Mode Expansions, Regge Trajectories, and Bosonic String Quantization

After conformal gauge, the classical string looks deceptively simple. In flat spacetime the embedding coordinates obey the free wave equation

$$(\partial_\tau^2-\partial_\sigma^2)X^\mu=0,$$

but the Virasoro constraints remain:

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

This page explains what those equations actually buy us. We will first see that a classical open string rotating rigidly in a plane lies on a linear Regge trajectory. Then we solve the wave equation in normal modes, quantize the oscillators, impose the Virasoro constraints, and read off the first open- and closed-string states.

Throughout this page we use flat target-space metric

$$\eta_{\mu\nu}=\operatorname{diag}(-,+,\ldots,+),$$

and define the string tension by

$$T={1\over 2\pi\alpha'}.$$

The parameter $\alpha'$ has dimensions of length squared. It sets both the mass spacing of the string spectrum and the slope of Regge trajectories.

A classical rotating open string

Before quantizing, it is worth looking at the most important classical solution. Consider an open string with Neumann boundary conditions in all target-space directions,

$$X'^\mu(\tau,0)=X'^\mu(\tau,\pi)=0.$$

A rigidly rotating solution in the $(X^1,X^2)$ plane is

$$X^0=A\tau, \qquad X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau,$$

with all other coordinates set to zero. Equivalently,

$$X^1+iX^2=A\cos\sigma\,e^{i\tau}.$$

At a fixed target-space time, the string is a straight line segment. The midpoint $\sigma=\pi/2$ is at the origin, while the endpoints $\sigma=0,\pi$ lie at opposite ends of the segment. The segment rotates with angular velocity $1$ in worldsheet time. Since $X^0=A\tau$, the physical endpoint speed is

$$\left|{d\vec X\over dX^0}\right|_{\sigma=0,\pi}=1,$$

so the endpoints move at the speed of light. That may sound alarming, but it is precisely what an ideal relativistic string with no endpoint masses does.

Let us check the constraints. The derivatives are

$$\dot X^\mu=(A,-A\cos\sigma\sin\tau,A\cos\sigma\cos\tau),$$

and

$$X'^\mu=(0,-A\sin\sigma\cos\tau,-A\sin\sigma\sin\tau).$$

Therefore

$$\dot X^2=-A^2+A^2\cos^2\sigma=-A^2\sin^2\sigma,$$

while

$$X'^2=A^2\sin^2\sigma, \qquad \dot X\cdot X'=0.$$

Thus

$$\dot X^2+X'^2=0, \qquad \dot X\cdot X'=0,$$

which is equivalent to $T_{++}=T_{--}=0$. The wave equation is also immediate, because both $\partial_\tau^2$ and $\partial_\sigma^2$ acting on $\cos\sigma\,e^{i\tau}$ produce a minus sign.

A rotating open string gives the cleanest classical derivation of a linear Regge relation. The endpoints move at the speed of light, and the conserved charges obey $J=\alpha' E^2$.

Energy, angular momentum, and the Regge slope

The canonical momentum density is

$$\Pi_\mu=T\dot X_\mu.$$

The target-space energy is

$$E=P^0=T\int_0^\pi d\sigma\,\dot X^0 =T\int_0^\pi d\sigma\,A =\pi T A.$$

The angular momentum in the rotation plane is

$$J^{12}=T\int_0^\pi d\sigma\, \left(X^1\dot X^2-X^2\dot X^1\right).$$

For the solution above,

$$X^1\dot X^2-X^2\dot X^1=A^2\cos^2\sigma,$$

so

$$J\equiv J^{12}=T A^2\int_0^\pi d\sigma\,\cos^2\sigma ={\pi T A^2\over2}.$$

Eliminating $A$ gives

$$J={E^2\over2\pi T}=\alpha' E^2.$$

In the rest frame, $E=M$, so

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

This is the classical Regge relation. It says that a long string rotating with larger angular momentum also has larger mass, but in such a way that $J$ grows linearly with $M^2$. Quantum mechanically the line is shifted by an intercept, but the slope is still controlled by $\alpha'$.

Historically, this observation was one of the reasons strings were introduced in hadron physics: mesons appeared to organize themselves along approximately linear trajectories in the $(M^2,J)$ plane. Modern string theory no longer treats the bosonic string as a realistic theory of hadrons, but this calculation remains conceptually central. It shows how an extended relativistic object produces infinitely many higher-spin states with a single scale.

