McMahon D. String Theory Demystified

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String Theory Demystiﬁ ed

has an immediate physical interpretation. It is the momentum density of the

string. We integrate along the length of the string ﬁ xing

to get the total momentum

carried by the string, which we label

pdP

∫

(3.23)

The other conserved current associated with the global symmetries of the action

comes from invariance under Lorentz transformations. In this case

δω

µµ

XX=

We can show that the lagrangian in Eq. (3.19) is invariant under a Lorentz

transformation in the following way:

δηδ ηδη

µν α

∂∂ =∂ ∂ +∂ ∂

()

XX X X X X

νν

µν α

µν

δη δη

=∂ ∂ +∂ ∂

=∂

() ()

(

XX X X

))

()

∂+∂∂

=∂∂ +

µν α

µν

ρα

µν

ηωη

ωη

XXX

ωωη

ωω

λα

µν

νρ α

µλ α

∂∂

=∂∂ +∂∂

XX XX (lowwer indices with

ωω

µν

νρ α

ρλ α

)

=∂∂ +∂XX X∂∂→

=∂∂

νρ α

µρ

(relabel dummy indices )

XXXX

ρν α

ωλν

+∂∂ →(relabel dummy indices )

==∂∂ +∂∂ =

ωω ω

νρ α

XX XX0 (antisymmetry

βββα

=− )

So the lagrangian is invariant under a Lorentz transformation, but what are the

currents? This is easy to ﬁ nd, since

hh X X

=− − ∂ ∂

⇒

∂

∂∂

αβ

µν

hh X

hh P

αβ

µν

αβ

(

)

=− − ∂ =− −

Using

εδ

µν

µµ

JL XX

=∂ ∂∂[/( )]

where

µµ

XX=

together with the anti-

symmetry of

, we conclude that the Lorentz current is

JTXPXP

µν µ

νν

=−

()

(3.24)

CHAPTER 3 Symmetries and Worldsheet Currents

We have introduced the energy-momentum tensor and looked at some conserved

currents that arise due to symmetries of the lagrangian. The next major piece of

the dynamics puzzle is the hamiltonian which governs the time evolution of the

worldsheet. It can be written down simply using formulas from classical mechanics:

HdXPL

dX X

=−

(

)

′

(

)

∫∫

σσ

µτ

σσ



(3.25)

The Hamiltonian

In this chapter we have extended our classical analysis of the string. We did this by

introducing the energy-momentum tensor and by describing the symmetries of the

Polyakov action. Then we derived conserved currents of the worldsheet, and wrote

down the hamiltonian. In the next chapter, we will conclude our classical description

of the string by writing down mode expansions of the hamiltonian and energy-

momentum tensor, and describing the Virasoro algebra. After writing down a mass

formula for the string, we will proceed to quantize the theory.

Summary

1. Let

σεσ

ααα

→

′

+ (

)

be an inﬁ nitesimal reparameterization. Considering

only terms that are ﬁ rst order in

, ﬁ nd the variation of the worldsheet

metric

αβ

2. Assuming that you can move the derivative

dd/

inside the integral in

Eq. (3.23), explain conservation of momentum for the cases of the open

and closed string.

3. The energy-momentum tensor has zero trace. Show that this is a

consequence of Weyl invariance.

4. Consider light-cone coordinates and derive the Virasoro constraints for the

energy-momentum tensor.

5. Consider the Lorentz current

µν

in Eq. (3.24). What is the equation that

describes that the current is conserved? What are the conserved charges and

what do they describe?

Quiz

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String Quantization

At this point we have the classical physics of the string in place. The next step

toward a quantum theory is, of course, to quantize the string. We will start off by

looking at a quantum theory of a single string. From quantum ﬁ eld theory you recall

that this procedure is known as ﬁ rst quantization. This is opposed to the procedure

called second quantization, which is a viewpoint of quantizing ﬁ elds (see Quantum

Field Theory Demystiﬁ ed for a review). The difference in the two approaches will

be on how we view the X

(s, t ). If we take them to be ﬁ elds, then the quantization

procedure is second quantization. If we take them to be space-time coordinates, as

we have so far, then the process is ﬁ rst quantization. This is the procedure we will

apply in this chapter. There are several different approaches to quantization of the

string, each with their own accompanying difﬁ culties and problems. The main

approaches used are called covariant quantization, light-cone quantization, and

BRST quantization. We consider the ﬁ rst two approaches here, and will discuss

BRST quantization later.

