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For $[a,a^{†}]=1$, the occupation number operator $a^{†}a$ has eigenvalues $0,1,2,\dots$ on the bosonic oscillator Fock space.

For a one-particle Hilbert space $H$, the bosonic or fermionic Fock space is ${\mathcal{F}}_{\pm }(H)={⨁}_{n=0}^{\infty }{H}^{{\otimes }_{\pm }n}$, using symmetric or antisymmetric tensor powers. Its zeroth summand is the one-dimensional vacuum sector. Creation operators and annihilation operators change particle number. The bosonic Fock space and fermionic Fock space implement the corresponding exchange statistics.

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.

Time-coordinate oscillators produce negative norms in the covariant indefinite Fock space. A physical negative-norm state would violate probabilistic unitarity. Such ghosts differ from the anticommuting worldsheet ghost fields introduced by gauge fixing.

The operator formalism describes a quantum system using states in a Hilbert space and observables represented by self-adjoint operators with suitable domains on that Hilbert space. Evolution and correlation functions are computed from Hamiltonian operators and their time-ordered products. For free quantized fields, the oscillator construction gives a Fock space. The operator-path-integral equivalence relates these matrix elements to integrals over field histories.

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 $A_{ij}$.

The old covariant quantization of a free-ended bosonic string retains oscillators in all target directions. In mostly-plus Minkowski spacetime, set ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ andTime-coordinate excitations have negative norms in this covariant Fock space. The string ghost states in this discussion are unwanted negative-norm physical states, distinct from the anticommuting Faddeev-Popov ghost fields in the path integral. Constraints and the null-state quotient of a string remove unphysical polarizations while maintaining target-space Lorentz covariance.

Why only positive-mode constraints annihilate states. The quantum Virasoro algebra isThe physical conditions arewhere $a$ is the string intercept. If both positive and negative modes annihilated states, $[L_1,L_{-1}]=2L_0$ would force $a=0$. Then $[L_2,L_{-2}]=4L_0+D/2$ would force $D=0$. Thus the Virasoro central extension and shifted zero mode prevent imposing every classical constraint strongly in a nontrivial string. As in Gupta-Bleuler quantization, negative-mode conditions act on physical bras, rather than also annihilating physical kets. Physical null string states are quotiented because their inner products with all physical states vanish.

The intercept bound at level one. For $|ϵ;p⟩={ϵ}_m{\alpha }_{-1}^m|p⟩$, the constraints and norm areIf $a>1$, momentum is spacelike and its orthogonal complement contains a timelike negative-norm polarization. The level-one intercept bound in covariant string quantization is thereforeFor $a<1$, momentum is timelike and the $D-1$ orthogonal polarizations are positive. This avoids level-one ghosts but gives a massive vector with one more polarization than the transverse light-cone gauge in string theory spectrum. Ghost absence alone is weaker than equivalence.

For $a=1$, momentum is null. Its orthogonal complement contains $D-2$ positive directions and the null direction $p$. The state $p\cdot {\alpha }_{-1}|p⟩$, proportional to $L_{-1}|p⟩$, is physical, spurious and null. Removing it givesThe quotient has exactly the $D-2$ massless transverse vector polarizations. Thus equivalence at level one selects $a=1$ and the null-state quotient of a string, not just the inequality.

Level two and the dimension. Set $a=1$. At level two, $p^2=-1/{\alpha }^{\prime }$ and $k=\sqrt{2{\alpha }^{\prime }}p$ has $k^2=-2$. A general state isUsing $[L_m,{\alpha }_n^a]=-n{\alpha }_{m+n}^a$, the nontrivial positive-mode conditions areThere are $D-1$ vector null string states $L_{-1}(v\cdot {\alpha }_{-1}|p⟩)$ with $k\cdot v=0$. The parent has $L_0=0$, so the descendant norm is zero. Quotienting these leaves the massive symmetric traceless rank-two tensor plus one additional scalar.

The level-two scalar in covariant string quantization can be chosen, for every $D$, asOn its three displayed structures, $L_1$ gives respectively $2,-4,2$ times $k\cdot {\alpha }_{-1}|p⟩$, and $L_2$ gives $D,-2,-4$ times the vacuum. The coefficients make both combinations vanish. The first two structures have norms $2D,8$ and cross inner product $-4$; the mode-two structure has norm $-4$ and is orthogonal to them. For $B=(D+4)/10$, $C=(D-1)/5$, this yieldsAbove 26 this is a physical negative-norm string state. Below 26 it is an extra positive-norm scalar, which cannot be discarded just to force the ordinary light-cone state count. At 26 it becomes null and can be removed. Thus level-two ghost absence gives $D\le 26$, whereas equivalence to the ordinary transverse spectrum requires $D=26$.

The critical scalar is also a Virasoro descendant. For $L_0|p⟩=-|p⟩$, the level-two scalar Virasoro null state candidateobeysAt $D=26$ it is physical and null, with $|S_{26}⟩=2|Z⟩$. At other dimensions it is not physical, so its norm must not be treated as a physical ghost test; the already-physical $|S_D⟩$ gives the correct test.

Finally, the covariant level-two oscillator space has dimension $D(D+3)/2$. The $D$ conditions from $L_1$ and one from $L_2$ leave $(D-1)(D+2)/2$ physical components. Quotienting $D-1$ vector null states and the critical scalar null state leavesthe massive spin-two field count and the level-two light-cone gauge in string theory count. The vector null descendants change the components of $\zeta$ orthogonal to $k$, and the critical scalar null descendant changes its component along $k$. Thus $\zeta$ can be set to zero. A representative then has $\zeta =0$, $k^m{ϵ}_{mn}=0$ and ${ϵ}^m{}_m=0$. The two approaches consequently agree on the massless level-one vector and massive level-two spin-two tensor for $a=1,D=26$.

Write $O=\int d[\bar{\psi },\psi ]e^{-\sum \bar{\psi }\psi }|\psi ⟩⟨\psi |$. For bosons the measure is ${\prod }_{n}d\mathrm{Re}{\psi }_{n}d\mathrm{Im}{\psi }_n/\pi$ over $\mathbb{C}$ in each mode, with ${\bar{\psi }}_n={\psi }_n^{\ast }$. For fermions it is an ordered Berezin integral over independent ${\bar{\psi }}_n,{\psi }_n$; choose $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$, so $\int d\bar{\psi }d\psi e^{-\bar{\psi }\psi }=1$.

For one bosonic mode, $a|\psi ⟩=\psi |\psi ⟩$ and $⟨\psi |a={\partial }_{\bar{\psi }}⟨\psi |$. Integration by parts in the Gaussian measure gives $aO=Oa$; the conjugate argument gives $a^{†}O=O{a}^{†}$. Boundary terms vanish because of the Gaussian weight.

For one fermionic mode, put $n=a^{†}a$ and move Grassmann coefficients to the left. The weighted projector isIts ordinary commutators are $[a,W]=-a+\psi I$ and $[a^{†},W]=a^{†}-\bar{\psi }I$. Their Berezin integrals vanish, so again $[a,O]=[a^{†},O]=0$. Equivalently the sole surviving coefficient in $\int W$ is $-\bar{\psi }\psi I$, giving $I$ directly.

The modes factorize. In the irreducible Fock space representation, commuting with every creation and annihilation operator makes $O$ a scalar multiple of the identity. Its vacuum matrix element is the normalized Gaussian integral, equal to one. Thus the coherent-state resolution of identity isFor infinitely many modes, this argument first uses a finite-mode regulator.

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.