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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.

Use a mostly-plus target metric tensor and the standard open-string mode expansion convention ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$. The integer string oscillator level is $N={\sum }_{n\ge 1}{\alpha }_{-n}\cdot {\alpha }_n$, whose eigenvalues are nonnegative integers. Then $L_0=N+{\alpha }^{\prime }{p}^2$, so the string mass-shell condition givesThe ground state is a tachyon, and the level-one state is massless. The physical string state conditions in old covariant string quantization areOnly the positive-mode Virasoro constraints annihilate a ket; their negative-mode partners annihilate the corresponding bra. Imposing both signs on every ket would contradict the nonzero Virasoro central extension.

The central charge comes from normal ordering the quadratic oscillator expressions. Classically their Poisson brackets have no central term. Quantum mechanically, moving annihilation operators through creation operators produces double contractions. From the oscillator commutator one obtains $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$, which gives the noncentral part $(m-n)L_{m+n}$.

The coefficient is fixed by the free-boson Virasoro central term. On a formal zero-momentum vacuum, for $m>0$,The two contraction pairings giveHere ${\eta }_{\mu \nu }{\eta }^{\mu \nu }=D$; the time coordinate contributes one, just as each spatial coordinate does. Comparing with the Virasoro algebra provesThis counts covariant target coordinates. It is distinct from the $D-2$ physical transverse polarizations. Including the central charge of reparameterization ghosts gives $c_{\mathrm{total}}=D-26$; cancellation of the worldsheet Weyl anomaly gives the usual critical dimension of the bosonic string $D=26$.

A general level-one state isThe zero-mode constraint requires $k^2=0$. Since $[L_1,{\alpha }_{-1}^{\mu }]={\alpha }_0^{\mu }$, the remaining nontrivial positive-mode constraint is $k\cdot ϵ=0$; all $L_m$ with $m\ge 2$ act trivially at this level. Thus the polarization vector is transverse to a nonzero null momentum.

There is also a null string state $L_{-1}|0;k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|0;k⟩$. It is physical when $k^2=0$ and orthogonal to every physical polarization. The null-state quotient of a string therefore identifiesChoose $k=(E,E,0,\dots ,0)$. Transversality sets ${ϵ}^1={ϵ}^0$, and the null-state identification removes this common component. The remaining $D-2$ spatial components have positive norm. Hence the physical level-one states form a massless vector with $D-2$ polarizations, or 24 at the critical dimension. The removed component is gauge redundancy. States violating the constraints, including independent timelike negative-norm polarizations, are unphysical; physical null states are removed by the quotient rather than counted as additional particles.

Take $0\le \sigma \le \pi$ and split the coordinates into $a=0,\dots ,p$ along the branes and $I=p+1,\dots ,25$ transverse to them. In conformal gauge the wave equation permits a separated mode $e^{-i\omega \tau }f(\sigma )$ with $f^{\prime \prime }+{\omega }^{2}f=0$. For Neumann boundary conditions at both ends, $f^{\prime }(0)=f^{\prime }(\pi )=0$ selects $\cos m\sigma$ and integer $m$. The zero-frequency solution is independent of $\sigma$, with a constant position and a term linear in time. Therefore the open-string mode expansion isThe canonical momentum density integrated over this interval is $k^a$.

For Dirichlet boundary conditions, subtract the static interpolation between fixed endpoint positions $y_0^I$ and $y_{\pi }^I$. The remaining field vanishes at both ends, selecting $\sin m\sigma$ with integer $m$ and excluding a momentum worldsheet zero mode. The expansion isHere ${\alpha }_m^{I†}={\alpha }_{-m}^I$. There is no factor $i$ in the sine series with this reality convention: changing $m$ to $-m$ reverses both the sine and the denominator. These are the integer-mode expansion with parallel D-brane endpoints; they are not the half-integer modes of a coordinate with a Neumann condition at one endpoint and a Dirichlet condition at the other.

