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The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.

For an ordinary open string, the boundary conditions relate the two chiral sectors. With ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{k}^{\mu }$, their independent Fourier generators areThe classical oscillator Poisson brackets $\{{\alpha }_m^{\mu },{\alpha }_n^{\nu }\}=-im{\eta }^{\mu \nu }{\delta }_{m+n,0}$ give the classical Virasoro constraint algebraThus they are first-class constraints. In the quantum theory one applies normal ordering to the product and uses $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$. The oscillator commutator givesFor example, for $m>0$ the extra vacuum term is $(D/2){\sum }_{r=1}^{m-1}r(m-r)=D(m^3-m)/12$. This exhibits the Virasoro central extension rather than concealing it in a classical constraint equation. The matter central charge is $D$; the worldsheet ghost fields have charge $-26$. Quantum consistency sets $D=26$ and the string intercept to one.

In old covariant string quantization the physical-state Virasoro conditions for an open string arePhysical string states are then identified modulo physical null string states. One does not impose both positive and negative generators as annihilation conditions: their central extension would obstruct that prescription. The resulting null-state quotient of a string agrees with the transverse light-cone spectrum at nonzero momentum.

At $N=0$, the oscillator vacuum $|k⟩$ is a single spacetime scalar field with ${\alpha }^{\prime }{M}^2=-1$: the bosonic tachyon. At $N=1$, write $|\Psi ⟩={ϵ}_{\mu }{\alpha }_{-1}^{\mu }|k⟩$. The zero-mode equation gives $k^2=0$, and $L_1$ gives $k\cdot ϵ=0$. The null descendant $L_{-1}|k⟩=\sqrt{2{\alpha }^{\prime }}k\cdot {\alpha }_{-1}|k⟩$ identifies $ϵ\sim ϵ+\lambda k$. This leaves 24 physical polarizations of a massless gauge vector.

At $N=2$, use the level-two open-string polarization decompositionThe mass is $M^2=1/{\alpha }^{\prime }$. The oscillator commutators give all the nontrivial positive-mode conditions:Higher positive modes annihilate this level. To see the quotient explicitly, the 25 transverse-parameter null states $L_{-1}(\xi \cdot {\alpha }_{-1}|k⟩)$, with $k\cdot \xi =0$, shiftThe additional level-two scalar Virasoro null state $(L_{-2}+\frac{3}{2}{L}_{-1}^2)|k⟩$ shiftsSubstitution into the two physical equations verifies the null-state conditions; the second residual is $(D-26)/2$. Because $k$ is timelike, these null shifts span all components of $b$, so choose a representative with $b=0$. The physical conditions then say $k^{\mu }{h}_{\mu \nu }=0$ and $h^{\mu }{}_{\mu }=0$. In the rest frame this is a traceless second-rank tensor in 25 spatial dimensions, with $25\cdot 26/2-1=324$ polarizations. The light-cone basis independently gives $24$ states ${\alpha }_{-2}^i|k⟩$ and $24\cdot 25/2=300$ symmetric states ${\alpha }_{-1}^i{\alpha }_{-1}^j|k⟩$. Hence

For the fermionic extension, an explicit convention prevents ambiguity in the signs. Keep worldsheet signature $(-,+)$ and useTake $\psi =({\psi }_1,{\psi }_2{)}^T$ and the constant parameter $ϵ=({ϵ}_1,{ϵ}_2{)}^T$ to be real Grassmann-odd Majorana components, with $\bar{\psi }={\psi }^T{\rho }^0$. Put ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$. Apart from the common factor $1/(4\pi {\alpha }^{\prime })$, the combined Lagrangian isThe rigid transformations can be writtenor, in components,For the ${ϵ}_2$ part, the bosonic variation is $i{ϵ}_2({\partial }_{+}{\psi }_1\cdot {\partial }_{-}X+{\partial }_{+}X\cdot {\partial }_{-}{\psi }_1)$, while the fermionic variation is $-i{ϵ}_2{\partial }_{-}X\cdot {\partial }_{+}{\psi }_1+i{ϵ}_2{\psi }_1\cdot {\partial }_{+}{\partial }_{-}X$. Anticommuting the constant parameter past the fermion is essential in the second term. The first terms cancel. The other chiral component works in the same way, giving the chiral proof of rigid worldsheet supersymmetryThus the action is invariant off shell up to a boundary term. The transformation is a genuine supersymmetry: on $X$ its commutator for parameters $ϵ,\zeta$ is the translationOn $\psi$ the same translation follows using ${\partial }_{+}{\psi }_1=0$ and ${\partial }_{-}{\psi }_2=0$; the remaining cross terms are proportional to these Dirac equations. This is on-shell closure of rigid worldsheet supersymmetry. On a surface with boundary, compatible endpoint conditions must make the boundary flux vanish. On a compact worldsheet, a constant parameter must preserve the spin structure; the local identity alone does not make a constant transformation compatible with every antiperiodic sector.

The fermions contribute to the worldsheet stress-energy tensor and produce a fermionic superconformal current. In holomorphic normalization with ${\psi }^{\mu }(z){\psi }^{\nu }(w)\sim {\eta }^{\mu \nu }/(z-w)$,The free-field contractions give $G(z)G(w)\sim (2c/3)/(z-w{)}^3+2T(w)/(z-w)$ and conformal weight $3/2$ for $G$, where $c=D+D/2=3D/2$. Consequently the modes obey the N=1 super-Virasoro algebraDropping the central terms gives the classical graded constraint algebra. Integer $r$ gives the Ramond sector; half-integer $r$ gives the Neveu–Schwarz sector. In the displayed conformal-plane convention $G_0^2=L_0-c/24$ in the Ramond sector. Shifting to the Ramond vacuum-normal-ordered zero mode gives the usual Ramond supercurrent zero-mode square $G_0^2=L_0^{(R)}$.