Open-string normal modes

Now solve the conformal-gauge wave equation on the open-string strip

$$0\leq \sigma\leq \pi.$$

For Neumann boundary conditions, the normal modes are cosines:

$$\partial_\sigma X^\mu(\tau,0)=\partial_\sigma X^\mu(\tau,\pi)=0 \quad\Longrightarrow\quad \cos n\sigma, \qquad n=0,1,2,\ldots .$$

With the standard normalization adapted to $T=1/(2\pi\alpha')$, the open-string expansion is

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

The zero modes have a direct spacetime interpretation. The constant $x^\mu$ is the center-of-mass position, and $p^\mu$ is the total spacetime momentum:

$$P^\mu=T\int_0^\pi d\sigma\,\dot X^\mu=p^\mu.$$

The oscillators $\alpha_n^\mu$ describe string vibrations. Reality of $X^\mu$ requires

$$(\alpha_n^\mu)^\dagger=\alpha_{-n}^\mu.$$

The positive-frequency modes $\alpha_n^\mu$ with $n>0$ will become annihilation operators, and the negative-frequency modes $\alpha_{-n}^\mu$ will become creation operators.

Open strings with Neumann boundary conditions have standing waves and a single set of oscillators. Closed strings are periodic and have independent left-moving and right-moving oscillators.

Canonical quantization of the oscillators

The equal-time canonical commutator is

$$[X^\mu(\tau,\sigma),\Pi_\nu(\tau,\sigma')] =i\delta^\mu{}_\nu\delta(\sigma-\sigma'), \qquad \Pi_\nu=T\dot X_\nu.$$

This gives

$$[x^\mu,p^\nu]=i\eta^{\mu\nu},$$

and

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

Thus for $n>0$, the normalized operators

$$a_n^\mu={\alpha_n^\mu\over\sqrt n}, \qquad (a_n^\mu)^\dagger={\alpha_{-n}^\mu\over\sqrt n}$$

obey the oscillator algebra

$$[a_m^\mu,(a_n^\nu)^\dagger]=\delta_{mn}\eta^{\mu\nu}.$$

The sign of $\eta^{00}$ is a warning: covariant quantization contains negative-norm oscillator states before imposing the constraints. The Virasoro constraints are not optional; they remove the unphysical timelike and longitudinal excitations. A fully clean covariant treatment uses ghosts and BRST cohomology. At this stage, one can either impose the Virasoro constraints directly or work in light-cone gauge, where only the $D-2$ transverse oscillators are manifestly physical.

Virasoro generators for the open string

It is convenient to define

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

The holomorphic stress tensor modes of the matter CFT are

$$L_m={1\over2}\sum_{n=-\infty}^{\infty}:\alpha_{m-n}\cdot\alpha_n:.$$

The colons denote normal ordering: annihilation operators are moved to the right of creation operators. For $m\neq0$, this is essentially the Fourier transform of the classical constraint $T_{++}=0$. For $m=0$, the result is

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

where the number operator is

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

In light-cone gauge the physical number operator is more transparently

$$N=\sum_{n=1}^{\infty}\alpha_{-n}^i\alpha_n^i, \qquad i=1, \ldots,D-2,$$

with only transverse directions included.

The quantum physical-state conditions are

$$(L_0-a)|\psi\rangle=0, \qquad L_m|\psi\rangle=0\quad(m>0),$$

modulo null states. The constant $a$ is the normal-ordering intercept. For the critical bosonic string,

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

The value $D=26$ is not an aesthetic choice. It is required by quantum Lorentz invariance, or equivalently by vanishing of the total conformal anomaly after adding the reparametrization ghosts. The quick light-cone way to remember the intercept is that each transverse oscillator contributes a zero-point energy

$${1\over2}\sum_{n=1}^{\infty}n={1\over2}\zeta(-1)=-{1\over24}.$$

With $D-2$ transverse directions, the total zero-point energy is

$$-{D-2\over24},$$

so the open-string intercept is

$$a={D-2\over24}.$$

When $D=26$, this gives $a=1$.

The open-string mass formula

The mass is

$$M^2=-p^2.$$

Using $L_0-a=0$,

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

Therefore

$$M^2={1\over\alpha'}(N-a).$$

In the critical bosonic string this becomes

$$M^2={1\over\alpha'}(N-1).$$

The first few levels are as follows.

