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Use Minkowski spacetime with signature $(-,+,\dots ,+)$, and write the string tension as $T=1/(2\pi {\alpha }^{\prime })$. If $X^{\mu }(\tau ,\sigma )$ is the string embedding map, its induced worldsheet metric is ${\gamma }_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$. The Nambu–Goto action isThe first variation uses $\delta {\gamma }_{ab}=2{\partial }_{(a}{X}_{\mu }{\partial }_{b)}\delta X^{\mu }$ and therefore givesAn integration by parts yields the bulk Nambu–Goto equations of motion and the endpoint term:For a closed string, periodicity cancels the endpoint term. For a freely moving open string, arbitrary endpoint variations require zero momentum flux, which is the free-end string boundary condition.

Choose orthogonal conformal coordinates on the nondegenerate part of the worldsheet, so that ${\gamma }_{\tau \sigma }=0$ and ${\gamma }_{\tau \tau }+{\gamma }_{\sigma \sigma }=0$. The Nambu–Goto equations of motion then become a wave equation, with the accompanying Virasoro constraints:Here a dot and a prime mean ${\partial }_{\tau }$ and ${\partial }_{\sigma }$. The last two equations are essential: a solution of the wave equation alone need not be a relativistic fundamental string motion.

For a free-ended open string, take $0\le \sigma \le \pi$ and $a>0$, and setwith the other coordinates constant. Every component satisfies the wave equation, andMoreover $X^{\prime }=0$ at both endpoints, giving the Neumann boundary condition. At fixed target time $t=a\tau$, the spatial image is the straight segmentThus its angular velocity is $1/a$. The speed at coordinate $r$ is $|r|/a$, so the endpoints have the null motion of a free string endpoint. The induced worldsheet metric degenerates there; the conserved densities below have finite limits, so the endpoint degeneracy causes no divergent charge.

In these coordinates ${\Pi }^{\tau \mu }=T{\dot{X}}^{\mu }$. The target-space Noether charges of a string give the energy, momentum and angular momentum:Since this is the centre-of-momentum frame, the rest mass is $M=E$. Eliminating $a$ proves the free-ended straight-string Regge relation:A spacetime Lorentz transformation of the solution changes its frame, not this rest-frame relation.

There is also a folded closed string version: use the same formula over $0\le \sigma \le 2\pi$. The spatial segment is then covered twice, with folds at $\sigma =0,\pi$, and the closed-string mode expansion is periodic. Each charge is twice its open string value, soThe folds also have degenerate induced metric and are understood through the finite canonical densities or a limiting smooth motion. The two classical Regge trajectories have different slopes because the closed string has two coincident branches. No quantum string intercept enters this classical calculation.

Take $0\le \sigma <2\pi$. Variation of the conformal-gauge action, with periodic variations, gives $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Hence $X^{\mu }=X_R^{\mu }(\tau -\sigma )+X_L^{\mu }(\tau +\sigma )$. In an uncompactified direction, periodicity allows equal linear slopes and independent periodic parts. Expanding the periodic parts in a Fourier series yields the closed-string mode expansionHere ${\alpha }_m$ labels the right-moving sector and ${\tilde{\alpha }}_m$ the left-moving sector; exchanging the names is harmless if done consistently. Integration of the canonical momentum density $(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}^{\mu }$ over the string gives $p^{\mu }$, fixing the zero-mode coefficient. Reality gives ${\alpha }_m^{†}={\alpha }_{-m}$ and the analogous relation for tildes, with ${\alpha }_0={\tilde{\alpha }}_0=\sqrt{{\alpha }^{\prime }/2}p$.

