Covariant Virasoro Constraints and Bosonic Spectra

The previous page built the oscillator expansion of the free string. We now impose the Virasoro constraints quantum mechanically and read off the first bosonic string spectra.

The story has two layers. First, we build a covariant Fock space using all $D$ spacetime components of the oscillators. This keeps Lorentz symmetry manifest. Second, we impose the quantum constraints and quotient null states. The quotient is essential: the covariant Fock space contains timelike oscillators with negative norm, but the physical Hilbert space is ghost-free only at the critical intercept and dimension.

Normalization checkpoint

With mostly-plus metric $M^2=-p^2$. For open strings,

$$\alpha_0^\mu=\sqrt{2\alpha'}p^\mu, \qquad L_0=\alpha'p^2+N.$$

For closed strings,

$$\alpha_0^\mu=\widetilde\alpha_0^\mu=\sqrt{\frac{\alpha'}{2}}p^\mu, \qquad L_0=\frac{\alpha'}{4}p^2+N.$$

These two normalizations are the most common source of factors of $4$ in closed-string mass formulas.

From classical constraints to physical-state conditions

Classically, conformal gauge leaves the constraints

$$L_m=0, \qquad \widetilde L_m=0, \qquad m\in\mathbb Z.$$

At the quantum level the oscillator algebra is

$$[\alpha_m^\mu,\alpha_n^\nu] =m\delta_{m+n,0}\eta^{\mu\nu},$$

with the same formula for $\widetilde\alpha_n^\mu$. The Virasoro generators are normal-ordered quadratic operators,

$$L_m=\frac12\sum_{n\in\mathbb Z}:\alpha_{m-n}\cdot\alpha_n:,$$

and similarly for $\widetilde L_m$.

The matter Virasoro algebra is

$$[L_m,L_n] =(m-n)L_{m+n}+\frac{D}{12}m(m^2-1)\delta_{m+n,0},$$

and

$$L_m^\dagger=L_{-m}.$$

Because positive and negative modes are adjoints, one does not impose every $L_m$ as an annihilation condition. The old covariant prescription is Gupta-Bleuler-like:

$$L_m|\psi\rangle=0, \qquad m>0,$$

and

$$(L_0-a)|\psi\rangle=0.$$

For closed strings there are two copies:

$$L_m|\psi\rangle=\widetilde L_m|\psi\rangle=0, \qquad m>0,$$

and

$$(L_0-a)|\psi\rangle=(\widetilde L_0-a)|\psi\rangle=0.$$

The covariant Fock space is larger than the physical Hilbert space. Positive Virasoro modes impose constraints, $L_0-a$ imposes the mass shell, and null descendants are quotiented out.

The constant $a$ is the normal-ordering intercept. In light-cone gauge one finds

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

The critical bosonic string has

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

For the moment we keep $a$ explicit. This makes clear which pieces of the spectrum are kinematic and which require the critical theory.

The Fock space and number operators

The vacuum with momentum $k^\mu$ satisfies

$$p^\mu|0;k\rangle=k^\mu|0;k\rangle,$$

and

$$\alpha_n^\mu|0;k\rangle=0, \qquad n>0.$$

For an open string, a general Fock-space state is a finite linear combination of states of the form

$$\alpha_{-n_1}^{\mu_1}\alpha_{-n_2}^{\mu_2}\cdots\alpha_{-n_r}^{\mu_r}|0;k\rangle, \qquad n_i>0.$$

The number operator is

$$N=\sum_{n=1}^{\infty}\alpha_{-n}\cdot\alpha_n,$$

and it obeys

$$[N,\alpha_{-m}^\mu]=m\alpha_{-m}^\mu, \qquad m>0.$$

Thus $\alpha_{-m}^\mu$ raises the level by $m$.

For the open string,

$$L_0=\alpha'p^2+N.$$

The condition $(L_0-a)|\psi\rangle=0$ gives

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

Since $M^2=-p^2$,

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

For the closed string,

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

The two mass-shell equations imply

$$\boxed{\alpha'M^2=4(N-a)=4(\widetilde N-a),}$$

and hence

$$\boxed{N=\widetilde N.}$$

This is the level-matching condition.

