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With 24 physical half-integer open-string oscillators, the first three string level operator eigenvalues are $0,1/2,1$. Their states are the vacuum, $a_{1/2}^{i†}|0⟩$, and $a_{1/2}^{i†}{a}_{1/2}^{j†}|0⟩$. The creation operators commute, so the third level is the symmetric square of the 24-dimensional transverse vector, of dimension $24\cdot 25/2=300$. Its trace is a scalar and its trace-free part has dimension 299. Including the vacuum shift gives ${\alpha }^{\prime }{M}^2=1/2,1,3/2$ when the two longitudinal directions are common directions with Neumann boundary conditions. The representation labels refer to the surviving ND rotation group, not full unbroken target Lorentz symmetry.

The string level operator for a free-ended open string is $N={\sum }_{k>0}{\alpha }_{-k}\cdot {\alpha }_k$. The quantum condition is ${\alpha }^{\prime }{M}^2=N-a$, where $a$ is the normal-ordering constant of a string. Lorentz-consistent bosonic string theory has $a=1$.

The Nambu–Goto action is invariant under worldsheet diffeomorphisms. Its canonical Nambu–Goto phase-space constraints areIn conformal gauge, $\Pi =T\dot{X}$, so these become $(\dot{X}\pm X^{\prime }{)}^2=0$. Their smeared generators implement the residual conformal transformations rather than independent propagating degrees of freedom. Classically their Fourier modes form the classical Virasoro constraint algebra. Quantization promotes them to the Virasoro algebra, and its constraints select the physical string states from the indefinite covariant Fock space.

For clarity, start with a Neumann boundary condition at both ends of an open string, $0\le \sigma \le \pi$. Its open-string mode expansion isCanonical quantization gives the string oscillator relationsWith ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$, the worldsheet stress-energy tensor obeysSubstituting the open-string mode expansion into this quadratic expression derives the Virasoro generators:The colons mean normal ordering, placing the negative-mode creation operators to the left. The string level operator $N$ increases by $r$ when an oscillator ${\alpha }_{-r}$ is added.

Using the commutator once givesApplying this to the quadratic expression for $L_n$ produces $(m-n)L_{m+n}$, together with the double-contraction Virasoro central extension. To determine its coefficient, take $m>0$ and evaluate $[L_m,L_{-m}]$ on a zero-momentum oscillator vacuum. Only the creation pairs with $1\le r\le m-1$ contribute, givingHere $D$ is the target spacetime dimension; the timelike free boson also contributes one to the central charge, since a double contraction contains two metric factors. ThusThis derives both the Virasoro generators and their quantum anomaly.

In old covariant string quantization, the physical-state Virasoro conditions for an open string arewhere $a$ is the normal-ordering constant of a string. Requiring all positive and negative modes to annihilate a state would contradict the nonzero Virasoro central extension; the positive-mode prescription is the appropriate covariant subsidiary condition. The zero-mode equation gives the open bosonic string mass spectrum,For a closed string, there are two commuting Virasoro algebras. With uncompactified zero modes, $L_0={\alpha }^{\prime }{p}^2/4+N_R$ and ${\tilde{L}}_0={\alpha }^{\prime }{p}^2/4+N_L$. Imposing both positive-mode conditions and both zero-mode conditions gives the closed-string level matching $N_L=N_R$ and $M^2=4(N_L-a)/{\alpha }^{\prime }$.

Covariance initially retains the timelike string oscillators. For example, ${\alpha }_{-1}^0|0;p⟩$ has negative norm because ${\eta }^{00}=-1$. Its existence in the unconstrained space is not an inconsistency if it is absent from the physical quotient. The no-ghost theorem for the critical bosonic string states that, at $D=26$ and $a=1$, the inner product on physical string states at ordinary nonzero momentum is positive semidefinite. After quotienting physical null string states, this physical space is represented by the 24 positive-norm transverse string oscillators.