CHAPTER 4

String Theory Demystiﬁ ed

Covariant Quantization

The procedure known as covariant quantization will be familiar to you from your

studies of ordinary quantum mechanics. In a nutshell, this is the imposition of

commutation relations on position and momentum. So, using this procedure we

continue with the notion that

στ

)

are space-time coordinates, but we need to say

what we are taking as the momentum. This can be done in the standard way using

lagrangian dynamics. Let

στ

(, )

be the momentum carried by the string. Given a

lagrangian density L we can calculate the momentum from the

στ

)

using

πστ

(, )=

∂

∂∂

()

This is easy to calculate using the Polyakov action as written in Eq. (2.31):

dXXXX

=∂∂−∂∂

()

⇒= ∂ ∂ −∂

∫

τµ σ

σµ

τµ σσ

σµ

XX∂

()

So we see that the conjugate momentum is

πστ

τµ

(, )=

∂

∂∂

()

=∂

With this deﬁ nition in hand, we are in a position to quantize the theory. To do this

we take the approach used in quantum ﬁ eld theory, namely, impose equal time

commutation relations on the position and momenta. In ordinary quantum mechanics

the position and momentum coordinates satisfy

[, ] [, ] [, ]

[, ] [, ] [,

xp yp zp i

xx yy z

xyz

===

==zzxyxzyz

pp pp

xx yy

][,][,][,]

[, ][, ]

====

======[, ][, ][, ][, ]pp pp pp pp

zz xy xz yz

where we have set

? =

. If we denote the coordinates as x

where i = 1, 2, 3 then

these relations can be written compactly as

[, ]

[, ][, ]

xp i

xx pp

ij ij

ij i j

CHAPTER 4 String Quantization

Here, d

is the Kronecker delta. Now we apply these relations to the string, with two

crucial differences. First, we have a relativistic theory and hence we let

δη

µν

→

Moreover, we expect that position and momentum will commute when they are

taken to be at different spatial locations of the string. That is, let s and s ′ be two

different locations on the string. Since we are taking equal time commutation

relations, we let the time coordinate be τ for both position and momentum. Then

[ (,), ( ,)]X

µν

στ π σ τ σ σ

′

=≠

′

0for

Now, to ensure that the position and momentum don’t commute at the same

spatial location, we use the Dirac delta function d ( s − s ′). So the equal time

commutation relations are:

µν µν

στ π σ τ ηδσ σ

στ

(, ), ( , ) ( )

(, )

′

⎡

⎣

⎤

⎦

=−

′

,, ( ,) (,), ( ,)X

νµν

στ πστπστ

′

⎡

⎣

⎤

⎦

′

⎡

⎣

⎤

⎦

= 0

(4.1)

To summarize, remember that

• t is the same for

στ

)

and

στ

(,)

′

• The presence of the Dirac delta function

σσ

()−

′

ensures that the

coordinates and momenta do commute at different points σ along the

string—the commutation relations are only nonzero when position and

momentum are evaluated at the same point on the string.

• The presence of

µν

comes from the fact we have a relativistic theory.

Ultimately, we want to write down commutation relations for the modes of the

string. This can be done most easily by transitioning to the light-cone coordinates

τσσ τσ

+−

=+ =−,

. First, using

∂= ∂+∂ ∂= ∂−∂

+−

12 12/( ) /(

)

τσ τσ

and

, we

can invert to write

∂+∂=∂ ∂−∂=∂

+− +−

τσ

(4.2)

In the case of

νν

στ πστ

(,) (,

)

′′

and

, we will have ∂′

= 1/2 (∂

+ ∂

σ′

) and

∂′

−

= 1/2 (∂

− ∂

σ′

). Using Eq. (4.2), we can write the commutation relation

[ (,), ( ,)] ( )Xi

µν µν

στ π σ τ ηδσ σ

′

=−

′

in a new way:

iX X

ηδσ σ στ πσ τ στ

µν µ ν µ

( )[ (,), (,)][ (,)−

′

= ,, ( , ) ]

[(,),( )(

TX X

∂

′

∂+

′

∂

′

+−

µν

στ

στ σ

,, ) ]

[ (,), ( ,)] [ (,)

στ σ τ στ

µν µ

′

∂

′

TX X TX ,, ( , ) ]

′

∂

′

−

στ

String Theory Demystiﬁ ed

Now we compute the derivative of

[ (,), ( ,)] ( )Xi

µν µν

στ π σ τ ηδσ σ

′

=−

′

with

respect to s. Since

στ

(,)

′

is not a function of s, only

στ

)

is affected.