Level	Representative state	Mass squared	Interpretation
$N=0$	$	0;k\rangle$	$-1/\alpha'$
$N=1$	$\zeta_\mu\alpha_{-1}^\mu	0;k\rangle$	$0$
$N=2$	$\alpha_{-2}^\mu	0;k\rangle$,$\alpha_{-1}^\mu\alpha_{-1}^\nu	0;k\rangle$

The ground state is tachyonic. This is one of the reasons the purely bosonic string is not a stable theory of nature. In superstring theory the GSO projection will remove the tachyon from the supersymmetric theories.

The first excited state is much more important. It has the form

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

The mass-shell condition gives

$$k^2=0.$$

The constraint $L_1|\zeta;k\rangle=0$ gives transversality,

$$k\cdot\zeta=0.$$

There is also a null state proportional to

$$k_\mu\alpha_{-1}^\mu|0;k\rangle,$$

which implements the gauge equivalence

$$\zeta_\mu\sim\zeta_\mu+\lambda k_\mu.$$

Thus the first open-string excitation is a massless spin-one gauge boson. Once Chan—Paton factors are attached to the endpoints, this state becomes a non-abelian gauge boson. That is the first major lesson of perturbative string theory:

$$\text{open strings contain gauge fields.}$$

Closed-string mode expansion

A closed string has periodicity

$$X^\mu(\tau,\sigma+2\pi)=X^\mu(\tau,\sigma).$$

The wave equation splits into independent left-moving and right-moving waves. With $0\leq\sigma<2\pi$, the standard expansion is

$$\begin{aligned} X^\mu(\tau,\sigma) &=x^\mu+\alpha'p^\mu\tau \\ &\quad +i\sqrt{\alpha'\over2}\sum_{n\neq0}{\alpha_n^\mu\over n}e^{-in(\tau-\sigma)} +i\sqrt{\alpha'\over2}\sum_{n\neq0}{\widetilde\alpha_n^\mu\over n}e^{-in(\tau+\sigma)}. \end{aligned}$$

The two sets of oscillators commute with each other:

$$[\alpha_m^\mu,\widetilde\alpha_n^\nu]=0,$$

and each separately obeys

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

The zero modes are

$$\alpha_0^\mu=\widetilde\alpha_0^\mu=\sqrt{\alpha'\over2}p^\mu.$$

The two Virasoro zero modes are

$$L_0={\alpha'p^2\over4}+N, \qquad \widetilde L_0={\alpha'p^2\over4}+\widetilde N.$$

The physical-state conditions are

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

and

$$L_m|\psi\rangle=\widetilde L_m|\psi\rangle=0 \qquad (m>0).$$

Subtracting the two zero-mode conditions gives level matching:

$$N=\widetilde N.$$

Adding them gives the closed-string mass formula

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

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

$$M^2={4\over\alpha'}(N-1), \qquad N=\widetilde N.$$

The first closed-string states

The closed-string ground state has

$$N=\widetilde N=0,$$

so

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

It is again tachyonic.

The first excited closed-string level has

$$N=\widetilde N=1.$$

A general state is

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

The mass formula gives

$$M^2=0.$$

The tensor $\epsilon_{\mu\nu}$ decomposes into three irreducible spacetime fields:

$$\epsilon_{\mu\nu} =\bigl(\epsilon_{(\mu\nu)}-{1\over D-2}\eta_{\mu\nu}\epsilon^\rho{}_{\rho}\bigr) +\epsilon_{[\mu\nu]} +{1\over D-2}\eta_{\mu\nu}\epsilon^\rho{}_{\rho}.$$

These are interpreted as

$$\begin{array}{ccl} \epsilon_{(\mu\nu)}\text{ traceless} &\longrightarrow& \text{graviton }G_{\mu\nu},\\ \epsilon_{[\mu\nu]} &\longrightarrow& \text{antisymmetric two-form }B_{\mu\nu},\\ \epsilon^\rho{}_{\rho} &\longrightarrow& \text{dilaton }\Phi. \end{array}$$

The Virasoro constraints impose transversality, and null states give the familiar gauge redundancies of a massless spin-two field and a two-form gauge field. The crucial conceptual point is already visible before writing an interaction:

$$\text{closed strings contain gravity.}$$

This is one of the deepest outputs of quantizing the string. A massless spin-two particle is not something that must be added by hand; it is part of the closed-string spectrum.

The critical bosonic string has a tachyonic ground state. At the first excited level, the open string gives a massless vector, while the closed string gives the graviton, antisymmetric two-form, and dilaton.

