Light-Cone Quantization and the Critical Dimension

The covariant quantization of the previous note is elegant because it keeps Lorentz invariance visible. Its price is that the Fock space initially contains unphysical timelike and longitudinal excitations. The physical-state conditions remove them, but the mechanism is indirect.

Light-cone quantization does the opposite. It solves the constraints before quantization, leaving only the genuinely propagating string coordinates. The Hilbert space is then manifestly positive-norm. The price is that Lorentz invariance becomes non-manifest and must be checked quantum mechanically. That check is where the famous values

$$D=26, \qquad a=1$$

first appear in their most concrete form.

Here $D$ is the target-space dimension and $a$ is the open-string normal-ordering constant, also called the intercept.

Conventions

The target-space metric is mostly plus,

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+), \qquad M^2=-p^2.$$

We define light-cone coordinates with the standard $\sqrt 2$ normalization,

$$X^\pm=\frac{X^0\pm X^{D-1}}{\sqrt2}, \qquad p^\pm=\frac{p^0\pm p^{D-1}}{\sqrt2}.$$

Then

$$V\cdot W=-V^+W^- - V^-W^+ + V^iW^i, \qquad i=1,\ldots,D-2,$$

and

$$V^2=-2V^+V^-+V^iV^i.$$

For the open string,

$$\alpha_0^\mu=\sqrt{2\alpha'}\,p^\mu,$$

while for the closed string,

$$\alpha_0^\mu=\widetilde\alpha_0^\mu=\sqrt{\frac{\alpha'}2}\,p^\mu.$$

The remaining gauge freedom in conformal gauge

In conformal gauge the Polyakov action becomes a free two-dimensional field theory, but the equations are accompanied by constraints:

$$\partial_+\partial_-X^\mu=0, \qquad T_{++}=0, \qquad T_{--}=0.$$

Here

$$\sigma^\pm=\tau\pm\sigma.$$

Conformal gauge has not fixed all gauge freedom. There are still residual reparametrizations

$$\sigma^+\to f^+(\sigma^+), \qquad \sigma^-\to f^-(\sigma^-),$$

which preserve the conformal form of the worldsheet metric. Light-cone gauge uses this residual freedom to choose one target-space coordinate, $X^+$, to be proportional to the worldsheet time.

For the open string, whose Neumann mode expansion is

$$X^\mu(\tau,\sigma) =x^\mu+2\alpha'p^\mu\tau +i\sqrt{2\alpha'}\sum_{n\neq0}\frac{\alpha_n^\mu}{n}e^{-in\tau}\cos n\sigma,$$

the light-cone gauge choice is

$$\boxed{X^+(\tau,\sigma)=x^+ + 2\alpha'p^+\tau.}$$

Equivalently,

$$\alpha_n^+=0, \qquad n\neq0,$$

and

$$\alpha_0^+=\sqrt{2\alpha'}\,p^+.$$

This gauge is valid in a sector with $p^+\neq0$. Intuitively, the string is being described by snapshots at fixed target-space light-cone time $X^+$.

After conformal gauge, residual transformations $\sigma^\pm\to f^\pm(\sigma^\pm)$ remain. Light-cone gauge uses them to set $X^+=x^+ +2\alpha'p^+\tau$ for an open string, removing all nonzero $X^+$ oscillators.

This is very different from merely choosing a Lorentz frame. It is a gauge condition on the worldsheet fields. Once $X^+$ is fixed, the Virasoro constraints determine $X^-$ in terms of the transverse coordinates $X^i$.

Solving the constraints for $X^-$

For the open string the Virasoro generators are

$$L_n=\frac12\sum_{m\in\mathbb Z}\alpha_{n-m}\cdot\alpha_m.$$

Split the dot product into light-cone and transverse pieces:

$$\alpha_{n-m}\cdot\alpha_m = -\alpha_{n-m}^+\alpha_m^- -\alpha_{n-m}^-\alpha_m^+ +\alpha_{n-m}^i\alpha_m^i.$$

Because $\alpha_n^+=0$ for $n\neq0$, the constraint $L_n=0$ for $n\neq0$ becomes

$$0=-\alpha_0^+\alpha_n^-+\frac12\sum_{m\in\mathbb Z}:\alpha_{n-m}^i\alpha_m^i:.$$

