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A first-class constraint generates an infinitesimal canonical transformation $\delta F=\{F,{ϵ}^i{\phi }_i\}$. For the phase-space action $I=\int (p\dot{q}-{\lambda }^i{\phi }_i)dt$, the multiplier transformation $\delta {\lambda }^k={\dot{ϵ}}^k-{\lambda }^i{ϵ}^j{C}_{ij}{}^k$ cancels the bulk variation when $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The remaining temporal boundary term vanishes for parameters that vanish at the endpoints.

An equation of motion determines how a physical state evolves. A variational action gives Euler-Lagrange equations; a phase-space action instead gives canonical Hamiltonian mechanics equations together with any imposed constraints.

Use the mostly-plus Minkowski metric, and write $\approx$ for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra $C_{ij}{}^k$ may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.

To see the gauge invariance directly, let $G={ϵ}^i(t){\phi }_i$ generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking ${ϵ}^i$ to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions ${ϵ}^i(t)$ therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.

For a closed string, choose $0\le \sigma <2\pi$ and periodic fields. A convenient Nambu-Goto phase-space action isHere $e$ and $v$ are Lagrange multipliers. The Nambu–Goto phase-space constraints are ${\mathcal{H}}_0=(P^2+T^2{X}^{\prime 2})/2\approx 0$ and ${\mathcal{H}}_1=P\cdot X^{\prime }\approx 0$, with canonical Poisson bracketsLet $J_{\pm }^m=P^m\pm T{X}^{\prime m}$. Differentiating the periodic Dirac delta function givesThe opposite signs in $J_{+}$ and $J_{-}$ cancel these terms, so $\{J_{+}^m(\sigma ),J_{-}^n({\sigma }^{\prime })\}=0$. Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields ${\ell }_n=i{e}^{in\sigma }{\partial }_{\sigma }$ on a circle satisfy $[{\ell }_m,{\ell }_n]=(m-n){\ell }_{m+n}$. Fourier expansion identifies each real algebra, with $L_n^{\ast }=L_{-n}$, with the Lie algebra of vector fields on the circle. The two commuting copies give ${\mathrm{Diff}}_1\oplus {\mathrm{Diff}}_1$, not a quantum central extension.

For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string $X^0=t$, take a boundary-adapted parametrization with $v=0$ at the ends. Fixing $X^i(t,a)=0$ gives $\delta X^i(t,a)=0$, a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero $e$ these require $X^{\prime i}(t,b)=0$, a Neumann boundary condition. Also $X^{\prime 0}=0$ in this temporal gauge for a string, so $X^{\prime m}(t,b)=0$. The constraint at this free-end string boundary condition reduces to $P^2=0$. Hamilton's equation $\dot{X}=eP+v{X}^{\prime }$ consequently gives ${\dot{X}}^2=0$ there. Since ${\dot{X}}^0=1$, the free endpoint has spatial speed one. This is the null motion of a free string endpoint.

A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set $e=1/T$, $v=0$, andTake $P_m=T{\dot{X}}_m$. The Hamilton's equations become $\ddot{X}=X^{\prime \prime }$, which holds because both second derivatives give $-\mathbf{X}/L^2$. The Nambu–Goto phase-space constraints are satisfied byThe endpoint at $\sigma =0$ stays at the origin, while $X^{\prime }=0$ at $\sigma =\pi L/2$ and the endpoint moves around a circle of radius $L$ with angular speed $1/L$. At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.

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 Schrödinger 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 $A_{ij}$.

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.

For the Monge-gauge Hamiltonian of a three-dimensional string, in Monge gauge, $X^{\prime m}=(0,1,{\Phi }^{\prime })$ and $P_m=(P_0,P_1,\Pi )$. The momentum constraint gives $P_1=-{\Phi }^{\prime }\Pi$. The other Virasoro constraint then readsChoosing the positive-energy branch,Substituting into the phase-space action leaves $I=\int dtd\sigma (\Pi \dot{\Phi }-\mathcal{H})$, soFor an infinite string this is an energy-density expression; even the undeformed string has infinite total energy. One may regulate the length or subtract its constant reference energy when a finite total energy is needed. The positive branch is unambiguous at the density level.