A spurious string state has the form ${\sum }_{n>0}{L}_{-n}|{\chi }_n⟩$. Its inner product with every physical string state vanishes, by $L_n|\psi ⟩=0$ and $L_n^{†}=L_{-n}$. Whenever such a state is itself physical, it is a null string state and represents a gauge redundancy, not an extra polarization. Consequently the physical quotient has no negative-norm propagating states, and agrees with light-cone gauge in string theory. This is the significance of the theorem; a proof is not needed here. It does not eliminate the bosonic tachyon, whose negative mass squared is distinct from negative norm, nor does it remove the auxiliary worldsheet ghost fields used in gauge fixing.

Distinguish target-space light-cone coordinates from the worldsheet $\pm$ labels. In target space take $X^{\pm }=(X^0\pm X^{25})/\sqrt{2}$, so the Minkowski metric is $d{s}^2=-2d{X}^{+}d{X}^{-}+{\sum }_i(d{X}^i{)}^2$. After conformal gauge has been chosen, the residual transformations of $\tau +\sigma$ and $\tau -\sigma$ may be used, on a patch with $p^{+}\ne 0$, to makeFor an open string on $0\le \sigma \le \pi$, the conventional normalization is $\kappa =2{\alpha }^{\prime }{p}^{+}$. For the closed-string convention of the preceding solution, $\kappa ={\alpha }^{\prime }{p}^{+}$. In either case, the two Virasoro constraints readSince ${\partial }_{\pm }{X}^{+}=\kappa /2$, they determine the longitudinal coordinate:Its oscillator content is fixed by the transverse coordinates, leaving $d-2$ independent fields. This is the light-cone gauge in string theory; it solves the constraints rather than imposing them afresh on each transverse state.

For Neumann boundary conditions at both endpoints, the open-string expansion isThe transverse string oscillators satisfy $[{\alpha }_m^i,{\alpha }_n^j]=m{\delta }^{ij}{\delta }_{m+n,0}$, $i,j=1,\dots ,24$, and ${\alpha }_{n>0}^i|0,p⟩=0$. Define $L_n^{\perp }=\frac{1}{2}{\sum }_m:{\alpha }_{n-m}^i{\alpha }_m^i:$ and ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}$. The solved longitudinal modes, including the zero-mode constraint, areThe critical bosonic string theory has 24 transverse bosons. Their regulated oscillator vacuum energy is $24\cdot \frac{1}{2}{\sum }_{n\ge 1}n=-1$, since the finite term of ${\sum }_{n\ge 1}n{e}^{-\epsilon n}={\epsilon }^{-2}-1/12+O({\epsilon }^2)$ is $-1/12$. Thus the normal-ordering constant of a string is $a=1$. A normalized transverse Fock space basis for general excited physical string states iswith finitely many nonzero occupation numbers. Arbitrary physical states are superpositions of this basis, with each momentum on its corresponding mass shell. The string level operator and open bosonic string mass spectrum giveAll spatial directions have free endpoints, so this string ends on a space-filling D25-brane. More generally, a D-brane has Neumann boundary conditions along its worldvolume and Dirichlet boundary conditions in its transverse directions; a D$p$-brane has $p$ spatial worldvolume directions. A stack may additionally supply Chan-Paton factors, without changing the oscillator construction above.

To see T-duality of D-brane boundary conditions directly, decompose the compact coordinate as $X=X_L(\tau +\sigma )+X_R(\tau -\sigma )$ and define its dual by $\tilde{X}=X_L-X_R$. Differentiation givesTherefore the original Neumann boundary condition ${\partial }_{\sigma }X=0$ becomes ${\partial }_{\tau }\tilde{X}=0$ at each endpoint. The dual endpoint coordinate is time-independent: the dual condition is Dirichlet. Conversely a fixed endpoint in $X$ becomes Neumann in the dual coordinate. For a circle, the T-duality radius is $\tilde{R}={\alpha }^{\prime }/R$. A wrapped Neumann brane loses one spatial dimension in the dual picture, for example D25 becomes D24. The two dual endpoint constants need not be identical; their positions encode the original Wilson-line data.

