McMahon D. String Theory Demystified

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126

String Theory Demystiﬁ ed

Quiz

1. Consider the Virasoro generators and calculate

[ , ], [ , ], [ , ],LL L LL L L L L

ij k jk i ki j

[]

[]]

2. Suppose that the BRST current

jz czT z cz czbz

zz z

() () () () ()()=+∂

is written as

a normal ordered expression. Find

{, }QQ

3. Looking at your answer to question 2, how many scalar ﬁ elds does the

theory contain if we require that

4. Use

0↓=,

to ﬁ nd

CHAPTER 7

RNS Superstrings

The real world as we know it described by the standard model of particle physics

contains two general classes of particles. These are deﬁ ned in terms of their spin

angular momentum as follows. Those particles with whole integer spin are called

bosons, while those with half-integer spin are called fermions. At the level of

fundamental particles—electrons, neutrinos, quarks, photons, gauge bosons, and

gluons—matter particles are fermions while force-mediating particles are bosons.

A symmetry which relates fermions and bosons is called a supersymmetry.

The string theory we have described so far in the book consists only of bosons.

Obviously, this cannot be a realistic theory that describes our universe since we see

fermions in the everyday world. This tells us that the theory we have developed so

far, starting with the Polyakov action, is not the whole story. The theory must be

extended to include fermions. In addition, recall that when we quantized the theory,

the ground state was a tachyon—a particle that travels faster than the speed of light.

These states with negative mass squared are physically unrealistic. More importantly,

a quantum theory with a tachyon has an unstable vacuum.

The remedy to this situation is to introduce supersymmetry into the theory. This

will allow us to develop string theory so that fermions are included in our description

of nature. We will also see that the unwanted tachyon state goes away. An interesting

128

String Theory Demystiﬁ ed

side effect of this effort will be that the critical dimension drops from 26 to 10. If

you aren’t familiar with the description of fermionic ﬁ elds (and supersymmetry) in

quantum theory, try reviewing your favorite quantum ﬁ eld theory book before

tackling this chapter (Quantum Field Theory Demystiﬁ ed provides a relatively

painless introduction).

The Superstring Action

We can proceed with a straightforward modiﬁ cation of the theory to include

fermions using an approach called the Ramond-Neveu-Schwarz (RNS) formalism.

This approach is supersymmetric on the worldsheet. Later we consider the Green-

Schwarz formalism, which is supersymmetric in space-time. When the number of

space-time dimensions is 10, these two approaches are equivalent.

The program we will follow can be done using basically the same which was

applied in the bosonic case: introduce an action, ﬁ nd the equations of motion, and

quantize the theory. However, this time we are going to include fermionic ﬁ elds on

the worldsheet. We start with the Polyakov action, ﬁ rst described in Eq. (2.27) and

reproduced here in the conformal gauge:

dXX=− ∂ ∂

∫

µα

(7.1)

To include free fermions in the theory using the RNS formalism, we add a kinetic

energy term for a Dirac ﬁ eld to the lagrangian. That is, we include D free fermionic

ﬁ elds

to the action, so that it assumes the form

dXXi=− ∂ ∂ − ∂

()

∫

σψρψ

µα

αµ

(7.2)

Again, if you are not familiar with Dirac ﬁ elds, consult Quantum Field Theory

Demystiﬁ ed or your own favorite quantum ﬁ eld theory text. The

are Dirac

matrices on the worldsheet. Since the worldsheet has 1 ⫹ 1 dimensions, the

are Dirac matrices in 1 ⫹ 1 dimensions. Hence there are two such

22×

matrices,

which can be written in the form

ρρ

−

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

(7.3)

using an appropriate choice of basis.

CHAPTER 7 RNS Superstrings

129

EXAMPLE 7.1

Show that the Dirac matrices on the worldsheet obey an anticommutation relation

known as the Dirac algebra

{,}

ρρ η

αβ αβ

=−2

(7.4)

by explicit computation.