Rigid symmetry of matter already supplies these currents, but promoting their vanishing to new gauge constraints requires coupling to worldsheet supergravity and varying its worldsheet gravitino. In superconformal gauge one retains $T=G=0$. The fermionic current constraints and their physical-state conditions extend the bosonic Virasoro conditions. The ghost charges are $-26+11=-15$, so cancellation with $3D/2$ gives the critical dimension of the RNS superstring $D=10$; the NS and Ramond intercepts are respectively $1/2$ and zero. This distinguishes the global supersymmetry established above from its locally gauged string completion.

In conformal gauge, a free closed bosonic string has $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$, periodicity in $\sigma$, and the two Virasoro constraints $(\dot{X}\pm X^{\prime }{)}^2=0$. Its left- and right-moving string oscillator families have independent Virasoro generators $L_n$ and ${\tilde{L}}_n$. In old covariant string quantization, a physical string state obeyswith $a=1$ in the critical bosonic string theory. The two zero-mode conditions also impose closed-string level matching. For an open string, keep just one string oscillator family. Negative-mode constraints are adjoints on bras, not additional annihilation conditions on physical string states.

A spurious string state is a Virasoro descendant orthogonal to every physical string state: if $|s⟩={\sum }_{n>0}{L}_{-n}|{\chi }_n⟩$, then $⟨\Phi |s⟩=0$ follows from $L_n|\Phi ⟩=0$ and $L_n^{†}=L_{-n}$. A null string state is a spurious state that is itself physical. It consequently has zero norm and is removed in the null-state quotient of a string. The calculations below concern one string oscillator family and use

For $|n_1⟩=L_{-1}|{\chi }_1⟩$ with $L_0|{\chi }_1⟩=0$ and all positive modes zero, $L_0|n_1⟩=|n_1⟩$. AlsoIt is therefore physical for string intercept one and spurious by construction. In particular $\Vert n_1{\Vert }^2=⟨{\chi }_1|2L_0|{\chi }_1⟩=0$. This proves the requested level-one descendant is null.

For the scalar descendant, write $|v_b⟩=(L_{-2}+b{L}_{-1}^2)|{\chi }_2⟩$ with $L_0|{\chi }_2⟩=-|{\chi }_2⟩$ and positive modes zero. Direct commutation givesFor instance $[L_1,L_{-1}^2]|{\chi }_2⟩=-2L_{-1}|{\chi }_2⟩$ and $L_2{L}_{-1}^2|{\chi }_2⟩=-6|{\chi }_2⟩$. Modes $m\ge 3$ annihilate $|v_b⟩$ because every remaining positive mode annihilates its parent. Hence the physical scalar descendant requires $b=3/2$ and $d=26$:It is spurious, so physicality also proves orthogonality and zero norm without an extra positivity assumption.

The printed coefficient $3/4$ does not give this null string state. It instead gives $L_1|v_{3/4}⟩=\frac{3}{2}{L}_{-1}|{\chi }_2⟩$, generally nonzero. For a normalized highest-weight parent with weight $-1$ at $d=26$, the descendant Gram matrix isThus the printed vector has norm $9/4$, whereas the corrected vector has norm zero. This is a genuine PDF coefficient error. Away from 26 dimensions even the corrected coefficient does not give the claimed physical null string state; the remaining condition is $L_2|v_{3/2}⟩=(d-26)|{\chi }_2⟩/2$.

The level-two open string state also needs a source correction: its vector string oscillator must be ${\alpha }_{-2}^{\mu }$, not the printed ${\alpha }_{-1}^{\mu }$. A single ${\alpha }_{-1}$ has level one and cannot share the stated mass-shell condition with the two-${\alpha }_{-1}$ term. Put $q^{\mu }={\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$, and write the genuine level-two state asThe mass-shell condition $L_0=1$ gives ${\alpha }^{\prime }{p}^2+2=1$, so $q^2=-2$. The commutators above giveFor the second formula, the contraction from the two ${\alpha }_{-1}$ operators is exactly $a^{\mu }{}_{\mu }$, not twice that value. Modes $m>2$ vanish automatically. We have derivedThe coefficient formulas in the question are consequently consistent after the string oscillator correction and with $d=26$.

To prove the full level-two open-string polarization decomposition, not merely check one candidate, defineThe constraints give $q\cdot b=-t/2$ and therefore $q\cdot A=0$. Contracting $c$ with $q$ givesIts trace is $t-t(26+3q^2)/20-2q\cdot A=0$. Thus $c$ is symmetric, traceless and transverse. Conversely, arbitrary $t$, a vector $A$ with $q\cdot A=0$, and such a tensor $c$ giveand direct contraction verifies both constraints. The number $20$ uses $d=26$; it would not reproduce the same unrestricted trace in arbitrary dimension.

Finally the two unphysical pieces are precisely the descendants already proved null. The vector part iswhose parent has $L_0=0$ and all positive modes zero because $q\cdot A=0$. For the scalar part useThenhas exactly the scalar tensor $t(3q_{\mu }{q}_{\nu }+{\eta }_{\mu \nu })/20$ and vector $t{q}_{\mu }/4$. The momentum vacuum has $L_0=-1$ and positive modes zero, as required. HenceThe physical quotient contains only the transverse traceless tensor. Its massive little-group space has dimension $25\cdot 26/2-1=324$, agreeing with the light-cone gauge in string theory count $24\cdot 25/2+24=324$ at level two.