closed string (β = 1) since the equation (2.47) becomes
Ẋ − ± X ′− =
1 1
(Ẋ l ± X ′l )2
α′ 2P +
(2.57)
And as like equation (2.56) we can also write for closed strings
l
′l 2
(Ẋ + X ) = 4α
′
X
1X l l
ᾱ ᾱ
2 p∈Z p n−p
X
1X l l
αα
2 p∈Z p n−p
n∈Z
(Ẋ l − X ′l )2 = 4α′
n∈Z
!
e−in(τ +σ) ≡ 4α′
X
!
e−in(τ −σ) ≡ 4α′
X
So we define the Verasoro mode for closed string
L̄⊥
n =
−in(τ +σ)
L̄⊥
ne
(2.58)
−in(τ −σ)
L⊥
ne
(2.59)
n∈Z
n∈Z
1X l l
1X l l
ᾱp ᾱn−p , L⊥
αα
n =
2 p∈Z
2 p∈Z p n−p
(2.60)
Now using equations (2.58), (2.59) and equations (2.57), (2.41) we can easily finds
√
2α′ αn− =
2 ⊥ √ ′ −
2
Ln , 2α ᾱn = + L̄⊥
+
P
P n
(2.61)
From this equation (2.61) we cal also write
ᾱn− =
1 X l
1
√
ᾱ ᾱl
2P + 2α′ p∈Z n−p p
And for n = 0 then ᾱ0− = ᾱ0− for equation (2.61), then, we have
⊥
L̄⊥
0 = L0
(2.62)
This equality is called level-matching and it means that one in the terms of right-moving
oscillators and one in terms of left-moving oscillators are equal as level matching.
In the quantum theory level matching implies that the states of right-moving excitations
of a string are equal to the states of its left-moving excitations [12]. We will explain
this later.
18
2.6
Quantization and Commutations relations
In the order to quantize our classical results we need to replace the Poisson brackets
by commutation relations and the fields by operators and then will solve the Virasoro
constraint equations, and will describe our theory in a Fock space that describes the
physical degrees of freedom [13]. Before we have written the classical equations of
motion in the light cone gauge. Now we will use (to quantization) the results of our
light cone analysis of the classical relativistic strings. Our Hamiltonian for open strings,
as we have X + = 2α′ p+ τ so from this we can write ∂τ = 2α′ p+ ∂X + and in addition, as
p− generates X + translations so from this we can write Hamiltonian as
H = 2α′ p+ p−
For quantization and commutation relations, we will start from a set of operators.
τl
+
(X l (τ, σ) , x−
0 , P (τ, σ) , P (τ ))
(2.63)
From this, we have the full set of basic operators of string theory because the above list
has the collection of all zero modes plus the infinite set of the annihilation and creation
operators [3]. First we will deal the open string and assume the space-filling D-brane.
Let postulate some commutation relations from the above set of operators
And
X l (τ, σ), P τ j (τ, σ ′ ) = iη lj δ(σ − σ ′ )
(2.64)
+
= −i
x−
0 ,P
(2.65)
And all the other commutation relations are equal to zero i.e.
X l (τ, σ), X j (τ, σ ′ ) = 0, P τ l (τ, σ), P τ j (τ, σ ′ ) = 0
− τl
l
x−
0 , X (τ, σ) = 0, x0 , P (τ, σ) = 0
P + , X l (τ, σ) = 0 , P + , P τ l (τ, σ) = 0
(2.66)
19
From the commutation relation in equation (2.64) we can also write as
h
l
j
i
X (τ, σ), Ẋ (τ, σ ) = i2πα′ η lj δ(σ − σ ′ )
′
(2.67)
Similarly the other commutation relations from the above relations are
h
h
j
i
X (τ, σ), Ẋ (τ, σ ) = i2πα′ η lj
′l
′
d
δ(σ − σ ′ )
dσ
i
Ẋ l (τ, σ), Ẋ j (τ, σ ′ ) = 0 , X ′l (τ, σ), X ′j (τ, σ ′ ) = 0
(2.68)
(2.69)
Now by solving these equations we get
h
i
d
(Ẋ l ± X ′l )(τ, σ), (Ẋ j ± X ′j )(τ, σ ′ ) = ±i4πα′ η lj δ(σ − σ ′ )
dσ
(2.70)
This commutation relation in equation (2.70) is useful to write down the commutation
relations for the oscillators. The classical modes αnl will become the quantum operators
with a nontrivial commutation relation. For commutation relation of oscillators, let
recall the equation (2.27), we can write
(Ẋ l + X ′l )(τ, σ) =
√
2α′
X
αnl e−in(τ +σ) , σ ∈ [0, π]
(2.71)
X
αnl e−in(τ +σ) , σ ∈ [−π, 0]
(2.72)
n∈Z
(Ẋ l − X ′l )(τ, −σ) =
l
Now let define an operator A (τ, σ)
Al (τ, σ) ≡
√
2α′
X
√
2α′
n∈Z
αnl e−in(τ +σ) , Al (τ, σ + 2π) = Al (τ, σ)
(2.73)
n∈Z
The periodicity is due to the definition of Al (τ, σ) and from this definition we can write
Al (τ, σ) =
(Ẋ l + X ′l )(τ, σ), σ ∈ [0, π]
(Ẋ l − X ′l )(τ, −σ), σ ∈ [−π, 0].
(2.74)
So from this set of equation (2.74), there are four possibilities of commutation relations,
we can summarized as
20
d
Al (τ, σ), Aj (τ, σ ′ ) = 4πα′ η lj δ(σ − σ ′ ) , σ, σ ′ ∈ [−π, π]
dσ
(2.75)
Now solving this equation (2.75) by using equation (2.73), we find
l
αm
, αnj = m η lj δm+n,0
(2.76)
This shows the commutation relation between the α modes and also α0l commutes with
others oscillators and these are equivalent to an infinite set of annihilation and creation
operators. To see this let start from our defining oscillators i.e.
√
√
µ
= aµ∗
αnµ = aµn n , α−n
n,n≥1
n
(2.77)
As both (a, α) are classical variables and now they become operators. As those classical
variables which are complex conjugate of each other will now become the Hermitian
operators in quantum theory i.e. the operators xl0 and pl are Hermitian as
(xl0 )† = xl0 , (pl )† = pl
(2.78)
Similarly the above oscillators and modes are now operators and we will take these like
√
√
l
= al†
αnl = aln n and α−n
n n, n ≥ 1
From this we have
l
(αnl )† = α−n
,n∈Z
(2.79)
From these conditions we can replace equation (2.76) as
And similarly
j
l
lj
, α−n
αm
= m η lj δm,n , alm , aj†
n = δm,n η
(2.80)
l† j†
l
αm
, αnj = 0, αm
, αn = 0
l
These commutation relations shows (al†
m , am ) satisfy the commutation relations of canon-
ical creation and annihilation operators of the quantum simple harmonic oscillator. So
21
l
are creation operators for
from this, we have αnl are annihilation operators while α−n
n ≥ 1. Using these Hermiticity conditions (2.78) and (2.79) we can write
(X l (τ, σ))† = X l (τ, σ)
Now to find the commutation relation between α0l and xln , for this as α0l =
√
2α′ P l and
using equation (2.64) and by simplification we get
xl0 , P j = i η lj
(2.81)
By quantization, replacing the classical variables to operators, our mode expansion then
become
∞
X
√
inτ cos nσ
√
X l (τ, σ) = xl0 + 2α′ pl τ + i 2α′
aµn e−inτ − al†
ne
n
n=1
(2.82)
Here replace the α modes by corresponding oscillators. And this is in the term of
annihilation and creation operators, expansion of the coordinate operator. Similarly
now quantize the closed string, the Poisson brackets will be replaced by commutation
relations and the classical variables by quantum operators. For commutation relations,
as the operators content of closed strings can be treated as the two commuting copies
the open string operators i.e. left moving and right moving. Similarly we can use the
almost same techniques to derive the commutation relations for closed strings. The
commutations relations for closed strings are
l
l
, ᾱnj = m η lj δm+n,0
ᾱm
, ᾱnj = m η lj δm+n,0 , ᾱm
l
l
l
ᾱm
, αnj = 0, ᾱm
, ᾱnj† = δm,n η lj
xl0 , p = iη lj →
xl0 , α0j
= iη lj
r
α′ l j
, x0 , ᾱ0 = iη lj
2
r
α′
2
(2.83)
Our Hamiltonian for closed strings in light cone coordinates is, as we know from our
light cone gauge, as p− generates X + translation and also we have X + = α′ p+ τ so from
22
this we can write ∂τ = α′ p+ ∂X + , so our Hamiltonian will be
H = α ′ p+ p−
There is only the factor of two in the open strings and closed strings Hamiltonian due
to the value of β, which is two for open string while one for closed string [3, 14].
2.7
Transverse Verasoro operators
As before we have write down the classical solution of motion of both open and closed
strings in light cone coordinates and then we solved for X − in the term of transverse
coordinates by using the constraints equation (2.47) i.e. αn− the in the term of αnl modes
as shown (for open string) in equation (2.53) so
√
2α′ αn− =
1 ⊥ ⊥ 1X l
L ,L ≡
α αl
P+ n n
2 p∈Z n−p p
(2.84)
Where l (repeated index) is summed over transverse light cone direction. As before L⊥
n
were transverse Virasoro modes but now it is called transverse Virasoro operators due
to the modes become operators. As the two α operators (inL⊥
n ) fail to commute for
n = 0, so L⊥
0 is the only operator which needs a Normal Ordering. From the definition
of L⊥
0 as
L⊥
0 =
∞
1X l l
α α
2 p∈Z −p p
(2.85)
∞
1
1X l l 1X l l
= α0l α0l +
α α +
αα
2
2 p=1 −p p 2 p=1 p −p
Since the last term is not Normal Ordered, taking the last term and making it as a
normal order by using the commutation relation, such that
∞
∞
∞
X
1 X l l 1
1 X l l
αα =
α α + (D − 2)
p
2 p = 1 p −p 2 p = 1 −p p 2
p =1
23
So L⊥
0 then become
L⊥
0
∞
∞
X
1 l l X l l 1
α−p αp + (D − 2)
p
= α0 α0 +
2
2
p=1
p=1
(2.86)
And also from the definition of L⊥
0 (with normal ordering constant a ) in the term of
p− is as
1
L⊥
n +a
+
p
2α′ P − ≡
(2.87)
From equation (2.86) and (2.87), we can say that
∞
X
1
a = (D − 2)
p
2
p =1
By Zeta function we find the value of
∞
P
(2.88)
p and so
p =1
a=−
1
(D − 2)
24
(2.89)
The Normal order L⊥
0 is then become
L⊥
0
′ l l
=αpp +
∞
X
p =1
l
α−p
αpl −
1
(D − 2)
24
(2.90)
This is all about the open string transverse Virasoro operators. Similarly for closed
strings, the transverse Virasoro operators are
√
2α′ αn− =
√
2α′ ᾱn− =
1 ⊥ ⊥ 1X l
L ,L ≡
α αl
P+ n n
2 p∈Z n−p p
1 ⊥ ⊥ 1X l
L̄ , L̄ ≡
ᾱ ᾱl
P+ n n
2 p∈Z n−p p
(2.91)
†
⊥
⊥ †
⊥
⊥ †
⊥
l †
l
l †
l
With (L⊥
0 ) = L0 , (Ln ) = L−n and (L̄n ) = L̄−n due to (αn ) = α−n and (ᾱn ) = ᾱ−n .
−
l
. As we shown that αm
and αnl commutes only when
Similarly we have (αn− )† = α−n
(m + n) equal to zero but the case is totally change for αn− i.e. two Virasoro operators
⊥
L⊥
m and Ln never commute for (m 6= n) . Now to find the commutation relations for
24
Virasoro operators lets begin from the commutation relation of Virasoro operator and
oscillator, which is
1 X l
j
αn−p αpl , αnj
L⊥
m , αn =
2 p∈Z
1
j
j
)
− nαm+n
= (−nαm+n
2
So we find
j
j
L⊥
m , αn = −nαm+n
(2.92)
This commutation relation is also valid for m = 0 . Similarly
√
l
′ l
L⊥
m , x0 = −i 2α αm
(2.93)
⊥
Now let us consider the two Virasoro operators L⊥
m and Ln , and the commutation
relation between them is not quit easy. So let the sum split as
L⊥
m =
1X l l
1X l
αm−k αkl +
α α
2 k≥0
2 k<0 k m−k
We made the above L⊥
m as normal ordered for any value of m. So we can now evaluate
the commutation relation like
1 X l l
1 X l
⊥
αm−k αkl , L⊥
αk αm−k , L⊥
L⊥
n +
n
m , Ln =
2 k≥0
2 k<0
Now evaluating then we get
1X
1X
⊥
l
L⊥
(m − k)αlm+n−k αkl +
(m − k)αkl αm+n−k
m , Ln =
2 k≥0
2 k<0
+
1X l
1X l
l
kαm−k αk+n
+
kα αl
2 k≥0
2 k<0 k+n m−k
(2.94)
Here we have now deal with two different cases m + n 6= 0, m + n = 0 and . If m + n 6= 0
, the two in each term commute then their order is irrelevant so in that case, we will
switch the order in the last two terms of equation (2.94) and then replace the variable
k by k − n. So we get
⊥
⊥
L⊥
m , Ln = (m − n)Lm+n , m + n 6= 0
(2.95)
25
A mathematical set of operators L⊥
m with n ∈ Z, satisfying the above equation (2.95)
defines the Lie algebra. This algebra is called the Witt algebra or the Virasoro algebra
without central extension. Now the second case m + n = 0, let n = −m in equation
(2.94) then an extra contribution arises due to insisting the normal order. Such that
1X
1X
l
⊥
(m − k)αl−k αkl +
(m − k)αlk α−k
L⊥
m , Ln =
2 k=0
2 k<0
+
1X l
1X l
l
l
kαm−k αk−m
+
kαk−m αm−k
2 k=0
2 k<0
(2.96)
For normal ordering, let replace k → −k in the second terms, k → m + k in the 3rd
terms, and k → m − k in the last terms
1X
l
1X
l
⊥
+
(m
−
k)α
α
(m + k)αl−k αkl
=
L⊥
,
L
m
n
−k k
2 k=0
2 k=1
+
1 X
1 X
(m + k)αl−k αkl +
(m − k)αl−k αkl
2 k=−m
2 k=m+1
(2.97)
Let assume that m > 0, (this argument goes the same way as in the other case) by
this all the terms are now normal ordered except the 3rd term for which −m ≤ k ≤ 0.
