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Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Continue with metric signature $+---$. The Pauli matrices obey $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication in the chiral gamma-matrix representation givesandHence $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }{I}_4$, the required Clifford algebra. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ yieldsThe antisymmetric part of ${\gamma }^{\nu }{\gamma }^{\mu }$ drops out because partial derivatives commute. Therefore every component of the Dirac spinor satisfies the Klein-Gordon equation. The converse is not true: four arbitrary scalar solutions need not obey the first-order Dirac equation.

The Lorentz group consists of real invertible matrices $\Lambda$ preserving the Minkowski metric: ${\Lambda }^{T}g\Lambda =g$. Writing $\Lambda =I+\omega +O({\omega }^2)$ gives ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$, so there are six continuous parameters: three spatial rotation angles and three boost rapidities. The Proper orthochronous Lorentz group is the connected component with determinant $+1$ that preserves time orientation. Parity and time reversal are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.

For the real-generator convention used here, vector generators can be writtenTake ${\Omega }_{\rho \sigma }=-{\Omega }_{\sigma \rho }$ and use the same parameters in the spinor representation. The Lorentz-spinor generators from a Clifford algebra areTo verify the Lorentz algebra, first commute a generator with one gamma matrix:For example, this follows by moving ${\gamma }^{\tau }$ through each product using the Clifford anticommutator. Applying the commutator derivation rule to $\frac{1}{4}[{\gamma }^{\tau },{\gamma }^{\nu }]$ then givesNo extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.

The same gamma commutator provesIndeed, to first order $S^{-1}{\gamma }^{\mu }S={\gamma }^{\mu }+\frac{1}{2}{\Omega }_{\rho \sigma }[{\gamma }^{\mu },M_{\mathrm{sp}}^{\rho \sigma }]$, exactly the infinitesimal vector action defined above. Exponentiating proves the finite relation. The spinor field transforms as ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$ with $x^{\prime }=\Lambda x$; combining this relation with ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$ preserves the Dirac equation. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover spin representation, rather than a single-valued ordinary representation of the Lorentz group itself.

For a boost in direction $i$, $M_{\mathrm{sp}}^{0i}=\frac{1}{2}{\gamma }^0{\gamma }^i$ is Hermitian, with eigenvalues $\pm \frac{1}{2}$. A real rapidity $\eta$ therefore produces eigenvalues $e^{\pm \eta /2}$ in $S$, whose moduli are not one. The finite-component spinor representation is not unitary for boosts, although the rotation matrices are unitary. The Lorentz group is noncompact; this does not forbid the unitary infinite-dimensional action on the physical Hilbert space of states.

Nevertheless, since $({\gamma }^{\mu }{)}^{†}={\gamma }^0{\gamma }^{\mu }{\gamma }^0$,This Dirac spinor pseudo-unitarity implies ${\bar{\psi }}^{\prime }(x^{\prime })=\bar{\psi }(x)S^{-1}$, and consequentlyThe Dirac adjoint is precisely the adjoint needed to form this Lorentz scalar; ordinary ${\psi }^{†}\psi$ alone is not a scalar.

Let ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\mathrm{diag}(-I_2,I_2)$. It anticommutes with every gamma matrix and commutes with every $M_{\mathrm{sp}}^{\rho \sigma }$, so $S^{-1}{\gamma }^{5}S={\gamma }^5$. Thus the axial current transforms as a four-vector under proper Lorentz transformations:Under parity it is an axial vector, acquiring the additional pseudovector sign. Contracting two axial currents cancels that sign and uses the invariant metric. Therefore $j_{5\mu }{j}_5^{\mu }$ is compatible with Lorentz invariance, including parity invariance of this contraction. Lorentz invariance of the interaction does not require the axial current to be conserved.

