Mode Expansions and Canonical Quantization

The previous page derived the classical Regge slope from a rotating open string. We now turn to the basic quantum-mechanical data of a free string: the Fourier modes of $X^\mu(\tau,\sigma)$ and their canonical commutation relations.

The central fact is simple but powerful. In conformal gauge the embedding fields obey the free two-dimensional wave equation, so every coordinate of the string decomposes into oscillator modes. Quantizing the string means quantizing those infinitely many oscillators. The Virasoro constraints then select the physical states from the resulting Fock space.

Conventions

For closed strings we use

$$0\leq \sigma <2\pi, \qquad X^\mu(\tau,\sigma+2\pi)=X^\mu(\tau,\sigma),$$

while for open strings we use

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

The target-space metric is mostly plus,

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

and the tension is

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

We keep $\alpha'$ explicit.

Closed strings on the cylinder

In conformal gauge, the Lorentzian Polyakov action for a closed string in flat spacetime becomes

$$S=\frac{1}{4\pi\alpha'}\int d\tau\,d\sigma\, \left(\dot X^2-X^{\prime 2}\right),$$

where

$$\dot X^\mu=\partial_\tau X^\mu, \qquad X^{\prime\mu}=\partial_\sigma X^\mu.$$

The equation of motion is the two-dimensional wave equation

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

For a closed string the worldsheet is a cylinder: at fixed $\tau$, the coordinate $\sigma$ parametrizes a circle.

A closed string sweeps out a cylinder. The periodic identification $\sigma\sim\sigma+2\pi$ means that a constant-$\tau$ slice is a closed loop.

Introduce worldsheet light-cone coordinates

$$\sigma^- = \tau-\sigma, \qquad \sigma^+ = \tau+\sigma.$$

Then

$$\partial_+\partial_-X^\mu=0,$$

so the general local solution separates into two chiral pieces,

$$X^\mu(\tau,\sigma)=X_R^\mu(\tau-\sigma)+X_L^\mu(\tau+\sigma).$$

A function of $\tau-\sigma$ moves toward increasing $\sigma$, while a function of $\tau+\sigma$ moves toward decreasing $\sigma$. On a closed string the two sectors are independent.

On the unwrapped cylinder, the two null families carry independent right- and left-moving waves. Periodicity identifies the two vertical sides.

The closed-string mode expansion

For a noncompact target-space coordinate, closed-string periodicity forbids a term linear in $\sigma$. The zero mode consists of the center-of-mass position and momentum,

$$x^\mu, \qquad p^\mu.$$

The standard closed-string mode expansion is

$$X^\mu(\tau,\sigma) =x^\mu+\alpha' p^\mu\tau +i\sqrt{\frac{\alpha'}{2}} \sum_{n\neq 0}\frac{1}{n} \left( \alpha_n^\mu e^{-in(\tau-\sigma)} +\widetilde\alpha_n^\mu e^{-in(\tau+\sigma)} \right).$$

Equivalently,

$$X_R^\mu(\tau-\sigma) =\frac12 x^\mu+\frac{\alpha'}{2}p^\mu(\tau-\sigma) +i\sqrt{\frac{\alpha'}{2}} \sum_{n\neq0}\frac{\alpha_n^\mu}{n}e^{-in(\tau-\sigma)},$$

and

$$X_L^\mu(\tau+\sigma) =\frac12 x^\mu+\frac{\alpha'}{2}p^\mu(\tau+\sigma) +i\sqrt{\frac{\alpha'}{2}} \sum_{n\neq0}\frac{\widetilde\alpha_n^\mu}{n}e^{-in(\tau+\sigma)}.$$

The factors of $\alpha'$ are fixed by two requirements: the total spacetime momentum should be $p^\mu$, and the oscillator commutators should have their canonical simple form.

Reality of $X^\mu$ implies

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

Thus modes with $n>0$ will become annihilation operators and modes with $n<0$ creation operators.