The endpoints can move on $(p+1)$-dimensional worldvolumes but are fixed in the transverse directions. Thus these are open strings attached to D-branes, or stretched between parallel D-branes at $y_0$ and $y_{\pi }$. The tangential momenta describe propagation on the brane; the transverse positions describe its embedding. Let $L=|y_{\pi }-y_0|$. The linear spatial worldsheet zero mode contributes $L^2/(4{\pi }^2{\alpha }^{\prime })$ to $L_0$. Quantization with string intercept one givesThe first term is also the square of the classical stretched-string energy $TL$, since the string tension is $T=1/(2\pi {\alpha }^{\prime })$.

For two coincident D3-branes, label the ends of an oriented open string by $i,j\in \{1,2\}$. The four endpoint combinations give Chan-Paton factors ${\lambda }_{ij}$, which form the full algebra of $2\times 2$ matrices. Under a change of brane basis these transform as $\lambda \mapsto U\lambda U^{-1}$, the adjoint action of $U(2)$. At $L=0$, the first oscillator level is massless. Its tangential polarization supplies a four-dimensional vector $A_a$, while the 22 transverse polarizations supply real adjoint scalar fields ${\Phi }^I$. This is the bosonic D3-brane low-energy field content.

The endpoint matrices multiply in their boundary order in disk amplitudes. The difference between the two orders in the three-vector interaction gives the matrix commutator; the kinematic factor is the one-derivative Yang-Mills theory cubic vertex. Factorization and the vector gauge redundancy then supply the quartic interaction with the same commutator. Equivalently, the leading low-energy action is the dimensional reduction of 26-dimensional Yang-Mills theory, withFor Hermitian scalar fields the last displayed term corresponds to a nonnegative potential, because their commutators are anti-Hermitian. Higher-derivative terms are suppressed at energies well below $1/\sqrt{{\alpha }^{\prime }}$. Ignoring the tachyon, the gauge sector is four-dimensional $U(2)$ Yang-Mills theory, accompanied by 22 adjoint scalar fields. It is not pure Yang-Mills theory, nor the supersymmetric six-scalar D3-brane theory of a ten-dimensional superstring.

Separate the two branes and use the D-brane scalar-position normalizationA Yang-Mills gauge transformation preserves this expectation value only when it is diagonal if $y_1\ne y_2$. Thus the unbroken group is $U(1)\times U(1)$. The scalar field kinetic term supplies a mass for the off-diagonal vectors, sinceSumming over $I$ gives $m_{12}^2=L^2/(4{\pi }^2{\alpha }^{\prime 2})$, exactly the mass of the stretched $N=1$ string. The two diagonal gauge vectors remain massless. This identifies brane separation with an adjoint Higgs mechanism, rather than an explicit deletion of the off-diagonal strings.

For the ground state of an open string joining the two different branes, $N=0$, so the bosonic stretched-string tachyon threshold isAt equality it is massless, and above it the stretched ground state has positive mass squared. The same-brane ground strings still have $M_0^2=-1/{\alpha }^{\prime }$; this calculation removes only the stretched-string instability, not the original instability of the bosonic branes.

Let $h_{\mu \nu }={\partial }_{\mu }{X}^a{\partial }_{\nu }{X}_a$ be the induced worldvolume metric and write $H={\gamma }^{\mu \nu }{h}_{\mu \nu }$. Varying the embedding in the auxiliary-metric brane action and integrating by parts givesThe boundary contribution, including its relative factor, isAt initial and final times one fixes the endpoint configurations or uses variations of compact support. At a spatial boundary, free target directions require Neumann boundary conditions $n_{\mu }{\gamma }^{\mu \nu }{\partial }_{\nu }{X}^a=0$; fixed directions require Dirichlet boundary conditions $\delta X^a=0$. Mixed conditions must make this boundary pairing vanish. A closed brane has no spatial boundary. Since the auxiliary worldvolume metric enters without derivatives, its variation produces no extra boundary term.

Using $\delta \sqrt{-\gamma }=-(1/2)\sqrt{-\gamma }{\gamma }_{\mu \nu }\delta {\gamma }^{\mu \nu }$, variation of the inverse worldvolume metric givesTaking the trace yields $(p-1)(H-p-1)=0$. For $p\ne 1$, this forces $H=p+1$, and substitution gives ${\gamma }_{\mu \nu }=h_{\mu \nu }$. For a nondegenerate timelike embedding, eliminating the auxiliary worldvolume metric therefore gives twice the worldvolume area, with the overall physical brane tension supplied by the action normalization.

