i i

N= ±’n ±n . (2.136)

i=1 n=1

Let us now construct the physical spectrum of the bosonic string in the

light-cone gauge.

In the light-cone gauge all the excitations are generated by acting with the

i i

transverse modes ±n . The ¬rst excited state, given by ±’1 |0; p , belongs to

a (D ’ 2)-component vector representation of the rotation group SO(D ’ 2)

in the transverse space. As a general rule, Lorentz invariance implies that

physical states form representations of SO(D ’ 1) for massive states and

SO(D ’ 2) for massless states. Therefore, the bosonic string theory in the

i

light-cone gauge can only be Lorentz invariant if the vector state ±’1 |0; p

is massless. This immediately implies that a = 1.

Having ¬xed the value of a, the next goal is to determine the space-

time dimension D. A heuristic approach is to compute the normal-ordering

constant appearing in the de¬nition of L0 directly. This constant can be

determined from

∞

D’2 +∞ D’2 +∞

1 1 1

i i i i

: + (D ’ 2)

±’n ±n = : ±’n ±n n. (2.137)

2 2 2

i=1 n=’∞ i=1 n=’∞ n=1

The second sum on the right-hand side is divergent and needs to be reg-

ularized. This can be achieved using ζ-function regularization. First, one

considers the general sum

∞

n’s ,

ζ(s) = (2.138)

n=1

which is de¬ned for any complex number s. For Re(s) > 1, this sum con-

verges to the Riemann zeta function ζ(s). This zeta function has a unique

analytic continuation to s = ’1, where it takes the value ζ(’1) = ’1/12.

Therefore, after inserting the value of ζ(’1) in Eq. (2.137), the result for

the additional term is

∞

D’2

1

(D ’ 2) n=’ . (2.139)

2 24

n=1

2.5 Light-cone gauge quantization 51

Using the earlier result that the normal-ordering constant a should be equal

to 1, one gets the condition

D’2

= 1, (2.140)

24

which implies D = 26. Though it is not very rigorous, this is the quickest

way to determined the values of a and D. The earlier analysis of the no-

negative-norm states theorem also singled out D = 26. Another approach

is to verify that the Lorentz generators satisfy the Lorentz algebra, which is

not manifest in the light-cone gauge. The nontrivial requirement is

[J i’ , J j’ ] = 0. (2.141)

’

Once the ±n oscillators are eliminated, J i’ becomes cubic in transverse

oscillators. The algebra is rather complicated, but the bottom line is that

the commutator only vanishes for a = 1 and D = 26. Other derivations of

the critical dimension are presented in the next chapter.

Analysis of the spectrum

Having determined the preferred values a = 1 and D = 26, one can now

determine the spectrum of the bosonic string.

The open string

At the ¬rst few mass levels the physical states of the open string are as

follows:

• For N = 0 there is a tachyon |0; k , whose mass is given by ± M 2 = ’1.

i

• For N = 1 there is a vector boson ±’1 |0; k . As was explained in the

previous section, Lorentz invariance requires that it is massless. This

state gives a vector representation of SO(24).

• N = 2 gives the ¬rst states with positive (mass)2 . They are

j

i i

±’2 |0; k ±’1 ±’1 |0; k ,

and (2.142)

with ± M 2 = 1. These have 24 and 24 · 25/2 states, respectively. The

total number of states is 324, which is the dimensionality of the symmetric

traceless second-rank tensor representation of SO(25), since 25·26/2’1 =

324. So, in this sense, the spectrum consists of a single massive spin-two

state at this mass level.

All of these states have a positive norm, since they are built entirely from

the transverse modes, which describe a positive-de¬nite Hilbert space. In

the light-cone gauge the fact that the negative-norm states have decoupled

52 The bosonic string

is made manifest. All of the massive representations can be rearranged in

complete SO(25) multiplets, as was just demonstrated for the ¬rst massive

level. Lorentz invariance of the spectrum is guaranteed, because the Lorentz

algebra is realized on the Hilbert space of transverse oscillators.

The number of states

The total number of physical states of a given mass is easily computed. For

example, in the case of open strings, it follows from Eqs (2.135) and (2.136)

with a = 1 that the number of physical states dn whose mass is given by

± M 2 = n ’ 1 is the coe¬cient of w n in the power-series expansion of

∞ ∞

24

±i ±i

(1 ’ wn )’24 .

tr wN = tr w = (2.143)

’n n

n=1 i=1 n=1

This number can be written in the form

tr wN

1

dn = dw. (2.144)

wn+1

2πi

The number of physical states dn can be estimated for large n by a saddle-

point evaluation. Since the saddle point occurs close to w = 1, one can use

the approximation

∞

4π 2

n ’24

N

(1 ’ w ) ∼ exp

tr w = . (2.145)

1’w

n=1

This is an approximation to the modular transformation formula

·(’1/„ ) = (’i„ )1/2 ·(„ ) (2.146)

for the Dedekind eta function

∞

iπ„ /12

1 ’ e2πin„ ,

·(„ ) = e (2.147)

n=1

as one sees by setting w = e2πi„ . Then one ¬nds that, for large n,

√

dn ∼ const. n’27/4 exp(4π n). (2.148)

The exponential factor can be rewritten in the form exp(M/M0 ) with

√

M0 = (4π ± )’1 . (2.149)

The quantity M0 is called the Hagedorn temperature. Depending on de-

tails that go beyond present considerations, it is either a maximum possible