Retain zeroth and first spatial-gradient orders in the long-wavelength approximation in cosmology. Spatial Ricci curvature, $D_i{D}_{j}N$, $q^2$ and scalar anisotropic stress are of second gradient order. Put $A^i{}_j={\tilde{K}}^i{}_j$ and $A^2=A^i{}_j{A}^j{}_i$. The supplied Einstein field equations then reduce toThe scalar-field stress projections show that the momentum constraint is a first-gradient equation and must be retained. The square $A^2$ is not a spatial derivative and cannot yet be dropped merely by invoking the long-wavelength approximation in cosmology. The time-derivative dot on the trace-free evolution equation is visible in the PDF and missing from the TeX aid.

Using $NH=\dot{a}/a$, the last equation becomes ${\partial }_t(a^3{A}^i{}_j)\simeq 0$. HenceThe time-independent spatial tensor must also satisfy the momentum constraint and the symmetry inherited from $K_{ij}$. This is inflationary shear damping: expanding regions lose anisotropic expansion, without requiring the frozen spatial metric shape to vanish. The matter evolution, useful below, is $\dot{\varphi }=N\Pi$ and $\dot{\Pi }\simeq -N(3H\Pi +V_{,\varphi })$, the leading Klein-Gordon equation in each local region.

Use natural units and the Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. For the real scalar field, choose ${\mathcal{L}}_0=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. The conjugate momentum is $\pi =\dot{\varphi }$. Under an active translation $\delta \varphi =\epsilon a^{\nu }{\partial }_{\nu }\varphi$, the density changes by $\delta {\mathcal{L}}_0=\epsilon {\partial }_{\mu }(a^{\mu }{\mathcal{L}}_0)$. The Noether theorem therefore gives the canonical stress-energy tensorIts divergence is $(\square \varphi +m^2\varphi ){\partial }^{\nu }\varphi$, which vanishes on the Klein-Gordon equation. With vanishing flux at spatial infinity, the charges are conserved. In particular,The minus sign is required because $P^i=\int T^{0i}$ and ${\partial }^i=-{\partial }_i$. These are the four-momentum of a free real scalar field.

For canonical quantization, impose $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=i{\delta }^3(\mathbf{x}-\mathbf{y})$, with both equal-time field-field commutators zero. Inverting the mode expansion givesThe equal-time canonical commutation relation then yieldsFor example, the two mixed field-momentum terms in the first commutator have coefficients $\frac{1}{2}\sqrt{E_{\mathbf{p}}/E_{\mathbf{q}}}$ and $\frac{1}{2}\sqrt{E_{\mathbf{q}}/E_{\mathbf{p}}}$; the delta function sets their sum to one.

Insert the mode expansion into the quadratic energy. Spatial integration supplies $(2\pi {)}^3{\delta }^3(\mathbf{p}\pm \mathbf{q})$. In the $aa$ and $a^{†}{a}^{†}$ terms, the coefficient is proportional to $-E_{\mathbf{p}}^2+{\mathbf{p}}^2+m^2=0$. The remaining terms giveHere $E_0$ is the divergent zero-point energy, formally $\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}{\delta }^3(0)$. Thus the vacuum-free energy formula requires normal ordering, or equivalently subtraction of this constant. In a finite box with a cutoff this is the ordinary sum $\frac{1}{2}{\sum }_{\mathbf{p}}{E}_{\mathbf{p}}$, so the subtraction is explicit before passing to the continuum.

