Mode Expansions and Canonical Quantization

A free string has infinitely many normal modes. Quantizing these modes gives the oscillator algebra from which string states are built. The boundary conditions determine whether there are two independent families of traveling waves or one family of standing waves; the worldsheet constraints determine which combinations can describe physical states.

This page develops the canonical quantization of the embedding fields in flat spacetime. It builds on conformal gauge and boundary conditions and uses the global conventions. The Fock space constructed here includes negative-norm states. Selecting the physical space is a further step, developed in Virasoro constraints and the spectrum.

Closed-string waves and their zero modes

The conformal-gauge action and its canonical momentum density are

$$S=\frac{T}{2}\int d\tau\,d\sigma\, \bigl(\dot X^2-X^{\prime 2}\bigr), \qquad \Pi_\mu=T\dot X_\mu.$$

The equation of motion is $\partial_+\partial_-X^\mu=0$. On a closed string, its solutions separate into waves of the two null coordinates:

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

A fixed phase of a function of $\sigma^+=\tau+\sigma$ moves toward decreasing $\sigma$; a fixed phase of a function of $\sigma^-=\tau-\sigma$ moves toward increasing $\sigma$. These descriptions concern propagation along the string, not motion in target space.

For a noncompact target coordinate, periodicity forbids a net term linear in $\sigma$. The expansion is

$$\begin{aligned} X^\mu(\tau,\sigma) ={}&x^\mu+\alpha'p^\mu\tau\\ &+i\sqrt{\frac{\alpha'}2} \sum_{n\ne0}\frac{1}{n} \left(\alpha_n^\mu e^{-in\sigma^+} +\widetilde\alpha_n^\mu e^{-in\sigma^-}\right). \end{aligned}$$

The label $\alpha_n$ belongs to $X_L$, and $\widetilde\alpha_n$ belongs to $X_R$. The zero modes can be split as

$$\begin{aligned} X_L^\mu&=\frac{x^\mu}{2}+\frac{\alpha'p^\mu}{2}\sigma^+ +i\sqrt{\frac{\alpha'}2}\sum_{n\ne0}\frac{\alpha_n^\mu}{n}e^{-in\sigma^+},\\ X_R^\mu&=\frac{x^\mu}{2}+\frac{\alpha'p^\mu}{2}\sigma^- +i\sqrt{\frac{\alpha'}2}\sum_{n\ne0}\frac{\widetilde\alpha_n^\mu}{n}e^{-in\sigma^-}. \end{aligned}$$

Assigning half of $x^\mu$ to each expression is convenient; it does not introduce two independent center-of-mass positions. Likewise, the noncompact left and right zero-mode momenta are equal. Compact directions admit winding and require a different zero-mode split, worked out in Exercise 3.

The total target-space momentum checks the coefficient of $\tau$:

$$\int_0^{2\pi}d\sigma\,\Pi^\mu =T(2\pi\alpha')p^\mu=p^\mu.$$

Reality of the classical field, or Hermiticity after quantization, requires

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

It is useful to include the momentum in the oscillator notation:

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

All $\alpha_n$ are dimensionless. The two nonzero-mode families are independent before the physical constraints are imposed.

Constant-phase characteristics on a local unwrapped-cylinder patch. Future worldsheet time points upward. The left-moving $\alpha_n$ family depends on $\sigma^+$, whereas the right-moving $\widetilde\alpha_n$ family depends on $\sigma^-$. The Wick-rotation dictionary shown below the diagram is derived in the cylinder-to-plane discussion. The worldsheet drawing is schematic.