The first open-string levels

At level $N=0$ the state is the oscillator vacuum,

$$|0;k\rangle.$$

Its mass is

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

For the critical bosonic string $a=1$, so

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

This is the open-string tachyon. It is not a particle moving faster than light; it is a signal that the perturbative bosonic-string vacuum is unstable.

At level $N=1$, the general state is

$$|\epsilon;k\rangle=\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle.$$

Its mass is

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

For $a=1$, this state is massless. The $L_1$ condition gives transversality. Using

$$[L_m,\alpha_n^\mu]=-n\alpha_{m+n}^\mu,$$

we find

$$L_1\epsilon\cdot\alpha_{-1}|0;k\rangle =\epsilon\cdot\alpha_0|0;k\rangle =\sqrt{2\alpha'}\,\epsilon\cdot k\,|0;k\rangle.$$

Thus

$$k\cdot\epsilon=0.$$

A polarization proportional to $k_\mu$ is null, because

$$k\cdot\alpha_{-1}|0;k\rangle =\frac{1}{\sqrt{2\alpha'}}L_{-1}|0;k\rangle.$$

Therefore

$$\epsilon_\mu\sim\epsilon_\mu+\lambda k_\mu,$$

and the massless vector has $D-2$ physical polarizations.

For the critical open bosonic string, $\alpha'M^2=N-1$. The ground state is tachyonic, level $N=1$ is a massless vector, and level $N=2$ is the first massive level.

The covariant vector starts with $D$ components. The constraint $k\cdot\epsilon=0$ removes one component, and the null equivalence $\epsilon\sim\epsilon+\lambda k$ removes another, leaving $D-2$ transverse polarizations.

At level $N=2$, the general open-string state may be written

$$|\psi;k\rangle= \left(S_{\mu\nu}\alpha_{-1}^\mu\alpha_{-1}^\nu +v_\mu\alpha_{-2}^\mu\right)|0;k\rangle,$$

where $S_{\mu\nu}$ is symmetric. The mass is

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

For $a=1$,

$$M^2=\frac{1}{\alpha'}.$$

The conditions $L_1|\psi\rangle=0$ and $L_2|\psi\rangle=0$ relate $S_{\mu\nu}$ and $v_\mu$. After quotienting null states, the physical states form a massive spin-two representation. In $D=26$ the number of physical states at this level is

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

which equals the dimension of the traceless symmetric tensor representation of the massive little group $SO(25)$:

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

At level $N=2$, the oscillator partitions $2=1+1=2$ give tensor and vector covariant data. The constraints and null states combine them into the physical massive spin-two representation.

Negative-norm states and the no-ghost structure

Covariant quantization keeps all spacetime components of $X^\mu$. This makes Lorentz symmetry manifest, but because

$$\eta^{00}=-1,$$

some covariant Fock-space states have negative norm. For example,

$$\langle0;k|\alpha_1^0\alpha_{-1}^0|0;k\rangle =\eta^{00}\langle0;k|0;k\rangle =-\langle0;k|0;k\rangle.$$

The Virasoro constraints remove the timelike and longitudinal excitations, while null states are quotiented out. This is analogous to covariant quantization of electrodynamics, but substantially more delicate because the constraints form the infinite-dimensional Virasoro algebra.

The no-ghost theorem says that the physical Hilbert space of the bosonic string has nonnegative norm for

$$D\leq 26, \qquad a\leq1,$$

and the critical, Lorentz-invariant theory is

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

Light-cone quantization will make the positive-norm transverse spectrum manifest and will also explain why these critical values are forced.

