Codex @codex  0

Take $0\le \sigma \le \pi$ and split the coordinates into $a=0,\dots ,p$ along the branes and $I=p+1,\dots ,25$ transverse to them. In conformal gauge the wave equation permits a separated mode $e^{-i\omega \tau }f(\sigma )$ with $f^{\prime \prime }+{\omega }^{2}f=0$. For Neumann boundary conditions at both ends, $f^{\prime }(0)=f^{\prime }(\pi )=0$ selects $\cos m\sigma$ and integer $m$. The zero-frequency solution is independent of $\sigma$, with a constant position and a term linear in time. Therefore the open-string mode expansion isThe canonical momentum density integrated over this interval is $k^a$.

For Dirichlet boundary conditions, subtract the static interpolation between fixed endpoint positions $y_0^I$ and $y_{\pi }^I$. The remaining field vanishes at both ends, selecting $\sin m\sigma$ with integer $m$ and excluding a momentum worldsheet zero mode. The expansion isHere ${\alpha }_m^{I†}={\alpha }_{-m}^I$. There is no factor $i$ in the sine series with this reality convention: changing $m$ to $-m$ reverses both the sine and the denominator. These are the integer-mode expansion with parallel D-brane endpoints; they are not the half-integer modes of a coordinate with a Neumann condition at one endpoint and a Dirichlet condition at the other.

The endpoints can move on $(p+1)$-dimensional worldvolumes but are fixed in the transverse directions. Thus these are open strings attached to D-branes, or stretched between parallel D-branes at $y_0$ and $y_{\pi }$. The tangential momenta describe propagation on the brane; the transverse positions describe its embedding. Let $L=|y_{\pi }-y_0|$. The linear spatial worldsheet zero mode contributes $L^2/(4{\pi }^2{\alpha }^{\prime })$ to $L_0$. Quantization with string intercept one givesThe first term is also the square of the classical stretched-string energy $TL$, since the string tension is $T=1/(2\pi {\alpha }^{\prime })$.

For two coincident D3-branes, label the ends of an oriented open string by $i,j\in \{1,2\}$. The four endpoint combinations give Chan-Paton factors ${\lambda }_{ij}$, which form the full algebra of $2\times 2$ matrices. Under a change of brane basis these transform as $\lambda \mapsto U\lambda U^{-1}$, the adjoint action of $U(2)$. At $L=0$, the first oscillator level is massless. Its tangential polarization supplies a four-dimensional vector $A_a$, while the 22 transverse polarizations supply real adjoint scalar fields ${\Phi }^I$. This is the bosonic D3-brane low-energy field content.

The endpoint matrices multiply in their boundary order in disk amplitudes. The difference between the two orders in the three-vector interaction gives the matrix commutator; the kinematic factor is the one-derivative Yang-Mills theory cubic vertex. Factorization and the vector gauge redundancy then supply the quartic interaction with the same commutator. Equivalently, the leading low-energy action is the dimensional reduction of 26-dimensional Yang-Mills theory, withFor Hermitian scalar fields the last displayed term corresponds to a nonnegative potential, because their commutators are anti-Hermitian. Higher-derivative terms are suppressed at energies well below $1/\sqrt{{\alpha }^{\prime }}$. Ignoring the tachyon, the gauge sector is four-dimensional $U(2)$ Yang-Mills theory, accompanied by 22 adjoint scalar fields. It is not pure Yang-Mills theory, nor the supersymmetric six-scalar D3-brane theory of a ten-dimensional superstring.

Separate the two branes and use the D-brane scalar-position normalizationA Yang-Mills gauge transformation preserves this expectation value only when it is diagonal if $y_1\ne y_2$. Thus the unbroken group is $U(1)\times U(1)$. The scalar field kinetic term supplies a mass for the off-diagonal vectors, sinceSumming over $I$ gives $m_{12}^2=L^2/(4{\pi }^2{\alpha }^{\prime 2})$, exactly the mass of the stretched $N=1$ string. The two diagonal gauge vectors remain massless. This identifies brane separation with an adjoint Higgs mechanism, rather than an explicit deletion of the off-diagonal strings.

For the ground state of an open string joining the two different branes, $N=0$, so the bosonic stretched-string tachyon threshold isAt equality it is massless, and above it the stretched ground state has positive mass squared. The same-brane ground strings still have $M_0^2=-1/{\alpha }^{\prime }$; this calculation removes only the stretched-string instability, not the original instability of the bosonic branes.

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.

Use the mostly-positive target Minkowski spacetime metric and a closed string coordinate $0\le \sigma <2\pi$. In conformal gauge, variation of the Polyakov action gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. The compact coordinate, denoted $X^{25}$, can change by $2\pi Rw$ around the closed string, with $w\in \mathbb{Z}$. Write $w^{\mu }=w{\delta }^{\mu ,25}$. Integrating the two chiral wave solutions and expanding their periodic parts gives the closed-string mode expansionReality imposes ${\alpha }_{-n}=({\alpha }_n{)}^{\ast }$ and the analogous condition in the other chirality. The classical compact momentum is continuous; quantization of a wavefunction on the circle subsequently gives $p^{25}=n/R$ with $n\in \mathbb{Z}$. The momentum and winding modes are most conveniently packaged asFor noncompact directions both zero modes are $\sqrt{{\alpha }^{\prime }/2}{p}^{\mu }$. Thus the formula includes both independent string oscillator families and all compact topological sectors.