Deriving the closed-string oscillator algebra

Canonical quantization begins with the equal-time relations

$$[X^\mu(\sigma),\Pi_\nu(\sigma')] =i\delta^\mu{}_{\nu}\,\delta_{2\pi}(\sigma-\sigma'), \qquad [X^\mu,X^\nu]=[\Pi_\mu,\Pi_\nu]=0,$$

where the periodic delta distribution is

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

To derive the oscillator normalization rather than guess it, take Fourier coefficients at $\tau=0$:

$$Q_n^\mu=\frac{1}{2\pi}\int_0^{2\pi}d\sigma\,e^{in\sigma}X^\mu, \qquad P_{n\mu}=\int_0^{2\pi}d\sigma\,e^{in\sigma}\Pi_\mu.$$

The different prefactors ensure

$$[Q_m^\mu,P_{n\nu}] =i\delta^\mu{}_{\nu}\delta_{m+n,0}.$$

In particular, $Q_0=x$ and $P_0=p$, so $[x^\mu,p_\nu]=i\delta^\mu{}_{\nu}$. The Kronecker delta involves $m+n$ because both Fourier transforms used the same sign in the exponential.

For $n\ne0$, the field expansion gives

$$Q_n^\mu=\frac{i}{n}\sqrt{\frac{\alpha'}2} \bigl(\alpha_n^\mu-\widetilde\alpha_{-n}^\mu\bigr), \qquad P_n^\mu=\frac{\alpha_n^\mu+\widetilde\alpha_{-n}^\mu}{\sqrt{2\alpha'}}.$$

Solving these two equations gives the inverse transformation:

$$\begin{aligned} \alpha_n^\mu &=\sqrt{\frac{\alpha'}2}\,P_n^\mu -\frac{in}{\sqrt{2\alpha'}}Q_n^\mu,\\ \widetilde\alpha_n^\mu &=\sqrt{\frac{\alpha'}2}\,P_{-n}^\mu -\frac{in}{\sqrt{2\alpha'}}Q_{-n}^\mu. \end{aligned}$$

Only the cross commutators between $Q$ and $P$ contribute. For example,

$$[\alpha_m^\mu,\alpha_n^\nu] =\frac{m-n}{2}\,\eta^{\mu\nu}\delta_{m+n,0} =m\eta^{\mu\nu}\delta_{m+n,0}.$$

The same calculation for the other family, and cancellation between its mixed terms, yield

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

The nonzero oscillators commute with $x$ and $p$. The appearance of $\eta^{\mu\nu}$ follows from raising the momentum index with the target metric; replacing it by a positive Euclidean metric would change the theory.

Oscillator levels and the indefinite Fock space

For each $n>0$, define $a_n^\mu=\alpha_n^\mu/\sqrt n$. Then

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

The oscillator vacuum at momentum $p$ obeys

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

Products of negative-index oscillators generate the matter Fock space. For a spacelike coordinate this is the familiar harmonic-oscillator construction. For the timelike coordinate, however,

$$\bigl\|\alpha_{-1}^{0}|0;p\rangle\bigr\|^2 =\langle0;p|\alpha_1^0\alpha_{-1}^0|0;p\rangle =-\langle0;p|0;p\rangle.$$

Here one may use wave packets to remove the delta-function normalization of momentum eigenstates. This negative norm is a feature of the covariant, unconstrained state space. Quantizing free fields has not yet established a unitary physical string theory.

The level operators are

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

They are weighted occupation operators, not simply the number of creation operators. Indeed,

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

For spacelike oscillators, $N=\sum_{n,\,i}n(a_n^i)^\dagger a_n^i$. The covariant operator continues to grade every creation operator by its mode number, even though the inner product is indefinite. Thus $\alpha_{-2}^i|0;p\rangle$ and $\alpha_{-1}^i\alpha_{-1}^j|0;p\rangle$ both have level two.

Neumann open strings and the interval delta function

An open string with Neumann conditions at both ends has standing waves. The chiral derivative matching derived on the previous boundary-condition page leaves one periodic oscillator family, giving

$$X^\mu(\tau,\sigma) =x^\mu+2\alpha'p^\mu\tau +i\sqrt{2\alpha'}\sum_{n\ne0}\frac{\alpha_n^\mu}{n} e^{-in\tau}\cos n\sigma.$$

The cosine modes make $X^{\prime\mu}=0$ at $\sigma=0,\pi$. The shorter integration interval explains the factor two in the zero mode:

$$T\int_0^\pi d\sigma\,\dot X^\mu=p^\mu, \qquad \alpha_0^\mu=\sqrt{2\alpha'}\,p^\mu.$$

