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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.

The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.

For an ordinary open string, the boundary conditions relate the two chiral sectors. With ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{k}^{\mu }$, their independent Fourier generators areThe classical oscillator Poisson brackets $\{{\alpha }_m^{\mu },{\alpha }_n^{\nu }\}=-im{\eta }^{\mu \nu }{\delta }_{m+n,0}$ give the classical Virasoro constraint algebraThus they are first-class constraints. In the quantum theory one applies normal ordering to the product and uses $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$. The oscillator commutator givesFor example, for $m>0$ the extra vacuum term is $(D/2){\sum }_{r=1}^{m-1}r(m-r)=D(m^3-m)/12$. This exhibits the Virasoro central extension rather than concealing it in a classical constraint equation. The matter central charge is $D$; the worldsheet ghost fields have charge $-26$. Quantum consistency sets $D=26$ and the string intercept to one.

In old covariant string quantization the physical-state Virasoro conditions for an open string arePhysical string states are then identified modulo physical null string states. One does not impose both positive and negative generators as annihilation conditions: their central extension would obstruct that prescription. The resulting null-state quotient of a string agrees with the transverse light-cone spectrum at nonzero momentum.

At $N=0$, the oscillator vacuum $|k⟩$ is a single spacetime scalar field with ${\alpha }^{\prime }{M}^2=-1$: the bosonic tachyon. At $N=1$, write $|\Psi ⟩={ϵ}_{\mu }{\alpha }_{-1}^{\mu }|k⟩$. The zero-mode equation gives $k^2=0$, and $L_1$ gives $k\cdot ϵ=0$. The null descendant $L_{-1}|k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|k⟩$ identifies $ϵ\sim ϵ+\lambda k$. This leaves 24 physical polarizations of a massless gauge vector.

At $N=2$, use the level-two open-string polarization decompositionThe mass is $M^2=1/{\alpha }^{\prime }$. The oscillator commutators give all the nontrivial positive-mode conditions:Higher positive modes annihilate this level. To see the quotient explicitly, the 25 transverse-parameter null states $L_{-1}(\xi \cdot {\alpha }_{-1}|k⟩)$, with $k\cdot \xi =0$, shiftThe additional level-two scalar Virasoro null state $(L_{-2}+\frac{3}{2}{L}_{-1}^2)|k⟩$ shiftsSubstitution into the two physical equations verifies the null-state conditions; the second residual is $(D-26)/2$. Because $k$ is timelike, these null shifts span all components of $b$, so choose a representative with $b=0$. The physical conditions then say $k^{\mu }{h}_{\mu \nu }=0$ and $h^{\mu }{}_{\mu }=0$. In the rest frame this is a traceless second-rank tensor in 25 spatial dimensions, with $25\cdot 26/2-1=324$ polarizations. The light-cone basis independently gives $24$ states ${\alpha }_{-2}^i|k⟩$ and $24\cdot 25/2=300$ symmetric states ${\alpha }_{-1}^i{\alpha }_{-1}^j|k⟩$. Hence

For the fermionic extension, an explicit convention prevents ambiguity in the signs. Keep worldsheet signature $(-,+)$ and useTake $\psi =({\psi }_1,{\psi }_2{)}^T$ and the constant parameter $ϵ=({ϵ}_1,{ϵ}_2{)}^T$ to be real Grassmann-odd Majorana components, with $\bar{\psi }={\psi }^T{\rho }^0$. Put ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$. Apart from the common factor $1/(4\pi {\alpha }^{\prime })$, the combined Lagrangian isThe rigid transformations can be writtenor, in components,For the ${ϵ}_2$ part, the bosonic variation is $i{ϵ}_2({\partial }_{+}{\psi }_1\cdot {\partial }_{-}X+{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1)$, while the fermionic variation is $-i{ϵ}_2{\partial }_{-}X\cdot {\partial }_{+}{\psi }_1+i{ϵ}_2{\psi }_1\cdot {\partial }_{+}{\partial }_{-}X$. Anticommuting the constant parameter past the fermion is essential in the second term. The first terms cancel. The other chiral component works in the same way, giving the chiral proof of rigid worldsheet supersymmetryThus the action is invariant off shell up to a boundary term. The transformation is a genuine supersymmetry: on $X$ its commutator for parameters $ϵ,\zeta$ is the translationOn $\psi$ the same translation follows using ${\partial }_{+}{\psi }_1=0$ and ${\partial }_{-}{\psi }_2=0$; the remaining cross terms are proportional to these Dirac equations. This is on-shell closure of rigid worldsheet supersymmetry. On a surface with boundary, compatible endpoint conditions must make the boundary flux vanish. On a compact worldsheet, a constant parameter must preserve the spin structure; the local identity alone does not make a constant transformation compatible with every antiperiodic sector.

The fermions contribute to the worldsheet stress-energy tensor and produce a fermionic superconformal current. In holomorphic normalization with ${\psi }^{\mu }(z){\psi }^{\nu }(w)\sim {\eta }^{\mu \nu }/(z-w)$,The free-field contractions give $G(z)G(w)\sim (2c/3)/(z-w{)}^3+2T(w)/(z-w)$ and conformal weight $3/2$ for $G$, where $c=D+D/2=3D/2$. Consequently the modes obey the N=1 super-Virasoro algebraDropping the central terms gives the classical graded constraint algebra. Integer $r$ gives the Ramond sector; half-integer $r$ gives the Neveu–Schwarz sector. In the displayed conformal-plane convention $G_0^2=L_0-c/24$ in the Ramond sector. Shifting to the Ramond vacuum-normal-ordered zero mode gives the usual Ramond supercurrent zero-mode square $G_0^2=L_0^{(R)}$.

Rigid symmetry of matter already supplies these currents, but promoting their vanishing to new gauge constraints requires coupling to worldsheet supergravity and varying its worldsheet gravitino. In superconformal gauge one retains $T=G=0$. The fermionic current constraints and their physical-state conditions extend the bosonic Virasoro conditions. The ghost charges are $-26+11=-15$, so cancellation with $3D/2$ gives the critical dimension of the RNS superstring $D=10$; the NS and Ramond intercepts are respectively $1/2$ and zero. This distinguishes the global supersymmetry established above from its locally gauged string completion.

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.