The closed-string physical-state Virasoro conditions arewhere $L_0={\alpha }^{\prime }{p}^2/4+N$ and ${\tilde{L}}_0={\alpha }^{\prime }{p}^2/4+\tilde{N}$. Thus closed-string level matching gives $N=\tilde{N}$, and $M^2=-p^2=4(N-1)/{\alpha }^{\prime }$. The massless level is $N=\tilde{N}=1$. Its general state isThe string oscillator commutators give $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$: the two terms obtained by commuting through the quadratic generator coincide and cancel its factor $1/2$. Therefore $L_1$ replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$, while $L_{m\ge 2}$ produces positive string oscillators that annihilate this level. The same holds independently for tildes. The complete polarization conditions areNull string states identify ${\zeta }_{\mu \nu }$ with ${\zeta }_{\mu \nu }+p_{\mu }{\lambda }_{\nu }+{\tilde{\lambda }}_{\mu }{p}_{\nu }$, with transverse $\lambda ,\tilde{\lambda }$. At nonzero momentum this leaves $24\cdot 24=576$ physical polarizations. Their symmetric traceless, antisymmetric and transverse-trace parts are the graviton, Kalb–Ramond field and dilaton.

The original PDF's condition ${\alpha }^{\prime }{p}^2=4$ is inconsistent with calling the displayed excited states massless. It is the on-shell condition of the string oscillator vacuum as a physical tachyon. For the level-one state, that value would give $L_0={\tilde{L}}_0=2$, violating the physical conditions. The ket $|0;p⟩$ used as the basis for a massless excitation must instead carry the excitation's momentum $p^2=0$; by itself it is then not an on-shell physical vacuum state.

For the compact coordinate, $X^1(\tau ,\sigma +2\pi )=X^1(\tau ,\sigma )+2\pi wR$, $w\in \mathbb{Z}$. Its momentum is $n/R$, $n\in \mathbb{Z}$, since the centre-of-mass wavefunction is single-valued on the circle. Thus its zero modes becomewithThese are momentum and winding modes. Let $k$ be the 25-dimensional noncompact momentum and $M^2=-k^2$. The two zero-mode constraints now giveTheir difference and average give, respectively,This also displays T-duality $R\leftrightarrow {\alpha }^{\prime }/R$, $n\leftrightarrow w$, with one chiral momentum reversed.

At a generic radius, the massless solutions are only $n=w=0$, $N=\tilde{N}=1$. Their fields in 25 dimensions are a graviton (275 polarizations), antisymmetric two-form (253), dilaton (one), two abelian vectors from $G_{a1}$ and $B_{a1}$ (23 each), and the radius scalar (one). Hence the generic massless spectrum has $275+253+1+46+1=576$ polarizations, with gauge group $U(1{)}_L\times U(1{)}_R$. Here generic excludes isolated radii where exceptional massless ground states of a bosonic circle occur. Indeed $N=\tilde{N}=0$ forces $nw=0$, and becomes massless at $R=|n|\sqrt{{\alpha }^{\prime }}/2$ or $R=2\sqrt{{\alpha }^{\prime }}/|w|$. These additional thresholds matter in the bosonic theory.

At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, masslessness and level matching readSince $N,\tilde{N}\ge 0$, their sum is at most two. For sum two, $n=w=0$ and level matching gives the ordinary $(1,1)$ family. For sum one the four possibilities areThey have one right-moving ${\alpha }_{-1}$ or one left-moving ${\tilde{\alpha }}_{-1}$ excitation, as their levels specify. Each family has 24 physical polarizations: a 25-dimensional vector with 23 polarizations and one compact-oscillator scalar. Thus these families add 96 states.

For sum zero, level matching forces $nw=0$, while the mass equation requires $n^2+w^2=4$. It gives four additional scalar states with charges $(\pm 2,0)$ and $(0,\pm 2)$, without any string oscillators. They must not be omitted. The massless spectrum at the bosonic self-dual circle therefore has 100 additional polarizations, 676 in total. The four new vector fields enhance the gauge group to $SU(2{)}_L\times SU(2{)}_R$. Including the original radius scalar, the nine non-dilaton scalars form its $(3,3)$ representation. Equivalently, the full count is $275+253+1+6\cdot 23+9=676$.