Regge trajectories from oscillator states

The classical rotating string relation becomes a quantum statement about the leading trajectory. For the open string, the highest-spin states at level $N$ are obtained by acting repeatedly with a transverse oscillator of one helicity, schematically

$$(\alpha_{-1}^{+})^N|0;k\rangle,$$

where $\alpha_{-1}^{+}$ denotes a complex transverse polarization such as

$$\alpha_{-1}^{+}=\alpha_{-1}^1+i\alpha_{-1}^2.$$

This state has spin

$$J=N,$$

while the mass formula gives

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

Therefore the leading open-string Regge trajectory is

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

The $+1$ is the quantum intercept. It is the same normal-ordering constant that made the open-string ground state tachyonic and the first excited state massless.

For the closed string, the leading states use both left- and right-moving oscillators:

$$(\alpha_{-1}^{+})^N(\widetilde\alpha_{-1}^{+})^N|0;k\rangle.$$

They have

$$J=2N,$$

and

$$M^2={4\over\alpha'}(N-1).$$

Thus

$$J={\alpha'\over2}M^2+2.$$

The closed-string Regge slope is one half of the open-string slope. This is a useful rule of thumb: closed-string states are made from a left-moving open-string-like excitation and a right-moving open-string-like excitation glued together.

Open strings, closed strings, and spacetime fields

The first few levels already summarize much of perturbative string theory:

$$\begin{array}{ccl} \text{open strings} &\Rightarrow& \text{spin-one gauge bosons},\\ \text{closed strings} &\Rightarrow& \text{spin-two gravitons plus }B_{\mu\nu}\text{ and }\Phi. \end{array}$$

This distinction will reappear throughout the course. Open-string diagrams have boundaries, and Chan—Paton labels on those boundaries become gauge indices. Closed strings have no endpoints and include the universal gravitational sector. D-branes will later bring these two worlds together: open strings end on D-branes, while closed strings propagate in the bulk and mediate gravity between branes.

There is also a useful large-$N$ gauge-theory intuition. In an old hadronic language, open strings resemble mesonic flux tubes, while closed strings resemble glueball-like flux loops. In modern gauge/gravity duality, this intuition is sharpened: closed strings in a curved bulk encode single-trace operators of a large-$N$ gauge theory, while open strings appear when flavor branes or other defects are added.

What to remember

The open-string mode expansion with Neumann boundary conditions is

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

The oscillator algebra is

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

The open-string mass formula is

$$M^2={1\over\alpha'}(N-a), \qquad a=1\quad\text{for the critical bosonic string}.$$

The closed-string mode expansion has independent left- and right-moving oscillators, with level matching

$$N=\widetilde N,$$

and mass formula

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

The first excited open-string state is a massless vector. The first excited closed-string states are the graviton, antisymmetric two-form, and dilaton. The bosonic string also has a tachyon, signaling instability of the bosonic vacuum and motivating the superstring.

Exercises

Exercise 1: verify the rotating-string solution

Show explicitly that

$$X^0=A\tau, \qquad X^1=A\cos\sigma\cos\tau, \qquad X^2=A\cos\sigma\sin\tau$$

solves the open-string equations of motion, satisfies Neumann boundary conditions, and obeys the Virasoro constraints.

Solution

The equation of motion is

$$(\partial_\tau^2-\partial_\sigma^2)X^\mu=0.$$

For $X^0=A\tau$, both second derivatives vanish. For the complex combination

$$X^1+iX^2=A\cos\sigma e^{i\tau},$$

we have

$$\partial_\tau^2(\cos\sigma e^{i\tau})=-\cos\sigma e^{i\tau}, \qquad \partial_\sigma^2(\cos\sigma e^{i\tau})=-\cos\sigma e^{i\tau},$$

so the wave equation holds.

The boundary condition is $X'^\mu=0$ at $\sigma=0,\pi$. Since

$$\partial_\sigma(\cos\sigma)=-\sin\sigma,$$

we have $X'^\mu=0$ at both endpoints.

Finally,

$$\dot X^\mu=(A,-A\cos\sigma\sin\tau,A\cos\sigma\cos\tau),$$

and

$$X'^\mu=(0,-A\sin\sigma\cos\tau,-A\sin\sigma\sin\tau).$$

Then

$$\dot X^2=-A^2+A^2\cos^2\sigma=-A^2\sin^2\sigma,$$ $$X'^2=A^2\sin^2\sigma,$$

and

$$\dot X\cdot X'=0.$$

Hence

$$\dot X^2+X'^2=0, \qquad \dot X\cdot X'=0,$$

which is equivalent to $T_{++}=T_{--}=0$.