Splitting the summation of 3rd term in above equation (2.97) and then simplifying, we
get
⊥
L⊥
m , Ln
∞
∞
1X
1X
=
(m − k)αl−k αkl +
(m + k)αl−k αkl + (D − 2)A(m)
2 k=0
2 k=1
(2.98)
Where A(m) is
m
m
m
1 X
1X
1X 2
A(m) =
k(m − k) = m
k−
k
2 k=0
2 k=1
2 k=1
(2.99)
By using the Mathematical induction and Zeta functions. We get
A(m) =
1
(m3 − m)
12
(2.100)
26
Now putting the value of A(m) from equation (2.100) into equation (2.98) and solving,
we find
1
⊥
⊥
(D − 2)(m3 − m)
L⊥
m , Ln = 2m L0 +
12
This completes the commutation relation of Virasoro operators for m + n = 0. Now
generalizing these two cases, we get
1
⊥
⊥
L⊥
,
L
(D − 2)(m3 − m)δm+n,0
m
n = 2m Lm+n +
12
(2.101)
The second term of right-hand side of the above equation (2.101) is called the central
extension and a mathematical set of operators L⊥
m with n ∈ Z, which satisfying the
above equation (2.101) defines the centrally extended Virasoro algebra. As the term is
said to be central because it is a constant and commute with all other operators in the
algebra. There is no central term for m = 0 and m = ±1.
These are the complete commutation relations, here i, j are the transverse indices.
Closed strings Virasoro operators and their commutation relations have the same techniques as we have done for open strings [12, 15].
27
2.8
Lorentz generators and critical dimensions
Lorentz invariance of a string action allows us to find a set of conserve word sheet
α
currents, Mµν
and the resulting word-sheet charges Mµν for open strings with σ ∈ [0, π],
are
Mµν =
Now putting the value of
Zπ
0
Pµτ
α
Mµν
(τ, σ)dσ =
Zπ
(Xµ Pντ − Xν Pµτ )dσ
(2.102)
0
from equation (2.15) so we get
Mµν
1
=
2πα′
Zπ
(Xµ Ẋν − Xν Ẋµ )dσ
(2.103)
0
Now using the explicit mode expansion for and from equation (2.26), by simplifying we
get
M
µν
=
xµ0 pν
−
xν0 pµ
∞
X
1 µ ν
ν
−i
αnµ )
α−n αn − α−n
n
n=1
(2.104)
where the 1st two terms are due to orbital angular momentum while the summation
terms are due to the angular momentum due to excited oscillator modes [12]. This
equation (2.104) is the classical Lorentz generators of open strings in the terms of
oscillators. Similarly for closed strings, the Lorentz generator becomes
M
µν
=
xµ0 pν
−
xν0 pµ
∞
∞
X
X
1 µ ν
1 µ ν
ν
ν
µ
−i
ᾱnµ )
(α−n αn − α−n αn ) − i
ᾱ−n ᾱn − ᾱ−n
n
n
n=1
n=1
(2.105)
M −l is the most delicate quantum Lorentz generators in light cone gauge because, X −
coordinate is a non-trivial function of the transverse coordinates X l and a consistent
M −l should be generates the Lorentz transformations on the strings coordinates. As
Lorentz algebra is not generally reproduces by generators , M µν which implies that the
theory is not Lorentz invariant, so to make it invariant, we should10
10
M −l , M −j = 0
As for point particle [J µ , J ν ] = 0, as J is the Lorentz generator for point particle.
(2.106)
28
Since from above equation (2.104), we have
M
−l
∼
l
x−
0p
−
xl0 p−
∞
X
1 − l
l
(α−n αn − α−n
αn− )
−i
n
n=1
(2.107)
As we know that the Lorentz generators should be Hermitian and normal ordered but
as from equation (2.107) we write
(M −l )† 6= M −l
So making it Hermitian by writing 21 (xl0 p− + p− xl0 ) instead of xl0 p− , we find
M
−l
∼
l
x−
0p
∞
X
1 l −
1 − l
− l
l
− (x0 p + p x0 ) − i
(α−n αn − α−n
αn− )
2
n
n=1
(2.108)
Now it’s fully Hermitian and it’s also normal ordered because of α− are normal ordered.
To make it complete Lorentz generator, we should put the definition of P − and α− from
equation (2.87) and (2.84), so our Lorentz charge M −l becomes
M
−l
=
l
x−
0p
∞
X
i
1
1 ⊥ l
l
⊥
⊥
l
l
− ′ + x0 (Ln + a) + (Ln + a)x0 − √
(L−n αn − α−n
L⊥
n)
′
+
4α p
2α p n=1 n
(2.109)
The above equation (2.109) is our Lorentz charges in light cone gauge, in the order
to make it Lorentz invariant, we should calculate the above commutation relation in
equation (2.106) and then simplifying, we got
With ∆m which is
−l
M ,M
∆m =
−j
∞
1 X
l
l
l
l
= − ′ +2
∆m (α−m
αm
− α−m
αm
)
α p m=1
1 1
1
(D − 2) + a
m 1 − (D − 2) +
24
m 24
(2.110)
(2.111)
As for Lorentz invariance, the above equation (2.110) should be equal to zero, so this
means
1
1
1
m 1 − (D − 2) +
(D − 2) + a = 0, ∀ m ∈ Z +
24
m 24
29
By solving this we get
D = 26
And a = −1. This shows that the relativistic strings can be properly quantized in the
flat Minkowski space if the number of space-time dimensions is 26 [3, 13, 16, 17, 18].
A similar calculation can fix the dimensionality of space time to the value D = 10 in
super strings.
2.9
State space and mass spectrum
The mass of relativistic string (that perform an arbitrary motion) can be calculated
from the mass operators, which is the relativistic equation in light cone coordinates, as
M 2 = −p2 = 2p+ p− − pl pl
Now writing the mass operator in the term of Virasoro operators (open strings), as from
equation (2.55) putting the value of 2p+ p− , with normal ordering constant (a = −1)
we find
1
M2 = ′
α
−1 +
∞
X
l
nal†
n an
n=1
!
(2.112)
As the sum inside the bracket in the Number operator, like
⊥
N ≡
∞
X
l
nal†
n an
(2.113)
n=1
So the mass operator then becomes
M2 =
1
(−1 + N ⊥ )
α′
(2.114)
And we can easily calculate the commutation relations of number operator with oscillators, which are
⊥ l
l
l†
N ⊥ , al†
n = nan , N , an = −nan
(2.115)
30
Now let us define our ground state of quantum strings. We have started the quantization
from our basic operators as in equation (2.63) and from this we have canonical pairs
+
(xl0 , pl ) and (x−
0 , p ) so we can define the ground state from this pairs, as it is usually
convenience to work in momentum space so the ground states is
p+ , p~T
(2.116)
These are the ground states of string for all values of momenta indicated by the labels
and also called the vacuum states for oscillators in string theory. We can create states
from the ground states by simply acting the creation operator on the ground states. As
we have an infinite numbers of creations operator, for which we can write the general
basis state |λ i of the state space, so
|λ i =
∞ Y
25
Y
al†
n
n=1 l=2
λn,l
p+ , p~T
(2.117)
Here λn,l is the positive integer and is define as the number of times that al†
n (creation
operators) appears. And as the number operator acts on the basis state, so their Eigen
value will be
⊥
⊥
⊥
N |λ i = Nλ |λ i , with Nλ =
Similarly for closed strings, the states becomes
λ, λ̄ =
∞ Y
25
Y
al†
n
n=1 l=2
λn,l
p+ , p~T ×
∞ Y
25
Y
∞ Y
25
Y
nλn,l
n=1 l=2
āj†
m
m=1 j=2
λ̄m,j
p+ , p~T
(2.118)
And the number operators are
⊥
N =
∞
X
n=1
l
nal†
n an
⊥
And N̄ =
∞
X
j
maj†
m am
(2.119)
m=1
From the zero modes of transverse Virasoro operators, the level matching condition is
N ⊥ = N̄ ⊥ . Some of states and the mass spectrum of both open and closed strings are
given in the below tables.
31
List of some open string states
List of some closed string states
Since from the closed string spectrum, we get a particle named graviton which shows
that the gravity emerge into string theory [3].
SUPERSTRINGS
The bosonic string theory has some unsatisfactory features, like the spectrum of the
closed-string contains a tachyon particle. Open strings spectrum also contains tachyons
which are unphysical because of implying instability of the vacuum. As the eliminations
of the open string tachyons has been well understood in the term of D-branes’s decay,
while, the closed-strings tachyon has not been understood yet [7].
And also the spectrum of the open as well as closed strings does not contain fermions.
With out fermions, the theory is unrealistic. To make a realistic theory which describes
the nature and all particles (Bosons and Fermions), we required a supersymmetry, which
relates the bosons and fermions, and the resultant theory is called superstring theory.
There are two basic approaches to develop the super string theories [3].
1. The Ramond-Neveu-Schwarz (RNS) formalism, which is the supersymmetric on
the string world sheet.
2. The Green-Schwarz (GS) formalism, which is supersymmetric in ten-dimensional
Minkowski space-time.
These approaches are equivalent at least for ten-dimensional Minkowski spacetime. This
chapter will describe the RNS formalism.
This version of string theory will be constructed along the same line as for the bosonic
theory i.e. we will consider the classical action, the solutions to the equation of motion
and their mode expansions and then we will apply the canonical and light cone gauge
quantization procedures [19].
32
33
3.1
The super string Action
To get fermions in our theory we introduce a new dynamical world-sheet variables
ψ1µ (τ, σ) , and ψ2µ (τ, σ) as like for bosons the dynamical word-sheet variable is X µ (τ, σ).
These classical variables are ψαµ (τ, σ) with (α = 1, 2) are anti-commuting variables,
rather than commuting variables.
In light cone gauge we set X + proportional to τ and X − was solved for, in terms of
other quantities. While in the case of superstrings, this remains the same but in additionally the light cone gauge condition also set ψα+ = 0 and then allow us to solve for
ψα− . As X − and ψα− , both gets contributions from transverse X l and ψαl , So this means
that both contributes into the light cone Lorentz Generator M −l , And from this we
can fixed the space time dimensions, which is D = 10 for super string.
Now let us start from the classical action that describes the full set of degree of freedom.
As we will use the light cone gauge which concern with transverse field, so we can write
the action in the term of transverse field
1
S=
4πα′
With
1
Sψ =
2π
Z
dτ
Z
Zπ
0
Zπ
dσ(Ẋ l Ẋ l − X ′l X ′l ) + Sψ
(3.1)
dσ ψ1l (∂τ + ∂σ )ψ1l + ψ2l (∂τ − ∂σ )ψ2l
(3.2)
dτ
0
Here Sψ action is the Dirac action for fermions, which live in the two dimension worldsheet [7, 20].
3.1.1
Equations of motion and Boundary conditions
In the order to find the equations of motion and the boundary conditions, we will vary
the fields ψαµ in action Sψ . So we have
1
δSψ =
2π
Z
dτ
Zπ
0
dσ δψ1l (∂τ + ∂σ )ψ1l + ψ1l (∂τ + ∂σ )δψ1l
34
+ δψ2l (∂τ − ∂σ )ψ2l + ψ2l (∂τ − ∂σ )δψ2l
(3.3)
By simplifying we can get the equations of motion and the boundary conditions, as
δSψ = 0 which gives the equations of motion
(∂τ + ∂σ )ψ1l = 0, (∂τ − ∂σ )ψ2l = 0
(3.4)
This implies that ψ1l is the right moving while ψ2l is left moving such as
ψ1l (τ, σ) = ψ1l (τ − σ)
ψ2l (τ, σ) = ψ2l (τ + σ)
(3.5)
Similarly from δSψ = 0, the boundary conditions becomes
ψ1l (τ, σ∗ )δψ1l (τ, σ∗ ) − ψ2l (τ, σ∗ )δψ2l (τ, σ∗ ) = 0
(3.6)
This should hold for both the end points, σ∗ = 0 and σ∗ = π for all τ . As from above
equation (3.6) we can also write ψ1l (τ, σ∗ ) = ± ψ2l (τ, σ∗ ) for each end point and by this,
with out loss of generality, we take
ψ1l (τ, 0) = ψ2l (τ, 0)
(3.7)
And the relative sign between ψ1l and ψ2l become important for other end of string. So
we have
ψ1l (τ, π) = ±ψ2l (τ, π)
(3.8)
Let we have a fermion field, ψ l which is define over an interval σ ∈ [−π, π] then the
boundary conditions can be written as
ψ1l (τ, σ) σ ∈ [0, π]
l
ψ ≡
ψ l (τ, −σ) σ ∈ [−π, 0].
2
(3.9)
And finally, from the relative sign, we have the two conditions i.e. periodic and anti
periodic fermions ψ l . which are
ψ l (τ, π) = +ψ l (τ, −π)
(3.10)
35
ψ l (τ, π) = −ψ l (τ, −π)
(3.11)
These are the two sectors or boundary conditions for fermion field and we will call
them, Ramond (R) sector for periodic as in equation (3.10) and Neveu Schwarz (NS)
sector for anti periodic fermion ψ l as in equation (3.11).
3.2
Neveu Schwarz sector
As Neveu Schwarz fermion is the function of τ − σ and has changed the sign when
σ → σ + 2π, so the mode expansion can be written as
ψ l (τ, σ) ∼
X
blr e−ir(τ −σ)
(3.12)
r∈Z+1/2
Here ψ l is an anti periodic, for any r = n +
1
2
with n is an integer.
As ψ l is an anti commuting ‘field which means that the expansion coefficients blr will be
then anti commuting operators, and for the negatively r like bl−1/2 , bl−3/2 , bl−5/2 ,..., are
the creation operators, while for positively like bl1/2 , bl3/2 , bl5/2 ,..., are the annihilation
operators. And these operators will also satisfy the anti commuting relation, like
blr , bjs = δr+s,0 δ lj
(3.13)
Similarly these operators will act on the ground or vacuum which we called Neveu
Schwarz vacuum or simply |NSi. And the states in the Neveu Schwarz sectors are
|λi =
9
∞
Y
Y
l=2
l
(α−n
)λn,l
n=1
9
Y
j=2
∞
Y
r=1/2,3/2..
