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Take ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ on the strip $0\le \sigma \le \pi$. Varying the Grassmann-odd worldsheet Majorana fermions gives ${\partial }_{-}{\psi }_1=0$, ${\partial }_{+}{\psi }_2=0$ and, up to the orientation of the strip boundary, the surface termAt each endpoint it vanishes if ${\psi }_1=\eta {\psi }_2$ with $\eta =\pm 1$, with the same relation on variations. A common sign can be absorbed by redefining one chiral field, so set ${\eta }_0=+1$. Since ${\psi }_1=f(\tau +\sigma )$ and ${\psi }_2=f(\tau -\sigma )$, the second endpoint imposes $f(u+2\pi )={\eta }_{\pi }f(u)$. Equal endpoint signs give the periodic Ramond sector, and opposite endpoint signs give the antiperiodic Neveu–Schwarz sector:These gluing choices preserve the Neumann-endpoint worldsheet supersymmetry conditions and give integer or half-integer worldsheet Majorana fermion modes, respectively.

First fix a normalization issue in the original PDF. With its stated canonical string oscillator algebra, the standard superconformal current mode iswithout the printed factor $1/2$. To check the Ramond supercurrent zero-mode square directly, pair positive and negative integer modes:The zero-mode Clifford relations give $({\alpha }_0\cdot {\psi }_0{)}^2={\alpha }_0^2/2$. Different $Q_m$ anticommute, and so do $Q_m$ with the zero-mode term. The two same-sign terms in $Q_m^2$ vanish by antisymmetry of the fermions and symmetry of the bosonic product. The cross terms giveThe second term comes from commuting ${\alpha }_m$ through ${\alpha }_{-m}$, so the frequency factor $m$ is essential. ThereforeHere $L_0^R$ uses vacuum-annihilating Ramond normal ordering; its string intercept is zero. A plane-to-cylinder central shift must be included consistently if a different zero-mode convention is used.

For the mode $F_r=G_r/2$ actually printed, the algebra instead gives $\{F_0,F_0\}=L_0^R/2$. Already the oscillator-free term gives $\{F_0,F_0\}={\alpha }_0^2/4$, whereas $2L_0^R={\alpha }_0^2$. Thus the requested identity is not valid with both that prefactor and the supplied string oscillator algebra. Removing the prefactor gives the identity just derived. Renaming $L_0^R/4$ as $L_0$ would fix this isolated equality but would spoil the usual Virasoro normalization and the subsequent mass formula. We use the standard normalized modes $G_r$ in the remaining physical constraints; their vanishing is unaffected by a common nonzero rescaling.

The RNS string physical-state conditions in the two sectors areThe last condition is the fermionic zero-mode constraint; its square implies the Ramond mass-shell condition. Physical null string states are quotiented. Matter central charge is $3d/2$, while the reparameterization worldsheet ghost fields and superconformal ghosts contribute $-26+11=-15$. Thus $d=10$ is the critical dimension of the RNS superstring, with $a_{NS}=1/2$ and $a_R=0$.

To solve the constraints, choose $X^{\pm }=(X^0\pm X^9)/\sqrt{2}$, so ${\eta }_{+-}=-1$, and work on a patch with ${\alpha }_0^{+}=\sqrt{2{\alpha }^{\prime }}{p}^{+}\ne 0$. In light-cone gauge in string theory fix all nonzero ${\alpha }_n^{+}$ and all ${\psi }_r^{+}$ to zero. For the Ramond sector, the printed restriction to nonzero fermion modes is not by itself complete: one must also reduce the longitudinal fermionic zero modes using the zero-mode supersymmetry constraint. The usual reduced light-cone description sets the remaining ${\psi }_0^{+}$ to zero as well and retains the transverse Clifford algebra. Covariantly this reduction is the Dirac equation on the ground-state particle polarization; it is not an extra unconstrained canonical string oscillator.