Use target signature $(-,+,\dots ,+)$ and a closed-string spatial coordinate $0\le \sigma \le 2\pi$. Write $Y$ for the compact coordinate and $X^A$ for the noncompact coordinates. In conformal gauge, the wave equation factorizes, so every coordinate is a sum of a function of $\tau +\sigma$ and a function of $\tau -\sigma$. The closed-string boundary conditions areThus the derivatives of each chiral part are periodic and have integer Fourier modes; only their worldsheet zero modes can produce the winding displacement. Integrating these Fourier series gives the closed-string mode expansionThe oscillator reality conditions are ${\alpha }_m^{†}={\alpha }_{-m}$ and similarly for the tilded modes. The coefficient of $\tau$ follows by integrating the canonical momentum density $\dot{Y}/(2\pi {\alpha }^{\prime })$. Single-valued wavefunctions of the center coordinate $y\sim y+2\pi R$ have momentum $n/R$, with $n\in \mathbb{Z}$. Equivalently, the compact chiral worldsheet zero modes are ${\alpha }^{\prime }{p}_L(\tau +\sigma )/2$ and ${\alpha }^{\prime }{p}_R(\tau -\sigma )/2$, whereHere tildes denote the left-moving sector. This convention will also fix the sign in compact-circle closed-string level matching.

Choose a noncompact light-cone pair and impose light-cone gauge in string theory $X^{+}=x^{+}+{\alpha }^{\prime }{p}^{+}\tau$, with $p^{+}\ne 0$. The Virasoro constraints $({\partial }_{\pm }X{)}^2=0$ then solve for ${\partial }_{\pm }{X}^{-}$ in terms of the transverse coordinates. For example, with ${\partial }_{\pm }={\partial }_{\tau }\pm {\partial }_{\sigma }$, one has ${\partial }_{\pm }{X}^{-}=({\partial }_{\pm }{X}^i{)}^2/(2{\alpha }^{\prime }{p}^{+})$. Thus there are two independent transverse oscillator Fock spaces, including $Y$ among the transverse coordinates. Quantization gives $[{\alpha }_m^i,{\alpha }_k^j]=m{\delta }^{ij}{\delta }_{m+k,0}$, and the same relation for the tilded oscillators, with the two sectors commuting. Lorentz consistency gives the critical dimension of the bosonic string $D=26$ and string intercept one in each sector. Accordinglycount 24 transverse oscillator species. The two zero-mode constraints give the lower-dimensional bosonic string mass spectrumTaking their difference and average yieldsEvery choice of transverse string oscillators and integer charges satisfying this closed-string level matching is a physical string state in light-cone gauge. The eliminated longitudinal oscillators do not supply additional polarizations.

To classify all massless states, positivity of the two compact terms implies $N+\tilde{N}\le 2$. When the sum is two, masslessness forces $n=w=0$, and closed-string level matching then forces $N=\tilde{N}=1$. The states ${\alpha }_{-1}^i{\tilde{\alpha }}_{-1}^j|0⟩$ have $24^2=576$ polarizations. In 25 noncompact dimensions, their decomposition gives a graviton, a Kalb–Ramond field, a dilaton, two Abelian gauge vectors from the metric and two-form with one compact index, and the radius scalar field. Their physical polarization counts areThe first two numbers are the symmetric traceless second-rank tensor and antisymmetric second-rank tensor of the massless little group $SO(23)$.

When $N+\tilde{N}=1$, closed-string level matching requires $|nw|=1$. The compact contribution obeysand equality occurs precisely at $R=\sqrt{{\alpha }^{\prime }}$. The charges $n=w=\pm 1$ have $(N,\tilde{N})=(1,0)$; the charges $n=-w=\pm 1$ have $(N,\tilde{N})=(0,1)$. Each charge has 24 polarizations supplied by its single excited sector. At a general radius these have

Finally, $N=\tilde{N}=0$ requires $nw=0$. The exceptional massless ground states of a bosonic circle occur atEach charge is a scalar field. These special radii must be included if “general radius” is to mean a complete classification, rather than only a generic radius. At a generic radius the massless spectrum is the neutral 576-state spectrum above; at exceptional radii one adds exactly the states just listed. The neutral ground state remains tachyonic.