(bj−r )ρr,j |NSi ⊗
p+ , p~T
(3.14)
This is the full ground state which is in the product, ⊗ of the ground state |p+ , p~T i for
l
α−n
and |NSi for bj−r operators.
For the NS sector the mass squared operator with out normal ordering, is given by
M2 =
1
α′ 2
1X
p6=0
l
α−p
αpl +
1
2
X
r∈Z+1/2
rbl−r blr
(3.15)
36
Now to find the normal ordering constant, we use the same procedure as we have done
1
for Bosonic string theory. As for bosonic oscillators, αl each coordinate contribute − 24
to the normal ordering constant, ” a ”. So we take this as
ab = −
1
24
Similarly for NS sector fermions the contribution term in mass squared operator M 2
will be
1
2
X
rbl−r blr =
r=−1/2,−3/2,..
And as
∞
X
X
r=
r=1
1
2
X
r=1/2,3/2,..
X
r+
+
r∈Zodd
rbl−r blr − (D − 2)
r=
+
r∈ZEven
r+2
r
(3.16)
r∈Z + +1/2
∞
X
r
r=1
+
r∈Zodd
From this we write
X
X
X
∞
+
r∈Zodd
+1/2
1
1X
r=
r=−
2 r=1
24
(3.17)
So the above equation (3.16) then becomes
1
2
X
r=−1/2,−3/2,..
rbl−r blr =
1
2
X
r=1/2,3/2,..
rbl−r blr −
1
(D − 2)
48
(3.18)
above equation (3.18) gives the idea that the contribution term for NS fermions is
aN S = −
1
48
And the full normal ordering constant for M 2 is written as
a = (D − 2)(aB + aN S ) = −(D − 2)
1
16
(3.19)
For D = 10 then from above equation (3.19), a = − 21 and hence the mass squared
operator becomes
1
M = ′
α
2
∞
X
1
⊥
l
− +N
, with N ⊥ =
α−p
αpl +
2
p=1
X
r=1/2,3/2,..
rbl−r blr
(3.20)
37
As the fermionic oscillators are contributing half integers N ⊥ to so by this we get some
of the first few states in this NS sector are as
1
, N ⊥ = 0 : |NSi ⊗ p+ , p~T
2
1
α′ M 2 = 0 , N ⊥ = : bl−1/2 |NSi ⊗ p+ , p~T
2
o
n
1
l
, bl−1/2 bj−1/2 |NSi ⊗ p+ , p~T
α′ M 2 = , N ⊥ = 1 : α−1
2
o
3 n l j
b−1/2 , bl−3/2 , bl−1/2 bj−1/2 bk−1/2 |NSi ⊗ p+ , p~T
α′ M 2 = 1 , N ⊥ = : α−1
2
α′ M 2 = −
(3.21)
The ground state (N ⊥ = 0) of the NS sector is a tachyon. The first excited states
(N ⊥ = 12 ) are massless and are eight states labeled by transverse index l.
As from this list we have both boson as well as fermions. As from above list (3.21)
States which have an even fermion number are bosonic while odd fermion number are
fermionic. So let us define an operator which distinguishes bosons and fermions from
one another. This operator will be called as (−1)F , where F stand for fermion number.
The operator has plus one value on the bosonic state while minus one value for fermionic
state. The fermions ground state is then written as
(−1)F |NSi ⊗
p+ , p~T = − |NSi ⊗
p+ , p~T
(3.22)
Similarly the operator (−1)F acts on the state, as in equation (3.14)
(−1)F |λi = −(−)
P
r,j
ρr,j
|λi
(3.23)
Hence this result allow us to take (−1)F operator as an anti commuting with all other
fermionic operators, like
(−1)F , blr = 0
(3.24)
The fermionic or bosonic character of the states is so far restricted to the (τ, σ) worldsheet [7, 21]. We will discuss later that these states are fermions or bosons in spacetime.
38
3.3
Ramond sector
Now the Ramond boundary conditions (3.10) in which field ψ l is a periodic and can be
expanded in the terms of oscillators
ψ l (τ, σ) ∼
X
dln e−ir(τ −σ)
(3.25)
n∈Z
As ψ l is an anti commuting, so the Ramond oscillators dln will be also anti commuting
operators. And the negatively like dl−1 , dl−2 , dl−3 ,..., are the creation operators, while
for positively like dl1 , dl2 , dl3 ,..., are the annihilation operators. Similarly the Ramond
oscillators dln will satisfy the anti commutation relations
dlm , djn = δm+n δ l j
(3.26)
As from this (3.26) the zero modes, dl0 should be treated with care. It turns out the
idea that these eight operators will have to organize by the simple linear combination
of the four creation and four annihilation operators. Let us we have the four creation
operators out of these eight operators
ξ 1 , ξ2 , ξ3 , ξ4
(3.27)
As being the zero modes, these creation operators will act with out changing the mass
squared state. Let we have a vacuum |0i, since the creation operators in (3.27) can
give the 24 degenerate Ramond ground states. Since we have total 16 ground states in
which eight of them will have an even number of ξs acting on |0i, while the other eight
will have an odd number of ξs acting on |0i. Let the eight states which have an even
number of creation operators is |Ra i with a = 1, 2, ..., 8 then
|0i ,
|Ra i :
ξ1 ξ2 |0i , ξ1 ξ3 |0i , ξ1 ξ4 |0i , ξ2 ξ3 |0i , ξ2 ξ4 |0i , ξ3 ξ4 |0i ,
ξ1 ξ2 ξ3 ξ4 |0i .
(3.28)
39
Similarly the other eight states which have an odd number of creation operators is |Rā i
with ā = 1̄, 2̄, ..., 8̄ then
|Rā i :
ξ1 |0i , ξ2 |0i , ξ3 |0i , ξ4 |0i ,
(3.29)
ξ ξ ξ |0i , ξ ξ ξ |0i , ξ ξ ξ |0i , ξ ξ ξ |0i .
1 2 3
1 2 4
1 3 4
2 3 4
Hence |Ra iand |Rā i states makes the complete set of the degenerate Ramond ground
states and can be denoted as |RA i with A = 1, 2, ..., 16 . The state space of Ramond
sector can be written as
|λi =
9
∞
Y
Y
l=2
l
(α−n
)λn,l
n=1
9
Y
∞
Y
j=2
r=1/2,3/2..
(dj−m )ρm,j |RA i ⊗
p+ , p~T
(3.30)
Just like as in NS sector, the Ramond sector has also an operator,(−1)F which is anti
commuting with all the other fermionic oscillators, including the zero modes
(−1)F , dln = 0
(3.31)
Conventionally, we declare |0i to be fermionic, like
(−1)F |0i = − |0i
(3.32)
The operator gives the idea that (−1)F states are fermionic while |Ra i states are bosonic.
In the R sector |Rā i, the mass squared operator without normal ordering can be written
as
1
M = ′
α
2
1X l l 1X l l
nd d
α α +
2 p6=0 −p p 2 n∈Z −n n
!
(3.33)
Similarly for R sector, the contribution term in M 2 is
1 X
1 X
1
ndl−n dln =
ndl−n dln + (D − 2)
2 n=−1,−2,..
2 n=1,2,..
24
(3.34)
Since the above equation (3.34) gives the idea that the contribution term for R fermions
is
aR =
1
24
40
Hence the contribution term for R sector aR is equal (with opposite sign) to the contribution term for bosonic aB . So the total normal ordering constant becomes zero, thus
we can write the mass squared operators as
M2 =
1 X l l
α−p αp + ndl−n dln
′
α n≥1
(3.35)
This implies that the Ramond ground states are massless and some of few excited states
in this R sector are as
α′ M 2 = 0 |Ra i k |Rā i
l
l
α′ M 2 = 1 α−1
|Ra i ,dl−1 |Rā i k α−1
|Rā i , dl−1 |Ra i
j
j
l
l
l
l
α′ M 2 = 2 α−2
, α−1
α−1
, dl−1 dj−1 |Ra i , k {α−2
, α−1
α−1
, dl−1 dj−1 } |Rā i ,
{αl−1 dj−1 ,dl−2 } |Rā i k {αl−1 dj−1 ,dl−2 } |Ra i
(3.36)
As we have separated the states into the two groups which have an identical number of
states, the left of the bars states gives (−1)F = −1 (fermionic states). While the right of
the bars states gives the bosonic states i.e. (−1)F = +1 . And hence for each fermionic
state, there is a corresponding bosonic state which is the supersymmetry [7, 21]. But
however, this supersymmetry is on the world sheet and the space time supersymmetry
will be arise by combining the states from both, Ramond and Neveu Schwarz sectors.
3.4
Super transverse Virasoro operators
For quantization of superstring theory, we needed the super Transverse Virasoro operators which are the generalization of the Transverse Virasoro operators. As before
we find the transverse Virasoro operators for bosonic string theory, the similar calculation will be required for the superstring [13, 22]. By including the supersymmetry, the
Virasoro operators will become the super Virasoro operators and can be written as
41
⊥(a)
L⊥
n = Ln
⊥(a)
L⊥
n = Ln
⊥(a)
Here Ln
⊥(b)
+ Ln
NS sector
⊥(d)
+ Ln
R sector
is the transverse Virasoro operator which can be written as
L⊥(a)
=
n
⊥(b)
And the Ln
1X l
α αl
2 p∈Z n−p p
⊥(d)
is terms is due to NS sector while Ln
term is for R sector, which can
be written as
⊥(b)
Ln
=
1
2
P
r∈Z+1/2
⊥(d)
(r + n2 )bl−r bln+r and Ln
=
1
2
These are helpful in quantizing the theory.
3.5
P
m∈Z
(m + n2 )dl−n dln+m
Counting states
In the order to count the number of states at any given mass squared operator. For
this we need a generating function which contains the information about the number of states. As for bosonic string theory (open string), we have the oscillators
(a†1 , a†2 , a†3 , ..., a†n ), then the generating function can be written as
f (x) =
∞
Y
n=1
1
1 − xn
(3.37)
As there are 24 transverse light cone directions for each oscillator, since each will have
a generating function. So the complete generating function for bosonic string theory
can be written as
f (x) =
∞
Y
n=1
1
(1 − xn )24
(3.38)
This is the generating function for the bosonic open string or simply for N ⊥ . Since the
mass squared α/ M 2 , is more physical than the number operator N ⊥ . So we need to
define the generating function by using α/ M 2 instead of N ⊥ . As for bosonic open string
α/ M 2 = N ⊥ − 1 then for this the generating function will be obtain by dividing the
42
above generating function (3.38) by the one power of x. Hence the generating function
for the bosonic open string theory, fos (x) can be written as
∞
1Y
1
fos (x) =
x n=1 (1 − xn )24
We can also expand it as
fos (x) =
1
+ 24 + 324x + 3200x2 + 25650x3 + ....
x
(3.39)
(3.40)
These are the states of the mass squared operators i.e. 24 massless states of photon
and so on.
Similarly the generating function for Neveu Schwarz is written as
fr (x) = 1 + xr
As there are 8 transverse light cone directions for each oscillators bl−1/2 , bl−3/2 , ... , since
each will have a generating function, and also the α′ M 2 = N ⊥ − 12 for NS sector, so the
complete generating function for NS sector can be written as
∞
1
1 Y (1 + xn− 2 )8
fN S (x) = √
x n=1 (1 − xn )8
(3.41)
√
√
1
fN S (x) = √ + 8 + 36 x + 128x + 402x x + ...
x
(3.42)
And by expanding we find
These are the states of the mass squared operators i.e. 8 massless states.
Now for Ramond sector, as α′ M 2 = N ⊥ , since the generating function becomes
∞
Y
(1 + xn )8
fR (x) = 16
(1 − xn )8
n=1
(3.43)
The overall 16 factor is due to each combination of the Ramond oscillator which gives
rise to the 16 states as by acting on each ground state. By expanding, we find
fR (x) = 16 + 256x + 2304x2 + ...
(3.44)
We notice that the Ramond coefficients (3.44) are actually double to the corresponding
Neveu Schwarz coefficients (3.42). We will discuss this later.
43
3.6
Open superstrings and the GSO projection
As from the Ramond sector, we see that the word sheet has a supersymmetry i.e. the
ground state which is divided into two groups of |Ra i and |Rā i, which built from the
zero modes dl0 , and has an equal number of fermionic and bosonic states having the
opposite value of (−1)F .
As the zero modes, dl0 carry the Lorentz index vector and can transform under Lorentz
transformation but the Ramond ground states (|Ra i , |Rā i) do not transform like a
vectors but transform as a spinors. Which shows that the index a, ā are the spinor
indices but are different spinors (and different fermions).
This means that there are two types of fermions in R sector, but both of them gives the
opposite values of (−1)F and with two types of fermions we cannot get any space time
supersymmetry because we identified |Rā i as a space time bosons, but bosons cannot
carry any kind of spinor index.
Now to solve this issue as, by projecting the spectrum of R sector in a specific way in
which we choose only those fermions which have (−1)F = −1, by this we can get space
time fermions, and this projection is called Gliozzi, Scherk, and Olive (GSO) projection
[3, 7, 21]. Hence the resultant projected R sector is then called the R− sector (R minus)
which have (−1)F = −1. Similarly, the , is the set of states for which (−1)F = +1. So
the generating function for R− sector then becomes as
∞
Y
(1 + xn )8
fR− (x) = 8
(1 − xn )8
n=1
(3.45)
By expanding, the power series of (3.45) is then
fR− (x) = 8 + 128x + 1152x2 + ...
(3.46)
So we see that there are eight fermionic massless states in R− sector. Similarly doing
the same thing for NS sector, hence, we find sector states for which (−1)F = −1 and
N S+ sector which have (−1)F = +1 . Since the N S− sector contain a tachyon, hence
it is useful to take the full open superstring by combining the set of states from the R−
44
sector and N S+ sector.
Now to find the generating function for N S+ sector fN S+ (x) by taking account of GSO
projection which eliminate the contribution of even number of fermions, so we get the
generating function for N S+ sector as
#
"∞
1
∞
Y (1 + xn− 12 )8 Y
1
(1 − xn− 2 )8
−
fN S+ (x) = √
(1 − xn )8
2 x n=1 (1 − xn )8
n=1
For supersymmetry, we need to have fN S+ (x) = fR− (x) so this means that
!