Thus

$$\boxed{ \alpha_n^-=\frac{1}{\sqrt{2\alpha'}\,p^+}L_n^\perp, \qquad n\neq0, }$$

where

$$L_n^\perp = \frac12\sum_{m\in\mathbb Z}:\alpha_{n-m}^i\alpha_m^i:.$$

The $n=0$ constraint gives the light-cone Hamiltonian. Including the normal-ordering constant,

$$0=L_0-a = -\alpha_0^+\alpha_0^- +\frac12\alpha_0^i\alpha_0^i +N-a.$$

Using

$$\alpha_0^\pm=\sqrt{2\alpha'}p^\pm, \qquad \alpha_0^i=\sqrt{2\alpha'}p^i,$$

we get

$$-2\alpha'p^+p^-+\alpha'p_i^2+N-a=0.$$

Therefore

$$\boxed{ p^-=\frac{p_i^2+(N-a)/\alpha'}{2p^+}. }$$

The invariant mass is

$$M^2=2p^+p^- - p_i^2,$$

so the open-string mass formula is

$$\boxed{ M^2=\frac{N-a}{\alpha'}. }$$

The gauge condition removes the nonzero $X^+$ oscillators. The Virasoro constraints then express every $\alpha_n^-$ in terms of transverse bilinears $L_n^\perp$ and determine $p^-$ from the mass-shell condition.

There is now no independent oscillator $\alpha_n^-$, and there is no independent oscillator $\alpha_n^+$. The only creation operators are

$$\alpha_{-n}^i, \qquad n>0, \qquad i=1,\ldots,D-2.$$

They obey

$$[\alpha_m^i,\alpha_n^j] = m\,\delta_{m+n,0}\delta^{ij}.$$

This Hilbert space is manifestly positive-norm. The string has $D-2$ transverse oscillators, just like a massless particle has $D-2$ physical polarizations.

Light-cone quantization removes $X^+$ by a gauge choice and determines $X^-$ by the constraints. The physical Fock space is built only from transverse oscillators $\alpha_{-n}^i$.

The intercept from zero-point energy

The number operator is

$$N=\sum_{n=1}^\infty \alpha_{-n}^i\alpha_n^i.$$

Because there are $D-2$ transverse bosons, normal ordering the light-cone Hamiltonian produces a zero-point energy. Formally each oscillator contributes $\frac12 n$, so

$$E_0 = \frac{D-2}{2}\sum_{n=1}^\infty n.$$

Using zeta-function regularization,

$$\sum_{n=1}^\infty n=\zeta(-1)=-\frac1{12},$$

we find

$$E_0=-\frac{D-2}{24}.$$

The mass formula is written as

$$M^2=\frac{N-a}{\alpha'},$$

so the intercept is

$$\boxed{ a=\frac{D-2}{24}. }$$

At this stage this is a statement about the light-cone zero-point energy. It does not yet say what $D$ must be. To find that, we must ask whether the quantized theory still represents the full Lorentz group.

The Lorentz-algebra test

Light-cone gauge makes the transverse rotation group $SO(D-2)$ manifest, but the full Lorentz group $SO(1,D-1)$ is hidden. Classically the Lorentz generators are

$$J^{\mu\nu} = \int d\sigma\,\left(X^\mu P^\nu-X^\nu P^\mu\right).$$

After light-cone gauge fixing, the dangerous generators are

$$J^{-i}.$$

They contain $X^-$ and therefore become nonlinear functions of the transverse oscillators. Schematically,

$$J^{-i} = x^-p^i-x^ip^- -i\sum_{n=1}^\infty\frac1n \left( \alpha_{-n}^-\alpha_n^i-\alpha_{-n}^i\alpha_n^- \right),$$

with $\alpha_n^-$ replaced by $L_n^\perp/(\sqrt{2\alpha'}p^+)$. Operator ordering now matters.