The canonical delta function must now reproduce functions on the interval. The orthonormal Neumann eigenfunctions are

$$\phi_0(\sigma)=\frac{1}{\sqrt\pi}, \qquad \phi_n(\sigma)=\sqrt{\frac{2}{\pi}}\cos n\sigma \quad(n\ge1).$$

Consequently,

$$\begin{aligned} \delta_N(\sigma,\sigma') &=\sum_{n=0}^{\infty}\phi_n(\sigma)\phi_n(\sigma')\\ &=\frac1\pi+\frac2\pi\sum_{n=1}^{\infty}\cos n\sigma\cos n\sigma',\\ [X^\mu(\sigma),\Pi_\nu(\sigma')] &=i\delta^\mu{}_{\nu}\delta_N(\sigma,\sigma'). \end{aligned}$$

The equality is distributional: integration against a suitable Neumann mode expansion reproduces that function. The spectral definition fixes the endpoint prescription, with no extra factor of one-half inserted by hand. Exercise 2 relates this kernel to reflection across the ends of the interval.

Define canonical mode coordinates and momenta at $\tau=0$ by

$$q_n^\mu=\int_0^\pi d\sigma\,\phi_nX^\mu, \qquad p_{n\mu}=\int_0^\pi d\sigma\,\phi_n\Pi_\mu.$$

Orthonormality gives $[q_m^\mu,p_{n\nu}]=i\delta^\mu{}_{\nu}\delta_{mn}$. Here $p_n$ denotes a mode momentum; the total target momentum is $p$ without a mode index. In particular, $q_0=\sqrt\pi\,x$ and $p_0=p/\sqrt\pi$.

For $n>0$, direct projection of the open-string expansion gives

$$q_n^\mu=\frac{i\sqrt{\pi\alpha'}}{n} (\alpha_n^\mu-\alpha_{-n}^\mu), \qquad p_n^\mu=\frac{\alpha_n^\mu+\alpha_{-n}^\mu}{2\sqrt{\pi\alpha'}}.$$

Thus

$$\alpha_n^\mu=\sqrt{\pi\alpha'}\,p_n^\mu -\frac{in}{2\sqrt{\pi\alpha'}}q_n^\mu, \qquad \alpha_{-n}^\mu=(\alpha_n^\mu)^\dagger,$$

and the canonical brackets imply

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

The full open-string oscillator algebra therefore has the same form as one closed-string chiral algebra. Its zero-mode normalization is different, and there is no independent tilde family.

From local constraints to Virasoro modes

The oscillator expansion turns the local conditions $C_{++}=C_{--}=0$ into algebraic conditions on infinitely many Fourier coefficients. For the closed string,

$$\partial_+X^\mu=\sqrt{\frac{\alpha'}2}\sum_{n\in\mathbb Z}\alpha_n^\mu e^{-in\sigma^+}, \qquad \partial_-X^\mu=\sqrt{\frac{\alpha'}2}\sum_{n\in\mathbb Z}\widetilde\alpha_n^\mu e^{-in\sigma^-}.$$

Squaring the series defines the classical generators:

$$\begin{aligned} C_{++}&=\alpha'\sum_m L_m^{\mathrm{cl}}e^{-im\sigma^+}, & L_m^{\mathrm{cl}}&=\frac12\sum_n\alpha_{m-n}\cdot\alpha_n,\\ C_{--}&=\alpha'\sum_m\widetilde L_m^{\mathrm{cl}}e^{-im\sigma^-}, & \widetilde L_m^{\mathrm{cl}}&=\frac12\sum_n\widetilde\alpha_{m-n}\cdot\widetilde\alpha_n. \end{aligned}$$

The $C_{ab}$ have dimensions of length squared in these dimensionless worldsheet coordinates; $L_m^{\mathrm{cl}}$ are dimensionless. Fourier inversion, including the tension, reads

$$L_m^{\mathrm{cl}} =T\int_0^{2\pi}d\sigma\,e^{im\sigma^+}C_{++}, \qquad \widetilde L_m^{\mathrm{cl}} =T\int_0^{2\pi}d\sigma\,e^{im\sigma^-}C_{--}.$$