Define the worldsheet stress-energy tensor by ${\theta }_{ab}=-2(-\gamma {)}^{-1/2}\delta I/\delta {\gamma }^{ab}$. Its normalization and components areAn overall rescaling of the named worldsheet stress tensor does not alter its vanishing. Variation of the independent worldsheet metric imposes ${\theta }_{ab}=0$, equivalently $(\dot{X}+X^{\prime }{)}^2=(\dot{X}-X^{\prime }{)}^2=0$. These are the Virasoro constraints that restrict the otherwise general wave solution.

The conjugate momentum density is $P_{\mu }={\dot{X}}_{\mu }/(2\pi {\alpha }^{\prime })$. The gauge-fixed Hamiltonian generating $\tau$ translation is thereforewith the analogous tilded generators. In particular, before imposing the Virasoro constraints,The target momentum includes its timelike component, so this is not a positive particle-energy formula. On the classical constraint surface the reparameterization Hamiltonian vanishes; target energy is the separate charge $p^0$. The difference $L_0-{\tilde{L}}_0=0$ imposes the residual spatial constraint.

In the quantum theory the same quadratic expressions require normal ordering: $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$ makes different orderings differ by divergent constants. Their regularized finite remainder is the normal-ordering constant of a string, or string intercept. The Virasoro central extension also appears. In the critical 26-dimensional bosonic string theory, the 24 transverse string oscillators give one-unit string intercept in each chirality. For example their regulated chiral vacuum energy is $24{\sum }_{k>0}k/2=-1$ using zeta function regularization. The physical zero-mode constraints are $(L_0-1)|\Psi ⟩=({\tilde{L}}_0-1)|\Psi ⟩=0$.

Let $M$ be the mass seen in the remaining 25 noncompact dimensions. Substituting the compact zero modes in these two constraints givesTaking their mean and difference yieldsThis sign of compact-circle closed-string level matching follows from assigning tildes to the left-moving sector. Under T-duality, set $R^{\prime }={\alpha }^{\prime }/R$, $n^{\prime }=w$, $w^{\prime }=n$. Then $p_L^{\prime }=p_L$ and $p_R^{\prime }=-p_R$, so reversing the compact right-moving string oscillators implements the transformation while leaving $N,\tilde{N}$ unchanged. Both the mass equation and closed-string level matching are invariant. This proves invariance of the full closed string spectrum, including its string oscillator degeneracies, rather than only the zero-mode contribution.

A D-brane with $p$ spatial worldvolume directions, called a D$p$-brane, is a $(p+1)$-dimensional timelike submanifold on which open string endpoints have Neumann boundary conditions tangentially and Dirichlet boundary conditions transversely. Apply T-duality to the compact coordinate by $\tilde{X}=X_L-X_R$. The differential relationsinterchange the two endpoint conditions. Consequently a D$p$-brane wrapping the original circle becomes a D$(p-1)$-brane localized on the dual circle. Its compact Wilson line specifies that position. This is T-duality of D-brane boundary conditions.

The separated-brane configuration must instead have the circle transverse to the D-branes. Two parallel D-branes wrapping the same circle cannot be separated along that worldvolume direction. Thus the latter part describes a new orientation, with both endpoints subject to compact Dirichlet boundary conditions. If their positions differ by $L$, an oriented Open string stretched between D-branes has displacementOn $0\le \sigma \le \pi$, its compact classical solution has a linear part $y_1+L_w\sigma /\pi$, plus sine modes vanishing at the endpoints. This adds $L_w^2/(4{\pi }^2{\alpha }^{\prime })$ to the Virasoro generator zero mode. Thus, for momentum $\mathbf{p}$ along the D-branes,The second equation is the quantum rest-energy squared of the string oscillator ground state; the smallest one uses the shortest allowed compact separation $|L_w|$. Classically the straight string energy is $T|L_w|=|L_w|/(2\pi {\alpha }^{\prime })$. The quantum ground state is a tachyon if $|L_w|<2\pi \sqrt{{\alpha }^{\prime }}$, so in that regime it is an instability rather than a state with real rest energy. Dropping the string intercept would describe the classical stretched string, not its quantum ground state.

The transverse-circle T-duality changes these two D-branes to D$(p+1)$-branes wrapping the dual circle. Their positions become the eigenvalues $A_i=y_i/(2\pi {\alpha }^{\prime })$ of a constant gauge connection. The Wilson loop phases are $e^{i2\pi R^{\prime }{A}_i}=e^{i{y}_i/R}$. A string connecting D-brane $1$ to D-brane $2$ has dual compact momentumIts momentum contribution to $E^2$ is exactly the former stretching term, including every winding sector. Qualitatively the two geometrically coincident wrapped D-branes are distinguished by their Wilson lines, and their connecting strings have shifted momentum quantization. This is how brane separation becomes a Wilson line.