The mapsends $p_L$ to $p_L$ and $p_R$ to $-p_R$. Reversing all right-moving compact string oscillators, ${\alpha }_m^Y\mapsto -{\alpha }_m^Y$, preserves their commutators and oscillator number; the left-moving oscillators are unchanged. Both the mass formula and compact-circle closed-string level matching are therefore unchanged, with a one-to-one map of the complete oscillator basis. This proves T-duality of the full spectrum, including winding, massive and tachyonic states.

At the self-dual circle the four single-oscillator charge families give $4\cdot 24=96$ new massless states. Each family decomposes into a 25-dimensional gauge vector with 23 polarizations and one scalar field. There are also four massless ground states, with charges $(\pm 2,0)$ and $(0,\pm 2)$. Thus the massless spectrum at the bosonic self-dual circle hasThe gauge interpretation follows directly from the chiral compact-boson operators. With $⟨Y_L(z)Y_L(0)⟩=-{\alpha }^{\prime }\log z/2$, the exponential $e^{ik{Y}_L}$ has conformal weight ${\alpha }^{\prime }{k}^2/4$. At the self-dual radius,have weight one and satisfy $J^3(z)J^3(0)\sim 1/(2z^2)$, $J^3(z)J^{\pm }(0)\sim \pm J^{\pm }(0)/z$, and $J^{+}(z)J^{-}(0)\sim z^{-2}+2J^3(0)/z$. These are the level-one $SU(2)$ current relations; the right sector supplies another copy. Multiplying these currents by a noncompact oscillator produces six gauge vectors in total. The nine scalar string vertex operators $J_L^a{J}_R^b$ form the $(3,3)$ representation, accounting for the radius scalar field and eight new scalar fields. The remaining neutral fields are the graviton, two-form and dilaton.

A radius deformation as a Higgs mechanism is the expectation value of the Cartan–Cartan scalar string vertex operator $J_L^3{J}_R^3\propto \partial Y_L\bar{\partial }{Y}_R$. Its stabilizer is $U(1{)}_L\times U(1{)}_R$, soThe four compact single-oscillator scalar field polarizations become longitudinal polarizations of the four massive vectors, exactly as in the Higgs mechanism. The other four extra scalar fields have masses $4/R^2-4/{\alpha }^{\prime }$ or $4R^2/{\alpha }^{\prime 2}-4/{\alpha }^{\prime }$ and need not remain massless; one pair becomes tachyonic on either side of the self-dual point. Together with the original bosonic tachyon, this means that the gauge-symmetry interpretation is formal and does not assert a stable bosonic-string vacuum.

Use target-space signature $(-,+,\dots ,+)$ and let a dot between target vectors denote contraction with the Minkowski metric. Varying the string embedding map in the Polyakov action, with periodic variations around the closed string, givesThus the embedding equation of motion isThe independent worldsheet metric variation uses $\delta \sqrt{-\gamma }=-\frac{1}{2}\sqrt{-\gamma }{\gamma }_{ab}\delta {\gamma }^{ab}$. It sets the worldsheet stress tensor to zero:In conformal gauge, ${\gamma }_{ab}=e^{2\omega }\mathrm{diag}(-1,1)$, the Weyl factor cancels. Define worldsheet derivatives ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$. The equations and Virasoro constraints reduce toEquivalently, $\ddot{X}-X^{\prime \prime }=0$, $\dot{X}\cdot X^{\prime }=0$ and ${\dot{X}}^2+X^{\prime 2}=0$. The free wave equation permits independent periodic left- and right-moving fields.