1
1
∞
∞
∞
Y
Y
(1 + xn− 2 )8 Y (1 − xn− 2 )8
(1 + xn )8
1
√
−
=
8
(1 − xn )8
(1 − xn )8
2 x n=1 (1 − xn )8
n=1
n=1
(3.47)
(3.48)
This identity is being proved by Carl Gustav Jacob Jacobi in his work on elliptic
functions, which is published in 1829. While in our case, for the basis of supersymmetric,
it is a key equation.
3.7
Closed string theories
As we know that the closed (bosonic) strings can be obtained by the combining the
left moving, and the right moving copies of the open strings. Similarly the closed
superstrings can be obtained by combining the open superstrings. Since there are two
sectors (NS and R) for open superstring, so there are four possibilities for the closed
string sectors, like
(NS, NS), (NS, R), (R, NS), (R, R)
(3.49)
As in the case of open superstrings, the space time bosons arises from NS sector while
the space time fermions arises from R sector. Similarly, in the case of closed superstring,
we can get the space time bosons from (NS, NS) as well as from the (R, R)1 sectors
while the space time fermions from (R, NS) and (NS, R) sectors.
1
The (R,R) sectors are ”doubly” fermionic and thus the spacetime bosons
45
3.7.1
Type IIA Superstring Theory
In the order to get the closed superstring theory with a supersymmetry, for this we
need to truncate the above four sectors in equation (3.49). Since we can truncate as
NS +
NS +
Left sector :
Right sector :
(3.50)
R−
R+
By this we find the four sectors which is called the type IIA superstring and these
sectors are
(NS + , NS + ), (NS + , R + ), (R − , NS + ), (R − , R + )
(3.51)
This is the type IIA superstring and for this, the mass squared is written as
1 ′ 2
α M = α′ ML2 + α′ MR2
2
(3.52)
Here ML2 and MR2 are denoting the mass squared operators of the left and right sectors
respectively. Now listing some of the massless states of the varies sectors, which are
(NS + , NS + ) : b̄l−1/2 |NSiL ⊗ bj−1/2 |NSiR ⊗
p+ , p~T
(3.53)
(NS + , R + ) : b̄l−1/2 |NSiL ⊗ |Rb̄ iR ⊗
p+ , p~T
(3.54)
(R − , NS + ) : |Ra iL ⊗ bl−1/2 |NSiR ⊗
p+ , p~T
(3.55)
(R − , R + ) : |Ra iL ⊗ |Rb̄ iR ⊗
p+ , p~T
(3.56)
Hence, there are 64 bosonic states in (3.53) due to the index and having eight values.
These are just like the bosonic closed string theory massless states and carry the two
indices. So we get the graviton (35 states), the Kalb-Ramond field (28 states), and the
dilaton (one state).
(NS + , NS + ) massless fields are : gµν , Bµν , φ
(3.57)
There are 64 fermionic states in each of the states (3.54) and (3.55) due to Both the
states given in (3.54) and (3.55) have included a Ramond ground state, so these states
46
are space time fermions. And give the total of 2 x 8 x 8 = 128 fermionic states. Similarly,
the states given (3.56) having the two R ground states, which have doubly fermionic,
so these will be the space time bosons which have 8 x 8 = 64 massless bosonic states
and together with the bosonic states in (3.53) gives the total massless, bosonic states
as 64 + 64 = 128 of the type IIA superstring. The space time bosonic states match
with the space time fermionic states, which is the supersymmetry of the theory.
3.7.2
Type IIB Superstring Theory
A different superstring theory arises by truncating the four sectors which is given in
(3.49) such that
Left sector :
NS +
Right sector :
R−
NS +
(3.58)
R −
By this we find the four sectors which is called the type IIB superstring and these
sectors are
(NS + , NS + ), (NS + , R − ), (R − , NS + ), (R − , R − )
(3.59)
Now listing the massless states of this type IIB superstring theory, which are
(NS + , NS + ) : b̄l−1/2 |NSiL ⊗ bj−1/2 |NSiR ⊗
p+ , p~T
(3.60)
(NS + , R − ) : b̄l−1/2 |NSiL ⊗ |Rb iR ⊗
p+ , p~T
(3.61)
(R − , NS + ) : |Ra iL ⊗ bl−1/2 |NSiR ⊗
p+ , p~T
(3.62)
(R − , R − ) : |Ra iL ⊗ |Rb iR ⊗
p+ , p~T
(3.63)
(NS - NS ) sector: This sector is same for both the superstrings, type IIA and type IIB.
(NS - R ) and (R - NS ) sectors: these sectors give the space time fermions and contain
a spin 3/2 gravitino (56 states) and a spin 1/2 fermion called the dilatino (eight states).
In the case of type IIB, the two gravitinos will have the same chirality while for the
type IIA superstring, they will have opposite chirality.
47
(R - R ) sector: This sector gives the space time bosons and in the case of type IIA
superstring theory, the massless bosons contain the Maxwell field (eight states) and the
anti-symmetric gauge field with three index (56 states), while the type IIB superstring
theory, the massless bosons contain the scalar field (one state), the Kalb Ramond field
(28 states), and the totally anti-symmetric gauge field with four index (35 states) [3, 7].
(R - R) the massless fields of type IIA : Aµ , Aµνρ
(3.64)
(R - R) the massless fields of type IIB : A, Aµν , Aµνρσ
(3.65)
These are the some massless field of R sectors.
3.7.3
Heterotic superstring theories
There are two types of Heterotic superstring theories. These are the closed superstring
theories. in the Heterotic string, we will combine the left moving bosonic open string
with the right moving open superstring as like the type II closed superstring theories
which arises by the combination of both left and right moving copies of the open superstrings. As there are 26 space time dimension of open bosonic string theory, ten of them
are matched by the right moving open bosonic coordinates with the open superstring
[3]. The extra 16 left moving coordinates can be described a torus with a very special
properties to gives a consistent superstring theory. There are precisely two distinct tori
which have the require properties and they will corresponds to a Lie algebras SO(32)2
and E8 × E8 3 . By this we can get a consistent theory which lives in a 10 dimensional
space time. Heterotic superstrings theory comes into two versions, E8 × E8 type and
SO(32) type. These groups characterize the symmetries which exist in the theories.
2
The group SO(32) is the group generated by 32-by-32 matrices which are the orthogonal and
having a unit determinant.
3
Es is the largest exceptional group, here E is for the exceptional.
48
3.8
Type I
We have discussed the oriented string theory
4
in which the operator X l (τ, σ) involves
a parameter σ ∈ [0, π] i.e. both the type II superstring theories and the Heterotic
superstring theories are the theories of oriented closed strings. We can also construct a
theory that will be an unoriented string theory. For this, we define an operator Ω which
can reverse the orientation of the strings [3, 7]. The unoriented strings can be obtained
by restricting the oriented strings spectrum to the set of the states which are invariant
under the action of Ω. Unoriented strings are not the strings with out the orientation
i.e. they viewed like the quantum superposition of states which as a whole are invariant
under the action of Ω. Hence, we can imagine the unoriented state as a superposition of
a string states and the same states with opposite orientation. A supersymmetric theory
of both open and closed unoriented strings is called the Type I superstring theory.
These are the five different superstrings theories which can be relate by the dualities.
3.9
Critical dimensions
For bosonic string theory, we have fixed the space time dimension by using the commutation relation of the Lorentz generators, M −l as
M −l , M −j = 0
The same idea will be used here as we used before in bosonic theory [7]. Now will take
the super Generators, by the same phenomenology we can construct the super Lorentz
generators. For NS sector the super Lorentz generator becomes
M
−l
=
l
x−
◦p
∞
X
1
i
1 ⊥ l
l
⊥
⊥
l
l
− ′ + x◦ (L0 + a) + ((L0 + a)x◦ − √
(L−n αn − α−n
L⊥
n)
′
+
4α p
n
2α p n=1
∞
X
i
l
l
⊥
r(L⊥
−√
−r br − b−r Lr )
2α′ p+ r=1/2
4
The orientation is defined as the direction of increasing of σ.
49
And similarly, the super Lorentz generator for R sector can be written as
l
M −l = x−
◦p −
∞
X
i
1 ⊥ l
1
l
l
⊥
⊥
l
√
x
(L
+
a)
+
((L
+
a)x
(L−n αn − α−n
L⊥
−
◦
0
0
◦
n)
′
+
′
+
4α p
2α p n=1 n
−√
∞
X
i
l
l
⊥
m(L⊥
−m dm − d−m Lm )
′
+
2α p m=1
By calculating the above commutation relation for N S sector the super Lorentz generator, we get the non vanishing result which is
with ∆m is
−l
M ,M
−j
∞
1 X
l
l
l
l
∆m (α−m
αm
− α−m
αm
)
= − ′ +2
α p m=1
1
1
1
− (D − 2) − 2a
∆m = m 1 − (D − 2) +
8
m
8
As for Lorentz invariance, the above equation should be equal to zero and for this
the dimension of space time, D = 10 and the normal ordering constant, NS sector
aN S = − 12 . and by the similarly method, for the super Lorentz generator R sector
aR = 0.
D-BRANES
A Dp-brane (with p spatial dimensions) is an extended object. The letter D stands
for Dirichlet. When p is equal to the total number of spatial dimensions, then this
type of Dp-brane is called space filling brane. In bosonic string theory, D25-brane is a
space filling brane because p is equal to the total number of spatial dimensions. The
endpoints of open strings must lie on the D-brane.
Similarly, when the Dirichlet conditions applied to the subset of the d indices of X µ ,
i.e. to the indices p + 1, ..., d, this means that the endpoints of the string are restricted
to move on a p-dimensional hyper plane. This higher dimensional object is then calling
a Dp-brane, where p indicates its (spatial) dimensions and D stands for Dirichlet.
It turns out that these objects should be considered to be dynamical as well.
4.1
Tachyons and D-brane decay
In bosonic string theory there is a particle having imaginary mass, called tachyon. The
open string is attached to a D-brane. The tachyons make the D-brane unstable [7, 23].
In the order to study the tachyons and D-brane decay, let start from the scalar field
because the field associated with the tachyons are the scalar field and the Lagrangian
for a scalar field can be written as
1
L = − η µν ∂µ φ ∂ν φ − V (φ)
2
50
51
Here V (φ) is the potential of scalar field. From scalar field Lagrangian, we can find the
equation of motion
∂ 2φ
− 2 + ∇2 φ − V ′ (φ) = 0
∂t
As the scalar field potential can be written in the form of mass, as
1
V (φ) = M 2 φ2
2
As when M 2 > 0 then the potential V (φ) will have the stable minimum at φ = 0 while
when M 2 < 0 then the potential V (φ) will have the unstable maximum at φ = 0. For
simplicity, let us consider the field only depends on time, then the equation of motion
becomes as
∂ 2 φ(t)
+ M 2 φ(t) = 0
∂t2
Now when the mass squared is greater than zero then the solution of the above equation
become
φ = A sin(M t + a0 )
As due to the stable point, the scalar field could be sitting at φ = 0 forever and will be
simply oscillate whenever it is displaced [24]. Now consider the imaginary mass squared,
like M 2 = −β 2 , where β is positive, in this case the equation of motion becomes as
∂ 2 φ(t)
− β 2 φ(t) = 0
2
∂t
The solution of this equation become as
φ(t) = A cosh(βt) + B sinh(βt)
Let the solution φ(t) = sinh(βt). As when time is zero, φ(0) = 0 but when time goes
to infinite then φ(∞) = ∞. This can be imagining as the field φ is rolled up. By
using the trivial solution, we can say that the tachyon will stay at φ = 0 but any small
perturbation could make it to roll-off. The point φ = 0 is an unstable point for tachyon
and tachyon cannot stay here for some definite time, which shows the instability. As
the mass squared of tachyon in an open string theory is
M2 = −
1
α′
52
Since the potential of the free tachyon becomes
f ree
Vtach
(φ) = −
1 2
φ
2α′
As the presence of the tachyon is the signal of instability of the open string theory. As
the open string is attached to the D-brane, since we can say that there is instability in
the background of space filling D-branes. In bosonic open string theory, the space filling
D-brane is D25-brane. As it is a physical object and its have some energy density T25 [7].
Tachyons are the states of an open string which is attached to D-brane this means that
the tachyons are the excited states of D-brane and by this, we can say the tachyons
states will lower the energy of D-brane. Explicitly, the existence of tachyons means
that the space filling brane, D25-brane is unstable.
4.2
Quantization of open strings in the presence of
various kinds of D-branes
4.2.1
Dp-branes and boundary conditions
Considering a Dp-brane and let introduces the spacetime coordinates xµ , with µ =
0, 1, 2, .., 25 for bosonic string theory, splitting these coordinates into two groups i.e.
tangent to the brane and normal to the brane. As the tangential coordinates includes
the time coordinate as well as let some spatial coordinates p , then the normal to the
brane includes (D − p) coordinates. We can write it as
x0 , x1 , x2 , .., xp
|
{z
}
DP tangential coordinates
, xp+1 , xp+2 , xp+3 , .., xd
|
{z
}
(4.1)
DP normal coordinates
For simplicity we let xa = x̄a , with (a = p + 1, ..., d). We can also write the above
equation (4.1) for string coordinates X µ (τ, σ) in similar fashion, like
X 0 , X 1 , X 2 , .., X p , X p+1 , X p+2 , X p+3 , .., X d
|
|
{z
}
{z
}
DP tangential coordinates
DP normal coordinates
(4.2)
53
Since the open string attached with the Dp-brane and the end points must lies on this
Dp-brane then the normal coordinates of string to the brane will satisfy the Dirichlet
boundary conditions such that
X a (τ, σ)|σ=0 = X a (τ, σ)|σ=π = x̄a , a = p + 1, ..., d
(4.3)
The coordinates X α will call DD coordinates due to these satisfying the Dirichlet boundary condition on both the end points. The end points of the open string can move everywhere along the tangent direction to the brane. By this the open string coordinates
which is tangent to the D-brane will satisfy the Neumann boundary conditions, as
X ′m (τ, σ)|σ=0 = X ′m (τ, σ)|σ=π = 0 , m = 0, 1, ..., p
(4.4)
These call NN coordinates due to both of the open string endpoints satisfying the
Neumann boundary condition. Hence we can summarize this as
X 0 , X 1 , X 2 , .., X p , X p+1 , X p+2 , X p+3 , .., X d
|
{z
} |
{z
}
NN coordinates
(4.5)
DD coordinates
Now in the order to use the light cone coordinates we need to have at least one of the
spatial NN coordinate from which we define X ± coordinates. So this means that we
will assume p ≥ 1 and our analysis will not apply to the strings which will attach to a
D0-brane. Hence in light cone coordinates, we may write
X + , X − , {X i } , {X a } , i = 2, ..., p and a = p + 1, ..., d
|
{z
} | {z }
NN
4.2.2
(4.6)
DD
Quantizing open strings on Dp-branes
As before we have quantized the open string in the presence of the space filling Dbrane and now we will quantize the open string in the presence of the Dp-brane and
to study the effects on string spectrum states due to the presence of the Dp-brane. As
the NN coordinates X i (τ, σ) will also satisfy the same conditions that were satisfied
by the light-cone coordinates X i (τ, σ) of the open strings which were attached to the
54
D25-brane. Since our previous results are useful for the coordinates X i . Let start from
equation (2.47) and let l → (i, a) then we finds
−
Ẋ ±X
′−
1 1 i
′i 2
a
′a 2
= ′ + (Ẋ ± X ) + (Ẋ ± X )
2α 2P
(4.7)
As X i (τ, σ) are satisfying exactly the same conditions which satisfied by X l (τ, σ) so by
this we can write
√
Ẋ i ± X ′i =
2α′
X
αni e−in(τ ±σ)
(4.8)
n∈Z
Now to investigate the normal coordinates X i (τ, σ), as these coordinates are normal to
the brane will satisfy the wave equation. Now solving the wave equation for X α , we
get1
√
X a (τ, σ) = x̄a +
2α′
X1
n6=0
And by this we can easily finds
X ′a ± Ẋ a =
√
2α′
X
n
αna e−inτ sin nσ
αna e−in(τ ±σ)
(4.9)
(4.10)
n6=0
This is same as equation (4.8) but differ by a minus sign and a zero mode which is
absent here in equation (4.10).