The canonical Hamiltonian field equations areConsequently $\Pi ={\Phi }^{\prime }$ impliesThe second equation also becomes $\dot{\Pi }={\Phi }^{\prime \prime }$, consistent with differentiating the first. These are one-direction traveling profiles along the string.

For the linear profile, the spatial string is the straight line $X^2=k{X}^1+kt$. It is tilted, and its coordinate intercept moves with speed $k$. This is also a uniformly moving straight string: the physical transverse velocity of a tilted string is the velocity normal to the line. With unit tangent $(1,k)/\sqrt{1+k^2}$ and coordinate velocity $(0,k)$, the normal speed isThus $k>1$ does not imply superluminal physical motion. Fixed-$\sigma$ labels have spacelike trajectories when $k>1$, but labels on a string can slide tangentially and are not identifiable material particles. Choosing $\dot{\sigma }=-k^2/(1+k^2)$ removes the tangential velocity and gives the subluminal velocity $(-k^2,k)/(1+k^2)$.

The induced worldsheet metric supplies a coordinate-invariant check:The string worldsheet remains Lorentzian for every finite $k$, including $k=1$ and $k>1$. Its constant energy density is $1+k^2$. This linear profile is a tilted boosted straight string, rather than a localized finite-energy wave.

The canonical momentum equation is $\dot{X}=eP+u{X}^{\prime }$. Eliminating $P$ from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is $-1$, as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so $e,u$ encode the metric modulo its conformal factor.

Define the induced worldsheet metric by $g_{\mu \nu }={\eta }_{mn}{\partial }_{\mu }{X}^m{\partial }_{\nu }{X}^n$. Varying the independent inverse metric tensor in the Polyakov action givesConsequently $g_{\mu \nu }=f{\gamma }_{\mu \nu }$, where $f=\frac{1}{2}{\gamma }^{\rho \sigma }{g}_{\rho \sigma }$. For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take $f>0$, or ${\gamma }_{\mu \nu }=e^{2\omega }{g}_{\mu \nu }$. The undetermined function is precisely Weyl invariance. Since $\sqrt{-\gamma }{\gamma }^{\mu \nu }{g}_{\mu \nu }=2\sqrt{-g}$, elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

The Jacobi identity for the Poisson bracket impliesFor first-class constraints with linearly independent differentials this gives $f_{[ij}{}^l{f}_{k]l}{}^m=0$, the Jacobi identity for the structure constants of the constraint algebra. Independence is an implicit assumption: for constraints obeying identities or vanishing identically, only the contracted identity follows, and arbitrary coefficients multiplying such constraints need not satisfy a Lie algebra identity.

With $G={ϵ}^i{\phi }_i$ and $\delta F=\{F,G\}$, the canonical variables transform asTo fix the sign convention, vary the phase-space action directly:Thus action invariance requires $\delta {\lambda }^i={\dot{ϵ}}^i+{ϵ}^j{\lambda }^k{f}_{jk}{}^i$. For $F^i={\lambda }^i-{\bar{\lambda }}^i$, the Faddeev-Popov determinant is that ofUsing anticommuting Faddeev-Popov ghost fields, the invariant-convention result isThe original PDF has a sign error in the stated multiplier transformation; the TeX transcription has the consistent plus sign. Keeping the PDF's displayed minus sign mechanically would instead give $i\int dt{b}_i({\dot{c}}^i-c^j{\bar{\lambda }}^k{f}_{jk}{}^i)$. That expression exponentiates the determinant of the printed transformation, but that transformation does not preserve the stated action with the canonical convention above. It cannot be used as the invariant result without changing another convention consistently.

As for the particle, constant multiplier moduli and any residual zero mode in field theory must be handled separately; the constant gauge fixing is understood locally on the gauge orbit.