SOLUTION

This result is easy to verify. Since

αβ

−

⎛

⎝

⎜

⎞

⎠

⎟

The Dirac algebra will be satisﬁ ed by the

if the following relations hold:

{,}

ρ ρ ρρ ρρ ρρ η

ρρ ρρ

0 0 00 00 00 00

11 1

222=+= =−=

1111 11 11

0 1 01 10

222

+ = =− =−

=+=−

ρρ ρρ η

ρρ ρρ ρρ

{,}

ηη

and likewise for

{, }

ρρ

. Now,

ρρ

00 0

−

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

⋅+−

()

⋅⋅−

(

ii i

))

+−

()

⋅

⋅+⋅ ⋅−

()

+⋅

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

iiii

00 00

== I

Hence the ﬁ rst relation

{,}

ρρ

2= I

is satisﬁ ed. We verify that the second

relation

{,}

ρρ

2=− I

is also satisﬁ ed:

ρρ

00 0 0

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

⋅+⋅ ⋅+⋅

⋅

ii i i

i ++⋅ ⋅+⋅

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

=−

000

01iii

Finally, noting that

ρρ

00 0

−

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟

⋅+−

()

⋅⋅+−

ii i i

(()

⋅

⋅+⋅ ⋅+⋅

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

00 00

iiii

ρρ

ii i i

00 0 0

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

⋅+⋅ ⋅−

()

+⋅

⋅+0000

⋅⋅−

()

+⋅

⎛

⎝

⎜

⎞

⎠

⎟

−

⎛

⎝

⎜

⎞

⎠

⎟

ii i

we see that

{,}{,}

ρρ ρρ

01 10

0==

130

String Theory Demystiﬁ ed

MAJORANA SPINORS

The ﬁ elds introduced in the action,

ψστ

µµ

= (, )

, are two-component Majorana

spinors on the worldsheet. Given that they have two components, they are

sometimes written with two indices

, where

=−01 1,, ,… D

is the space-time

index and

A =±

is the spinor index. We can write

as a column vector in the

following way (suppressing the space-time index):

⎛

⎝

⎜

⎞

⎠

⎟

−

Under Lorentz transformations, these ﬁ elds transform as vectors in space-time

[recall that a contravariant vector ﬁ eld

(

)

is one that transforms as

Vx V x Vx

µµµ

() ( ) ()→

′′

=Λ

under

xx x

µµµ

→

′

=Λ

where

is a Lorentz

transformation].

Following the convention used with Dirac spinors in quantum ﬁ eld theory, we

have the deﬁ nition:

ψψρ

µµ

= ()

†0

Note that the deﬁ nitions used here depend on the basis used to write down the Dirac

matrices Eq. (7.3), and that other conventions are possible. We can also introduce a

third Dirac matrix analogous to the

matrix you’re familiar with from studies of

the Dirac equation, which in this context we denote by

ρρρ

301

−

⎛

⎝

⎜

⎞

⎠

⎟

It will be of interest to make left movers and right movers manifest. This can be

done by recalling the following deﬁ nitions:

στσ

=±

(7.5)

∂= ∂±∂

()

τσ

(7.6)

∂=∂+∂ ∂=∂−∂

+− +−

τσ

(7.7)

EXAMPLE 7.2

Show that

ψρ ψ ψ ψ ψ ψ

µα

αµ

∂= ⋅∂+⋅∂

−+− +−+

2( )

CHAPTER 7 RNS Superstrings

131

SOLUTION

We can rewrite

ρψ

µα

αµ

∂

in a more enlightening way by expanding out the sum

explicitly:

ψρ ψ ψ ρ ρ ψ

µα

αµ

∂= ∂+∂()

Now

∂=

−

⎛

⎝

⎜

⎞

⎠

⎟

∂

∂=

⎛

⎝

⎜

⎞

⎠

⎟

∂

So, the summation is

ρρ

τσ

∂+ ∂=

−

⎛

⎝

⎜

⎞

⎠

⎟

∂+

⎛

⎝

⎜

⎞

⎠

⎟

∂

−∂

i( −−∂

∂+∂

⎛

⎝

⎜

⎞

⎠

⎟

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

−

τσ

)