Likewise, use the Hermitian momentum expression $-\frac{1}{2}\int (\pi \nabla \varphi +\nabla \varphi \pi )d^{3}x$. Terms containing two annihilators or two creators vanish by antisymmetry under $\mathbf{p}\mapsto -\mathbf{p}$, leaving the symmetric number-operator expression. Its vacuum term is zero with an inversion-symmetric regulator. Therefore the normal-ordered free scalar four-momentum isThe creation operator commutators follow directly from $[a_{\mathbf{q}}^{†}{a}_{\mathbf{q}},a_{\mathbf{p}}^{†}]=a_{\mathbf{q}}^{†}(2\pi {)}^3{\delta }^3(\mathbf{q}-\mathbf{p})$:Starting from a vacuum annihilated by every $a_{\mathbf{p}}$, each creation adds a particle of mass $m$, energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$. The real scalar field has one spin-zero species: its antiparticle is the same species. Products of creation operators commute, so multiparticle states are invariant under exchange of their labels. In a normalized discrete mode, $(a^{†}{)}^r|0⟩/\sqrt{r!}$ exists for every $r\ge 0$; there is no exclusion restriction. These are bosonic statistics from commuting creation operators and prove Bose–Einstein statistics. For completeness, the single-mode thermal sum ${\sum }_{r\ge 0}{e}^{-\beta Er}$ gives $⟨r⟩=(e^{\beta E}-1{)}^{-1}$, the Bose-Einstein distribution at zero chemical potential.

Particle gauge fixing and BRST. Put $C=(p^2+{\mu }^2)/2$. The classical constraint generates ${\delta }_{\epsilon }{x}^m=\epsilon p^m$, ${\delta }_{\epsilon }{p}_m=0$, ${\delta }_{\epsilon }e=\dot{\epsilon }$. Consequently the variation of $e-\bar{e}$ along a gauge orbit is the operator ${\partial }_t$ on $\epsilon$. A path integral restricted to that gauge must include its Faddeev-Popov determinant; anticommuting FP ghosts represent the determinant, rather than its inverse. Constant gauge zero modes and any proper-time modulus must be treated separately, so the relevant determinant is $\det {}^{\prime }{\partial }_t$.

A convenient point-particle Faddeev–Popov ghost action at $\bar{e}=1$ isThe FP ghost normalization has been chosen to give the graded Poisson bracket $\{b,c\}=-i$, with the bracket symmetric on two odd variables. The particle BRST charge and Klein–Gordon constraint arewhere the odd constant parameter $\Lambda$ is placed on the left. The product $\Lambda Q$ is even. Accounting for the odd parameter in this convention givesThese are canonical BRST transformations. Direct variation checks the sign: the bosonic kinetic term changes by $\Lambda (\dot{c}{p}^2+cp\cdot \dot{p})$, while the FP ghost term changes by $-\Lambda C\dot{c}$. Henceand the action is invariant for the corresponding boundary conditions. The BRST charge is conserved because $C$ commutes with the gauge-fixed Hamiltonian and $c$ is constant on the FP ghost equation of motion.

Quantization gives $\{\hat{b},\hat{c}{\}}_{+}=1$ and ${\hat{c}}^2=0$. Since $C$ commutes with the FP ghosts,It generates the same variations by $\delta \hat{F}=-i[\hat{F},\Lambda \hat{Q}]$, with an ordinary commutator against the even generator. Represent $\hat{c}$ by multiplication and $\hat{b}$ by differentiation. On a ghost-number-zero wavefunction $\Psi (x)$, $\hat{Q}\Psi =0$ is exactly $\hat{C}\Psi =0$, namelythe Klein-Gordon equation for the mostly-plus metric. The specified FP ghost sector matters: for an unrestricted wavefunction ${\Psi }_0+c{\Psi }_1$, BRST closure only constrains ${\Psi }_0$, not every component independently. The complete physical prescription uses BRST cohomology, not an assertion that every vector in the entire ghost-extended kernel is a new Klein–Gordon particle.

String oscillator algebra and FP ghost Virasoro generators. Use the same bosonic brackets as above and the odd bracketsQuantization turns them intoThe $b,c$ here are worldsheet ghost fields, not the matter fermion oscillators used in the NS question.