For the open string, differentiation of the cosine expansion gives the same two derivative formulas with a single $\alpha_n$ family and the open-string value of $\alpha_0$. Both components of the constraint therefore involve the same $L_m^{\mathrm{cl}}$. A correct interval inversion combines both halves of the doubled string:

$$L_m^{\mathrm{cl}}=T\int_0^\pi d\sigma\, \left(e^{im\sigma^+}C_{++}+e^{im\sigma^-}C_{--}\right).$$

One chiral integral over half the circle would not isolate an arbitrary integer Fourier mode. Classically all these constraint modes vanish.

Normal ordering and the central term

The quantum products require normal ordering. Moving positive-index oscillators to the right gives the plane-vacuum prescription for this free theory:

$$L_m=\frac12\sum_{n\in\mathbb Z} :\alpha_{m-n}\cdot\alpha_n:, \qquad L_m^\dagger=L_{-m}.$$

In particular, the two halves of the nonzero-mode sum combine into $N$, so

String sector	Zero mode	Plane generator
Closed, holomorphic	$\alpha_0=\sqrt{\alpha'/2}\,p$	$L_0=\alpha'p^2/4+N$
Closed, antiholomorphic	$\widetilde\alpha_0=\sqrt{\alpha'/2}\,p$	$\widetilde L_0=\alpha'p^2/4+\widetilde N$
Open, Neumann	$\alpha_0=\sqrt{2\alpha'}\,p$	$L_0=\alpha'p^2+N$

No intercept has been subtracted in these definitions. The oscillator algebra gives

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

as derived in Exercise 4. Applying this relation twice shows that $[L_m,L_n]-(m-n)L_{m+n}$ commutes with every oscillator. Its possible vacuum contribution is the central term.

For $m>0$, evaluate the commutator in the formal $p=0$ oscillator vacuum. The only terms in $L_{-m}|0;0\rangle$ contain two creation operators, with mode numbers $r$ and $m-r$. Contracting these with $L_m$ gives

$$\langle0;0|L_mL_{-m}|0;0\rangle =\frac D2\sum_{r=1}^{m-1}r(m-r) =\frac D{12}(m^3-m),$$

with unit oscillator-vacuum normalization. The double target-index contraction is $\eta_{\mu\nu}\eta^{\mu\nu}=D$: the timelike boson contributes $+1$ to the central charge despite its negative-norm excitations. The result is

$$[L_m,L_n]=(m-n)L_{m+n} +\frac D{12}(m^3-m)\delta_{m+n,0}.$$

Closed strings have a commuting tilde copy. Open strings have one copy. This is the matter Virasoro algebra, with $c_X=D$. It already shows why one cannot require every $L_m$, positive and negative, to annihilate every physical state: the central term obstructs that prescription. Constraints must be imposed with the quantum state-space construction taken into account.

The Euclidean cylinder and the complex plane

The chirality labels are tied to the analytic continuation. Under

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

we have the exact mode and derivative dictionary

$$\begin{aligned} e^{-in\sigma^+}&\longrightarrow e^{-nw}=z^{-n}, &\partial_+&\longrightarrow i\partial_w,\\ e^{-in\sigma^-}&\longrightarrow e^{-n\bar w}=\bar z^{-n}, &\partial_-&\longrightarrow i\partial_{\bar w}. \end{aligned}$$

Thus $\alpha_n$ are holomorphic and $\widetilde\alpha_n$ antiholomorphic. Choosing the opposite Lorentzian assignment without changing the complex coordinate would exchange these labels inconsistently.

The cylinder is mapped to the punctured plane. A constant-$\tau_E$ circle has radius $e^{\tau_E}$, and increasing $\sigma$ runs counterclockwise. The infinite past maps to the origin; the infinite future maps to infinity. This turns Euclidean time ordering into radial ordering.

The map $z=e^{\tau_E+i\sigma}$ sends closed-string spatial circles to concentric circles with $|z|=e^{\tau_E}$. The three labeled radii are exact; the cylinder drawing is schematic. The origin is a limiting puncture at $\tau_E\to-\infty$, not a finite-time point on the cylinder.