For the separated-brane problem, use two static parallel D$p$-branes and no relative angle or background field. Let their displacement in transverse directions be $d^I$, with ${\sum }_I(d^I{)}^2=d^2$. A Dirichlet coordinate contains the classical termThe spatial gradient contributes to the worldsheet HamiltonianParallel branes have integer modes in their NN and DD directions, so their oscillator vacuum energy is still $-1$. The zero-mode Virasoro constraint is consequentlyWith string tension $T=1/(2\pi {\alpha }^{\prime })$, the Open string stretched between D-branes hasThe classical stretching energy is $Td$, while the quantum ground-state energy, when real, isThe vacuum correction is an additive term in energy squared, not a constant subtraction from the classical stretching energy. It follows that the bosonic stretched-string tachyon threshold isBelow this threshold the lowest state is a tachyon: small-momentum modes have imaginary frequency and the background is unstable, rather than having an ordinary real negative ground-state energy. It is massless at the threshold. The assumption of parallel branes is needed for the integer moding used here; relative angles change the vacuum contribution.

Use target-space signature $(-,+,+,+)$ and conformal gauge $g_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$. Dots and primes denote differentiation with respect to $\tau$ and $\sigma$. Varying the Polyakov action with respect to the string embedding map gives the wave equation; varying the worldsheet metric sets the worldsheet stress tensor to zero. Explicitly,With ${\sigma }^{\pm }=\tau \pm \sigma$ and ${\partial }_{\pm }=\frac{1}{2}({\partial }_{\tau }\pm {\partial }_{\sigma })$, the two Virasoro constraints are $({\partial }_{+}X{)}^2=({\partial }_{-}X{)}^2=0$.

A fixed spatial circle would sweep out a cylindrical string worldsheet. In static coordinates its string embedding map is $X=(t,R\cos \theta ,R\sin \theta ,0)$ and its induced worldsheet metric is $\mathrm{diag}(-1,R^2)$. The radial part of the Nambu–Goto equations of motion is proportional to $R^{-2}{\partial }_{\theta }^2(R\cos \theta ,R\sin \theta )=-(\cos \theta ,\sin \theta )/R$, which is nonzero. Thus a nonzero-radius cylinder cannot solve the equations. Motion of labels around this same cylinder is a worldsheet diffeomorphism and cannot alter this conclusion.

The familiar pulsating circular string makes the obstruction particularly explicit. Put $X^0=\kappa \tau$ and $X^1+i{X}^2=r(\tau )e^{i\sigma }$. The wave equation and Virasoro constraints reduce toA constant $r=R\ne 0$ violates the first equation. Even rotating the labels, $X^1+i{X}^2=R{e}^{i(\sigma +\Omega \tau )}$, cannot help: the wave equation requires ${\Omega }^2=1$, whereas $\dot{X}\cdot X^{\prime }=R^2\Omega$ requires $\Omega =0$.

Nor does a rigid rotation of the circular plane rescue the usual circular ansatz in three spatial dimensions. Write its spatial part as $A(\tau )\cos \sigma +B(\tau )\sin \sigma$, with $A^2=B^2=R^2$ and $A\cdot B=0$. The wave equation gives $\ddot{A}=-A$ and $\ddot{B}=-B$. The constant norms give $A\cdot \dot{A}=B\cdot \dot{B}=0$ and ${\dot{A}}^2={\dot{B}}^2=R^2$. The mixed Virasoro constraint, together with the derivative of $A\cdot B=0$, gives $\dot{A}\cdot B=A\cdot \dot{B}=0$. Differentiating $A\cdot B$ twice gives $\dot{A}\cdot \dot{B}=0$. This would require four mutually orthogonal nonzero spatial vectors in ${\mathbb{R}}^3$. This obstruction is specific to the stated spatial dimension; a rigid circular string in four spatial dimensions has more room.

To express the Virasoro constraints in modes, extend the string oscillators byDifferentiating the two chiral expansions gives ${\partial }_{-}{X}^{\mu }=\sqrt{{\alpha }^{\prime }/2}{\sum }_n{\alpha }_n^{\mu }{e}^{-in{\sigma }^{-}}$ and the corresponding expression with tildes for ${\partial }_{+}X$. Equating every Fourier coefficient of their squares to zero yieldsReality of the string embedding map means ${\alpha }_{-n}={\alpha }_n^{\ast }$ and ${\tilde{\alpha }}_{-n}={\tilde{\alpha }}_n^{\ast }$. In particular, define the classical string level operators by $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$ and $\tilde{N}={\sum }_{n>0}{\tilde{\alpha }}_{-n}\cdot {\tilde{\alpha }}_n$. The zero-mode Virasoro constraints becomeConsequently the classical bosonic string mass spectrum isThe equality of the two levels is closed-string level matching. There is no string intercept here: that shift comes from quantum normal ordering, not from these classical equations.