We can easily quantize this string which is attached with a Dp-Brane. As P τ a =
1
Ẋ a
2πα′
then the non vanishing commutation relations are
h
i
X a (τ, σ), Ẋ b (τ, σ ′ ) = i2πα′ η ab δ(σ − σ ′ )
(4.11)
Similarly we can easily find that
a
αm
, αnb = m η ab δm+n,0
(4.12)
The mass squared operators becomes as
1
M2 = ′
α
1
−1 +
∞
X
n=1
Here x̄a is not an operator but a number.
i
nai†
n an +
∞
X
m=1
a
maa†
m am
!
(4.13)
55
Here the repeated index will be summed, the normal ordering constant is (a = −1),
and the critical dimensions are same as for D25-brane. As l → (i, a) since our ground
state will not remain the same as D25-brane. Since for Dp-brane, the ground state will
be labeled by p+ , pi and then becomes | p+ , p~i as with p~ = (p2 , ..pp ). Now the state
space are written as
|λ i =
"
p
∞ Y
Y
n=1 i=2
λn,i
ai†
n
#"
∞
d
Y
Y
m=1 a=p+1
λm,a
aa†
m
#
p+ , p~
(4.14)
The associated wave functions take the form
ψi1.... ip a1.... aq. (τ, p+ , p~)
Since some of the fields that satisfy M 2 ≤ 0 associated with Dp-branes are, the scalar
+
field (tachyons), the massless Maxwell field (photons) ai†
~i and similarly the os1 |p , p
cillator that acts on the ground states from those coordinates normal to the Dp-branes
+
are ai†
~i , these are the massless the Lorentz scalar (due to the index ) which are
1 |p , p
normal to the branes.
4.2.3
Open string between parallel Dp-branes
Now considering the open strings which is extends between the two parallel Dp-branes.
Let the first Dp-brane is located at xa = x̄a1 ’while the second one is at xa = x̄a2 . The
classes of the open string which are supported on the particular configurations of the
D-branes are called sectors. In the presence of a two parallel Dp-branes, an open string
can begin from one brane and end on the other brane. There are four sectors in our
case of two parallel Dp-branes.
Consider the sector, in which open strings begins on the brane one and end on the
brane two. For this, the boundary conditions for the DD strings coordinates are
X a (τ, σ)|σ=0 = x̄a1 , X a (τ, σ)|σ=π = x̄a2 , a = p + 1, ..., d
(4.15)
56
Similarly for X a which are normal to the branes and will satisfy the wave equation. By
solving the wave equation for X a , we find
X a (τ, σ) = x̄a1 + (x̄a2 − x̄a1 )
σ √ ′ X 1 a −inτ
+ 2α
α e
sin nσ
π
n n
n6=0
(4.16)
From this we get
X ′a ± Ẋ a =
√
2α′
X
αna e−in(τ ±σ) and
√
2α′ α0a =
n∈Z
1 a
(x̄2 − x̄a1 )
π
Similarly the mass squared will becomes
2
a
x̄2 − x̄a1
1
2
M =
+ ′ −1 + N ⊥
′
2πα
α
Where
⊥
N =
p
∞ X
X
i
nai†
n an
+
d
∞ X
X
a
maa†
m am
(4.17)
(4.18)
(4.19)
m=1 a=p+1
n=1 i=2
To develop the ground state for our parallel Dp-branes, we need additional labels which
distinguish the various sectors [7]. These additional labels will be two numbers. The
first number will denote the brane on which the open string end point σ = 0 lies while
the second number will denote the brane on which the other end point σ = π lies. Since
the ground states can be written as | p+ , p~; [ij]i and the four sectors are
p+ , p~; [11] , p+ , p~; [22] , p+ , p~; [12] , p+ , p~; [21] .
(4.20)
Here the first two sectors are those which we discuss earlier i.e. x̄a2 = x̄a1 . Let us
first discuss the [12] sector, the mass squared operators for the lowest value of number
operators becomes
+
p , p~; [12]
1
,M =− ′ +
α
2
x̄a2 − x̄a1
2πα′
2
When the separations between the two branes vanished then the field associated is a
scalar tachyonic field. From this we can find the critical separation which is
√
|x̄a2 − x̄a1 | = 2π α′
57
Hence the ground states represents the massless scalar field for the zero separation while
for large separation, the ground states represents the massive scalar fields. The next
states (due to the number operators) are
a†
a
1
p , p~; [12] , M = − ′ +
α
+
2
x̄a2 − x̄a1
2πα′
2
These are the (d − p) Lorentz scalar which is normal to the branes.
i†
a
1
p , p~; [12] , M = − ′ +
α
+
2
x̄a2 − x̄a1
2πα′
2
These are the (p + 1) − 2 = p + 1 massive states. As we know that the massive gauge
fields has more degree of freedom than the massless gauge field. Hence from this we
have one massive vector as well as (d − p − 1) massive scalars.
We have obtained a very interesting situation, as the separation between the Dp-branes
goes to zero and coincide but still they are distinguishable and we have four open string
sectors. The massless string states that represents a strings extending from Dp-brane
one to Dp-brane two, includes the massless gauge field and (d − p) states of a massless
scalars. This is same content as that of a sector in which strings begin and end on
the same Dp-brane. Whenever the two D-branes coincides, we get four massless gauge
fields. From the world-volume of two coincide D-branes we indeed to get a U(2) YangMills theory.
Let we have N Dp-branes, for this the sectors will be labeled by the pairs [i, j]2 , here
i = 1, 2, ..N . The [i,j] sector will consist of the open strings which will start on ith and
end on jth brane and we have N 2 sectors. The interaction of strings can be visualized
as The first open string will join with a second open string and will form another open
string. For this, the end point of first string (σ = π) will join the beginning point of the
second string (σ = 0). And the new string (became from these two) will begin at the
beginning point of the first string and will end on the end point of the second string
[7, 25]. If the strings stretched between the D-branes, then the first string will from
2
The discrete labels i, j are used to label the branes, and the various open string sectors are
sometimes called Chan-Paton indices.
58
the sector will have to join by the second string from sector to gives the product open
string in the sector. The interaction can be written as
[i, j] ∗ [j, k] = [i, k], here j is not summed
If the N Dp-branes are coincides, then the N 2 sectors results in the N 2 interacting
massless gauge fields. Since from this N coinciding D-branes will carry U(N) massless
gauge fields.
4.2.4
Strings between parallel Dp and Dq-branes
Now considering the two parallel D-branes, but having different dimension. Let we have
two D-branes which are Dp-brane and Dq-brane and assume that p > q. From this,
there will be a p-dimensional hyperplane parallel to the Dp-brane which will contain
the Dq-brane. This means that we will have some common tangent as well as normal
directions for both branes. For strings coordinates X µ , we can write for parallel Dpbrane and Dq-brane with p > q
X 0 , X 1 , .., X q
|
{z
}
common tangential coordinates
, X q+1 , X q+2 , .., X p ,
|
{z
}
mixed coordinates
X p+1 , X p+2 , .., X d
|
{z
}
(4.21)
common normal coordinates
Here the mixed coordinates will satisfy the Neumann boundary conditions on the starting Dp-brane while the Dirichlet boundary conditions on the ending of the Dq-brane
and these coordinates will call ND coordinates. Similarly for the strings coordinates,
we can also write
X 0 , X 1 , .., X q , X q+1 , X q+2 , .., X p , X p+1 , X p+2 , .., X d
|
{z
} |
{z
} |
{z
}
(4.22)
X + , X − , {X i } , {X r } , {X a }
{z
} | {z } | {z }
|
(4.23)
NN coordinates
ND coordinates
DD coordinates
In light cone coordinates, we can write these as
NN
ND
DD
Where i = 2, ..., q , r = q + 1, ..., p and a = p + 1, ..., d . Now let for the position of the
Dp-brane is while for position of Dq-brane is x̄r2 and x̄a2 . The boundary conditions for
59
the ND coordinates X r are
X ′r (τ, σ)|σ=0 = 0 , X r (τ, σ)|σ=π = x̄r2
(4.24)
The expansion of X r (τ, σ), by using the same procedure, can be written as
nσ
X 2
√
n
X r (τ, σ) = x̄r2 + i 2α′
αrn e−iτ 2 cos
n 2
2
n∈Z
(4.25)
odd
r
The Hermiticity of the above expansion, X r gives (αrn )† = α−
n . And by the above
2
2
expansion (4.16) we can find
Ẋ r ± X ′r =
√
2α′
X
n
αrn e−i 2 (τ ±σ)
2
(4.26)
n∈Zodd
The non vanishing commutation relations are
h
r
s
i
X (τ, σ), Ẋ (τ, σ ) = i(2πα′ )δ(σ − σ ′ )δ rs
′
(4.27)
And by the similar method as we have done for space filling D-Brane, we can also find
as
h
r
s
i
αm , αn =
2
2
m rs
δ δm+n,0
2
(4.28)
Now as we have three types of coordinates like NN, DD, and ND or DN. The normal
ordering contribution term for both NN and DD coordinates is same because of all
oscillators are integrally moded
aDD = aNN = −
1
24
(4.29)
However, for the ND or DN coordinates, the contribution term can be calculated as
1 X r
1
1 X r
α− m αrm + (p − q)
α− m αrm =
2
2
2
2
2 m∈Z
2
48
+
odd
(4.30)
m∈Zodd
Since this means that the contribution term for ND or DN is
aDN = aND =
1
48
(4.31)
60
Similarly we can easily find the total normal order constant, which is
a=−
1
1
[q + 25 − (p + 1)] + (p − q)
24
48
1
= −1 + (p − q)
16
Now the mass squared operator then becomes
a
2
x̄2 − x̄a1
1
1
2
⊥
M =
+ ′ N − 1 + (p − q)
2πα′
α
16
With
⊥
N =
p
∞ X
X
n=1 i=2
i
nai†
n an
p
∞ X
d
X X
X
k r† r
a
αk αk +
+
maa†
m am
2
2
2
+ r=q+1
m=1 a=p+1
(4.32)
(4.33)
(4.34)
k∈Zodd
And the ground state are now labeled as
p+ , p~; [12]
, p~ = (p2 , ....., pq )
(4.35)
The labels of the ground state indicates that the corresponding fields will be living in
a (q+1 ) dimensions of the space time i.e. the fields will live on the Dq-brane worldvolume, which has the lower dimensions. The ground states having N ⊥ = 0 corresponds
to a scalar field on Dq-brane. This scalar depends upon the separation of the branes.
4.3
4.3.1
String charge and electric charge
Fundamental string charge
As we know that, for the point particle, the word line is one dimensional and Maxwell
gauge field Aµ carrying one index and the point particle carries electric charges whenever
it interacts with Maxwell field, in which the particle couples to the Maxwell field.
Similar idea used for string charge, the string should couples to a new kind of gauge
field. This gauge field is the Kalb Ramond anti symmetric tensor (of rank two)Bµν
which is the massless field and arisen in closed string. The complete dynamics of the
string coupling to Kalb Ramond field can be written in the term of action is
Z
Z
1
1
µ
ν
ν
µ
S = Sstr −
dτ dσBµν (X) (∂τ X ∂σ X − ∂τ X ∂σ X ) − 2 dD x Hµνρ H µνρ (4.36)
2
6k
61
Here Sstr is the string action the second term is the action term due to string charge
while the last term is the action for the Kalb Ramond. And k is the dimensionful
constant which makes the action dimensionless i.e. ([k 2 ] = M 6−D ). The Hµνρ is the
field strength of Bµν which can be written as
Hµνρ ≡ ∂µ Bνρ + ∂ν Bµρ + ∂ρ Bµν
(4.37)
In the second term of the above action (4.36), Bµν (X(τ, σ)) can be written for the
general space time Bµν (X) as
Bµν (X(τ, σ)) =
Z
dD xδ D (x − X(τ, σ)) Bµν (x)
Now the variation of above action (4.36) gives us
Z
1
µνρ
µν
D
∂ρ H
−j
δS = d x δBµν (x)
k2
Here j µν is the (anti symmetric) current
Z
1
µν
j =
dτ dσδ D (x − X(τ, σ)) (∂τ X µ ∂σ X ν − ∂τ X ν ∂σ X µ )
2
(4.38)
(4.39)
(4.40)
By the variation principle, we find
1
∂ρ H µνρ = j µν
2
k
(4.41)
The tensor j µν is conserved quantity like
1 ∂ 2 H µνρ
∂j µν
=
=0
∂xµ
k 2 ∂xµ ∂xρ
(4.42)
As the above equation give the idea of conservation of j µν but there is a ν free index,
which means that j µν is the set of conserved current labeled by (ν) a free index. And
the charge density, as the zeroth component of j µν , since the charge density (j̄ 0 ) of the
Kalb Ramond field is then j 0k , here k will be running over the spatial value. Similarly
from the above equation (4.42), we find ∇.~j 0 = 0 and from the charge density we can
easily find the charge of string, Q̄ as
~ =
Q
Z
dd x ~j 0
(4.43)
62
To understand it more, using the static gauge X 0 = τ = t and then simplifying the
equation (4.40), we get
1
j (~x, t) =
2
~0
Z
~ ′ (t, σ)
~ σ) X
dσδ ~x − X(t,
(4.44)
From this, we get the idea that the charge density is tangent to the every point on
string and lies along orientation of the string.