The continued closed-string field becomes

$$\begin{aligned} X^\mu(z,\bar z) ={}&x^\mu-\frac{i\alpha'}2p^\mu\log|z|^2\\ &+i\sqrt{\frac{\alpha'}2}\sum_{n\ne0}\frac1n \left(\alpha_n^\mu z^{-n} +\widetilde\alpha_n^\mu\bar z^{-n}\right). \end{aligned}$$

Differentiating, including the momentum as the $n=0$ term, yields

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

Substitution in the normalized plane stress tensor reproduces precisely the oscillator generators:

$$T_X(z)=-\frac1{\alpha'}:\partial X\cdot\partial X: =\sum_{m\in\mathbb Z}L_m z^{-m-2}, \qquad L_m=\oint_{|z|=r}\frac{dz}{2\pi i}\,z^{m+1}T_X(z).$$

The contour is counterclockwise. The factor of $-i$ in $\partial X$ is necessary: its square cancels the minus sign in $T_X$. Before quantum normal ordering, the Lorentzian quantity $C_{++}/\alpha'$ continues to $-(\partial_wX)^2/\alpha'$. The quantum stress tensor on the cylinder also contains a vacuum shift, discussed below.

For the open string, the same exponential maps the strip $0\le\sigma\le\pi$ to the upper half-plane. Its two edges map to the positive and negative real axes. Neumann reflection extends the holomorphic derivative through the boundary, so the doubled description has one chiral oscillator family. This is a relation between derivatives; it does not equate open- and closed-string center-of-mass normalizations.

Under $z=e^{\tau_E+i\sigma}$, a constant-time open-string slice becomes an upper semicircle. The dashed lower semicircle is its reflected continuation, not an additional physical string. Neumann derivative gluing leaves one $\alpha_n$ family and one Virasoro algebra. The geometry is schematic and the open zero-mode coefficient remains $\alpha_0=\sqrt{2\alpha'}\,p$.

Plane ordering, cylinder energy, and the physical intercept

Three quantities that can contain constant shifts must be distinguished. First, the plane generator defined above has $L_0=\alpha'p^2/4+N$ in a closed-string chiral sector. Its $p=0$ oscillator vacuum has $L_0=0$.

Second, the cylinder stress tensor transforms with the Schwarzian derivative:

$$T_{\mathrm{cyl}}(w) =\left(\frac{dz}{dw}\right)^2T_{\mathrm{plane}}(z) +\frac c{12}\{z,w\}, \qquad \{z,w\}=\frac{z'''}{z'}-\frac32\left(\frac{z''}{z'}\right)^2.$$

For $z=e^w$, $\{z,w\}=-1/2$, and hence

$$T_{\mathrm{cyl}}(w)=\sum_nL_ne^{-nw}-\frac c{24}.$$

The matter contribution to the dimensionless worldsheet-time translation generator is therefore

$$H_{\mathrm{cyl}}^{X} =\begin{cases} L_0+\widetilde L_0-D/12,&\text{closed circle of period }2\pi,\\ L_0-D/24,&\text{Neumann strip of width }\pi. \end{cases}$$

These are worldsheet generators, not the target-space energy $p^0$. The negative constant is the matter Casimir energy in these coordinate units. The anomalous transformation and its relation to vacuum energy are developed in Tong, Section 4.4.1; Exercise 6 evaluates the shift directly for our complex coordinate.

Third, the physical intercept appears when imposing the quantum string constraints, conventionally written

$$(L_0-a)|\mathrm{phys}\rangle=0, \qquad L_n|\mathrm{phys}\rangle=0\quad(n>0),$$

with another set for closed strings. For the critical bosonic string, consistency gives $D=26$ and $a=1$. One cannot infer this by setting $a=D/24$: at $D=26$, that matter-cylinder shift is $13/12$, not $1$. The physical construction removes longitudinal excitations and includes the gauge constraints; in light-cone quantization the corresponding transverse zero-point sum gives $a=(D-2)/24$. Deriving why these constructions agree belongs to the subsequent treatment of physical states and the critical dimension, not to the canonical brackets alone.

