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The oscillator vacuum is annihilated by all annihilation operators. Acting with creation operators constructs Fock states. In single-string quantization this is the unexcited string, not the empty spacetime state.

For a Neumann-Dirichlet open-string boundary condition, the transverse modes have frequencies $r\in \mathbb{Z}+1/2$. The canonical commutation relations give $[{\alpha }_r^i,{\alpha }_s^j]=r{\delta }^{ij}{\delta }_{r+s,0}$ and ${\alpha }_r^{†}={\alpha }_{-r}$. One derivation expands $X-y={\sum }_{r>0}{q}_r\cos (r\sigma )$: spatial orthogonality reduces the Polyakov action to $(8{\alpha }^{\prime }{)}^{-1}\int {\sum }_r({\dot{q}}_r^2-r^2{q}_r^2)d\tau$. These are harmonic oscillators of mass $1/(4{\alpha }^{\prime })$, whose normalized creation operators satisfy ${\alpha }_{-r}=i\sqrt{r}{a}_r^{†}$ in the conventional phased expansion.

With 24 physical half-integer open-string oscillators, the first three string level operator eigenvalues are $0,1/2,1$. Their states are the vacuum, $a_{1/2}^{i†}|0⟩$, and $a_{1/2}^{i†}{a}_{1/2}^{j†}|0⟩$. The creation operators commute, so the third level is the symmetric square of the 24-dimensional transverse vector, of dimension $24\cdot 25/2=300$. Its trace is a scalar and its trace-free part has dimension 299. Including the vacuum shift gives ${\alpha }^{\prime }{M}^2=1/2,1,3/2$ when the two longitudinal directions are common directions with Neumann boundary conditions. The representation labels refer to the surviving ND rotation group, not full unbroken target Lorentz symmetry.

The oscillator-generated covariant photon Fock space has an indefinite Hermitian form, not a positive Hilbert space inner product. For a normalizable one-photon wave packet of polarization $\lambda$, its squared norm is proportional to $-{\eta }^{\lambda \lambda }\int d^{3}p|f(\mathbf{p}){|}^2$. Thus $\lambda =0$ gives a negative-norm photon state. The divergent ${\delta }^3(0)$ of an unsmeared momentum eigenstate is a separate normalization issue, avoided by the wave packet.

Choose contravariant polarization vectors ${ϵ}^{0\mu }=(1,0)$, ${ϵ}^{3\mu }=(0,\hat{\mathbf{p}})$, and two ${ϵ}^{r\mu }=(0,{\mathbf{e}}_r)$ with ${\mathbf{e}}_r\cdot \mathbf{p}=0$ for $r=1,2$. The third spatial polarization is longitudinal polarization; the first two are transverse polarization. The timelike photon polarization is distinct from the longitudinal one.

The Gupta-Bleuler quantization condition sets the divergence of the positive-frequency part of a quantum field to zero on physical states:Restore the factors $e^{-i|\mathbf{p}|t}$ to the annihilation terms. Since $p_{\mu }{ϵ}^{0\mu }=|\mathbf{p}|$ and $p_{\mu }{ϵ}^{3\mu }=-|\mathbf{p}|$, Fourier transform gives the equivalent conditionIts sign depends on the chosen sign of the longitudinal polarization vector; the covariant condition does not.

To see its content for a general Fock state, temporarily discretize momentum and decompose one unphysical oscillator sector as $|\Psi ⟩={\sum }_{n,m\ge 0}|n,m{⟩}_{0,3}\otimes |{\chi }_{nm}{⟩}_{\perp }$, where $|n,m⟩=(a^{0†}{)}^n(a^{3†}{)}^m|0⟩/\sqrt{n!m!}$. The temporal oscillator obeys $a^0|n,m⟩=-\sqrt{n}|n-1,m⟩$, while $a^3|n,m⟩=\sqrt{m}|n,m-1⟩$. The condition therefore becomesEquivalently, the allowed finite-particle states use the transverse creation operators and only $b^{†}=a^{0†}-a^{3†}$ in the unphysical sector. Indeed the constraint acts on a polynomial of the two unphysical creation operators as $-{\partial }_{a^{0†}}-{\partial }_{a^{3†}}$, whose kernel consists of polynomials in their difference. Also $[b,b^{†}]=-1+1=0$, and a state containing $b^{†}$ is orthogonal to every constrained state because $b$ annihilates every such state. This argument applies mode by mode and extends by smearing to continuum momentum.

For one photon, $c_0{a}^{0†}|0⟩+c_3{a}^{3†}|0⟩$ is constrained only when $c_3=-c_0$, producing a null state. The Gupta-Bleuler null-state quotient removes these null directions. The condition excludes negative-norm physical states; quotienting its null states leaves the two positive-norm transverse photon polarizations. The condition alone gives a positive semidefinite Hermitian form, not yet a positive definite Hilbert space.

Set $\hslash =1$. Work on the strip $0\le \sigma \le \pi$ in conformal gauge, with worldsheet signature $(-,+)$ and target signature $(-,+,\dots ,+)$. In light-cone gauge in string theory, the independent fields are the $d=D-2=24$ transverse coordinates. Put the Neumann boundary condition at $\sigma =0$ and the Dirichlet boundary condition at $\sigma =\pi$; reversing the endpoints exchanges cosine and sine modes without changing the spectrum. The fixed endpoint is $X^i(\tau ,\pi )=y^i$.