Compact coordinates

If a spatial target coordinate is compact, $X\sim X+2\pi R$, a closed string can obey

$$X(\tau,\sigma+2\pi)=X(\tau,\sigma)+2\pi wR, \qquad w\in\mathbb Z.$$

The integer $w$ is the winding number. Winding changes the zero modes and is the seed of T-duality. For now we work in noncompact flat spacetime.

Canonical momentum and oscillator algebra

The momentum conjugate to $X^\mu$ is

$$P^\tau_\mu(\tau,\sigma) =\frac{\partial \mathcal L}{\partial \dot X^\mu} =\frac{1}{2\pi\alpha'}\dot X_\mu.$$

The total spacetime momentum is

$$p_\mu=\int_0^{2\pi}d\sigma\,P^\tau_\mu.$$

Indeed,

$$\int_0^{2\pi}d\sigma\,\dot X^\mu =2\pi\alpha' p^\mu,$$

because all nonzero Fourier modes average to zero.

Canonical quantization imposes

$$[X^\mu(\tau,\sigma),P^\tau_\nu(\tau,\sigma')] =i\delta^\mu{}_{\nu}\,\delta_{2\pi}(\sigma-\sigma'),$$

with periodic delta function

$$\delta_{2\pi}(\sigma-\sigma') =\frac{1}{2\pi}\sum_{n\in\mathbb Z}e^{in(\sigma-\sigma')}.$$

This is equivalent to

$$[x^\mu,p_\nu]=i\delta^\mu{}_{\nu},$$

and

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

The closed string therefore has two independent copies of the same oscillator algebra.

The zero modes describe the center of mass, while the nonzero modes form two independent oscillator towers. The right-moving and left-moving oscillators commute with each other.

It is useful to include the center-of-mass momentum as the zeroth oscillator:

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

With this convention, many closed-string formulas look uniform in the integer mode number $n$.

Virasoro modes from the classical constraints

The conformal-gauge action looks like $D$ free scalar fields, but the string is not merely a collection of free fields. The metric equation of motion still imposes

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

Equivalently,

$$(\dot X+X')^2=0, \qquad (\dot X-X')^2=0.$$

Using the mode expansion,

$$\dot X^\mu-X^{\prime\mu} =\sqrt{2\alpha'}\sum_{n\in\mathbb Z}\alpha_n^\mu e^{-in(\tau-\sigma)},$$

and

$$\dot X^\mu+X^{\prime\mu} =\sqrt{2\alpha'}\sum_{n\in\mathbb Z}\widetilde\alpha_n^\mu e^{-in(\tau+\sigma)}.$$

Thus the stress-tensor constraints become Fourier constraints. Define

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

and

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

Then

$$(\dot X-X')^2 =4\alpha'\sum_{m\in\mathbb Z}L_m e^{-im(\tau-\sigma)},$$

and

$$(\dot X+X')^2 =4\alpha'\sum_{m\in\mathbb Z}\widetilde L_m e^{-im(\tau+\sigma)}.$$

Classically,

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

The Virasoro generators are the quadratic Fourier modes of the two chiral stress tensors. Classically the metric equation of motion sets every Fourier coefficient to zero.

At the quantum level, the quadratic products in $L_m$ must be normal ordered. For $m\neq0$ this is essentially unambiguous. For $L_0$ it produces the intercept that determines the first string mass levels. That shift is the subject of the next page.

The matter oscillators satisfy the Virasoro algebra

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

with an identical formula for $\widetilde L_m$, and

$$[L_m,\widetilde L_n]=0.$$

The central charge of the $D$ free embedding coordinates is

$$c_X=D.$$

The cylinder-to-plane map

The cylinder is natural for canonical quantization. The complex plane is natural for conformal field theory.

After Wick rotation $\tau=-i\tau_E$, set

$$w=\tau_E+i\sigma, \qquad z=e^w.$$

The identification $\sigma\sim\sigma+2\pi$ is precisely what makes $z$ single-valued. Constant Euclidean-time slices become circles around the origin, so time ordering on the cylinder becomes radial ordering on the plane.