Proceeding and again using Eq. (4.2) we ﬁ nd

iTXX

δσ σ σ τ σ τ

µν

∂

−

′

=∂

′

∂

′

( ) [ (,), ( ,)]

[ (,), ( ,)]

[( ) (

+ ∂

′

∂

′

=∂−∂

−

+−

TX X

µν

στ σ τ

σστ σ τ

στ

,), ( ,)]

[( ) ( , ),

′

∂

′

+∂−∂

+−

TX (,)]

[ (,), ( ,)]

′

∂

′

=∂

′

∂

′

−

TX X

µν

στ

στ σ τ

−−∂

′

∂

′

+∂

−+

TX X

[ (,), ( ,)]

[(,),

µν

στ σ τ

στ

′′

∂

′

−∂

′

∂

′

−−−

XTXX

νµν

στ στ στ

(,)] [ (,), (,)]

Now we can utilize the commutation relation for the conjugate momenta in

Eq. (4.1). We have

0 =

′

=∂ ∂[ (,), ( ,)][ (,),

πστπστ στ

µν

TX TX((,)]

[ (,), ( ,)]

[

′

=∂ ∂

′

=∂

στ

στ σ τ

TX X

µµ

στ σ τ

( , ), ( , )]∂

′

(Since it equals zero,, we divide

by for later convenience)T

T= [(∂∂+∂

′

∂+

′

∂

′

=∂

+− +−

) (,),( ) ( ,)]

[

µν

στ σ τ

(( , ), ( , )] [ ( , ), (

στ σ τ στ

νµν

′

∂

′

+∂

′

∂

++−

XTXX

′′

+∂

′

∂

′

+∂

−+

στ

στ σ τ

µν

,)]

[ (,), ( ,)] [TX X T

−−−

′

∂

′

µν

στ σ τ

( , ), ( , )]

Now since

[ (,), ( ,)]

πστπστ

µν

′

= 0

, let’s form the sum [p

(s, t), p

(s ′, t)] +

(/ s ) d ( s - s′). We obtain

[ (,), ( ,)] ( )

[

πστπστ η

δσ σ

µν µν

′

∂

−

′

=∂

TX(( , ), ( , )] [ ( , ), (

στ σ τ στ

νµν

′

∂

′

+∂

′

∂

++−

XTXX

′′

+∂

′

∂

′

+∂

−+

στ

στ σ τ

µν

,)]

[ (,), ( ,)] [TX X T

−−−

′

∂

′

+∂

µν

στ σ τ

στ

( , ), ( , )]

[(,),

′′

∂

′

−∂

′

∂

′

+−+

XTXX

νµν

στ στ στ

( ,)] [ (,), ( ,)]

[

(

)

(

)

][

(

,+∂

′

∂

′

−∂

−−

TX X TX

µν µ

στ σ τ στ

))

(

)

]

′

∂

′

−

στ

CHAPTER 4 String Quantization

That is,

iTXX

δσ σ σ τ σ τ

µν µ ν

∂

−

′

=∂

′

∂

′

( ) [ (,), (,)]2 ++∂

′

∂

′

+−

2TX X[ (,), ( ,)]

µν

στ σ τ

It is also straightforward to show that

[ (,), ( ,)] ( )

[

πστπστ η

δσ σ

µν µν

′

−

∂

−

′

=∂

−

TX2

µµν µ

στ σ τ στ

( , ), ( , )] [ ( , ),

′

∂

′

+∂

′

∂

−−+

XTXX2

νν

στ

(,)]

′

In the chapter quiz, you will have the opportunity to show that [

(s, t ),



−

(s, t )] = [

−

(s, t ), 

(s′, t )] = 0. Therefore the commutation relations are

[ (,), ( ,)] ( )∂

′

∂

′

∂

−

′

µν

στ σ τ

δσ σ

(4.3)

[ (,), (,)] (∂

′

∂

′

=−

∂

−

′

−−

µν

στ σ τ

δσ σ

))

(4.4)