For the compact boson, closed-string periodicity permits a winding number $w\in \mathbb{Z}$, so $X^3(\sigma +2\pi )-X^3(\sigma )=2\pi wR$. Let $q$ be the momentum in this direction. Its zero-mode expansion and the corresponding chiral momenta areThus replace the equal left and right zero modes by ${\tilde{\alpha }}_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_L^3$ and ${\alpha }_0^3=\sqrt{{\alpha }^{\prime }/2}{p}_R^3$; the nonzero string oscillators remain integer-moded. Classically $q$ is continuous. Quantizing the compact boson makes momentum wavefunctions single-valued around its circle, giving $q=n/R$ with $n\in \mathbb{Z}$.

If $k^a$, $a=0,1,2$, is the noncompact momentum and $M^2=-k^a{k}_a$, the two zero-mode Virasoro constraints giveAveraging and subtracting gives the momentum and winding modes and compact-circle closed-string level matching formulasThe latter becomes $\tilde{N}-N+nw=0$ on imposing quantum momentum quantization. The sign follows from assigning tildes to the left mover ${\sigma }^{+}$.

An explicit winding-supported circular string, valid classically for every $R>0$, isThe noncompact projection is a circle of constant radius. Every coordinate obeys the wave equation. Its compact winding is $w=1$, and a direct calculation givesThus both Virasoro constraints hold and the induced worldsheet metric is $2R^2\mathrm{diag}(-1,1)$, not a degenerate cylinder. The compact motion cancels the mixed stress that spoiled rotation of labels in flat space. Here $q=-R/{\alpha }^{\prime }$, $N=0$ and $\tilde{N}=R^2/{\alpha }^{\prime }$ satisfy the classical compact-circle closed-string level matching condition. Requiring this particular classical momentum to equal an individual quantum eigenvalue would impose an additional radius-dependent condition; that is not required to construct the classical solution.

First-class reduction. A regular set of independent constraints is first class when their Poisson brackets vanish on the constraint surface, equivalently locally $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The coefficients can be functions on phase space. With the generator $G={\sum }_i{\xi }^i{\phi }_i$, an infinitesimal canonical gauge transformation isEach independent first-class condition removes one phase-space dimension, and quotienting its independent gauge orbit removes another. Thus the regular physical phase space hasThis regular first-class phase-space reduction requires independent constraints and gauge directions. The count can fail at singular strata or for reducible constraints; first-class closure alone does not guarantee independence. For a concrete counterexample to counting redundant equations, take two canonical pairs and ${\phi }_1=p_1$, ${\phi }_2=2p_1$. Both equations are first class, but there is only one independent condition and one gauge direction. The surviving pair $(q^2,p_2)$ has dimension two, whereas blindly substituting $N=n=2$ gives zero.

Oscillator symplectic form. The center-of-mass pair $(x^m,p_m)$ describes translations and total momentum. The nonzero Fourier coefficients ${\alpha }_k^m$ describe the standing-wave modes compatible with free-end Neumann boundary conditions; their reality condition is ${\alpha }_{-k}={\alpha }_k^{\ast }$, and ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ in the standard normalization. For $k>0$, regarding ${\alpha }_k^m$ as a complex coordinate makes its conjugate momentum $i{\alpha }_{-k,m}/k$. Equivalently write ${\alpha }_k=(q_k+i{s}_k)/\sqrt{2}$. Up to a total derivative, its kinetic term is $-q_k\cdot {\dot{s}}_k/k$, a real canonical form. Inverting this oscillator symplectic form of an open string givesThe zero oscillator commutes with nonzero oscillators but is not independent of $p$: $\{x^m,{\alpha }_0^n\}=\sqrt{2{\alpha }^{\prime }}{\eta }^{mn}$. Brackets between the center pair and independent nonzero oscillators vanish.