The exponential map $z=e^w$ sends the Euclidean cylinder to the punctured plane. Increasing Euclidean time corresponds to increasing radial distance from the origin.

On the plane the closed-string field has the expansion

$$X^\mu(z,\bar z) =x^\mu-i\frac{\alpha'}{2}p^\mu\ln(z\bar z) +i\sqrt{\frac{\alpha'}{2}} \sum_{n\neq0}\frac{1}{n} \left(\alpha_n^\mu z^{-n}+\widetilde\alpha_n^\mu\bar z^{-n}\right).$$

Taking derivatives removes the logarithmic ambiguity:

$$\partial X^\mu(z) =-i\sqrt{\frac{\alpha'}{2}} \sum_{n\in\mathbb Z}\frac{\alpha_n^\mu}{z^{n+1}},$$ $$\bar\partial X^\mu(\bar z) =-i\sqrt{\frac{\alpha'}{2}} \sum_{n\in\mathbb Z}\frac{\widetilde\alpha_n^\mu}{\bar z^{n+1}}.$$

The holomorphic matter stress tensor is

$$T(z)=-\frac{1}{\alpha'}:\partial X^\mu\partial X_\mu:(z),$$

and its Laurent expansion is

$$T(z)=\sum_{n\in\mathbb Z}\frac{L_n}{z^{n+2}}.$$

Equivalently,

$$L_n=\frac{1}{2\pi i}\oint dz\,z^{n+1}T(z).$$

This contour formula is the bridge between canonical string quantization and two-dimensional CFT.

Open-string comparison

For an open string with Neumann boundary conditions in all target-space directions,

$$X^{\prime\mu}(\tau,0)=X^{\prime\mu}(\tau,\pi)=0,$$

the two chiral halves are glued together at the boundaries. The open string has a single independent oscillator algebra.

The mode expansion is

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

where $0\leq\sigma\leq\pi$. The zero mode is normalized so that

$$p^\mu=\frac{1}{2\pi\alpha'}\int_0^\pi d\sigma\,\dot X^\mu.$$

For the open string one writes

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

and

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

For Neumann boundary conditions, the strip can be doubled across its boundaries. The left- and right-moving halves are identified, leaving one oscillator algebra and one Virasoro algebra.

Before the normal-ordering shift,

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

for the open string, while

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

for the closed string. These formulas are the starting point for the first quantum spectra.

Summary

The free string has been reduced to harmonic oscillators:

$$X^\mu(\tau,\sigma) \quad\longleftrightarrow\quad x^\mu,p^\mu,\alpha_n^\mu,\widetilde\alpha_n^\mu.$$

The constraints are encoded in Virasoro generators:

$$T_{\pm\pm}=0 \quad\longleftrightarrow\quad L_m=\widetilde L_m=0 \quad\text{classically}.$$

Quantization modifies the story in two ways: products of oscillators must be normal ordered, and the Virasoro constraints are imposed as physical-state conditions rather than as operator identities on the whole Fock space.

Exercises

Exercise 1: Periodic delta function

Show that

$$\delta_{2\pi}(\sigma-\sigma') =\frac{1}{2\pi}\sum_{n\in\mathbb Z}e^{in(\sigma-\sigma')}$$

has the defining property

$$\int_0^{2\pi}d\sigma'\,\delta_{2\pi}(\sigma-\sigma')f(\sigma')=f(\sigma)$$

for a periodic function $f$.