[ (,), ( ,)][ (,),∂

′

∂

′

=∂

′

∂

+− −+

XX X

µν µ

στ σ τ στ

στ

(,)]

′

= 0

(4.5)

Using Eqs. (4.3) to (4.5), deriving commutation relations for the modes, which will

help us get to the quantum physics, will be a much simpler matter. In order to derive

the commutation relations, we will need the following expression for the Dirac

delta function:

()=

=−∞

∞

∑

ikx

Recalling Eq. (2.47), which told us that we can write the equations for the string

in terms of left-moving and right-moving modes:

XXX

µµµ

στ στ στ

(, ) (, ) (, )=+

Notice the left-moving modes are functions of

only, while the right-moving

modes are functions of

−

only, so that

∂=∂ ∂=∂

++ −−

XX XX

µµ µµ

στ στ στ στ

(, ) (, ) (, ) (, ))

String Theory Demystiﬁ ed

COMMUTATION RELATIONS FOR THE CLOSED STRING

We will derive the commutation relation [Eq. (4.3)] for the modes explicitly, and

simply state the results in the other cases. Hence we write down the left-moving

modes [Eq. (2.55)] which are functions of

. The left-moving modes are restated

here in the case of the closed string

ssm

τσ

στ τ σ

(, ) ( )

()

=+ ++

−+

mm≠

∑

(4.6)

Now since

τσ

, the derivative is

∂=+ =

−+

≠

∑

Xp e

µµµτσ

στ α α

(, )

()

CC C

µτσ

−+

=−∞

∞

∑

()

(4.7)

To get the last step, we used

2= (/ )C

. Now let’s calculate the left-hand side

of Eq. (4.3). We have

[ (,), ( ,)]

()

∂

′

∂

′

−+

XX e

µν µτσ

στ σ τ α

==−∞

∞

−+

′

=−∞

∞

−

∑∑

⎡

⎣

⎢

⎤

⎦

⎥

()

ντσ

iim n

im n

()

=−∞

∞

−+

′

∑

⎡

⎣

⎤

⎦

τσσµν

αα

But, this must be proportional to

(/ ) ( )∂∂ −

′

σδσ σ

, and furthermore, the term on the

right-hand side of Eq. (4.3) does not depend on

. So, we have to remove the

dependence. We can do so by noting that

emn

im n−+

→=−

()

1 when

We will be able to enforce this condition by introducing the Kronecker delta

term

mn+ ,

, which is 1 when

mn=−

and 0 otherwise. Furthermore, we take note of

the following expression for the Dirac delta function:

δσ σ

σσ

()

−

′

−−

′

=−∞

∞

∑

(4.8)

Notice that

∂

−

′

=−

−−

′

=−∞

∞

∑

δσ σ

σσ

()

CHAPTER 4 String Quantization

Using Eqs. (4.3) and (4.8) together with our previous result, we have

im n

−+

=−∞

∞

−+

′

∑

()

[, ]

τσσµν

αα

δσ σ

µν

σσ

∂

−

′

=−

−−

′

=−∞

()

∞∞

−−

′

=−∞

∞

∑

⎛

⎝

⎜

⎞

⎠

⎟

µν

σσ

2 T

()

We have also used Eq. (2.50) to relate

and the string tension T. This gives us

the commutation relation for the modes

αα η δ

µν µν

mn mn

⎡

⎣

⎤

⎦

+ 0

Equation (4.5) can be used to show that the

αα

µν

and

commute. We can write all

of the commutation relations for the modes of the closed string as

αα ηδ αα ηδ

µν µν µν µν

mn mn mn m

mm,,

⎡

⎣

⎤

⎦

⎡

⎣

⎤

⎦

++0 nnmn,

αα

µν

⎡

⎣

⎤

⎦

(4.9)

In the chapter quiz you will also derive a commutation relation for the center-of-

mass position and momentum of the string

[, ]xp i

µν µν

COMMUTATION RELATIONS FOR THE OPEN STRING

In the case of the open string, it can be shown that together with

[, ]xp i

µν µ

, the

commutation relations are

αα ηδ

µν µν

mn mn

⎡

⎣

⎤

⎦

+ 0

(4.10)

THE OPEN STRING SPECTRUM

With the commutation relations in hand, we can proceed to ﬁ nd the states of the

string. Because the open string case is simpler, we consider this ﬁ rst. Notice that in

our quantization procedure where we have imposed commutation relations on the