The full Lorentz algebra requires

$$[J^{-i},J^{-j}]=0.$$

A direct computation gives an anomalous term proportional to

$$\sum_{n>0} \left[ \left(\frac{D-2}{12}-2\right)n + \frac1n\left(2a-\frac{D-2}{12}\right) \right] \left( \alpha_{-n}^i\alpha_n^j-\alpha_{-n}^j\alpha_n^i \right).$$

The two independent coefficients vanish only if

$$\frac{D-2}{12}=2, \qquad 2a=\frac{D-2}{12}.$$

Therefore

$$\boxed{ D=26, \qquad a=1. }$$

This is one of the cleanest derivations of the critical dimension of the bosonic string.

Light-cone quantization has a positive-norm Hilbert space, but Lorentz invariance is not automatic. Closure of the hidden generators $J^{-i}$ fixes $D=26$ and $a=1$.

Relation to the no-ghost theorem

Covariant quantization keeps Lorentz invariance manifest but must prove that negative-norm states decouple. Light-cone quantization keeps positivity manifest but must prove Lorentz invariance. The two descriptions agree precisely at

$$D=26, \qquad a=1.$$

This agreement is the physical content behind the no-ghost theorem for the bosonic string.

Open-string spectrum in light-cone gauge

Set now

$$D=26, \qquad a=1.$$

The open-string mass formula becomes

$$\boxed{ \alpha'M^2=N-1. }$$

The first few levels are especially instructive.

Level $N=0$

The vacuum state is

$$|0;p\rangle,$$

with

$$\alpha'M^2=-1.$$

This is the open-string tachyon. Its presence is a sign that the bosonic string vacuum is unstable. The superstring will remove this tachyon by the GSO projection.

Level $N=1$

The first excited states are

$$\alpha_{-1}^i|0;p\rangle, \qquad i=1,\ldots,24.$$

They have

$$M^2=0.$$

There are $24=D-2$ polarizations, exactly the number of physical polarizations of a massless vector in $26$ dimensions. Thus this level is interpreted as a massless gauge boson.

Level $N=2$

At the next level the independent light-cone states are

$$\alpha_{-2}^i|0;p\rangle,$$

and

$$\alpha_{-1}^i\alpha_{-1}^j|0;p\rangle.$$

The count is

$$24+\frac{24\cdot25}{2}=324.$$

This number is not random. A massive spin-two particle in $26$ dimensions transforms as a symmetric traceless tensor of the massive little group $SO(25)$, whose dimension is

$$\frac{25\cdot26}{2}-1=324.$$

Thus the light-cone states, although written in terms of $SO(24)$ transverse oscillators, recombine into a full $SO(25)$ representation. This is a concrete low-level check of Lorentz invariance.

For the open bosonic string, $\alpha'M^2=N-1$. The level $N=2$ states have degeneracy $324$, matching the symmetric traceless representation of the massive little group $SO(25)$.

The leading Regge trajectory is obtained by repeatedly applying the same transverse oscillator,

$$\alpha_{-1}^{i_1}\cdots\alpha_{-1}^{i_N}|0;p\rangle.$$

It contains states of spin growing linearly with $N$, so

$$J\sim N\sim \alpha'M^2.$$

This is the quantum continuation of the classical Regge relation derived from the rotating string.

Normalization checkpoint

We keep $\alpha'$ explicit. If one temporarily sets $\alpha'=2$, the open-string formula $\alpha' M^2=N-1$ becomes $M^2=(N-1)/2$.

Closed strings in light-cone gauge

The closed string has two independent transverse oscillator algebras,

$$\alpha_n^i, \qquad \widetilde\alpha_n^i,$$

corresponding to right- and left-moving modes. The zero-mode convention is

$$\alpha_0^\mu=\widetilde\alpha_0^\mu = \sqrt{\frac{\alpha'}2}\,p^\mu.$$

In light-cone gauge we set

$$X^+(\tau,\sigma)=x^+ + \alpha'p^+\tau.$$

Thus

$$\alpha_n^+=\widetilde\alpha_n^+=0, \qquad n\neq0.$$

The two Virasoro constraints solve both longitudinal oscillator families:

$$\alpha_n^-=\frac{1}{\sqrt{\alpha'/2}\,p^+}L_n^\perp, \qquad \widetilde\alpha_n^-=\frac{1}{\sqrt{\alpha'/2}\,p^+}\widetilde L_n^\perp, \qquad n\neq0.$$