Solution

Expand the periodic function as

$$f(\sigma')=\sum_{m\in\mathbb Z}f_m e^{im\sigma'}.$$

Then

$$\int_0^{2\pi}d\sigma'\,\delta_{2\pi}(\sigma-\sigma')f(\sigma') =\frac{1}{2\pi}\sum_{n,m}f_m e^{in\sigma}\int_0^{2\pi}d\sigma'\,e^{i(m-n)\sigma'}.$$

Using

$$\int_0^{2\pi}d\sigma'\,e^{i(m-n)\sigma'}=2\pi\delta_{mn},$$

the result becomes

$$\sum_m f_m e^{im\sigma}=f(\sigma).$$

Exercise 2: The zero-mode normalization

Insert the closed-string expansion into

$$p^\mu=\frac{1}{2\pi\alpha'}\int_0^{2\pi}d\sigma\,\dot X^\mu$$

and verify that the coefficient of the linear term in $\tau$ must be $\alpha'p^\mu$.

Solution

Differentiating the expansion gives

$$\dot X^\mu=\alpha'p^\mu+ \sqrt{\frac{\alpha'}{2}}\sum_{n\neq0} \left(\alpha_n^\mu e^{-in(\tau-\sigma)}+\widetilde\alpha_n^\mu e^{-in(\tau+\sigma)}\right).$$

The integral over $\sigma\in[0,2\pi]$ kills every nonzero Fourier mode. Therefore

$$\frac{1}{2\pi\alpha'}\int_0^{2\pi}d\sigma\,\dot X^\mu =\frac{1}{2\pi\alpha'}(2\pi\alpha'p^\mu)=p^\mu.$$

Exercise 3: The oscillator commutator from Hermiticity

Given

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

show that the Hermiticity condition $(\alpha_n^\mu)^\dagger=\alpha_{-n}^\mu$ is consistent with the commutator.

Solution

Taking the adjoint of the commutator gives

$$[\alpha_m^\mu,\alpha_n^\nu]^\dagger =[\alpha_n^{\nu\dagger},\alpha_m^{\mu\dagger}] =[\alpha_{-n}^\nu,\alpha_{-m}^\mu].$$

Using the algebra on the right-hand side,

$$[\alpha_{-n}^\nu,\alpha_{-m}^\mu] =(-n)\delta_{-n-m,0}\eta^{\nu\mu} =-n\delta_{m+n,0}\eta^{\mu\nu}.$$

On the support of $\delta_{m+n,0}$, one has $-n=m$, so this equals

$$m\delta_{m+n,0}\eta^{\mu\nu},$$

which is the adjoint of the original right-hand side.

Exercise 4: Deriving $L_m$

Use

$$\dot X^\mu-X^{\prime\mu} =\sqrt{2\alpha'}\sum_{n\in\mathbb Z}\alpha_n^\mu e^{-in(\tau-\sigma)}$$

to show that

$$(\dot X-X')^2=4\alpha'\sum_{m\in\mathbb Z}L_m e^{-im(\tau-\sigma)},$$

where

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

Solution

Multiplying two copies gives

$$(\dot X-X')^2 =2\alpha'\sum_{r,s\in\mathbb Z}\alpha_r\cdot\alpha_s\,e^{-i(r+s)(\tau-\sigma)}.$$

Set $m=r+s$, or $r=m-s$. Then

$$(\dot X-X')^2 =2\alpha'\sum_{m\in\mathbb Z}\left(\sum_{s\in\mathbb Z}\alpha_{m-s}\cdot\alpha_s\right)e^{-im(\tau-\sigma)}.$$

Since

$$L_m=\frac12\sum_s\alpha_{m-s}\cdot\alpha_s,$$

the expression becomes

$$(\dot X-X')^2=4\alpha'\sum_m L_m e^{-im(\tau-\sigma)}.$$

Exercise 5: Open-string zero mode

For the open-string expansion

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

verify that

$$p^\mu=\frac{1}{2\pi\alpha'}\int_0^\pi d\sigma\,\dot X^\mu.$$

Solution

Differentiating with respect to $\tau$ gives

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

For every nonzero integer $n$,

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

Therefore

$$\frac{1}{2\pi\alpha'}\int_0^\pi d\sigma\,\dot X^\mu =\frac{1}{2\pi\alpha'}(2\alpha'p^\mu\pi)=p^\mu.$$