()i

Hence,

ψρ ψ ψ ρ ρ ψ

µα

αµ

∂= ∂+∂

−∂

∂

⎛

⎝

⎜

⎞

⎠

−

()

⎟⎟

Now, we write out the components of the spinors. For simplicity, we suppress the

space-time index for a moment. First, note that

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

−

⎞⎞

⎠

⎟

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

−+

+−

Using

ψψρ

µµ

= ()

†0

, we have

ψψψψ

µµ

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

()

−

⎛

⎝

⎜

⎞

⎠

⎟

−

−+

−−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

()

∂

⎛

⎝

⎜

⎞

−+

+−

−+

+−

−+

ψψ

µµ

⎠⎠

⎟

=∂+∂

=⋅∂+⋅∂

−+− +−+

ψψ ψψ

ψψψψ

()

132

String Theory Demystiﬁ ed

The result obtained in Example 7.2 allows us to write the fermionic part of the

action in a relatively simple way. Denoting the fermionic action by

we have

=− − ∂

=− − ⋅∂

∫

−+−

σψρψ

σψψ

µα

αµ

()

()( ++⋅∂

=⋅∂+⋅∂

+−+

−+− +−+

∫

ψψ

σψ ψ ψ ψ

)

()iT d

It can be shown that by varying the fermionic action

one can obtain the free

ﬁ eld Dirac equations of motion:

∂=∂=

+− −+

ψψ

µµ

(7.8)

The Majorana ﬁ eld

−

describes right movers while the Majorana ﬁ eld

describes

left movers.

SUPERSYMMETRY TRANSFORMATIONS ON THE WORLDSHEET

Now, we introduce a supersymmetry (SUSY for short) transformation parameter

which is denoted by

. This inﬁ nitesimal object is also a Majorana spinor, which

has real, constant components given by

⎛

⎝

⎜

⎞

⎠

⎟

−

Since the components of

are taken to be constant, this represents a global symmetry

of the worldsheet. If it were a local symmetry, it would depend on the coordinates

)

στ

. Furthermore, the components of

are Grassman numbers. Two Grassmann

numbers

ab, anticommute such that ab ba+=

Now we use

to deﬁ ne our symmetry. The action which includes the fermionic

ﬁ elds is invariant under the supersymmetry transformations:

δεψ

δψ ρ ε

µµ

µα

=− ∂

(7.9)

Using

δψ

, we also ﬁ nd that

δψ ρ ε ερ

µα

=− ∂ = ∂iXi X

. Notice that this takes

the free boson ﬁ elds into fermionic ﬁ elds, and vice versa. We can relate individual

components as follows. First, we have

δεψεε

εψ εψ

µµ

X ==

⎛

⎝

⎜

⎞

⎠

⎟

−+

−

−− ++

()

(7.10)

CHAPTER 7 RNS Superstrings

133

In the second case, we have

δψ

⎛

⎝

⎜

⎞

⎠

⎟

−

and

ρερερε

ρρ ε

τσ

∂=∂+∂

=∂+∂

−

XXX

()

⎛

⎝

⎜

⎞

⎠

⎟

∂+

⎛

⎝

⎜

⎞

⎠

⎟

∂

⎡

⎣

⎢

⎤

⎦

⎥

−∂−∂

τσ

i()

()∂+∂

⎛

⎝

⎜

⎞

⎠

⎟

−∂

∂

⎛

⎝

⎜

⎞

⎠

⎟

−

τσ

Hence,

δψ ε

µµ

−−+

=− ∂2 X

(7.11)

δψ ε

µµ

++−

=∂2 X

(7.12)

Conserved Currents

At this point, we need to identify the conserved currents associated with the action

in Eq. (7.2). Before tackling supersymmetry, let’s review how to calculate a

conserved current by looking at momentum. You can practice in the Chapter Quiz

by looking at Lorentz transformations. Let’s start with a simple example to remind

ourselves of the method.