The two $L_0$ constraints are

$$L_0-a=0, \qquad \widetilde L_0-a=0.$$

Since

$$L_0=\frac{\alpha'}4p^2+N, \qquad \widetilde L_0=\frac{\alpha'}4p^2+\widetilde N,$$

we get

$$N=\widetilde N$$

and

$$\boxed{ M^2=\frac4{\alpha'}(N-a) = \frac4{\alpha'}(\widetilde N-a). }$$

At the critical value $a=1$,

$$\boxed{ M^2=\frac4{\alpha'}(N-1), \qquad N=\widetilde N. }$$

The condition

$$N=\widetilde N$$

is called level matching. It is the operator version of invariance under translations around the closed-string spatial circle.

The first closed-string levels

At

$$N=\widetilde N=0$$

we find the closed-string tachyon,

$$M^2=-\frac4{\alpha'}.$$

At

$$N=\widetilde N=1$$

the states are

$$\alpha_{-1}^i\widetilde\alpha_{-1}^j|0;p\rangle, \qquad i,j=1,\ldots,24.$$

They are massless and have

$$24\times24=576$$

physical polarizations. Decompose this tensor under the massless little group $SO(24)$:

$$24\otimes24 = \left(\text{symmetric traceless}\right) \oplus \left(\text{antisymmetric}\right) \oplus \left(\text{trace}\right).$$

The dimensions are

$$\frac{24\cdot25}{2}-1=299,$$ $$\frac{24\cdot23}{2}=276,$$

and

$$1.$$

These are the physical polarizations of the spacetime fields

$$G_{\mu\nu}, \qquad B_{\mu\nu}, \qquad \Phi,$$

namely the graviton, the antisymmetric two-form, and the dilaton.

Closed strings have independent left- and right-moving oscillator levels. Level matching forces $N=\widetilde N$. The first massless states form $24\otimes24$, decomposing into the graviton, two-form, and dilaton.

Why the graviton appears automatically

The closed-string massless state has two vector-like oscillator indices, one from each chiral half. The symmetric traceless part is a spin-two state. Interactions then force it to couple as a graviton. This is the technical reason why perturbative closed string theory always contains gravity.

Summary

Light-cone quantization gives the most economical description of the free string:

$$\text{physical Hilbert space} = \text{Fock space of }D-2\text{ transverse oscillators}.$$

For the bosonic string, the transverse zero-point energy gives

$$a=\frac{D-2}{24}.$$

Requiring the hidden Lorentz generators to close gives

$$D=26, \qquad a=1.$$

The open-string spectrum is

$$\alpha'M^2=N-1,$$

and the closed-string spectrum is

$$M^2=\frac4{\alpha'}(N-1), \qquad N=\widetilde N.$$

The next question is not just what the first few levels are, but how many states appear at very high level. That question leads to the string partition function and the Hagedorn temperature.

Exercises

Exercise 1. Light-cone scalar product

Using

$$X^\pm=\frac{X^0\pm X^{D-1}}{\sqrt2},$$

show that

$$V\cdot W=-V^+W^- -V^-W^+ + V^iW^i.$$

Solution

Invert the definitions:

$$V^0=\frac{V^+ + V^-}{\sqrt2}, \qquad V^{D-1}=\frac{V^+ - V^-}{\sqrt2},$$

and similarly for $W$. The contribution from the $0$ and $D-1$ directions is

$$-V^0W^0+V^{D-1}W^{D-1} = -\frac12(V^+ + V^-)(W^+ + W^-) +\frac12(V^+ - V^-)(W^+ - W^-).$$

Expanding gives

$$-V^+W^- - V^-W^+.$$

Adding the transverse directions gives

$$V\cdot W=-V^+W^- -V^-W^+ + V^iW^i.$$

Exercise 2. Derive $\alpha_n^-$ from $L_n=0$

For the open string in light-cone gauge, use

$$\alpha_n^+=0, \qquad n\neq0,$$

to derive

$$\alpha_n^-=\frac{1}{\sqrt{2\alpha'}p^+}L_n^\perp, \qquad n\neq0.$$

Solution

Start with

$$L_n=\frac12\sum_m\alpha_{n-m}\cdot\alpha_m.$$

In light-cone components,

$$\alpha_{n-m}\cdot\alpha_m = -\alpha_{n-m}^+\alpha_m^- -\alpha_{n-m}^-\alpha_m^+ +\alpha_{n-m}^i\alpha_m^i.$$