In the order to write this in more explicit form that the string density depends upon
the orientation, for this let a long static string stretched along the xl , then the string
can be describe as
X 1 (τ, σ) = f (σ) , X 2 = X 3 = .... = X d = 0
(4.45)
Here f (σ) is the function of σ with the range of −∞ to ∞, this function will be either
increasing or decreasing. Now using the above equation (4.45) and solving the equation
(4.44), then we get
j 01 (x1 , ..., xd ; t) =
1
1
sgn(f ′ )δ(x2 )δ(x3 )...δ(xd ) = sgn(f ′ )δ(x⊥ )
2
2
(4.46)
Here sgn(f ′ ) is for the sign of f ′ . This result show explicitly that the charge density ~j 0
depends upon the orientation or the sign of f ′ (σ).
4.3.2
Visualizing string charge
In the order to visualize the charge of string, let us consider we have a static string
for which j ik = 0 ( i,k are spatial components), since for a static string, the equation
(4.41) becomes as
∂H ikρ
=0
∂xρ
By this H are to be time independent so H ijk = 0 and the other equation is
(4.47)
∂H okl
= k 2 j ok
(4.48)
∂xl
~ H . This is called field strength dual to H and
Now let us introduce a new vector B
which can be define as
~H
H 0kl = εklm B
(4.49)
63
Here εijk is the anti symmetric Levi-civita symbol. Since the above equation (4.48)
then becomes
~ H )k = k 2 j 0k
(∇ × B
(4.50)
Now from this equation we can write
~ H = k 2~j 0
∇ × B
(4.51)
~ H is the magnetic field. Integrating
This equation is the Ampere’s equation, where B
the above equation (4.51) over a curve Γ of a two dimensional surface S. so we get
Z
I
1
~
~
(4.52)
BH .dl = ~j 0 .d~a
k2 Γ
S
By this equation, we get the idea that the curve Γ is link to a string and the strings
number N which is associated with Γ is define as
Z
I
1
1
~
~
N = 2
BH .dl = ~j 0 .d~a
2
k Γ
S
(4.53)
Here N is the number of strings which is linked by the curve Γ.
4.3.3
Strings ending on D-branes
As by the quantization of closed strings, there arises a massless field called Kalb Ramond
field which lives over the all spacetime and the string can couple electrically with it to
get charged. Now considering the charged string which attached to D-branes, for this,
the action for the string couple to Kalb Ramond field, is written as
Z
1
dτ dσ εαβ ∂α X µ ∂β X ν Bµν (X(τ, σ))
SB = −
2
(4.54)
Here εαβ is the two dimensional indices α, β = 0, 1, anti symmetric with ε01 = 1. Now
using the above action and calculating the variation in SB , which we find that
Z
δSB = − dτ dσ (∂τ (Λν ∂σ X ν ) − ∂σ (Λν ∂τ X µ ))
(4.55)
Here we use
δBµν (X) =
∂Λν
∂Λµ
−
µ
∂X
∂X ν
(4.56)
64
The arguments of Λ are the string coordinates X(τ, σ)due to the arguments of Bµν .
As from equation (4.55), the two total derivative in which the time derivative give no
contribution, since this means that Λ are vanishes at the end of time. For a closed string
both the derivative vanishes while for the open string, there introduces a boundary
contributions terms, which in not vanishes. In the order to find δBµν for an open
string, let the string coordinates along the brane are X m and the string coordinates
normal to the brane are X a . Since we can write this as
X µ = (X m , X a ) , µ = (m, a)
(4.57)
If this D-brane is Dp-brane, then (m = 0, 1, ...p) and solving the equation (4.55) for
X µ = (X m , X a ) coordinates, so the coordinates X a are DD for which ∂τ X a = 0 and
the limits we take the σ ∈ [0, π] then we get
Z
Z
m
δSB =
dτ Λm ∂τ X |σ=π − dτ Λm ∂τ X m |σ=0
(4.58)
Here we get the two boundary terms and hence, the gauge invariance has failed. From
this we got the idea that the string charge conservation failed at the end points of the
string. In the order to restore the gauge invariance, we should add some terms which
give the electric charges to the end points of the string. then the Action can be written
as
S = SB +
Z
m
dτ Am (X)∂τ X |σ=π −
Z
dτ Am (X)∂τ X m |σ=0
(4.59)
As whenever we vary Λµ (i.e. δBµν = ∂µ Λν − ∂ν Λµ ), then we have to also vary Maxwell
field Aµ on D-brane, like wise δAm = −Λm , and the two terms have opposite sign which
show that the end points of string, which lies on Dp-brane, are oppositely charged. This
action (4.59) restored the string charge conservation and we got the idea that the string
end points have the opposite charge due to the Maxwell gauge field Am and by this we
restore the conservation of charge invariance.
The Maxwell action is also not invariant, as it is proportional to F 2 , like
δFmn = ∂m δAn − ∂n δAm = −∂m Λn + ∂n Λm = −δBmn
(4.60)
65
To make it invariant, we should add Bmn into Maxwell field Fmn such that
Fmn ≡ Fmn + Bmn , δFmn = 0
(4.61)
This Fmn is called the field strength on D-brane and the Lagrangian density on D-brane
is proportional to − 14 F mn Fmn , so expanding it
1
1
1
1
− F mn Fmn = − B mn Bmn − F mn Fmn − F mn Bmn
4
4
4
2
(4.62)
As the last term has some interesting physical significance, expanding it as
1
− F mn Bmn = −F 0k B0k + ...
2
(4.63)
In the order to understand the physical significance of this term, let as the action for
string charge, second term in (4.36), can be written as
Z
≡ − dD x Bµν (x) j µν (x)
(4.64)
This equation tells us that string charge density j 0k couples to B0k . Since anything
couple to B0k will carry string charge so it means that F 0k will represent the string
charge on D-brane but as F 0k is equal to the electric field (F 0k = Ek ), which gives the
idea that the electric field lines will carry the string charge on D-branes.
4.4
D-brane charges and stable D-branes in Type
II
There are also other extended objects rather than strings that carry charge i.e. Dpbranes. To study the charge of a Dp-branes, let start from our previous concept of
charge that the string carry charge when they couple to Kalb Ramond gauge field, so
this coupling is written as
−
Z
dτ dσ ∂τ X µ ∂σ X ν Bµν (X(τ ,σ))
(4.65)
66
Now generalizing this idea like, a Dp-brane will be electrically charge if it coupled to a
massless anti symmetric tensor field of indices (p + 1 ). As Dp-brane has (p+ 1 ) dimensions and is parameterize τ by and the set of its coordinates (σ 1 , σ 2 , ..., σ p ). Let the
space time coordinates which describe the position of this brane is X µ (τ, σ 1 , σ 2 , ..., σ p )
with µ = 0, 1, ...d, and the anti symmetric tensor field is describe by Aµµ1 µ2 ...µp (x) then
the generalize coupling from (4.65) is written as
Z
Sp = − dτ dσ1 ..dσp ∂τ X µ ∂σ1 X µ1 ...∂σp X µp Aµµ1 µ2 ...µp (X µ (τ, σ 1 , σ 2 , ..., σ p ))
(4.66)
Kalb Ramond gauge field is the only massless anti symmetric tensor in bosonic closed
string theory and there is no other massless anti symmetric tensor which means that the
Dp-branes cannot be charged in bosonic string theory. While the type II superstring
theories have some additional anti symmetric massless tensor in (R - R ) sectors, as
listed in (3.64) and (3.65)
(R - R ) the massless fields of type IIA: Aµ , Aµνρ
(4.67)
(R - R ) the massless fields of type IIA: A, Aµν , Aµνρσ
(4.68)
This means that Aµ coupled to the D0-branes and Aµνρ coupled to the D2-branes while,
Aµν coupled to the D1-branes and Aµνρσ coupled to the D3-branes. And the field A does
not couple to any D-brane (it coupled electrically to the object called, D-instanton).
Summarizing these as
T ypeIIA : D0, D2,
(4.69)
T ypeIIB : D1, D3.
(4.70)
These branes are the stable charged and cannot decay into closed or open string states,
while the bosonic D-branes are unstable due to the existence of tachyons, and carry no
charge. Similarly in type IIA superstring theory, the Dp-branes with even are stable
but are unstable in type IIB superstring theory while, in type IIB superstring theory,
the Dp-branes with odd p are stable but are unstable in type IIA superstring theory.
All the stable D-branes of type II superstring theories are charged and the Dp-branes
67
which is not appeared in the above lists (4.69) and (4.70), like D4, D6, and D8 of type
IIA superstring theory while the D5, D7, and D9 of type IIB superstring theory, means
that these are carrying the magnetic charges for either (R - R ) gauge fields in listed
above in (4.67) and (4.68) or for the other (R - R ) states that we not includes in the
discussion [3, 7, 27, 28].
The (electric) charge of the Dp-brane has the simple physical significance, when the p
(spatial dimensions) are curled up to a circles and Dp-brane is wrapping around, the
resulting compact space. Since here, p compact space directions will lies on D-brane
and others space time directions, which defined here, the lower dimensional spacetime,
will be normal to the brane. Hence, the lower-dimensional observer which has only
access to the non compact directions and will see the brane as a point particle.
Let (x1 , ...., xp ) be the compact directions and (X 1 , ...., X p ) be their corresponding coordinates of brane. If compact directions are curled up to the circles of radii (R1 , ...., Rp )
and assuming the parameters σ k ∈ [0, 2π] then
X k (τ, σ 1 , ..., σ p ) = Rk σ k , k = 1, ..., p
(4.71)
Here the repeated index k is not summed. This represent the wrapped Dp-branes and
the coordinates X k are running from zero to 2πRk . Let the non compact dimensions
be the X m with m index is for the non compact directions, then
X m (τ, σ 1 , ..., σ p ) = xk (τ )
(4.72)
This equation shows that, Dp-brane appeared as the point particle to the observer
which has the lower-dimensions. These assumptions made the equation (4.66) as
Z
Sp = − dτ dσ1 ..dσp ∂τ X µ R1 ...Rp Aµ 12...p (X(τ, σ 1 , σ 2 , ..., σ p ))
(4.73)
Since the µ will take the values over the non compact directions, µ = m, then the above
equation (4.73) becomes
Z
Sp = − dτ dσ1 ..dσp ∂τ X m R1 ...Rp Am 12..p X m (τ ), X k (σ k )
(4.74)
68
Finally, let the part of A...field which is independent of compact coordinates i.e. Am 12..p (xm (τ )),
then the above equation (4.74) becomes
Z
1
p
Sp = −R ...R
dτ dσ1 ..dσp ∂τ xm Am 12..p (xm (τ ))
By solving this integral, we find
Z
Z
Vp
Vp
m
m
Ām dxm
Sp = − ′ p/2 dτ ∂τ x Ām (x (τ )) = −
(α )
(ls )p
(4.75)
(4.76)
Here Vp is the volume of compact space, Vp = (2πR1 )....(2πRp ) , And ls is the string
length ls = (α′ )1/2 . And the gauge field Ām as
1
Ā (xm (τ ))
(α′ )p/2 m
= Am 12...p (xm (τ )).
The equation (4.76) recognized that the coupling of the point particle to the Maxwell
field Ām which means that the Dp-brane appeared as a charged point particle. From
this the charge Q of the brane is
Q=
Vp
(ls )p
The charge Q depends upon the volume of the branes [7].
(4.77)
STRING DUALITIES
Duality is generally used for the relationship between the two systems which have very
different descriptions, but identical physics. There are two types of dualities in string
theory, T-duality and S-duality.1 In many cases, the T-duality implies that the two
different geometries of the extra dimensions are physically equivalent i.e. the circle of
radius (R) is equivalent to the circle of (α′ R−1 ) radius. As there are five superstring
theories which looks like the different theories from one another, but the T-duality
relates the two type II superstring theories and the two Heterotic superstring theories.
Similarly, S-duality relates, string’s coupling constant gs to (gs−1 ), in the same ways as
like the T-duality. S-duality relates the type I superstring theory to Heterotic SO (32)
string theory, and type IIB superstring theory relates to itself.
To deal with the T-duality, we should first discuss the effects on the string when one of
the spatial dimensions has curled up to compact space.
5.1
5.1.1
T-duality and closed strings
5.1.1 Mode expansions for compact dimension
Let us consider one of the spatial dimension be curled up,2 in free bosonic string theory
i.e. the X 25 dimension is curled up into a circle of radius R. Now we are going to check
the effects of this on closed string [7, 29], as before we have taken the closed string
1
2
Where some author writes T for target ‘while some use it for toroidal and S is for Strong coupling.