The quadratic constraints obeyFor example, the two terms in the bracket with $L_n=\frac{1}{2}{\sum }_j{\alpha }_j\cdot {\alpha }_{n-j}$ give equal contributions after relabelling $j$. Applying this identity to both factors of $L_m$ givesThis classical Virasoro constraint algebra has no central term and closes on the constraints, hence is first class. For $G={\sum }_n{\xi }_{-n}{L}_n$, the oscillator gauge transformation isReality is respected when ${\xi }_{-n}={\xi }_n^{\ast }$.

Light-cone reduction and mass. Choose light-cone coordinates $X^{\pm }=(X^0\pm X^{D-1})/\sqrt{2}$; a vector square is $-2{\alpha }^{+}{\alpha }^{-}+{\mathbf{\alpha }}^2$, with $D-2$ transverse components. On the proposed gauge slice ${\alpha }_{k\ne 0}^{+}=0$, the variation becomesFor ${\alpha }_0^{+}\ne 0$, every nonzero-mode gauge parameter has an invertible coefficient, so the conditions locally fix the corresponding gauge freedom. Globally this is the usual patch in which $X^{+}$ is an admissible worldsheet clock; it is not a claim about strings for which that coordinate has turning points. The zero-mode reparameterization is left over.

On that slice the nonzero Virasoro constraints are linear in the longitudinal oscillators:No nonzero longitudinal oscillator remains independent. The residual action isThis residual mass-shell action in light-cone string gauge displays the remaining zero-mode constraint. At the quantum level normal ordering replaces its oscillator term by $N-a$. One can further set $x^{+}$ equal to time and solve for the light-cone Hamiltonian $p^{-}=[{\mathbf{p}}^2+(N-a)/{\alpha }^{\prime }]/(2p^{+})$.

Canonical quantization gives the transverse relationsHere $\hslash =1$, ${\alpha }_{k>0}$ annihilate the oscillator vacuum, and ${\alpha }_{-k}$ create excitations. Define $a_k^I={\alpha }_k^I/\sqrt{k}$ for $k>0$. The string level operator isThus it counts oscillator number weighted by mode number, and has nonnegative integer eigenvalues. The residual mass-shell condition givesThe constant $a$ is the normal-ordering constant of a string, or intercept, not an extra classical tension. In the usual Lorentz-invariant critical bosonic string, level one carries the $D-2$ transverse polarizations of a massless vector. A massive vector would need $D-1$ polarizations; the longitudinal one is not present. Lorentz consistency therefore requires that this vector level be massless, giving $a=1$. Equivalently the regularized transverse zero-point value $a=(D-2)/24$ gives $D=26$. This critical-vector assumption in the string intercept argument concerns the usual critical quantum theory. In $D=3$ the lone transverse level-one polarization transforms trivially under the transverse rotation group; a massive scalar interpretation is not excluded by counting. Thus the stated masslessness conclusion is not a consequence of polarization counting for arbitrary $D$. An arbitrary intercept in a transverse oscillator model also need not define the usual covariant critical theory.

The full self-dual massless spectrum. In the closed-string sector, $N$ and $\tilde{N}$ are the independent nonnegative integer oscillator levels of the two chiral sectors. The integer $n$ quantizes center momentum around the circle, $p_{\mathrm{circle}}=n/R$, and $w$ counts how many times the string winds it. At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, zero mass requiresBoth levels are nonnegative, so their sum is at most two. Exhausting these possibilities gives the massless spectrum at the bosonic self-dual circle:

The last family is easy to miss because the uncompactified ground state is a tachyon; its positive compact energy cancels that negative contribution at these charges. There are no other possibilities: at level sum one, $n^2+w^2=2$ forces both charges to be $\pm 1$; at sum zero, $n^2+w^2=4$ gives exactly the four listed pairs. With $d=D-2$ transverse oscillators the number of independent massless polarizations is $d^2+4d+4=D^2$, hence 676 in the critical $D=26$ bosonic theory. The circle oscillator is included among the $d$ components; it should not be discarded when interpreting the lower-dimensional scalar states.