Contract $b_m$ with the two $c$ factors in the cubic FP ghost term of the given BRST operator. The first contraction contributes $-\frac{1}{2}{\sum }_q(m-q)c_{-q}{b}_{m+q}$ and the second contributes $+\frac{1}{2}{\sum }_p(p-m)c_{-p}{b}_{p+m}$. Their sum is ${\sum }_q(q-m)c_{-q}{b}_{m+q}$. Reordering and normal ordering therefore give the ghost oscillator Virasoro generatorsFor $m\ne 0$ these formulas have no additive ambiguity. At $m=0$, moving annihilators past creators produces infinite zero-point sums. Their regularized finite constant shifts $L_0$ and corresponds to a term proportional to $c_0$ in $Q$. The convention fixed below has a ghost oscillator vacuum weight $-1$, or equivalently the usual intercept-one shift. Whether that shift is written outside the normally ordered sum or incorporated in its definition is a convention; it must not be omitted twice or counted twice. In the chosen oscillator convention this is explicitlyEquivalently, the normally ordered BRST charge contains the intercept term $-c_0$; the formal un-ordered operator in the question includes that choice only after its ordering prescription is fixed.

The oscillator brackets directly giveIf $Q^2=0$, then $[L_n,Q]=[\{b_n,Q\},Q]=[b_n,Q^2]=0$. The graded Jacobi identity now yields Virasoro closure from BRST nilpotence:In particular a central extension cannot remain in the total generators if the quantum BRST charge is nilpotent.

Low-level descendant calculation. Let $|\Omega ;p⟩$ be the oscillator ground state, annihilated by ${\alpha }_{n>0}$, $b_{n\ge 0}$ and $c_{n>0}$. All $L_{n>0}$ annihilate it. The nonzero-mode FP ghost generators act asFor example $L_1(-b_{-1}{c}_0|\Omega ⟩)=-2b_0{c}_0|\Omega ⟩=-2|\Omega ⟩$. Similarly $L_2$ on the two level-two terms gives $-8|\Omega ⟩$ and $-9|\Omega ⟩$, hence $-17|\Omega ⟩$. These signs are essential: FP ghosts have an indefinite pairing.

The matter generators createContracting the level-one term with $L_1$ gives ${\alpha }_0^2$. At level two the first term gives $2{\alpha }_0^2$, and the double contraction of the second gives $D/2$; cross terms vanish because oscillator levels differ. Thus the normalized oscillator/Virasoro descendant coefficients, conventionally denoted by the requested norms, areComparing the first with $[L_1,L_{-1}]=2L_0$ gives the oscillator highest weightIn this normalization the question's $⟨0|L_0|0⟩$ is $h_0$. Comparing level two with $[L_2,L_{-2}]=4L_0$ givesThis is the bosonic-string critical dimension from ghost descendants. It checks cancellation of the matter and FP ghost Virasoro anomalies without replacing the FP ghost calculation by a remembered central-charge value.

There is a necessary ghost zero-mode pairing qualification to the word “norm”. If $b_0,c_0$ are Hermitian in the full FP ghost space and $b_0|\Omega ⟩=0$, thenA bare full-ghost inner product cannot at the same time normalize this ket to one. FP ghost zero modes need saturation in actual amplitudes. The above equations are the coefficients of $L_1{L}_{-1}|\Omega ⟩$ and $L_2{L}_{-2}|\Omega ⟩$, or the corresponding normalized contravariant Virasoro form. They are not positive-definite Hilbert norms. The algebraic coefficient calculation is sufficient for the requested critical-dimension argument and remains valid without that false normalization.

Finally, impose the stated ground-state conditions $Q|\Omega ;p⟩=b_0|\Omega ;p⟩=0$. They imply $L_0|\Omega ;p⟩=\{b_0,Q\}|\Omega ;p⟩=0$, so $h_0=0$ andThis is the bosonic tachyon. When using an oscillator vacuum at arbitrary momentum to establish the coefficients, it need not already be BRST closed; demanding closure selects this ground-state mass shell. The two uses should not be confused.

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 $L_0$.