= Solution
Use $x^\pm=(x^0\pm x^{D-1})/\sqrt2$, transverse coordinates $x^i$, $i=1,\ldots,d=D-2$, and the <Minkowski metric>
$$
ds^2=-2dx^+dx^-+dx^i dx^i,\qquad
p^2=-2p_+p_-+p_i p_i.
$$
The <relativistic particle phase-space action> becomes
$$
I=\int dt\left[\dot x^+p_++\dot x^-p_-+\dot x^i p_i
-\frac e2(-2p_+p_-+p_i^2+\mu^2)\right].
$$
In the <light-cone gauge> $x^+=t$, solve the <mass-shell condition> for $p_+$, assuming $p_-\ne0$. The reduced <phase-space action> is $\int dt(\dot x^-p_-+\dot x^ip_i-H)$, with
$$
\boxed{H=-p_+=-\frac{p_i^2+\mu^2}{2p_-}
=\frac{p_i^2+\mu^2}{2p^+},\qquad p^+=-p_->0.}
$$
The last equality selects the future-directed <momentum> sector and makes positivity transparent. With $\hbar=1$ and $p_a=-i\partial_a$, the <Schrodinger equation> is
$$
i\partial_+\Psi=-\frac{-\Delta_\perp+\mu^2}{2(-i\partial_-)}\Psi.
$$
The inverse acts only on <Fourier modes> with nonzero $p_-$. Multiplication by $2p_-$ gives $2\partial_-\partial_+\Psi=(\Delta_\perp-\mu^2)\Psi$. Therefore
$$
\boxed{(\square_D-\mu^2)\Psi=0,\qquad
\square_D=-2\partial_+\partial_-+\Delta_\perp.}
$$
The <light-cone Hamiltonian> thus gives the same <Klein-Gordon equation> as covariant quantization.
For the <massive two-form field>, take $\mu\ne0$. Apply $\partial^n$ to its field equation. Antisymmetry of $F_{mnp}$ makes $\partial^n\partial^m F_{mnp}=0$, so
$$
\partial^n A_{np}=0.
$$
Expanding $F=dA$, the other divergence terms vanish by this condition, leaving \b[$(\square_D-\mu^2)A_{np}=0$]. The <light-cone decomposition of a massive two-form> makes its dependent components explicit. The divergence equation is
$$
-\partial_- A_{+n}-\partial_+A_{-n}+\partial_iA_{in}=0.
$$
Taking $n=-$ and $n=i$, respectively, gives
$$
\boxed{A_{+-}=-\partial_-^{-1}\partial_iA_{-i},\qquad
A_{+i}=-\partial_-^{-1}\partial_+A_{-i}
+\partial_-^{-1}\partial_jA_{ji}.}
$$
The $n=+$ equation follows from these expressions: the two terms containing $\partial_+\partial_iA_{-i}$ cancel and $\partial_i\partial_jA_{ji}=0$. Consequently \b[$A_{-i}$ and $A_{ij}$ are independent], each satisfying the <Klein-Gordon equation> with mass $\mu$. The number of independent <particle polarizations> is
$$
\boxed{d+\binom d2=\binom{D-1}{2}.}
$$
This 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 $\binom{D-2}{2}$ 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 $\alpha_k^i,\widetilde\alpha_k^i$ are independent left- and right-moving transverse <string oscillators>. Their complex conjugates are $\alpha_{-k}^i,\widetilde\alpha_{-k}^i$. The two zero-mode <Lagrange multipliers> impose the remaining <mass-shell condition> and <closed-string level matching>. The <string level operators> are
$$
N=\sum_{k>0}\alpha_{-k}\cdot\alpha_k,\qquad
\widetilde N=\sum_{k>0}\widetilde\alpha_{-k}\cdot\widetilde\alpha_k.
$$
Their quantum definitions use <normal ordering>. The symplectic terms in the <phase-space action> give
$$
[x^m,p_n]=i\delta^m_n,\qquad
[\alpha_m^i,\alpha_n^j]=m\delta^{ij}\delta_{m+n,0},\qquad
[\widetilde\alpha_m^i,\widetilde\alpha_n^j]=m\delta^{ij}\delta_{m+n,0},
$$
with all brackets between distinct sectors zero. The nonzero-index <string oscillators> obey $\alpha_n^{i\dagger}=\alpha_{-n}^i$ and similarly for the right-moving sector. Define the momentum-labelled <oscillator vacuum> by
$$
\alpha_k^i|0;p\rangle=\widetilde\alpha_k^i|0;p\rangle=0\quad(k>0),\qquad
p_m|0;p\rangle=p_m^{\mathrm{label}}|0;p\rangle.
$$
For $k>0$, $a_k^i=\alpha_k^i/\sqrt{k}$ has $[a_k^i,a_l^{j\dagger}]=\delta_{kl}\delta^{ij}$. Hence
$$
N=\sum_{k,i}k\,a_k^{i\dagger}a_k^i,\qquad
[N,\alpha_{-k}^i]=k\alpha_{-k}^i.
$$
Starting with $N|0;p\rangle=0$, a finite product with $r_{ki}$ <creation operators> of mode $k$ has <eigenvalue> $\sum_{k,i}kr_{ki}$. The <Fock space> is generated by these products; \b[both level operators have nonnegative integer <eigenvalues>]. This establishes the <integer string oscillator level> property. Subtracting their physical zero-mode constraints enforces $N=\widetilde N$.
There is a distinction between the displayed classical zero modes and their quantum constraints. With the <normal-ordering constant of a string> $a$, these are
$$
\frac{p^2}{8\pi T}+N-a=0,\qquad
\frac{p^2}{8\pi T}+\widetilde N-a=0.
$$
At the <massless first closed-string level>, the states are
$$
\alpha_{-1}^i\widetilde\alpha_{-1}^j|0;p\rangle.
$$
Their 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> is
$$
\boxed{M_N^2=-p^2=8\pi T(N-1)=\frac4{\alpha'}(N-1),\qquad
\alpha'=\frac1{2\pi T},\quad N=\widetilde N.}
$$
The ground state has $M_0^2=-8\pi T$ and is a <tachyon>; level one is massless; for $N\geq2$ 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 $L_0$.
\b[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 are
$$
\alpha_{-2}^i|0\rangle,\qquad
\alpha_{-1}^i\alpha_{-1}^j|0\rangle.
$$
They have $d+d(d+1)/2=(D-1)D/2-1$ components and assemble into the <symmetric traceless square> $S^2_0V$ of the massive <little group> vector space $V=\mathbb R^{D-1}$. The full closed-string level is $S^2_0V\otimes S^2_0V$. For two symmetric trace-free matrices $S,\widetilde S$, the map
$$
(S,\widetilde S)\longmapsto [S,\widetilde S]_{ab}
=S_{ac}\widetilde S_{cb}-\widetilde S_{ac}S_{cb}
$$
is an equivariant map onto antisymmetric matrices. To verify surjectivity, take $S$ diagonal with distinct entries in positions $a,b$ and $\widetilde S$ 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 $\binom{D-1}{2}$ <particle polarizations> and is described by the massive field equation with \b[$\mu^2=8\pi T$]. At $D=26$ this gives 300 <particle polarizations>, consisting in <light-cone coordinates> of 24 components $A_{-i}$ and 276 components $A_{ij}$.
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