Use $x^{\pm }=(x^0\pm x^{D-1})/\sqrt{2}$, transverse coordinates $x^i$, $i=1,\dots ,d=D-2$, and the Minkowski metricThe relativistic particle phase-space action becomesIn the light cone gauge $x^{+}=t$, solve the mass-shell condition for $p_{+}$, assuming $p_{-}\ne 0$. The reduced phase-space action is $\int dt({\dot{x}}^{-}{p}_{-}+{\dot{x}}^i{p}_i-H)$, withThe last equality selects the future-directed momentum sector and makes positivity transparent. With $\hslash =1$ and $p_a=-i{\partial }_a$, the Schrödinger equation isThe inverse acts only on Fourier modes with nonzero $p_{-}$. Multiplication by $2p_{-}$ gives $2{\partial }_{-}{\partial }_{+}\Psi =({\Delta }_{\perp }-{\mu }^2)\Psi$. ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.

For the massive two-form field, take $\mu \ne 0$. Apply ${\partial }^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes ${\partial }^n{\partial }^m{F}_{mnp}=0$, soExpanding $F=dA$, the other divergence terms vanish by this condition, leaving $({\square }_D-{\mu }^2)A_{np}=0$. The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking $n=-$ and $n=i$, respectively, givesThe $n=+$ equation follows from these expressions: the two terms containing ${\partial }_{+}{\partial }_i{A}_{-i}$ cancel and ${\partial }_i{\partial }_j{A}_{ji}=0$. Consequently $A_{-i}$ and $A_{ij}$ are independent, each satisfying the Klein-Gordon equation with mass $\mu$. The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group $SO(D-1)$. In the analogous Proca equation, ${\partial }^m{A}_m=0$ determines $A_{+}$ from $A_{-}$ and $A_i$, leaving $D-1$ components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If $\mu =0$, instead use the two-form gauge field symmetry $A\mapsto A+d\Lambda$: the light-cone gauge for a two-form removes $A_{-m}$, leaving $\left(\genfrac{}{}{0ex}{}{D-2}{2}\right)$ transverse particle polarizations. The massive and massless counts are different.

In the closed-string mode expansion, $x^m,p_m$ are center-of-mass canonical variables, while ${\alpha }_k^i,{\tilde{\alpha }}_k^i$ are independent left- and right-moving transverse string oscillators. Their complex conjugates are ${\alpha }_{-k}^i,{\tilde{\alpha }}_{-k}^i$. The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey ${\alpha }_n^{i†}={\alpha }_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor $k>0$, $a_k^i={\alpha }_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j†}]={\delta }_{kl}{\delta }^{ij}$. HenceStarting with $N|0;p⟩=0$, a finite product with $r_{ki}$ creation operators of mode $k$ has eigenvalue ${\sum }_{k,i}k{r}_{ki}$. The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces $N=\tilde{N}$.

There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string $a$, these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts $d(d+1)/2-1$, $d(d-1)/2$, and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes $a=1$. Equivalently, regularized transverse zero-point energy gives $a=(D-2)/24$, and Lorentz consistency fixes the critical dimension of the bosonic string $D=26$.

It follows that the bosonic string mass spectrum isThe ground state has $M_0^2=-8\pi T$ and is a tachyon; level one is massless; for $N\ge 2$ the mass is $M_N=\sqrt{8\pi T(N-1)}$. The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical $L_0$.

A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the symmetric traceless square $S_0^{2}V$ of the massive little group vector space $V={\mathbb{R}}^{D-1}$. The full closed-string level is $S_0^{2}V\otimes S_0^{2}V$. For two symmetric trace-free matrices $S,\tilde{S}$, the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\tilde{S}$ with only its symmetric $ab$ entry nonzero. Their commutator gives the $ab$ antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly $\left(\genfrac{}{}{0ex}{}{D-1}{2}\right)$ particle polarizations and is described by the massive field equation with ${\mu }^2=8\pi T$. At $D=26$ this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components $A_{-i}$ and 276 components $A_{ij}$.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.