Take $0\le \sigma <2\pi$ and the uncompactified closed-string mode expansionThe Virasoro generators are the Fourier coefficients of the two quadratic worldsheet stress tensors. Their quantum expressions areThe canonical commutation relations are $[{\alpha }_m^{\mu },{\alpha }_n^{\nu }]=m{\eta }^{\mu \nu }{\delta }_{m+n,0}$, with an identical tilded algebra and commuting chiral sectors. Normal ordering places annihilation modes to the right. It gives $L_0={\alpha }^{\prime }{p}^2/4+N$, where $N={\sum }_{n>0}{\alpha }_{-n}\cdot {\alpha }_n$, and similarly on the other side. In old covariant string quantization, physical string states obeymodulo physical null string states. Negative generators are adjoint constraints on bras, not further annihilation conditions on the same ket. In critical bosonic string theory, $d=26$ and the string intercept is $a=1$. These closed-string physical-state Virasoro conditions implyThus the massless first closed-string level has $N=\tilde{N}=1$, and its general state isTo derive the polarization tensor constraints, the oscillator algebra gives $[L_n,{\alpha }_m^{\mu }]=-m{\alpha }_{n+m}^{\mu }$. Applying $L_1$ to this state replaces ${\alpha }_{-1}^{\mu }$ by ${\alpha }_0^{\mu }$; the independent tilded condition replaces the other oscillator by ${\tilde{\alpha }}_0^{\nu }$. All higher positive generators annihilate the state automatically. ThereforeThere is no condition that the tensor trace vanish: each chiral side has only one oscillator, so neither $L_2$ nor ${\tilde{L}}_2$ imposes a trace condition. The null-state quotient of a string also identifiesFor example, $L_{-1}({\xi }_{\nu }{\tilde{\alpha }}_{-1}^{\nu }|0,p⟩)$ is a momentum-longitudinal state with zero inner product against every transverse physical polarization. This explains the string-state gauge redundancy, rather than merely counting covariant components. For an ordinary nonzero null momentum, choose a frame adapted to $p$. The constraints remove one null direction in each index, and these null-state identifications remove the remaining longitudinal directions. The representatives are then ${\zeta }_{ij}$, $i,j=1,\dots ,24$.

Decompose that transverse matrix into its symmetric trace-free part, antisymmetric part and trace. They describe the graviton, Kalb–Ramond field and dilaton, respectively. Their particle polarization counts areThese are representations of the rotational part $SO(24)$ of the massless little group. The dilaton corresponds to the transverse trace; a covariant scalar polarization can be written proportional to ${\eta }_{\mu \nu }-p_{\mu }{\ell }_{\nu }-{\ell }_{\mu }{p}_{\nu }$, with ${\ell }^2=0$ and $p\cdot \ell =1$. It is transverse in both indices and has trace 24. The nonzero-momentum assumption is appropriate for propagating particle states; zero momentum requires separate treatment of gauge and background zero modes.

In a general background, these fields replace the flat-space free action by the string nonlinear sigma model. A convenient Euclidean convention isHere ${ϵ}^{ab}$ is the orientation density; reversing its convention changes the displayed sign of the two-form coupling. The metric tensor is $G_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }+\cdots$, $B$ is the antisymmetric Kalb–Ramond field, and $\Phi$ is the dilaton. Their worldsheet couplings reproduce the corresponding massless closed-string vertex operators when expanded about flat space. The Polyakov path integral integrates $e^{-S_E}$ over embedding fields and metrics modulo worldsheet diffeomorphisms and Weyl transformations; gauge fixing supplies ghost determinants and moduli integrations. Quantum consistency of a general background additionally requires vanishing sigma-model beta functions, not just insertion of arbitrary backgrounds into a free theory.

Finally, separate the constant dilaton mode, $\Phi ={\Phi }_0+\hat{\Phi }$. By the Gauss-Bonnet theorem, its action isFor a connected oriented surface of genus $g$, its Euler characteristic is $\chi =2-2g$. Defining the string coupling by $g_s=e^{{\Phi }_0}$ givesThus the string genus expansion is a topology sum with two extra powers of $g_s$ per handle. An amplitude with $n$ canonically normalized external closed-string vertices has the additional vertex normalization $g_s^n$, so its overall factor is $g_s^{2g-2+n}$. The curvature term, rather than a guessed coupling inserted at each internal interaction, supplies this dependence directly from the functional integral.

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.

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.