The early operator approach is developed in the original analysis of Goddard, Goldstone, Rebbi, and Thorn. Tong, Sections 1.4 and 2.1, provides a complementary account of the mode expansion and the covariant oscillator construction.

Exercises

1. Recover traveling waves from initial data

For one nonzero integer $m>0$, suppose the classical closed-string initial data are

$$\begin{aligned} X(0,\sigma)&=x+A\cos m\sigma+B\sin m\sigma,\\ \dot X(0,\sigma)&=\alpha'p+C\cos m\sigma+D\sin m\sigma. \end{aligned}$$

Here $A,B,C,D$ are real target-space vectors with dimensions of length. Determine $\alpha_m$ and $\widetilde\alpha_m$. Find the condition for a purely left-moving nonzero-mode wave and check it directly in the initial data. No Virasoro constraint is assumed in this exercise.

Solution

The positive-index Fourier coefficients are

$$Q_m=\frac{A+iB}{2}, \qquad P_m=\frac{C+iD}{2\alpha'}.$$

The second formula follows from $\Pi=T\dot X$ and $T\pi=1/(2\alpha')$. Inserting these coefficients into the canonical inverse transformation gives

$$\begin{aligned} \alpha_m&=\frac{C+mB+i(D-mA)}{2\sqrt{2\alpha'}},\\ \widetilde\alpha_m&=\frac{C-mB-i(D+mA)}{2\sqrt{2\alpha'}}. \end{aligned}$$

The negative-index coefficients are the complex conjugates. A purely left-moving wave has $\widetilde\alpha_m=0$, requiring

$$C=mB, \qquad D=-mA.$$

At $\tau=0$ these relations give

$$\dot X-\alpha'p =mB\cos m\sigma-mA\sin m\sigma =X'.$$

This is the correct relation for a function of $\tau+\sigma$. A purely right-moving wave instead has $C=-mB$, $D=mA$, so $\dot X-\alpha'p=-X'$. These are solutions of the wave equation; generic choices still need not satisfy $C_{\pm\pm}=0$.

2. Derive the reflection kernels

Show that the Neumann interval kernel can be written as

$$\delta_N(\sigma,\sigma') =\delta_{2\pi}(\sigma-\sigma')+\delta_{2\pi}(\sigma+\sigma').$$

Find the corresponding Dirichlet kernel using an odd reflection. Check Neumann reproduction at both endpoints for $f(\sigma)=1+2\cos\sigma+\tfrac12\cos2\sigma$.

Solution

Adding the two periodic Fourier expansions and using

$$\cos n(\sigma-\sigma')+\cos n(\sigma+\sigma') =2\cos n\sigma\cos n\sigma'$$

gives

$$\delta_N=\frac1\pi+\frac2\pi\sum_{n\ge1}\cos n\sigma\cos n\sigma'.$$

This is the even-reflection kernel. Subtracting instead yields the odd-reflection kernel

$$\begin{aligned} \delta_D(\sigma,\sigma') &=\delta_{2\pi}(\sigma-\sigma')-\delta_{2\pi}(\sigma+\sigma')\\ &=\frac2\pi\sum_{n\ge1}\sin n\sigma\sin n\sigma'. \end{aligned}$$

It reproduces the Dirichlet sine expansion and has no constant mode. It vanishes when either argument is an endpoint, as appropriate for fields whose endpoint values are fixed.

For the specified Neumann function, orthogonality on $[0,\pi]$ gives

$$\int_0^\pi d\sigma'\,\delta_N(\sigma,\sigma')f(\sigma') =1+2\cos\sigma+\frac12\cos2\sigma.$$

The endpoint values are $f(0)=7/2$ and $f(\pi)=-1/2$, with their full coefficients. In the image representation, the two coincident image contributions at a boundary implement this same spectral prescription. One must not apply another ad hoc endpoint factor after using the cosine completeness relation.

3. Include a compact direction and winding

Let a spatial target coordinate satisfy $Y\sim Y+2\pi R$, with integer momentum quantum number $k$ and winding number $\nu$. Derive its left and right zero-mode momenta, and find the level-matching condition when the left and right physical intercepts are equal.