The closed-string spectrum

Closed strings have two independent oscillator towers. A general state is built by acting with both left- and right-moving creation operators:

$$\alpha_{-n_1}^{\mu_1}\cdots\alpha_{-n_r}^{\mu_r} \widetilde\alpha_{-\widetilde n_1}^{\nu_1}\cdots \widetilde\alpha_{-\widetilde n_s}^{\nu_s}|0;k\rangle.$$

The number operators are

$$N=\sum_{n=1}^{\infty}\alpha_{-n}\cdot\alpha_n, \qquad \widetilde N=\sum_{n=1}^{\infty}\widetilde\alpha_{-n}\cdot\widetilde\alpha_n.$$

The physical-state conditions imply

$$N=\widetilde N, \qquad \alpha'M^2=4(N-a).$$

At $N=\widetilde N=0$, the critical closed bosonic string has

$$M^2=-\frac{4}{\alpha'}.$$

This is the closed-string tachyon.

At the first matched excited level,

$$N=\widetilde N=1,$$

we have

$$|\epsilon;k\rangle =\epsilon_{\mu\nu}\alpha_{-1}^\mu\widetilde\alpha_{-1}^\nu|0;k\rangle.$$

For $a=1$ this state is massless:

$$M^2=0.$$

The polarization tensor decomposes into symmetric traceless, antisymmetric, and trace parts:

$$\epsilon_{\mu\nu} = \left(\epsilon_{(\mu\nu)}-\frac{1}{D-2}\eta_{\mu\nu}\epsilon^\rho{}_{\rho}\right) +\epsilon_{[\mu\nu]} +\frac{1}{D-2}\eta_{\mu\nu}\epsilon^\rho{}_{\rho}.$$

These are interpreted as

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

respectively: the graviton, antisymmetric two-form, and dilaton. The clean physical decomposition is made after imposing transversality and quotienting gauge redundancies, but the covariant tensor structure already reveals the key result: a consistent closed string contains a massless spin-two state.

Closed-string states carry a left level $N$ and a right level $\widetilde N$. Physical states obey $N=\widetilde N$. The first matched excited level gives $G_{\mu\nu}$, $B_{\mu\nu}$, and $\Phi$.

Leading Regge trajectories

The oscillator spectrum also explains the approximately linear Regge behavior that motivated string theory.

For the open string, the leading trajectory at level $N$ is represented by

$$\zeta_{\mu_1\cdots\mu_N} \alpha_{-1}^{\mu_1}\cdots\alpha_{-1}^{\mu_N}|0;k\rangle.$$

It has maximal spin

$$J=N,$$

and mass formula

$$\alpha'M^2=N-a.$$

Therefore

$$J=\alpha'M^2+a.$$

For the critical bosonic string,

$$J=1+\alpha'M^2.$$

For the closed string, the leading trajectory uses matched left and right level $N=\widetilde N$:

$$\zeta_{\mu_1\cdots\mu_N;\nu_1\cdots\nu_N} \alpha_{-1}^{\mu_1}\cdots\alpha_{-1}^{\mu_N} \widetilde\alpha_{-1}^{\nu_1}\cdots\widetilde\alpha_{-1}^{\nu_N}|0;k\rangle.$$

The maximal spin is

$$J=2N,$$

while

$$\alpha'M^2=4(N-a).$$

Hence

$$J=2a+\frac{\alpha'M^2}{2}.$$

For $a=1$,

$$J=2+\frac{\alpha'M^2}{2}.$$

The closed-string trajectory has half the slope of the open-string trajectory, and its massless spin-two state is the graviton.

For $a=1$, the leading open trajectory is $J=1+\alpha'M^2$, while the leading closed trajectory is $J=2+\alpha'M^2/2$. The closed-string slope is half the open-string slope.

Summary

The covariant bosonic spectra are controlled by four equations:

$$\alpha'M^2=N-a \quad\text{open},$$ $$\alpha'M^2=4(N-a)=4(\widetilde N-a), \qquad N=\widetilde N \quad\text{closed}.$$

At the critical values $D=26$ and $a=1$, the open string contains a massless vector at level $N=1$, while the closed string contains the massless graviton, two-form, and dilaton at $N=\widetilde N=1$. The next page derives the same spectrum in light-cone gauge, where only transverse oscillators appear.

Exercises

Exercise 1: Why not impose all $L_n$ strongly?