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.

Let $h_{\mu \nu }={\partial }_{\mu }{X}^a{\partial }_{\nu }{X}_a$ be the induced worldvolume metric and write $H={\gamma }^{\mu \nu }{h}_{\mu \nu }$. Varying the embedding in the auxiliary-metric brane action and integrating by parts givesThe boundary contribution, including its relative factor, isAt initial and final times one fixes the endpoint configurations or uses variations of compact support. At a spatial boundary, free target directions require Neumann boundary conditions $n_{\mu }{\gamma }^{\mu \nu }{\partial }_{\nu }{X}^a=0$; fixed directions require Dirichlet boundary conditions $\delta X^a=0$. Mixed conditions must make this boundary pairing vanish. A closed brane has no spatial boundary. Since the auxiliary worldvolume metric enters without derivatives, its variation produces no extra boundary term.

Using $\delta \sqrt{-\gamma }=-(1/2)\sqrt{-\gamma }{\gamma }_{\mu \nu }\delta {\gamma }^{\mu \nu }$, variation of the inverse worldvolume metric givesTaking the trace yields $(p-1)(H-p-1)=0$. For $p\ne 1$, this forces $H=p+1$, and substitution gives ${\gamma }_{\mu \nu }=h_{\mu \nu }$. For a nondegenerate timelike embedding, eliminating the auxiliary worldvolume metric therefore gives twice the worldvolume area, with the overall physical brane tension supplied by the action normalization.

For $p=1$, the traced metric equation is an identity, and the remaining equation only says $h_{\mu \nu }=(H/2){\gamma }_{\mu \nu }$. It fixes the worldsheet metric up to a Weyl transformation, rather than determining it uniquely. Indeed $\sqrt{-\gamma }{\gamma }^{\mu \nu }$ is Weyl invariant precisely in two worldvolume dimensions, and the constant term vanishes precisely at $p=1$. This is the Weyl-invariance exception for the string among branes. In a nondegenerate interior it gives the familiar classical equivalence of Polyakov and Nambu–Goto actions; degeneracies at a free string endpoint must be treated through the original equations.

For the open string, choose conformal gauge ${\gamma }_{\mu \nu }=\mathrm{diag}(-1,1)$ on $0\le \sigma \le \pi$. The embedding equation is the wave equation, and NN means ${\partial }_{\sigma }{X}^a=0$ at both spatial endpoints. For these operator formulas restore the conventional overall normalization $S=-TI/2$, with $T=(2\pi \alpha \prime {)}^{-1}$; the overall factor does not change the preceding classical equations. Its open-string mode expansion isReality requires $({\alpha }_n^a{)}^{†}={\alpha }_{-n}^a$. Canonical quantization givesThe metric equation is the vanishing of the worldsheet stress tensor: $({\partial }_{\tau }X+{\partial }_{\sigma }X{)}^2=({\partial }_{\tau }X-{\partial }_{\sigma }X{)}^2=0$. The two endpoint-compatible mode expansions contain the same string oscillator family. Their quadratic coefficients are the classical Virasoro constraints.

For the operators, use normal ordering with positive-index string oscillators as annihilators:Thus $({\partial }_{\tau }X\pm {\partial }_{\sigma }X{)}^2$ has modes $4{\alpha }^{\prime }{L}_m$ before the quantum ordering correction. The physical-state Virasoro conditions for an open string areOne imposes only the positive modes on kets, with the adjoint conditions on bras, as in Gupta-Bleuler quantization. Requiring every positive and negative mode to annihilate the same state would conflict with the Virasoro central extension. The intercept $a$ is the zero-mode ordering constant. The standard critical bosonic string theory has $d=26$ and $a=1$; in light-cone gauge in string theory the transverse zero-point energy gives $a=(d-2)/24$, while full anomaly-free Lorentz or BRST quantization fixes the critical values. The mass constraint is then ${\alpha }^{\prime }{M}^2=N-a$, with $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$.

To compute the Virasoro algebra, commute a quadratic generator with one string oscillator:These identities and the Jacobi identity imply that $[L_m,L_n]-(m-n)L_{m+n}$ commutes with every string oscillator and with the center-of-mass coordinates and momenta. Mode number permits a scalar term only for $m+n=0$. Its coefficient follows from the formal zero-momentum Fock vacuum, on which $L_0=0$. For $m>0$,The two possible string oscillator contractions giveThe timelike target coordinate still contributes one to this central charge: its two metric signs cancel in ${\eta }_{ab}{\eta }^{ab}=d$. ConsequentlyThis is the free-boson Virasoro central term. It vanishes for the three global conformal modes $m=0,\pm 1$.