With this reduction the string oscillator generators have the formRepeated $i$ runs over the eight transverse directions. The coefficient in $L_n^{\perp }$ comes from the fermion worldsheet stress tensor $-\psi \partial \psi /2$; for $n=0$, positive and negative $r$ combine to give the occupation energy $r{\psi }_{-r}{\psi }_r$. Thus the light-cone super-Virasoro constraints are explicitly solved byHere $a$ is $1/2$ in the NS sector and zero in the R sector. The zero-mode equation says $2p^{+}{p}^{-}=p_i^2+(N-a)/{\alpha }^{\prime }$. All longitudinal excitations are consequently determined by transverse bilinears. The critical dimension and string intercepts are necessary for the resulting quantum theory to preserve the target Lorentz algebra.

The frequency-weighted Neveu–Schwarz level operator isThe PDF's displayed number operator omits the fermion frequency weights and the $r=1/2$ term. With the stated anticommutators that cannot be a level operator: it would assign no level to the first fermionic excitation. The bosonic term already includes its frequency through $[{\alpha }_n,{\alpha }_{-n}]=n$, whereas the fermionic term must include its frequency explicitly.

One independent way to obtain the string intercept is the transverse zero-point energy. Each periodic boson contributes $\frac{1}{2}{\sum }_{n\ge 1}n=-1/24$. Each antiperiodic real fermion contributes $-\frac{1}{2}{\sum }_{r\ge 1/2}r=-1/48$, with the minus sign due to fermionic normal ordering. For example a common exponential cutoff givesSubtracting the local divergent terms leaves $8(-1/24-1/48)=-1/2$, so the NS string intercept is $a=1/2$. In the Ramond sector the integer fermions instead contribute $+1/24$ per component and cancel the bosonic contribution, giving $a_R=0$.

The unprojected NS sector oscillator vacuum has $N=0$ andIt is a spacetime scalar tachyon. The first excitation ${\psi }_{-1/2}^i|0;p⟩$ has $N=1/2$ and is massless, with eight vector particle polarizations. Covariantly its particle polarization obeys $p\cdot ϵ=0$ modulo the longitudinal null particle polarization, leaving the same eight degrees of freedom.

The R sector oscillator vacuum is also massless, but it is degenerate because $\{{\psi }_0^{\mu },{\psi }_0^{\nu }\}={\eta }^{\mu \nu }$. The operators ${\Gamma }^{\mu }=\sqrt{2}{\psi }_0^{\mu }$ give the target Clifford algebra, and $G_0|u;p⟩=0$ becomes $p_{\mu }{\Gamma }^{\mu }u=0$. At nonzero null momentum this Dirac equation leaves sixteen unprojected on-shell spinor states. Equivalently the eight transverse zero modes give the sixteen-dimensional spin representation $8_s\oplus 8_c$ of $SO(8)$. Thus the Ramond ground state describes spacetime fermions, not a single scalar oscillator vacuum.

To define the GSO projection explicitly, let $f_{NS}$ count excited NS fermions and give its vacuum even parity. ThenIt retains odd fermion number, removes the tachyon, and retains the eight massless vector particle polarizations. In the Ramond sector let ${\Gamma }_{\ast }$ be transverse chirality and $f_R$ count nonzero-mode fermions. Choose one consistent chirality sign $s=\pm 1$ and useEach operator squares to itself. At the ground level $P_R$ retains one eight-dimensional spinor chirality, so the retained spectrum has eight massless bosonic and eight massless fermionic states. The parity convention can instead assign the NS vacuum odd parity and write a plus projector; the retained states are identical.

Equality continues at every mass level. Let $q$ count ${\alpha }^{\prime }{M}^2$, suppressing common momentum and endpoint-label factors, and set $D(q)={\prod }_{n\ge 1}(1-q^n{)}^8$. The projected string oscillator generating functions areThe Ramond factor eight follows because either chirality projector selects eight ground states for every nonzero-oscillator parity. The Jacobi abstruse identity makes the bracket equal to $16q^{1/2}{\prod }_{n\ge 1}(1+q^n{)}^8$, soThis proves level-by-level equality of bosonic and fermionic multiplicities. The appropriate spacetime supercharge from an RNS spin field relates the two sectors; the projection ensures that it acts within the retained state space. These sector and chirality conventions agree with the primary superstring lecture notes. Equal total counts alone are not the definition of the projection; they follow from the stated parity selection and the string oscillator identity.

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.