EXAMPLE 7.3

Consider the action

STd XXi=− ∂ ∂ − ∂

∫

/( )2

σψρψ

µα

αµ

and ﬁ nd the

conserved current associated with translational invariance.

134

String Theory Demystiﬁ ed

SOLUTION

Let’s write down the lagrangian, which is

XXi=− ∂ ∂ − ∂

()

µα

αµ

ψρ ψ

To examine translational invariance, we let

XXa

µµ

→+

where

is an

inﬁ nitesimal parameter. A key insight into the fact that

is inﬁ nitesimal is that we

can drop terms that are second order in

. Taking

XXa

µµ

→+

changes the

lagrangian as follows:

Xa Xa i

→−

∂+∂+− ∂

⎡

⎣

⎤

⎦

=−

µµα

µµ

µα

αµ

ψρ ψ

()()

(()()∂+∂ ∂+∂

⎡

⎣

⎤

⎦

=− ∂ ∂

µα

XaXaL

µµα

µα

+∂ ∂ +∂ ∂ +∂ ∂

⎡

⎣

⎤

⎦

=− ∂

Xa aX aa L

αα

µα

XX Xa aX L

∂+∂∂+∂∂

⎡

⎣

⎤

⎦

(drop

ssecond-order term )∂∂

=− ∂ ∂

µα

++∂ ∂

⎡

⎣

⎤

⎦

µα

(addd in to to get total lagrangian∂∂

µα

XXL

))

Note that the term

ψρ ψ

µα

αµ

∂∝

is unaffected by

XXa

µµ

→+

. Now, we

focus on the leftover extra term:

µα

Xa aX=∂∂+∂∂

⎡

⎣

⎤

⎦

We will manipulate this expression to get our conserved current, which will be a

term multiplying

∂

. We can ﬁ x up this expression doing some index gymnastics,

which is a good exercise for us to go through given the level of this book. For more

practice doing this, consult Relativity Demystiﬁ ed.

We want to ﬁ x up the second term so that it looks like the ﬁ rst term. We do this

by raising and lowering indices with the metric and using the fact that

ηη δ δ

µν

νλ λ

µαβ

βγ γ

==hh

CHAPTER 7 RNS Superstrings

135

Now, the ﬁ rst thing to notice is that the order of the derivatives doesn’t matter.

So, the ﬁ rst step is to write

∂∂ =∂∂

µα

aX Xa

Next, let’s raise the space-time index on

and lower it on

. Since these are

space-time indices, we use the Minkowski metric to do this:

∂∂=∂ ∂

µα

µν

µλ

ηη

Xa X a()()

The metric is not space-time dependent (in the ﬂ at space or Minkowski space-

time we are considering here), so we can pull the metric terms outside of the

derivatives. You might recognize that this is actually true in general—because the

derivatives are with respect to worldsheet coordinates, but the metric, if it depends

on coordinates, is space-time dependent. So,

∂∂=∂∂

=∂ ∂

µν

µλ

λµν

µλ α ν

αλ

λα ν

ηηηη

()()Xa Xa

XaaXa

αν

=∂ ∂

The index

is a repeated or dummy index, so we can call it what we like. Let’s

change it to match the ﬁ rst term in

µα

LT X a a X=∂∂+∂∂/[ ]2

∂∂=∂∂

αν

αµ

Xa Xa

Now, we repeat the process for the indices on the derivatives. This time, the

indices are worldsheet indices. So, we obtain

∂∂=∂ ∂

=∂∂

=∂

αµ

αβ

αγ

αβ

αγ β

µγ

Xah Xh a

hh X a

ββ

µγ

µβ

µα

Xa Xa

∂

=∂ ∂ =∂ ∂

And so, the variation in the lagrangian reduces to

µα

Xa aX TXa=∂ ∂ +∂∂ =∂∂

()