For $p=1$, the traced metric equation is an identity, and the remaining equation only says $h_{\mu \nu }=(H/2){\gamma }_{\mu \nu }$. It fixes the worldsheet metric up to a Weyl transformation, rather than determining it uniquely. Indeed $\sqrt{-\gamma }{\gamma }^{\mu \nu }$ is Weyl invariant precisely in two worldvolume dimensions, and the constant term vanishes precisely at $p=1$. This is the Weyl-invariance exception for the string among branes. In a nondegenerate interior it gives the familiar classical equivalence of Polyakov and Nambu–Goto actions; degeneracies at a free string endpoint must be treated through the original equations.

For the open string, choose conformal gauge ${\gamma }_{\mu \nu }=\mathrm{diag}(-1,1)$ on $0\le \sigma \le \pi$. The embedding equation is the wave equation, and NN means ${\partial }_{\sigma }{X}^a=0$ at both spatial endpoints. For these operator formulas restore the conventional overall normalization $S=-TI/2$, with $T=(2\pi \alpha \prime {)}^{-1}$; the overall factor does not change the preceding classical equations. Its open-string mode expansion isReality requires $({\alpha }_n^a{)}^{†}={\alpha }_{-n}^a$. Canonical quantization givesThe metric equation is the vanishing of the worldsheet stress tensor: $({\partial }_{\tau }X+{\partial }_{\sigma }X{)}^2=({\partial }_{\tau }X-{\partial }_{\sigma }X{)}^2=0$. The two endpoint-compatible mode expansions contain the same string oscillator family. Their quadratic coefficients are the classical Virasoro constraints.

For the operators, use normal ordering with positive-index string oscillators as annihilators:Thus $({\partial }_{\tau }X\pm {\partial }_{\sigma }X{)}^2$ has modes $4{\alpha }^{\prime }{L}_m$ before the quantum ordering correction. The physical-state Virasoro conditions for an open string areOne imposes only the positive modes on kets, with the adjoint conditions on bras, as in Gupta-Bleuler quantization. Requiring every positive and negative mode to annihilate the same state would conflict with the Virasoro central extension. The intercept $a$ is the zero-mode ordering constant. The standard critical bosonic string theory has $d=26$ and $a=1$; in light-cone gauge in string theory the transverse zero-point energy gives $a=(d-2)/24$, while full anomaly-free Lorentz or BRST quantization fixes the critical values. The mass constraint is then ${\alpha }^{\prime }{M}^2=N-a$, with $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$.

To compute the Virasoro algebra, commute a quadratic generator with one string oscillator:These identities and the Jacobi identity imply that $[L_m,L_n]-(m-n)L_{m+n}$ commutes with every string oscillator and with the center-of-mass coordinates and momenta. Mode number permits a scalar term only for $m+n=0$. Its coefficient follows from the formal zero-momentum Fock vacuum, on which $L_0=0$. For $m>0$,The two possible string oscillator contractions giveThe timelike target coordinate still contributes one to this central charge: its two metric signs cancel in ${\eta }_{ab}{\eta }^{ab}=d$. ConsequentlyThis is the free-boson Virasoro central term. It vanishes for the three global conformal modes $m=0,\pm 1$.

To keep the intercept convention separate, define ${\mathcal{L}}_m=L_m-a{\delta }_{m,0}$. The Virasoro zero-mode shift changes the displayed central term toThe matter central charge here is $d$, not $d-2$. Covariant worldsheet ghost fields contribute $-26$; their inclusion cancels the anomaly at $d=26$, while the intercept is handled by the appropriate zero-mode and physical-state convention.

