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The canonical commutation relations imply $[N,{\alpha }_{-k}^i]=k{\alpha }_{-k}^i$ for $N={\sum }_{k>0}{\alpha }_{-k}\cdot {\alpha }_k$. With zero level assigned to the oscillator vacuum, a Fock space basis state containing $r_{ki}$ creation operators of each mode has level ${\sum }_{k,i}k{r}_{ki}$. The level is therefore a nonnegative integer; it counts oscillator excitation weighted by frequency, not simply the number of creation operators.

The first levels are the oscillator vacuum, ${\alpha }_{-1}^I|0⟩$, and ${\alpha }_{-2}^I|0⟩,{\alpha }_{-1}^I{\alpha }_{-1}^J|0⟩$. At level two the transverse vector representation and symmetric square assemble into the massive symmetric traceless square of one extra rotation dimension.

Antiperiodic worldsheet Majorana fermions give transverse modes $b_r^I$, $r\in \mathbb{Z}+\frac{1}{2}$, with $\{b_r^I,b_s^J\}={\delta }^{IJ}{\delta }_{r+s,0}$. Positive modes annihilate the oscillator vacuum; negative modes are fermionic creation operators.

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 Schrodinger 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 $A_{ij}$.

For the relativistic particle phase-space action, the first-class constraint $\phi =(p^2+m^2)/2$ generatesIndeed the integrand varies by $\frac{1}{2}d[ϵ(p^2-m^2)]/dt$. The canonical gauge transformation is an invariance when the gauge parameter vanishes at fixed temporal endpoints, or when all fields and the parameter are periodic. The boundary restriction matters for the proper-time modulus.

Normalize the worldline interval to $[0,1]$. Thenis invariant because $\delta s=ϵ(1)-ϵ(0)=0$. Every allowed $e$ in its orbit can be written $e(t)=s+\dot{ϵ}(t)$: set $ϵ(t)={\int }_0^t[e(u)-s]du$. Thus $s$ remains a gauge-invariant integration variable, not another removable nonconstant mode. For a worldline circle the constant gauge parameter is a residual zero mode. Without the endpoint restriction, the assertion that $s$ is invariant would not hold.

The worldline gauge-orbit determinant is the Jacobian from gauge-orbit coordinates $ϵ$ to the nonconstant part of $e$ is the Faddeev-Popov determinant of ${\partial }_t$. Equivalently, the gauge-fixing identity has the formThe determinant is taken between the appropriate boundary-condition spaces, with the modulus removed; on a circle the prime also removes the constant parameter. Gauge fixing therefore leaves a factor $\det {}^{\prime }{\partial }_t$ and a modulus measure, after division by any residual gauge volume. Even though this determinant is field independent in the present Abelian example, it is the required change-of-variables Jacobian. A Grassmann integral over the Faddeev-Popov ghosts exponentiates it:The overall determinant phase depends on the integration convention and can be absorbed into normalization. Zero modes and the same endpoint restrictions must be treated separately rather than included in an invertible determinant.

For the free-ended open string, take $0\le \sigma \le \pi$. A canonical cosine expansion at a fixed time isIt implements the Neumann boundary conditions and has ${\alpha }_n^{†}={\alpha }_{-n}$. With ${\alpha }_0^m=p^m/\sqrt{\pi T}$, its Nambu-Goto phase-space action, up to a total time derivative, isReality requires ${\lambda }_{-n}={\lambda }_n^{\ast }$. Numerical factors can be absorbed into these Lagrange multipliers. In this covariant quantization of the bosonic string the oscillators retain all $D$ spacetime components, in contrast to the transverse oscillators in the preceding solution. The canonical commutation relations areThe oscillator vacuum $|0;p⟩$ is annihilated by ${\alpha }_n^m$ for $n>0$. Its momentum label will sometimes be suppressed.

Define the matter Virasoro algebra generators using normal ordering:No additive intercept is included in this definition of $L_0$. For $n>0$ the indices of the two factors in each term add to $n$; they cannot both be negative. After normal ordering there is a positive-mode annihilation operator on the right, possibly accompanied by the zero mode. Hence $L_n|0;p⟩=0$ for every $n>0$. In $L_0$, commuting positive modes past negative modes formally adds $\frac{D}{2}{\sum }_{k>0}k$. This divergent constant needs a prescription, and a finite shift is an ordering ambiguity. Our convention instead puts the physical string intercept into the constraint $L_0-a$.

With this convention the matter Virasoro algebra isA different additive constant in $L_0$ would change the linear-in-$m$ central term, so stating the convention is essential.

For the worldsheet ghost fields, useand choose a ghost oscillator vacuum withThen $c_0|0{⟩}_{\mathrm{gh}}\ne 0$ since $\{b_0,c_0\}=1$. This choice specifies the ghost zero-mode doublet; it is not a claim that both zero modes annihilate one state. With the printed ghost Virasoro zero-mode convention, $L_0^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ and $L_m^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ for $m>0$. The latter follows by putting positive ghost modes on the right; a possible contraction requires $m=0$ and is absent here.

Apply the supplied BRST charge to the matter state times this ghost oscillator vacuum. Terms with a rightmost $c_m$, $m>0$, vanish, as do the positive-mode ghost generators. ThusThese one-ghost states are independent, as can also be seen by applying $b_0,b_m$. Therefore the BRST physical-state constraints of an open string areFor the matter oscillator vacuum, $L_0={\alpha }^{\prime }{p}^2$, so $p^2=1/{\alpha }^{\prime }$ and $M^2=-1/{\alpha }^{\prime }=-2\pi T$: the physical ground state is a tachyon. The momentum must satisfy this equation; the zero-momentum oscillator vacuum by itself would not be BRST-closed.