For $n\neq0$, the first light-cone term contributes only when $n-m=0$, namely $m=n$, and the second contributes only when $m=0$. Thus the two light-cone terms together give

$$-\alpha_0^+\alpha_n^-.$$

The constraint $L_n=0$ becomes

$$0=-\alpha_0^+\alpha_n^-+\frac12\sum_m:\alpha_{n-m}^i\alpha_m^i:.$$

Since

$$\alpha_0^+=\sqrt{2\alpha'}p^+,$$

we obtain

$$\alpha_n^-=\frac{1}{\sqrt{2\alpha'}p^+}L_n^\perp.$$

Exercise 3. The intercept

Using zeta-function regularization, compute the open-string intercept for $D-2$ transverse bosons.

Solution

The zero-point energy is

$$E_0=\frac{D-2}{2}\sum_{n=1}^\infty n.$$

With

$$\sum_{n=1}^\infty n=\zeta(-1)=-\frac1{12},$$

we find

$$E_0=-\frac{D-2}{24}.$$

The light-cone mass formula is written as

$$M^2=\frac{N-a}{\alpha'},$$

so the intercept is the negative of the zero-point energy:

$$a=-E_0=\frac{D-2}{24}.$$

For $D=26$ this gives

$$a=1.$$

Exercise 4. Count the first massive open-string level

At level $N=2$, count the light-cone states

$$\alpha_{-2}^i|0;p\rangle, \qquad \alpha_{-1}^i\alpha_{-1}^j|0;p\rangle,$$

for $i,j=1,\ldots,24$. Show that the answer equals the dimension of a symmetric traceless tensor of $SO(25)$.

Solution

The states $\alpha_{-2}^i|0;p\rangle$ contribute

$$24$$

states. The states $\alpha_{-1}^i\alpha_{-1}^j|0;p\rangle$ are symmetric in $i,j$ because the oscillators commute, so they contribute

$$\frac{24\cdot25}{2}=300$$

states. Therefore

$$24+300=324.$$

A symmetric rank-two tensor of $SO(25)$ has dimension

$$\frac{25\cdot26}{2}=325.$$

Removing its trace gives

$$325-1=324.$$

Thus the level-$2$ light-cone states form precisely the massive spin-two representation.

Exercise 5. Decompose the massless closed-string state

Show that

$$24\otimes24$$

decomposes into $299+276+1$ states.

Solution

The tensor product of two $SO(24)$ vectors decomposes into symmetric traceless, antisymmetric, and trace parts:

$$V_i\widetilde V_j = \left(V_{(i}\widetilde V_{j)}-\frac1{24}\delta_{ij}V\cdot\widetilde V\right) + V_{[i}\widetilde V_{j]} + \frac1{24}\delta_{ij}V\cdot\widetilde V.$$

The symmetric tensors have dimension

$$\frac{24\cdot25}{2}=300.$$

Removing the trace gives

$$300-1=299.$$

The antisymmetric tensors have dimension

$$\frac{24\cdot23}{2}=276.$$

The trace gives one scalar. Hence

$$24\otimes24=299+276+1=576.$$

These are the graviton, two-form, and dilaton polarizations.

Exercise 6. Why the critical dimension follows from Lorentz closure

Use the anomalous coefficient

$$\left(\frac{D-2}{12}-2\right)n + \frac1n\left(2a-\frac{D-2}{12}\right)$$

to show that Lorentz closure requires $D=26$ and $a=1$.

Solution

The expression must vanish for every positive integer $n$. Since it contains an $n$ term and a $1/n$ term, each coefficient must vanish separately:

$$\frac{D-2}{12}-2=0,$$

and

$$2a-\frac{D-2}{12}=0.$$

The first equation gives

$$D-2=24, \qquad D=26.$$

Substituting this into the second equation gives

$$2a=2, \qquad a=1.$$