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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.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.

For covariant quantization of the bosonic string,All other commutators between independent canonical variables vanish. The covariant Fock vacuum obeys ${\alpha }_k^m|0;p⟩=0$ for every $k>0$ and every $m$. Normal ordering moves these annihilating string oscillators to the right and givesThen ${\hat{N}}_{\mathrm{cov}}|0;p⟩=0$. Its commutator $[{\hat{N}}_{\mathrm{cov}},{\alpha }_{-k}^m]=k{\alpha }_{-k}^m$ grades a Fock state basis by $N=\sum k{n}_{km}$, although its inner product is indefinite because of the timelike oscillator.

The quantum Virasoro algebra has central charge $D$:The Virasoro central extension comes from commutators needed to order infinite oscillator sums; it is a quantum effect absent from the classical Poisson brackets. Replacing a classical Poisson bracket by a commutator cannot recover that term without a regularized ordering calculation.

If a state were annihilated by every nonzero ${\hat{L}}_j$, the $j=1,k=-1$ commutator would imply ${\hat{L}}_0|\psi ⟩=0$. The $j=2,k=-2$ commutator would then imply $(D/2)|\psi ⟩=0$. For $D>0$, only the zero vector is annihilated by all nonzero Virasoro constraints. This explains why only positive modes annihilate a physical string state.

The vacuum is a physical string state when $p^2=2\pi Ta$. At level one all states have the form $|A;p⟩=A_m{\alpha }_{-1}^m|0;p⟩$. Since $[{\hat{L}}_j,{\alpha }_k^m]=-k{\alpha }_{j+k}^m$, the positive-mode Virasoro constraints giveThus the complete level-one conditions and norm arewith the common vacuum normalization suppressed.

For $a>1$, choose spacelike momentum $p^m=(0,k,0,\dots )$, $k^2=2\pi T(a-1)$. A purely timelike polarization has $A\cdot p=0$ and norm $-1$. Hence a negative-norm physical string state exists when $a>1$.

For $a<1$, the mass-shell condition gives $M^2=2\pi T(1-a)>0$. In a rest frame, $A\cdot p=0$ forces $A_0=0$, leaving $D-1$ positive-norm vector-particle polarizations, those of a massive vector.

For $a=1$ and nonzero null momentum, $A\cdot p=0$ leaves a null direction $A_m\propto p_m$. The corresponding null string state is $p\cdot {\alpha }_{-1}|0;p⟩=\sqrt{\pi T}{\hat{L}}_{-1}|0;p⟩$. It is orthogonal to every physical string state because ${\hat{L}}_1$ annihilates them. Quotienting by this gauge redundancy, $A_m\sim A_m+\zeta p_m$, leaves $D-2$ positive vector-particle polarizations. Thus the level-one spectrum agrees with light-cone gauge in string theory at $a=1$. This level-one argument alone does not establish consistency or absence of negative norms at all higher levels.

In the displayed version of light-cone gauge in string theory, the oscillators have $D-2$ transverse components andThe center-of-mass Poisson brackets remain $\{x^m,p_n\}={\delta }_n^m$, and the transverse oscillator Poisson brackets are $\{{\alpha }_j^a,{\alpha }_k^b\}=-ij{\delta }^{ab}{\delta }_{j+k,0}$. Other independent brackets vanish. With $\hslash =1$, canonical quantization gives the canonical commutation relationsThe oscillator Fock vacuum at momentum $p$ is defined by ${\alpha }_k^a|0;p⟩=0$ for $k>0$. It is the ground state of one string, rather than the empty spacetime vacuum. Set $a_k^a={\alpha }_k^a/\sqrt{k}$ for $k>0$. The normal ordering prescription gives the string level operatorA Fock state basis is obtained by applying ${\prod }_{k,a}((a_k^a{)}^{†}{)}^{n_{ka}}/\sqrt{n_{ka}!}$ to $|0;p⟩$, with string level operator eigenvalue $N={\sum }_{k,a}k{n}_{ka}$. In particular, its level-one states areThey transform as the transverse vector of the little group rotation subgroup $SO(D-2)$, precisely the vector-particle polarizations of a massless vector. A massive vector would instead require $D-1$ vector-particle polarizations. This conclusion uses a quantization compatible with the Lorentz group of the bosonic string theory.

The classical mass constraint alone has no quantum zero-point energy shift. Its quantum version includes the normal-ordering constant of a string $a$:Masslessness at level one fixes $a=1$, soThus the ground state is a tachyon. Without the quantum ordering shift, the displayed classical constraint would give $M_1^2=2\pi T$ and would not support the stated massless interpretation. In the usual transverse vacuum regularization, $a=(D-2)/24$; consistency with $a=1$ also gives the critical dimension of string theory $D=26$.

After imposing conformal gauge, transformationsremain, with a compensating Weyl transformation. For a closed string, the functions obey the corresponding periodicity conditions. Introduce target-space light-cone coordinates $X^{\pm }=(X^0\pm X^{D-1})/\sqrt{2}$. When $p^{+}\ne 0$, the equations make $X^{+}$ a sum of left- and right-moving functions, and the two residual reparameterizations can setThis light-cone gauge in string theory removes all nonzero oscillators of $X^{+}$. Any residual constant shifts merely choose the worldsheet origin, so the local conformal freedom is fixed.

In conformal gauge and light-cone gauge in string theory $X^{+}=\kappa \tau$, the transverse part of the string sigma model in a spacetime metric reduces toThe omitted longitudinal term is a total time derivative. Each transverse coordinate obeys a massive wave equation. If the spatial period is $L$, the longitudinal momentum is $p^{+}=T\kappa L$ in this normalization. Keeping the gauge slope explicit prevents missing a factor of $p^{+}$ when changing conventions.

Before the GSO projection, the Neveu–Schwarz sector has string oscillator levels $0,1/2,1,3/2$, with physical multiplicities $1,8,36,128$ in ten-dimensional light-cone gauge in string theory. The eight transverse worldsheet Majorana fermions supply eight single ${\psi }_{-1/2}$ states. Level one has eight ${\alpha }_{-1}$ states and $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$ pairs of distinct ${\psi }_{-1/2}$ modes. Level $3/2$ has $8$ single ${\psi }_{-3/2}$ modes, $8\cdot 8$ mixed boson–fermion modes and $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$ triples. Their squared masses are respectively $-1/(2{\alpha }^{\prime }),0,1/(2{\alpha }^{\prime }),1/{\alpha }^{\prime }$. In the Ramond sector, the Ramond zero-mode Clifford algebra gives a sixteen-dimensional ground space. At level one, ${\alpha }_{-1}^i$ and ${\psi }_{-1}^i$ each provide $8\cdot 16$ states, hence multiplicities $16,256$ at squared masses $0,1/{\alpha }^{\prime }$. No half-integer Ramond levels occur. The GSO projection keeps odd NS fermion number and one total Ramond chirality, giving $8$ bosons and $8$ fermions at zero mass, then $128$ of each at the first massive level.