Matter and ghost generators commute with one another. Add their two algebras and write ${\mathcal{L}}_m=L_m+L_m^{\mathrm{gh}}$. The given ghost constant must be retained:Define the shifted generators ${\hat{\mathcal{L}}}_m={\mathcal{L}}_m-{\delta }_{m,0}$. ThenBRST nilpotence requires cancellation of the anomalous central term in this shifted constraint algebra, together with the intercept one already present in the charge. At $m=2,n=-2$ the remaining anomalous coefficient is $(D-26)/2$, so $D=26$. At this value the shifted total generators obey the Witt algebra. The unshifted ${\mathcal{L}}_m$ still have the displayed linear zero-mode shift; it must not be silently discarded.

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 PDF's displayed zero-mode term has no derivative. Read literally, $d_0\cdot d_0$ vanishes for classical Grassmann variables and supplies no zero-mode symplectic structure. The subsequent canonical-algebra requests therefore require the standard kinetic term $\frac{i}{2}{d}_0\cdot {\dot{d}}_0$. We use that intended correction explicitly; the rest of the displayed action fixes the nonzero-mode normalization.

The Ramond level operator, with vacuum-annihilating normal ordering, isThe canonical oscillator relations, for transverse indices $I,J$, areThe hermiticity convention is ${\alpha }_k^{†}={\alpha }_{-k}$ and $d_k^{†}=d_{-k}$. The commuting bosonic zero mode is supplied by the center-of-mass momentum. For $k>0$, define $a_k^I={\alpha }_k^I/\sqrt{k}$ and $f_k^I=d_k^I$. Their bosonic occupation numbers are $0,1,2,\dots$ and their fermionic occupation numbers are $0,1$. Therefore, on the Fock space generated from an oscillator vacuum,Equivalently, creation operators raise the level by $k$, since $[N_R,{\alpha }_{-k}^I]=k{\alpha }_{-k}^I$ and $[N_R,d_{-k}^I]=k{d}_{-k}^I$. The zero modes commute with $N_R$ and do not change the level. The multiplier imposesso the $N=0$ states are massless. In the Ramond sector the bosonic and fermionic oscillator zero-point contributions cancel, consistently with the stated zero intercept. The massless ground states are spacetime spinors, as the Ramond zero-mode Clifford algebra now shows.

Normalize ${\gamma }^I=\sqrt{2}{\hat{d}}_0^I$. ThenLet $v=|0⟩\ne 0$. Each positive-frequency bosonic annihilator commutes with ${\gamma }^I$, while each fermionic annihilator anticommutes with it. Applying either annihilator to ${\gamma }^{I}v$ therefore gives zero. Thus all eight ${\gamma }^{I}v$ are oscillator vacua. For real $a_I$, the hermitian operator $A={\sum }_I{a}_I{\gamma }^I$ satisfies $A^2=({\sum }_I{a}_I^2)1$, soThis proves the real independence of Clifford-generated vectors, and hence their linear independence over $\mathbb{R}$.

The same argument applies to the nonzero vacuum ${\gamma }^{1}v$, because $({\gamma }^1{)}^2=1$. It gives eight real-linearly independent oscillator vacua ${\gamma }^I{\gamma }^{1}v$. They include $v$ itself, at $I=1$. Products of two zero modes preserve vacuum annihilation just as products of one do.

For the chirality matrix ${\gamma }_9={\gamma }^1{\gamma }^2\cdots {\gamma }^8$, reversing eight anticommuting factors introduces $(-1{)}^{8\cdot 7/2}=1$. HenceMoving any ${\gamma }^I$ through the other seven factors also gives ${\gamma }_9{\gamma }^I=-{\gamma }^I{\gamma }_9$. If ${\gamma }_{9}v=v$, thenThe first collection has negative chirality; the second has positive chirality. If $u_{+}$ and $u_{-}$ have these respective chiralities, hermiticity gives $⟨u_{+},u_{-}⟩=⟨{\gamma }_9{u}_{+},u_{-}⟩=⟨u_{+},{\gamma }_9{u}_{-}⟩=-⟨u_{+},u_{-}⟩$, so they are orthogonal. Combining the two real-independent collections therefore gives at least sixteen real-linearly independent oscillator vacua, eight in each chirality.

The real qualification in the question matters: the particular eight vectors generated from an arbitrary complex $v$ need not be independent over $\mathbb{C}$. Nevertheless the dimension bound from paired Clifford involutions also follows from the full Clifford algebra. Define four commuting hermitian involutions $J_j=i{\gamma }^{2j-1}{\gamma }^{2j}$, $j=1,\dots ,4$. Their joint spectral projections preserve the vacuum space, so it contains a nonzero common eigenvector $w$. Multiplication by ${\gamma }^{2j-1}$ flips the eigenvalue of $J_j$ and leaves the other three eigenvalues unchanged. The sixteen products obtained by independently choosing whether to apply these four odd-indexed Gamma matrices to $w$ consequently have distinct joint eigenvalue quadruples. They are nonzero, mutually orthogonal oscillator vacua. Thus the unprojected vacuum space also has complex dimension at least sixteen, with eight states of each chirality in the minimal representation. A further chiral projection is an additional physical restriction, not part of the oscillator-vacuum conditions here.