Solving the massive subsidiary conditions. In light-cone coordinates, use $d{s}^2=-2d{X}^{+}d{X}^{-}+{\delta }_{IJ}d{X}^{I}d{X}^J$ and write $d=D-2$. The divergence condition readsThus, when ${\partial }_{-}$ is invertible,First apply this with $n=-$, then with $n=I$, and finally with $n=+$; symmetry supplies the mixed components already determined. The trace condition becomesThe independent components are $h_{--}$, $h_{-I}$ and the trace-free part of $h_{IJ}$. Under transverse rotations they form a scalar representation, a vector representation, and a symmetric traceless rank-two tensor. ThereforeThis is the light-cone decomposition of a massive spin-two field for $D\ge 3$. In $D=2$, the trace and divergence give $h_{+-}=h_{++}=0$ and ${\partial }_{+}{h}_{--}=0$. The massive wave equation then forces $h_{--}=0$, so there are no polarizations, consistent with the zero value of the printed count. The massive particle little group is $SO(D-1)=SO(d+1)$, and its symmetric traceless square branches asThe mixed components with the extra direction give the vector; one independent trace combination gives the scalar. These are exactly the polarizations of a massive spin-two field. The remaining independent components retain the massive Klein-Gordon equation.

Transverse bosonic modes and mass levels. The variables ${\alpha }_k^I$ are Fourier amplitudes of the physical transverse open-string mode expansion. Classically, reality requires ${\alpha }_{-k}^I=({\alpha }_k^I{)}^{\ast }$. The symplectic term in the action fixes their quantum commutators:With string tension convention ${\alpha }^{\prime }=(2\pi T{)}^{-1}$, the zero mode of the constraint givesThe longitudinal nonzero modes have already been removed in light-cone gauge in string theory. Quantum normal ordering introduces the string intercept $a$, giving the open bosonic string mass spectrumEach bosonic occupation number is a nonnegative integer, so $N$ is a nonnegative integer weighted by oscillator frequency.

Suppressing the common momentum label, the lowest light-cone levels of an open bosonic string areThe oscillator vacuum is annihilated by every positive ${\alpha }_k^I$. At level one there are only $d$ vector polarizations. For a Lorentz-consistent vector, these are the transverse polarizations of a massless particle, transforming under the rotation part of its massless particle little group. A massive vector would need $D-1$ polarizations, including a scalar under $SO(d)$ that is absent here. Thus the first bosonic vector level must be massless, fixing $a=1$.

At level two the commuting creation operators give a symmetric square. Its scalar trace and symmetric traceless square, together with the mode-two vector, are the massive-spin-two decomposition above. In the consistent bosonic theory they form one massive spin-two field with $m^2=1/{\alpha }^{\prime }$. The covariant equations describe its propagation while eliminating the redundant components. For $D=26$ the transverse counts are $300+24=324$, the symmetric traceless rank-two tensor dimension of $SO(25)$.

Half-integer fermionic modes. The Neveu–Schwarz sector has antiperiodic worldsheet Majorana fermions. Its Neveu–Schwarz fermionic oscillators obeyThe oscillator vacuum satisfies ${\alpha }_k^I|0;p⟩=0$ for $k>0$ and $b_r^I|0;p⟩=0$ for $r>0$. A negative fermion mode is a fermionic creation operator for a transverse worldsheet excitation. The Neveu–Schwarz level operator and mass condition areThe smallest positive frequency is $1/2$, so the only first-excited states areThe same vector-polarization argument requires them to be massless in Lorentz-consistent quantization, fixing $a_{\mathrm{NS}}=1/2$.

At $N=1$, ${\alpha }_{-1}^I|0;p⟩$ is a vector, while $b_{-1/2}^I{b}_{-1/2}^J|0;p⟩$ is the exterior square: interchanging the indices changes the sign and equal indices give zero. Together they branch from an antisymmetric tensor:Thus the Neveu–Schwarz level-one massive tensor has $(D-1)(D-2)/2$ polarizations and mass squared $1/(2{\alpha }^{\prime })$. It differs from spin two because the two-fermion tensor is antisymmetric and has neither the symmetric trace-free representation nor its scalar trace. At $D=10$, the count is $8+28=36$, compared with $44$ for a massive spin-two field. The specified states are before the GSO projection; the usual tachyon-removing GSO projection also removes this integer level.

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

An infinitesimal Lorentz transformation is $\delta X^{\mu }={\omega }^{\mu }{}_{\nu }{X}^{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. Applying Noether theorem to this continuous symmetry gives the conserved Lorentz currentIts Noether charge is $M^{\mu \nu }={\int }_0^{\pi }d\sigma J_{\tau }^{\mu \nu }$. Substituting the open-string mode expansion and using orthogonality of the cosine modes yieldsThe first term is orbital angular momentum; the sum is the contribution of the string oscillators.