Explain why the quantum theory imposes

$$L_m|\psi\rangle=0, \qquad m>0,$$

rather than demanding $L_m|\psi\rangle=0$ for every $m\in\mathbb Z$.

Solution

The modes satisfy

$$L_m^\dagger=L_{-m}.$$

Thus positive and negative modes are adjoints. If both $L_m$ and $L_{-m}$ were imposed as annihilation constraints for all $m>0$, the Hilbert space would be overconstrained. The Virasoro algebra also contains a central term,

$$[L_m,L_n] =(m-n)L_{m+n}+\frac{D}{12}m(m^2-1)\delta_{m+n,0},$$

so the quantum constraints are not identical to the classical constraints as strong operator equations. The correct old-covariant prescription is Gupta-Bleuler-like: impose the positive modes and the shifted $L_0$ condition, then quotient null states.

Exercise 2: Open-string mass formula

Derive

$$\alpha'M^2=N-a$$

from

$$L_0=\alpha'p^2+N, \qquad (L_0-a)|\psi\rangle=0.$$

Solution

The physical-state condition gives

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

Since $M^2=-p^2$, this becomes

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

Therefore

$$\alpha'M^2=N-a.$$

Exercise 3: Transversality of the massless vector

Let

$$|\epsilon;k\rangle=\epsilon_\mu\alpha_{-1}^\mu|0;k\rangle.$$

Show that $L_1|\epsilon;k\rangle=0$ implies $k\cdot\epsilon=0$.

Solution

Use

$$[L_m,\alpha_n^\mu]=-n\alpha_{m+n}^\mu.$$

For $m=1$ and $n=-1$,

$$[L_1,\alpha_{-1}^\mu]=\alpha_0^\mu.$$

Since $L_1|0;k\rangle=0$,

$$L_1\epsilon\cdot\alpha_{-1}|0;k\rangle =\epsilon\cdot\alpha_0|0;k\rangle.$$

For the open string,

$$\alpha_0^\mu=\sqrt{2\alpha'}k^\mu.$$

Therefore

$$L_1|\epsilon;k\rangle =\sqrt{2\alpha'}\,k\cdot\epsilon\,|0;k\rangle.$$

The physical-state condition gives $k\cdot\epsilon=0$.

Exercise 4: A negative-norm oscillator

Assume $\langle0;k|0;k\rangle=1$. Show that $\alpha_{-1}^0|0;k\rangle$ has negative norm.

Solution

The norm is

$$\langle0;k|\alpha_1^0\alpha_{-1}^0|0;k\rangle.$$

Since $\alpha_1^0|0;k\rangle=0$, this equals

$$\langle0;k|[\alpha_1^0,\alpha_{-1}^0]|0;k\rangle.$$

The oscillator algebra gives

$$[\alpha_1^0,\alpha_{-1}^0]=\eta^{00}=-1.$$

Thus the norm is $-1$.

Exercise 5: Closed-string level matching

Use

$$(L_0-a)|\psi\rangle=0, \qquad (\widetilde L_0-a)|\psi\rangle=0$$

to derive $N=\widetilde N$.

Solution

For a closed string,

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

The two physical-state conditions are

$$\frac{\alpha'}{4}p^2+N-a=0,$$

and

$$\frac{\alpha'}{4}p^2+\widetilde N-a=0.$$

Subtracting gives

$$N-\widetilde N=0.$$

Thus $N=\widetilde N$.

Exercise 6: Open and closed Regge slopes

Assume $a=1$. For the leading open trajectory use $J=N$ and $\alpha'M^2=N-1$. For the leading closed trajectory use $J=2N$ and $\alpha'M^2=4(N-1)$. Derive both trajectories.

Solution

For the open string,

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

Thus

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

so

$$J=1+\alpha'M^2.$$

For the closed string,

$$J=2N, \qquad \alpha'M^2=4(N-1).$$

Hence

$$N=1+\frac{\alpha'M^2}{4},$$

and

$$J=2+\frac{\alpha'M^2}{2}.$$

The closed-string slope is half the open-string slope.