By identification we can compact a dimension i.e. x ∼ x + 2πR
69
70
periodicity condition as
X µ (τ, σ) = X µ (τ, σ + 2π)
(5.1)
This periodicity condition is used whenever the closed string moving in a non compact
dimensions. As usual we will use the light cone coordinates. When there is a compact
dimension, let say X 25 ≡ X and the light cone coordinate are
X + , X − , X 2 , X 3 , ..., X 24 , X
|
{z
}
(5.2)
X(τ, σ) = X(τ, σ + 2π) + m(2πR)
(5.3)
Xi
Then the periodicity condition becomes as
Here m is the number, called the winding number, and defined as, the number of times
that the string winds around the circle of compact dimension and its sign depend upon
the direction of winding. For other non compact dimensions, µ 6= 25 the equation (5.1)
holds. Let defines the winding w in the terms of the winding number m and the radius
of the compact space as
w≡
mR
α′
(5.4)
By this the above periodicity condition in equation (5.3) becomes as
X(τ, σ) = X(τ, σ + 2π) + 2πα′ w
(5.5)
The mode expansions for the this equation (5.5) can be similarly derived as we have
done before, the expansion become as
r
r
1 L
α′
α′ X 1
XL (u) = x0 +
ᾱ0 u + i
ᾱn e−inu
2
2
2 n6=0 n
1
XR (v) = xR
+
2 0
r
And from these two equations we get
r
α′
α′ X 1
ᾱ0 v + i
ᾱn e−inv
2
2 n6=0 n
ᾱ0 − α0 =
√
2α′
(5.6)
(5.7)
(5.8)
71
As we see that ᾱ0 is equal to α0 for zero winding. By calculating the momentum along
the compact direction like
1
P =
2πα′
Z2π
(ẊL + ẊR )dσ = √
0
1
(ᾱ0 +α0 )
2α′
Form solving these two equations (5.8) and (5.9), we get
r
α′
(p − w)
α0 =
2
r
α′
ᾱ0 =
(p + w)
2
Using these new definition by which we get
r
α′ X 1 −inτ
′
′
e
(αn einσ + ᾱn e−inσ )
X(τ, σ) = x0 + α pτ + α wσ + i
2 n6=0 n
(5.9)
(5.10)
(5.11)
This is the mode expansion of compact dimension and by this we can get easily
Ẋ + X ′ =
√
2α′
X
ᾱn e−in(τ +σ)
(5.12)
X
αn e−in(τ −σ)
(5.13)
n∈Z
Ẋ − X ′ =
5.1.2
√
2α′
n∈Z
Quantization and commutation relations
For quantization we adopted the similar method like the modes becomes operators and
starting from the commutation relation between the momentum P τ (τ, σ) and the string
compact dimension, X which is
[X(τ, σ), P τ (τ, σ ′ )] = iδ(σ − σ ′ )
(5.14)
Similarly the other commutation relations, will become by same method as we done
before for non compact dimensions, as
[ᾱm , ᾱn ] = [αm , αn ] = mδm+n,0 , [αm , ᾱn ] = 0
(5.15)
72
Now by the explicit form of ᾱ0 and α0 , we get
[p, w] = 0
(5.16)
Since ᾱ0 and α0 commuting with ᾱn , αn which means that
[p, ᾱn ] = [p, αn ] = [w, αn ] = [w, ᾱn ] = 0
(5.17)
And from equation (5.14) we easily find that
[x0 , α0 ] = [x0 , ᾱn ] = i
r
α′
2
(5.18)
And using the explicit form of ᾱ0 and α0 we find
[x0 , p] = i , [x0 , w] = 0
(5.19)
As we see that the winding w commutes with all the operators which appear in X.
This gives the idea that the winding w is a constant or just a constant number. More
physically the winding w is an operator which gives the Eigen values of w corresponding
to various possible windings of strings. The quantization is only possible for those closed
string sectors which has some fixed winding w. As along the compact dimension, the
string would act as like a point particle that is moving on a circle3 since the momentum
will be then quantize and can be written as
p=
n
,n∈Z
R
(5.20)
Here n is named as Kaluza Klein excitation number. Since both the operators p and w
have discrete values.
5.1.3
Constraint and mass spectrum
For the mass spectrum, we begin from the Virasoro operators in which the sum over
transverse l is splits into a sum over i and a term due to compact dimension, as
1 l l
1 i i
⊥
⊥
L̄⊥
0 = ᾱ0 ᾱ0 + N̄ = p p + ᾱ0 ᾱ0 + N̄
2
4
3
By winding, we loss and gain some states for string, while for particle, we just lost states when
we compact a dimension by identification because the particle cannot wrap around circle to gives new
states.
73
1 i i
1 l l
⊥
⊥
L⊥
0 = α0 α0 + N = p p + α0 α0 + N
2
4
(5.21)
All the oscillators of X l and X will contribute to the number operators N̄ ⊥ and N ⊥
[2]. Thus we find that
1
⊥
⊥
⊥
L⊥
0 − L̄0 = (α0 α0 − ᾱ0 ᾱ0 ) + N − N̄
2
= −α′ pw + N ⊥ − N̄ ⊥
(5.22)
⊥
Since L⊥
0 − L̄0 does not vanish here which imposed a constraint that
N ⊥ − N̄ ⊥ = α′ pw
(5.23)
As both the number operators having the numbers Eigen values, so by the quantization
of both p and w makes the above equation (5.23) more physical as
N ⊥ − N̄ ⊥ = n m
(5.24)
This gives the level matching condition and now the mass squared operators, M 2 =
2p+ p− − pi pi becomes as
M 2 = p2 + w 2 +
2
(N ⊥ + N̄ ⊥ − 2)
′
α
(5.25)
This means that both the momentum p and the winding w give the contribution to the
mass squared operators [3, 7].
5.1.4
State Space of compactified closed strings
As there are additional terms in mass squared operators so for the ground states we
will add the additional labels such that
p+ , p~T ; n, m
, n, m ∈ Z
(5.26)
And the state space can be constructed by applying the creation operators on the
ground states like
#" ∞
#" ∞
#
# " ∞ 24
" ∞ 24
Y
Y
Y
Y
YY
λ
λ̄
k
l
λ̄j,s
λi,r
āj†
a†k
ai†
ā†l
p+ , p~ T ; n, m
s
r
r=1 i=2
s=1 j=2
k=1
l=1
(5.27)
74
And the number operators which will act on above state space are
⊥
N =
∞ X
24
X
r λi,r +
r=1 i=2
∞
X
⊥
k λk , N̄ =
∞ X
24
X
s λ̄j,s +
s=1 j=2
k=1
∞
X
l λ̄l
(5.28)
l=1
The mass squared operators can also be written as
n 2 mR 2
2
2
+ ′ (N ⊥ + N̄ ⊥ − 2)
+
M =
′
R
α
α
(5.29)
This is the mass squared operators for both none zero momentum and winding for the
modified level matching condition N ⊥ − N̄ ⊥ = n m
5.2
T-Duality for Closed Strings
As the mass squared operators depend upon the compactified radius R and there is a
remarkable property that if we replace the radius R by a radius R̃ = α′ R−1 and n by
m then the mass squared operators remain the same, this symmetry is called T-duality
for the closed string and the radii R and R̃ are dual radii.
R ↔ α′ R−1 ≡ R̃
(5.30)
Hence the mass squared operators for both the dual radii are same as
2
M (R; n, m) = M
2
R̃; m, n
(5.31)
This gives the idea that if we start a theory having a small compactified radius R, we
can transform to a dual theory which have the large radius R̄ [7].now for a complete
dual theory, let us define a dual coordinate (compact coordinate) operator
X̃ ≡ XL (τ + σ) − XR (τ − σ)
The mode expansions for dual coordinate X̃ are
r
α′ X 1 −inτ
′
′
X̃(τ, σ) = q0 + α wτ + α pσ + i
e
(ᾱn e−inσ − αn einσ )
2 n6=0 n
(5.32)
(5.33)
75
By comparing this with equation (5.11), we find that
x0 → q0 p → w αn → −αn
(5.34)
q → x w → p ᾱ → ᾱ
0
0
n
n
The dual momentum P˜τ becomes as
P˜τ ≡
1
1
∂
(ẊL − ẊR )
X̃
=
τ
2πα′
2πα′
(5.35)
The commutation relations for (X̃ ,P˜τ ) can be calculated by the same manner as we
have done for (X , P τ ). The Hamiltonian for both (X , P τ ) and (X̃ ,P˜τ ) is same and
written as
H=
2 i i
(p p + p2 + w2 ) + N ⊥ + N̄ ⊥ − 2
α′
(5.36)
In the order to make a map between these two (X, P τ ) and (X̃, P˜τ ), the above (5.34)
transform as
x0 → q̃0 p → w̃ αn → −α̃n
(5.37)
q → x̃ w → p̃ ᾱ → ˜¯α
0
0
n
n
This map makes the physical equivalence of the two different theories and hence, Tduality is the symmetry, which exists between the different string theories. A complete
summery is given in the following table (5.1)
Table: (5.1) complete dual theory
76
5.2.1
Type II superstrings and T-duality
Let us consider X 9 coordinate is curled up into a circle of radius R in type II superstring
theories and the T-duality transformation carried out on this coordinate as
XL9 → XL9 and XR9 → −XR9
This interchanged the momentum and the winding numbers similarly, the word sheet
fermions will also transform under T-duality as
ψL9 → ψL9 and ψR9 → −ψR9
This means that the T-duality reversed the chirality of the right moving ground states of
Ramond sector. As the relative chirality of both left and right moving ground states is
the thing which distinguished the type IIA and type IIB theories. If one is compactified
on a circle of R, let say Type IIA, then the T-duality gives the type IIB which will be
compactified on R̃ [3].
5.3
5.3.1
T-duality of open strings
T-duality and open strings
In the case of open strings, the situation is a little bit different when we compact a
dimension because the open string cannot winds around the compact dimension. Let
we have a D25-brane and a compactified X 25 circle having the radius R, the open string
has quantize momentum p25 but no winding but after the T-duality transformation,
we will have a D24-brane and a compactified X 25 circle of radius R̃. The Dirichlet
boundary conditions will impose a zero momentum constraint but the open string will
have winding now. The open string mass squared of the two theories will coincide when
R̃ = α′ R−1 , because of the momentum states contribute to M 2 in the first theory in
the same way as the open string winding states contribute to M 2 in the second theory.
By permitting the duality conversion to change the D-brane, we can write it as
(D25; R) → (D24; R̃)
77
Now to find how the T-duality in the case of open strings, let us start from the assumption that we have space filling D25-brane i.e. the open strings have NN-type coordinates
and let the compactified dimension X 25 (τ, σ) ≡ X(τ, σ), for this the mode expansion
can be written as
X(τ, σ) = x0 +
√
X1
√
2α′ α0 τ + i 2α′
αn cosσe−inτ
n
n6=0
With
α0 =
√
2α′ p =
√
2α′
n
R
(5.38)
(5.39)
Now separating the above string coordinate (5.38) into left moving and right moving,
as
X(τ, σ) = XL (τ + σ) + XR (τ − σ)
Where
1
XL = (x0 + q0 ) +
2
r
1
XR = (x0 − q0 ) +
2
r
(5.40)
i√ ′X 1
α′
2α
α0 (τ + σ) +
αn e−inτ e−inσ
2
2
n
n6=0
i√ ′X 1
α′
α0 (τ − σ) +
αn e−inτ e+inσ
2α
2
2
n
n6=0
(5.41)
Here q0 is an arbitrary constant. For the T-duality transformation, let the dual coordinate is
X̃(τ, σ) ≡ XL (τ + σ) − XR (τ − σ)
(5.42)
For this X̃ the mode expansions are
X̃(τ, σ) = q0 +
√
X1
√
αn sin σe−inτ
2α′ α0 σ + i 2α′
n
n6=0
(5.43)
This equation is some thing like equation (4.16) in which a string is stretched from one
D-brane to another D-brane, and hence, there is a correspondences between the above
equation (5.43) and (4.16), and gives the idea that the coordinate X̃ is of DD type i.e.
The end points are fixed as ∂τ X̃ = 0 for the end points, σ = 0 and σ = π. The open
string stretches when σ goes from 0 to π such as
X̃(τ, π) − X̃(τ, 0) =
√
2α′ α0 (π − 0) = 2π R̃n
(5.44)
78
Since n will have any possible integer values, the physics behind this are that, infinite collections of D24-branes having uniform spacing 2π R̃ along the compactified X 25
direction. These configurations will be physically equivalent to the single D24-brane
which is at some fixed positions on the circle of radius R̃. We notice that the T-duality
maps the Neumann boundary conditions into Dirichlet boundary conditions as
∂σ X = XL′ (τ + σ) − XR′ (τ − σ) = ∂τ X̃
∂τ X = XL′ (τ + σ) + XR′ (τ − σ) = ∂σ X̃
(5.45)
T-duality has transformed the open string with Neumann boundary conditions, on a
circle of radius R to an open string with Dirichlet boundary conditions, on a circle
of radius R̃. Now let we have D25-brane in a word in which k spatial dimensions are
compactified into circles, Then the T-duality will transformation on each circle will give
a physically equivalent world in which we will have a D(25 - k )-brane and each circle
will be replaced by a circle with dual radius. And generally, if we have a Dp-brane
which is stretched around a compact dimension, T-duality along this direction will give
a D(p - 1 )-brane at some fixed position on a circle of dual radius [7].
5.3.2
Open strings and Wilson lines
To study the effects and its dual picture, of open strings on D-branes having the gauge
fields which are characterizing by holonomies. Let us start from a compact dimension
on circle and on this, a flat potential4 will have non trivial physical effects as like
Aharanov-Bohm effect. As when the component of the gauge potential along the circle
of compactified dimension takes none zero constant values, it gives the holonomy W or
Wilson line, which can be written as
I
W ≡ exp(iw) = exp iq dx Ax
Here Ax is the gauge potential along the compact direction x. As w = q
(5.46)
H
dx Ax lives on
H
a unit circle due to compact dimension, so it means that w ∈ [0, 2π] and w = q dx Ax
4
The potential which gives the vanishing field strength
79
is some thing like the angle θ such that
θ≡w=q
I
dx Ax
(5.47)
Since the Wilson line W ≡ exp(iθ) is a gauge invariant and the θ gauge equivalents
will gives same holonomy W . For a constant gauge potential Ax the above equation
becomes
q Ax =
θ
2πR
(5.48)
Now let us consider the open strings which have the opposite charges on the end points,
so that the string will act as a neutral and the Wilson line will be no effects. If a Dbrane wraps around the compact dimension then the mass-squared operators can be
written as
M 2 = p2 +
l
1
⊥
(N
−
1)
,
p
=
α′
R
(5.49)
In case of open string which having opposite charges on their end points and which
lies on the same Dp-brane then shifts in momentum
5
is as p − qAx + qAx = p. Now
lets us consider the two Dp-branes and the string stretching between these two, as each
Dp-branes has own Maxwell field. Let the negatively charged end points lies on first
Dp-brane and the positively charged end points lies on the second Dp-branes then the
momentum will be shifting from p to p − qA1 + qA2 and can be written explicitly as
l
l
θ2
θ1
→
−
+
R
R 2πR 2πR
The mass squared operator then becomes as
2
2πl − (θ2 − θ1 )
1
2
M =
+ ′ (N ⊥ − 1) , l ∈ Z
2πR
α
(5.50)
(5.51)
If θ1 = θ2 then as a result the effects of holonomies will be cancel out and the equation
will be reduces to equation (5.49). The T-duality picture of the two Dp-branes having
the different parameters of θ will be consisting the two D(p - 1 )-branes having the
different (angular) positions due to the values of θ [3, 7].