Varying the transverse Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^i=0$ and the spatial boundary contribution $-(2\pi {\alpha }^{\prime }{)}^{-1}\int d\tau [X^{\prime i}\delta X^i{]}_0^{\pi }$. At the free endpoint this vanishes precisely when $X^{\prime i}=0$; at the fixed endpoint $\delta X^i=0$. Separation of variables then gives $\cos (r\sigma )$ with $\cos (r\pi )=0$, hence $r=n+\frac{1}{2}$, $n=0,1,\dots$. These are Neumann-Dirichlet open-string boundary conditions. No dynamical transverse worldsheet zero mode survives: a constant solution must equal the prescribed $y^i$, and a linear-in-$\sigma$ solution violates the free-end condition.

Write $X^i-y^i={\sum }_{r>0}{q}_r^i(\tau )\cos (r\sigma )$. The orthogonality relation ${\int }_0^{\pi }\cos (r\sigma )\cos (s\sigma )d\sigma =(\pi /2){\delta }_{rs}$ reduces the action to independent harmonic oscillators:The canonical commutation relations $[q_r^i,P_s^j]=i{\delta }^{ij}{\delta }_{rs}$ determine normalized annihilation operators at $\tau =0$Thus $q_r^i=\sqrt{2{\alpha }^{\prime }/r}(a_r^i{e}^{-ir\tau }+a_r^{i†}{e}^{ir\tau })$. These formulas derive the quantization from the action rather than import integer-moded open-string rules.

For the usual phased string oscillators, set ${\alpha }_r^i=-i\sqrt{r}{a}_r^i$ and ${\alpha }_{-r}^i=i\sqrt{r}{a}_r^{i†}$ for $r>0$. Then the half-integer open-string oscillator expansion and algebra areThe corresponding field momentum density is ${\Pi }_i={\dot{X}}^i/(2\pi {\alpha }^{\prime })$. Completeness of the mixed-boundary eigenfunctions gives $[X^i(\sigma ),{\Pi }_j({\sigma }^{\prime })]=i{\delta }_j^i{\delta }_{\mathrm{ND}}(\sigma ,{\sigma }^{\prime })$, where ${\delta }_{\mathrm{ND}}=(2/\pi ){\sum }_{r>0}\cos (r\sigma )\cos (r{\sigma }^{\prime })$. This is a distribution identity on the mixed-boundary function space, not an unrestricted value at a fixed endpoint with a Dirichlet boundary condition.

Classically, the transverse Virasoro algebra zero-mode generator isThere is no transverse momentum term. If the two light-cone directions are common directions with Neumann boundary conditions, the full zero-mode constraint adds ${\alpha }^{\prime }{p}_{\parallel }^2=-2{\alpha }^{\prime }{p}^{+}{p}^{-}$: $L_{0,\mathrm{cl}}={\alpha }^{\prime }{p}_{\parallel }^2+L_{0,\mathrm{cl}}^{\perp }$. This assumption about the longitudinal directions is needed to interpret oscillator levels as target-space masses.

Quantizing the symmetrically ordered transverse generator gives $L_0^{\perp }=N+E_0$, where the string level operator is $N={\sum }_{i,r>0}r{a}_r^{i†}{a}_r^i$. Each harmonic oscillator contributes $r/2$ to the vacuum energy. Use zeta function regularization and the Riemann zeta function with the actual half-integer spectrum:Consequently the Neumann-Dirichlet string zero-point energy isIn the common convention $L_0={\alpha }^{\prime }{p}_{\parallel }^2+N-a$, the normal-ordering constant of a string is $a=-\frac{1}{2}$. This positive shift differs from the $-1$ vacuum energy of 24 integer-moded transverse bosons. The difference between those two vacuum energies is $3/2$. To distinguish zero-mode conventions, the ND twist conformal weight is $1/16$ per transverse boson. With 24 bosons, the plane matter generator is $L_0^{\mathrm{plane}}={\alpha }^{\prime }{p}_{\parallel }^2+N+3/2$; its physical open-string condition $L_0^{\mathrm{plane}}-1=0$ is exactly the strip/light-cone constraint ${\alpha }^{\prime }{p}_{\parallel }^2+N+1/2=0$ used here. The transverse plane and strip constants differ by the central charge shift $c_{\perp }/24=24/24=1$. A common exponential frequency cutoff independently gives ${\sum }_{r>0}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, confirming the finite part and avoiding invalid termwise manipulation of divergent sums.

Let $a_r^i|0⟩=0$. The lowest levels of a fully transverse ND bosonic string areThe third level has two $r=1/2$ excitations; there is no $r=1$ oscillator. Its indices are symmetric because the creation operators commute. With $L_0=0$ and the common longitudinal momentum convention, the rest energies and masses obeyAt fixed positive $p^{+}$ the corresponding light-cone energies are $p^{-}=(N+1/2)/(2{\alpha }^{\prime }{p}^{+})$; the table lists excitation levels, not a spectrum that remains discrete if longitudinal momentum is varied continuously.

The surviving transverse rotations form $SO(24)$. The ground state is a scalar, the next level its vector, and the third level the symmetric square of the vector. Separating its trace givesThe trace state is proportional to ${\sum }_i(a_{1/2}^{i†}{)}^2|0⟩$; subtracting this trace gives the 299-dimensional symmetric traceless square.

There is a qualification to the printed “little group”. The little group with mixed string boundary conditions must preserve the endpoints as well as momentum. These boundary conditions break the full 26-dimensional Lorentz group, so the massive states above cannot be classified as representations of the unbroken 26-dimensional massive little group $SO(25)$. In a D1–D25 realization the common worldvolume has Lorentz group $SO(1,1)$ and trivial connected massive little group; $SO(24)$ acts on the ND coordinates as an internal rotation group. The scalar, vector and symmetric trace-free decomposition is under the surviving transverse SO(24), with this boundary-background qualification.