Exercise 2: derive the classical Regge relation

Using the same rotating solution, compute $E=P^0$ and $J=J^{12}$, then show that

$$J=\alpha'E^2.$$

Solution

The energy is

$$E=P^0=T\int_0^\pi d\sigma\,\dot X^0.$$

Since $\dot X^0=A$,

$$E=\pi T A.$$

The angular momentum is

$$J=T\int_0^\pi d\sigma\,(X^1\dot X^2-X^2\dot X^1).$$

For the rotating solution,

$$X^1\dot X^2-X^2\dot X^1=A^2\cos^2\sigma.$$

Therefore

$$J=T A^2\int_0^\pi d\sigma\,\cos^2\sigma ={\pi T A^2\over2}.$$

Eliminate $A$ using $E=\pi T A$:

$$J={E^2\over2\pi T}.$$

Since

$$T={1\over2\pi\alpha'},$$

we obtain

$$J=\alpha'E^2.$$

In the rest frame $E=M$, so the classical trajectory is $J=\alpha'M^2$.

Exercise 3: normalize the open-string zero mode

For the open-string expansion

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

show that the total spacetime momentum is $P^\mu=p^\mu$.

Solution

The total momentum is

$$P^\mu=T\int_0^\pi d\sigma\,\dot X^\mu.$$

The derivative of the zero mode is

$$\dot X^\mu_{\rm zero}=2\alpha'p^\mu.$$

The oscillator part contains $\cos n\sigma$ with $n\geq1$, and

$$\int_0^\pi d\sigma\,\cos n\sigma=0.$$

Thus only the zero mode contributes:

$$P^\mu=T\int_0^\pi d\sigma\,2\alpha'p^\mu =2\pi T\alpha' p^\mu.$$

Using $T=1/(2\pi\alpha')$ gives

$$P^\mu=p^\mu.$$

Exercise 4: the open-string mass formula

Assume the open-string physical-state condition

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

Derive the mass formula and determine the masses of the first two levels in the critical bosonic string.

Solution

The condition gives

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

Since

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

we get

$$M^2={1\over\alpha'}(N-a).$$

For the critical bosonic string, $a=1$. Hence

$$M^2={1\over\alpha'}(N-1).$$

At $N=0$,

$$M^2=-{1\over\alpha'},$$

so the ground state is a tachyon. At $N=1$,

$$M^2=0,$$

so the first excited state is massless. Its representative state is

$$\zeta_\mu\alpha_{-1}^\mu|0;k\rangle,$$

and the remaining Virasoro constraints make it a massless vector.

Exercise 5: closed-string level matching

Starting from

$$L_0={\alpha'p^2\over4}+N, \qquad \widetilde L_0={\alpha'p^2\over4}+\widetilde N,$$

and the physical-state conditions

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

show that $N=\widetilde N$ and derive the closed-string mass formula.

Solution

Subtract the two equations:

$$\left({\alpha'p^2\over4}+N-a\right) - \left({\alpha'p^2\over4}+\widetilde N-a\right)=0.$$

This gives

$$N-\widetilde N=0,$$

or

$$N=\widetilde N.$$

Using either one of the two physical-state equations,

$${\alpha'p^2\over4}+N-a=0.$$

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

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

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

$$M^2={4\over\alpha'}(N-1), \qquad N=\widetilde N.$$

Exercise 6: the closed-string massless fields

At level $N=\widetilde N=1$, the closed-string state is

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

Decompose $\epsilon_{\mu\nu}$ into symmetric traceless, antisymmetric, and trace parts, and identify the corresponding spacetime fields.

Solution

Any rank-two tensor can be decomposed as

$$\epsilon_{\mu\nu} =\epsilon_{(\mu\nu)}+ \epsilon_{[\mu\nu]}.$$

The symmetric part can be further split into a trace and a traceless part:

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

After imposing the massless constraints and gauge equivalences, the physical interpretation is

$$\epsilon_{(\mu\nu)}\text{ traceless}\quad\longrightarrow\quad G_{\mu\nu},$$ $$\epsilon_{[\mu\nu]}\quad\longrightarrow\quad B_{\mu\nu},$$

and

$$\epsilon^\rho{}_{\rho}\quad\longrightarrow\quad \Phi.$$

Thus the first massless closed-string level contains the graviton, the antisymmetric two-form, and the dilaton.