5
For point particle the addition of Wilson line make a shift in p as p-qA.
80
5.4
5.4.1
Electromagnetic fields on D-branes and T-duality
Maxwell fields coupling to open strings
In the presence of background Maxwell fields, the string end points will couple to
Maxwell potential Am . As we have written this coupling terms in equation (4.59),
adding this to the string action
Z
Z
Z
′
m
S = dτ dσL(Ẋ, X ) + dτ Am (X)∂τ X |σ=π − dτ Am (X)∂τ X m |σ=0
(5.52)
Now let us consider those background fields which has constant electromagnetic field6
strength Fmn , for this let define the gauge potential
1
Am (x) = Fmn xm
2
(5.53)
Putting this gauge potential in the above action (5.52) and then by using the variation,
we find the boundary conditions as
σ
Pm
+ Fmn ∂τ X n = 0, σ = 0, π
(5.54)
We get this by taking the usual Dirichlet boundary condition δX a = 0 for the coordinates normal to the brane. There are the background electromagnetic fields which
σ
changed the boundary conditions. Now using the explicit form of the Pm
and simplify-
ing, we find
∂σ Xm − 2πα′ Fmn ∂τ X n = 0, σ = 0, π
(5.55)
This is the boundary condition for open string in the case of background field.
5.4.2
D-branes with Electric fields and T-dualities
Let us consider a D-brane which wraps around x25 compact dimension and carrying
constant electric fields along x25 such that
F25,0 = E25 ≡ E
6
The background field will be purely electric if F0i (= −Fi0 ) and purely magnetic if Fij .
(5.56)
81
Since the boundary conditions (5.55) becomes as
∂σ X0 − 2πα′ F25,0 ∂τ X 25 = 0
∂σ X2 5 − 2πα′ F25,0 ∂τ X 0 = 0
(5.57)
As X0 = −X 0 and let X 25 ≡ X then the above equations becomes
∂σ X 0 − E∂τ X = 0
∂σ X − E∂τ X 0 = 0
(5.58)
Here E is the dimensionless electric field
E ≡ 2πα′ E
(5.59)
Now writing the above equations (5.58) in a more beautiful way, for this let ∂± ≡
1
(∂
2 τ
± ∂σ ) and by simplifying, we get
∂+
X0
X
=
1+E 2
1−E 2
2E
1−E 2
2E
1−E 2
1+E 2
1−E 2
∂−
X0
X
(5.60)
This is the desired form of the boundary conditions (5.58) in the form of matrix. And
the duality relations becomes
∂+ X = ∂+ X̃ , ∂− X = −∂− X̃
(5.61)
In the order to find a dual description, let us consider a D(p - 1 )-brane which is
moving with a constant velocity along the compact dimension. Let we have two frame
of reference S and S ′ . The S ′ is the rest frame of D(p - 1 )-brane while the relative
parameter boost of these two are β = v/c, where is the speed of moving brane. Let the
string coordinates in S ′ are X ′0 and X̃ ′ then the Lorentz transformation can be written
as
X ′0 = γ(X 0 − β X̃)
X̃ ′ = γ(−βX 0 + X̃)
(5.62)
82
Now using the duality relations and simplifying, we find
2
0
2β
1+β
X0
X
2
2
1−β
1−β
∂−
∂+ =
2β
1+β 2
X
X
1−β 2
1−β 2
(5.63)
These are the boundary conditions for the dual D(p - 1 )-brane. Comparing these to
equation (5.60) then we get
E = 2πα′ E = β
(5.64)
This shows that a moving D(p - 1 )-brane with the boost parameter β = v/c is actually
the T-dual to the Dp-brane which is wrapped around on the dual circle having the
electric field E = β along in the direction of circle. Since the T-duality relates the
Dp-brane with electric field is physically equal to the moving D(p - 1 )-brane with no
electric field [7, 28].
5.4.3
D-branes with Magnetic fields and T-dualities
Now we will consider the magnetic fields in background. For this, let a Dp-brane having
its two directions lies on (x2 , x̃3 ) plane, x̃3 is compactified to a circle of radius R̃3 such
that both x2 and x̃3 give a cylinder of circumference of 2π R̃3 . Let the open string
coordinates are X 2 and X̃ 3 , which are Neumann. After T-duality on X̃ 3 , gives X 3 as a
Dirichlet, since the dual picture is now a D(p - 1 )-brane which is stretch along X 2 at
some fixed position X 3 = 0. Now the magnetic field on this Dp-brane is let F23 = B.
For this the boundary conditions (5.55) becomes as
∂σ X 2 − B∂τ X̃ 3 = 0
∂σ X̃ 3 + B∂τ X 2 = 0
(5.65)
Here B is the dimensionless electric field
B ≡ 2πα′ B
And similarly we can write these boundary conditions as
1−B2
2B
X2
X2
1+B2
1+B2
=
∂+
∂−
1−B2
2B
3
3
− 1+B2 1+B2
X̃
X̃
(5.66)
(5.67)
83
These are the boundary condition in the presence of background magnetic fields. In
the case of dual picture, let a D(p-1)-brane which have tilted on the cylinder, then the
boundary conditions can be written in the form of X ′2 and X
′3
that is rotated by an
angle α . The and are Neumann and Dirichlet and the proper rotation can be written
as
X ′2
X
′3
=
cos α
sin α
− sin α cos α
X2
X
3
Using the duality relation and simplifying, we get
X2
X2
cos 2α − sin 2α
∂−
=
∂+
3
3
sin 2α cos 2α
X̃
X̃
(5.68)
(5.69)
Now comparing this result with the boundary conditions (5.67), first let the diagonals
elements which gives
1 − B2
= cos 2α
1 + B2
(5.70)
B = 2πα′ B = − tan α
(5.71)
Simplifying this, we get B = ± tan α, we will take the negative sign such as
This equation gives that the zero magnetic field cannot rotate the D-brane and it will
required an infinite magnetic field to rotate it. But putting the B = − tan α, we can
easily confirm the off diagonals. In equation (5.71) the negative sign is necessary for the
confirmation of the off diagonals elements. Since the titled D-brane is the dual picture
of the D-brane which have magnetic fields [7].
5.5
String coupling and the dilaton
The massless scalar field dilaton has an interesting property such as its expectation
value can controls the coupling of string and this coupling is a dimensionless number
which can set the strength of the interactions of the strings.
In string theory, the string coupling is not a constant and can be written in the form
84
of dilaton field. The closed string coupling gs can be evaluated from the dilaton field
φ(x), such as
gs ∼ eφ
The string coupling gs is the not an adjustable parameter but a dynamical parameter in
string theory. The dynamical nature of coupling is an ideal property for the unification
of all the other interactions. Whenever the string coupling is small then the amplitudes
for the string interactions can be easily calculable by using the Riemann surfaces. The
Riemann surfaces can also allow us to understand the infinites which come in the
amplitudes of general relativity [3, ?].
5.6
S-duality
Another kind of duality which is called S-duality which relates the string coupling gs
to 1/gs as likes the T-duality which relates the radius R to α′ R−1 . The S-duality
relates the type I superstring theory to the SO(32) Heterotic superstring theory, and
the type IIB superstring theory to itself. As when the coupling gs is small then we
use the perturbation theory but whenever the coupling gs is large then we can use the
S-duality i.e. the large coupling gs of the type I superstring theory is equivalent to the
weak coupling gs of SO(32) Heterotic theory. S-duality tell us that how these three
superstring theories behave at strong coupling but when the coupling is large, in Type
IIA and E8 × E8 Heterotic, then both of them developed an eleventh dimension of size
the ls gs . The 11th dimension is a circle in the type IIA and a line interval in the
Heterotic7 .
7
Under the T-duality, one can get easily g̃s =
√
α′
R gs .
85
M-theory is a new type of quantum theory which lives in 11 dimensions of spacetime. At
the low energies, it is approximately equivalent to supergravity, a classical field theory
lives in 11-dimensions of space time, but the M-theory is much more than supergravity
[3, 7, 30, 31, 32].
Appendix A
Appendix A
The Massless States Of Closed
String
As the massless level of closed string, the general state of fixed momentum. We write
it as
X
j†
+
~T
Rlj al†
1 ā1 p , p
l,j
Here Rlj are the elements of an arbitrary square matrix of the size (D − 2). Any square
matrix can be decomposed into its symmetric part and its antisymmetric part as
1
1
Rlj = (Rlj + Rjl ) + (Rlj − Rjl ) ≡ Slj + Alj
2
2
And the symmetric part Slj can be written as
1
1
δlj S +
δlj S
Slj = Slj −
D−2
D−2
with S ≡ S ll = δ lj Slj The 1st term is traceless as
1
1
lj
Slj −
δ
δlj S = S −
δ lj δlj S = 0
D−2
D−2
86
Hence Rlj is decomposed into a traceless matrix plus a multiple of the unit matrix. Let
say the traceless part of Rlj is Ŝlj and S ′ =
S
D−2
then we have
Rlj = Ŝlj + Alj + S ′ δlj
Hence we decomposed the Matrix Rlj into a symmetric-traceless part, an antisymmetric and a traceless part. Since we split the total massless states into three linear
independent states as
X
j†
+
~T
Ŝlj al†
1 ā1 p , p
(A-1)
X
j†
+
Alj al†
~T
1 ā1 p , p
(A-2)
l,j
l,j
j†
+
~T
S ′ al†
1 ā1 p , p
(A-3)
The equation (A-1) is similar to the quantum theory of the free gravitational field states.
Hence equation (A-1) represents Graviton states. The equation (A-2) corresponds to
the one-particle states of the Kalb-Ramond field, an antisymmetric tensor field Bµν
with two indices. The equation (A-3) corresponds to a one-particle state of a massless
scalar field. This field is called the dilaton [7].
87
Appendix B
Appendix B
The Spinors Algebra In 2D
As the Clifford algebra satisfies
{Γµ , Γν } = 2ηµν
In 2D it gives
Γ21 = η11 , Γ22 = η22 , Γ1 Γ2 + Γ2 Γ1 = 0.
Lorentizian:
In this case we have
Γ21 = +1 , Γ22 = −1 , Γ1 Γ2 + Γ2 Γ1 = 0
Now we can choose the representation of these operators as a matrices such that
0 −1
0 1
.
, Γ2 = −iσ2 =
Γ 1 = σ1 =
1 0
1 0
And the generator of the spin(1, 1) is then given by
Γ12 =
1
2
0
1
1
1
,
[Γ1 , Γ2 ] = Γ1 Γ2 = σ3 =
1
4
2
2
0 −2
88
with
γ1
eγΓ12 = e 2 σ3 =
e
γ
2
0
γ
e− 2
0
.
Given a vector v = v1 Γ1 + v2 Γ2 the Lorentz transformation is then given by
v 7→ eγΓ12 ve−γΓ12 ,
By simplifying, we get
And hence
7→
v1
v2
0
e
e
−γ(v1 −v2 )
7→
+γ(v1 +v2 )
0
cosh γ sinh γ
sinh γ cosh γ
.
v1
v2
.
Thus the Clifford algebra generated by {1, σ1 , −iσ2 , σ3 } is the same as a 2 × 2 real
matrices and the even part of the Clifford algebra Cl(1, 1) is generated by 1 and σ3
both of which are the diagonal matrices and therefore Cl(1, 1)even ∼
= ℜ ⊗ ℜ. Because
the Γ12 is real and diagonal therefore the irreducible representations of spin(1, 1) are
one dimensional (Weyl) and real (Majorana). They transform as
γ
ψ + 7→ e+ 2 ψ (Positive chirality)
γ
ψ − 7→ e− 2 ψ(Negative chirality)
Euclidean:
In the case of Euclidean, we have
Γ21 = +1 , Γ22 = +1 , Γ1 Γ2 + Γ2 Γ1 = 0
Now by the similar way we can choose the representations of these operators as matrices
1 0
0 1
.
, Γ 2 = σ3 =
Γ 1 = σ1 =
0 −1
1 0
The generator of spin(2) is then given by
Γ12 =
0
1
1
−i
[Γ1 , Γ2 ] = Γ1 Γ2 =
σ2 =
1
4
2
2
89
2
− 12
0
,
With
γ1
cos( γ2 )
eγΓ12 = e−i 2 σ2 =
− sin( γ2 )
cos( γ2 )
sin( γ2 )
Given a vector v = v1 Γ1 + v2 Γ2 then the spin(2)
.
v 7→ eγΓ12 ve−γΓ12 ,
By simplifying, we get
v1
v2
cos γ
7→
sin γ
− sin γ cos γ
v1
.
v2
Thus the Clifford algebra generated by {1, σ1 , −iσ2 , σ3 } is the same as a 2 × 2 real
matrices and the even part of the Clifford algebra Cl(2) is generated by 1 and Γ1 Γ2
since (Γ1 Γ2 )2 = −1 therefore Cl(1, 1)even ∼
= C and the spinors transforms as
cos( γ2 ) − sin( γ2 )
ψ1
ψ1
.
7→
γ
γ
ψ2
sin( 2 ) cos( 2 )
ψ2
We can also form the irreducible representation by taking
ψ+ = ψ1 + iψ2 , ψ− = ψ1 − iψ2 .
ψ± transform under Spin(2) as
γ
ψ+ 7→ e+i 2 ψ+ (Positive chirality)
γ
ψ− 7→ e−i 2 ψ− (Negative chirality)
The chirality is the eigenvalue under
Γ 3 = Γ1 Γ 2 =
ψ1
ψ2
7→ Γ3
ψ1
ψ2
90
0 −1
1
0
=
−ψ2
+ψ1
and
ψ+ 7→ iψ+ , ψ− 7→ −iψ− .
Thus in both Lorentzian and Euclidean signature, we have chiral (Weyl) spinors. The
supersymmetry generators are spinors and in the two dimensions we can choose p
Weyl spinors of positive chirality and q Weyl of negative chirality. This gives us (p, q)
supersymmetry in the two dimensions [33, 34, 35].
91
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