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The eigenvalue of a mode number operator counts excitations in that mode. Bosonic occupation numbers range over all nonnegative integers; fermionic occupation numbers are zero or one.

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.