1
The Virasoro constraints
Consider the constraint
X, X,+ = 0
+
(1)
From the mode expansion we nd this constraint can be written as
Ln
1
2
nn n = 0
(2)
n
where we have chosen the normalization constant for later convenience.
Classically we would like to
1
The disc amplitude
Note: The treatment here follows Zwiebach.
Suppose we have two open strings scattering o each other. This will be a 4-point diagram, since
there are two incoming quanta and two outgoing quanta. We will take the open strings to be in
t
1
Path integrals
The string worldsheet was 2-dimensional. On this worldsheet we had the variables X (, ). Upon
quantization we have to do a path integral over these variables. We can also regard the theory in
a Hamiltonian language, where X will become lo
1
Fermions
Consider the string world sheet. We have bosons X (, ) on this world sheet. We will now also
1
put (, ) on the world sheet. These fermions are spin 2 objects on the worldsheet.
In higher dimensions, we can take a local orthonormal frame and spi
1
Light cone coordinates on the world sheet
On the world sheet we have used the coordinates , . We will see however that the physics is
simpler in light cone coordinates
+ = + ,
Then we have
=
(1)
=
+
+ +
+
(2)
=
+
+
(3)
2
2
[ ]X = 0
(4)
+ x = 0
1
Quantizing the open string
1.1
The action
The Polyakov action is
S = T
1
d2 g a X b X gab
2
(1)
In the gauge
gab = ab
we get
S=T
1
d d [ X X X X ]
2
(2)
(3)
We will let the range of be
0<
1.2
(4)
Dierent boundary conditions
Neumann boundary conditions (
1
Modular invariance
In string theory we have a 2-d world sheet. The path integral over this world sheet gives a quantity
Z . But we can compute Z in a Hamiltonian formulation as well, which needs that we choose
one direction as space and another as time.
1
The light cone gauge
We have seen in the previous set of notes how to make physical states at level 2. After writing
down all possible oscillator excitations at this level, we had to impose the physical constraints
Ln | = 0, n > 0
(1)
(L0 1)| = 0
(2)
an
1
Hermitian conjugation
Recall that the operators n were dened by
a
n
= ,
n
n
n>0
( )
a
n =
n
(1)
(2)
where an , a are annihilation and creation operators for the mode n. Thus we have
n
n = ( )
n
(3)
Now look at the Virasoro operators. We have
L = (n
1
The story of fermions
Consider a 1-d chain of lattice sites. At each site there is a fermion, represented by a Grassman
number
k , k l + l k = 0
(1)
so that these k are anticommuting objects. The path integral is performed with an action
S = i
i
k
k (k
1
The Classical String
A point particle sweeps out a 1-dimensional worldline in spacetime. A string sweeps out a 2dimensional worldsheet. The action of the point particle is given by the length of the worldline
S = m
where m is the mass of the particle, w
1
Integration by parts
Suppose we have an integral
d2 zz f (z, z )
(1)
D
where the integral ranges over a rectangular region called D :
D : a < x1 < b,
(2)
c < x2 < d
We should be able to integrate this by parts and write the result as an integral over th