Solution

A center-of-mass wavefunction must be periodic under $y\to y+2\pi R$. Thus $e^{ip_Yy}$ is single-valued only if $p_Y=k/R$, with $k\in\mathbb Z$. Winding means

$$Y(\tau,\sigma+2\pi)-Y(\tau,\sigma)=2\pi\nu R.$$

The zero-mode solution is therefore

$$Y_0=y+\alpha'\frac{k}{R}\tau+\nu R\sigma =y+\frac{\alpha'}2(p_L\sigma^++p_R\sigma^-).$$

Matching the coefficients of $\tau$ and $\sigma$ gives

$$p_L=\frac{k}{R}+\frac{\nu R}{\alpha'}, \qquad p_R=\frac{k}{R}-\frac{\nu R}{\alpha'}.$$

Consequently $\alpha_0^Y=\sqrt{\alpha'/2}\,p_L$ and $\widetilde\alpha_0^Y=\sqrt{\alpha'/2}\,p_R$. The remaining noncompact momenta cancel from the difference of the two generators:

$$L_0-\widetilde L_0 =\frac{\alpha'}4(p_L^2-p_R^2)+N-\widetilde N =k\nu+N-\widetilde N.$$

Equal physical intercepts require this difference to vanish, giving

$$N-\widetilde N=-k\nu.$$

For example, $k\nu=1$ requires $\widetilde N=N+1$. The sign follows from associating $\alpha$ with $\sigma^+$ and from the stated positive winding orientation. Reversing either naming convention changes the displayed sign but not the physical matching condition. When $k\nu=0$, the ordinary equality $N=\widetilde N$ is recovered.

4. Show how Virasoro generators act on oscillators

Derive $[L_m,\alpha_n^\mu]=-n\alpha_{m+n}^\mu$. Use it to compute the $L_0$ eigenvalue of the closed-string chiral state $\alpha_{-2}^i\alpha_{-1}^j|0;p\rangle$. Explain why this is a level statement rather than a physical-state test.

Solution

Normal-ordering subtractions are constants and do not affect a commutator with a single oscillator. Thus

$$\begin{aligned} [L_m,\alpha_n^\mu] =\frac12\sum_r\Bigl(&\alpha_{m-r,\nu}[\alpha_r^\nu,\alpha_n^\mu]\\ &+[\alpha_{m-r,\nu},\alpha_n^\mu]\alpha_r^\nu\Bigr). \end{aligned}$$

The first term survives at $r=-n$ and contributes $-n\alpha_{m+n}^\mu/2$. The second survives at $r=m+n$ and contributes the same amount. Therefore

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

For $m=0$ and negative oscillator indices,

$$[L_0,\alpha_{-2}^i]=2\alpha_{-2}^i, \qquad [L_0,\alpha_{-1}^j]=\alpha_{-1}^j.$$

Since $L_0|0;p\rangle=(\alpha'p^2/4)|0;p\rangle$, the state has eigenvalue

$$\frac{\alpha'p^2}{4}+3.$$

It contains two creation operators but has level three. It need not satisfy the positive-mode Virasoro conditions, a mass-shell condition, or closed-string level matching with an antiholomorphic sector. Those are additional requirements on physical states.

5. Express the rotating string in oscillator variables

For the classical rotating open string

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

with $A>0$, find the nonzero oscillators. Check all classical Virasoro constraints and relate $N^{\mathrm{cl}}=\sum_{n>0}\alpha_{-n}\cdot\alpha_n$ to the angular momentum found on the rotating-string page.

