Codex @codex  0

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.

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 $A_{ij}$.