Solution

The zero modes are $p^0=A/(2\alpha')$ and $\mathbf p=0$, so

$$\alpha_0^0=\frac{A}{\sqrt{2\alpha'}}, \qquad \boldsymbol\alpha_0=0.$$

Matching the coefficients of $e^{-i\tau}\cos\sigma$ gives

$$\alpha_1^1=-\frac{iA}{2\sqrt{2\alpha'}}, \qquad \alpha_1^2=\frac{A}{2\sqrt{2\alpha'}}, \qquad \alpha_1^0=0.$$

The $n=-1$ coefficients are their conjugates; all $|n|>1$ coefficients vanish. Therefore

$$N^{\mathrm{cl}}=\alpha_{-1}\cdot\alpha_1 =\frac{A^2}{4\alpha'}.$$

The zero-mode constraint is

$$L_0^{\mathrm{cl}}=\alpha'p^2+N^{\mathrm{cl}} =-\frac{A^2}{4\alpha'}+\frac{A^2}{4\alpha'}=0.$$

The remaining possible nonzero constraints are

$$L_{\pm1}^{\mathrm{cl}}=\alpha_0\cdot\alpha_{\pm1}=0, \qquad L_{\pm2}^{\mathrm{cl}}=\frac12\alpha_{\pm1}\cdot\alpha_{\pm1}=0.$$

The last equality uses the relative factor of $i$ between the two spatial oscillators: their squares cancel. Higher modes vanish because no pair of nonzero oscillator indices can sum to them. Thus all $C_{\pm\pm}$ vanish.

The classical angular momentum is $J^{12}=A^2/(4\alpha')$, so $N^{\mathrm{cl}}=J^{12}=\alpha'E^2$. This is a relation among classical amplitudes. It does not assert that the solution is an exact oscillator-number eigenstate or add a quantum intercept to the classical trajectory.

6. Compute the cylinder vacuum shift

Evaluate the Schwarzian derivative for $z=e^w$. Determine the matter vacuum value of the chiral cylinder generator and the change produced by $\alpha_{-2}^i$. Explain why setting $D=26$ does not identify the matter-cylinder shift with the bosonic physical intercept.

Solution

All derivatives of $e^w$ equal $e^w$, so

$$\{e^w,w\}=1-\frac32=-\frac12.$$

Consequently the chiral cylinder zero mode is $L_0-D/24$. In the plane oscillator vacuum with $p=0$, its value is $-D/24$; adding $\alpha_{-2}^i$ changes it to $2-D/24$. The excitation gap is still two. For a closed string with both sectors in their oscillator vacuum, the combined constant is $-D/12$. A single chiral excitation by itself is only a kinematic example and need not satisfy closed-string matching.

One can check the constant without invoking the transformation law. Point-split one free boson by a small separation $\epsilon$ in $w$. Transforming the plane two-point function gives

$$\langle\partial_wX(w+\epsilon/2)\partial_wX(w-\epsilon/2)\rangle =-\frac{\alpha'}2\frac{1}{4\sinh^2(\epsilon/2)}.$$

Subtract its local singular part $-\alpha'/(2\epsilon^2)$ and multiply by $-1/\alpha'$ from the stress tensor. Since

$$\frac{1}{4\sinh^2(\epsilon/2)} =\frac1{\epsilon^2}-\frac1{12}+O(\epsilon^2),$$

the finite expectation value is $-1/24$ per boson. A target-index contraction gives $-D/24$ for the covariant matter system.

At $D=26$, this number is $-13/12$. The physical condition involves $L_0-a$ with $a=1$, obtained after treating the worldsheet gauge constraints consistently. Matter-cylinder vacuum energy, plane ordering, and the physical intercept answer different questions and must not be substituted for one another.

References

• Goddard, Peter; Goldstone, Jeffrey; Rebbi, Claudio; and Thorn, Charles B. “Quantum Dynamics of a Massless Relativistic String.” Nuclear Physics B 56 (1973), 109–135. doi:10.1016/0550-3213(73)90223-X. The original systematic operator treatment of the relativistic string and its consistency conditions.
• Tong, David. String Theory. University of Cambridge Part III Mathematical Tripos lecture notes (2009), arXiv:0908.0333, version 3 (23 February 2012). Full text. Sections 1.4 and 2.1 cover mode expansions and covariant quantization; Section 4.4.1 explains the cylinder Casimir shift, and Sections 4.5.1–4.5.2 develop radial quantization and the Virasoro algebra. Complex-coordinate and chiral-label choices must be translated when comparing formulas.