Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 46 (String Theory)

Actions and equations. Use a mostly-plus Minkowski metric and define the induced worldsheet metric $h_{ab}={\partial }_{a}X\cdot {\partial }_{b}X$. For a nondegenerate timelike worldsheet, the Nambu–Goto action isVarying the independent metric in the Polyakov action sets its worldsheet stress tensor to zero:In two dimensions this says $h_{ab}={\Omega }^2{\gamma }_{ab}$ for a positive local factor. The factor drops out of $\sqrt{-\gamma }{\gamma }^{ab}$, and substitution gives the Nambu–Goto action. Conversely, any nondegenerate induced metric solves the auxiliary-metric equation up to a Weyl transformation. This classical equivalence of Polyakov and Nambu–Goto actions is a statement about classical embeddings; quantum equivalence additionally requires treatment of the metric measure and anomaly.

Variation of $X^m$ in the metric action, followed by elimination of $\gamma$, gives the Nambu–Goto equations of motionA closed string has periodic $X$ and no spatial endpoint variation.

For a background metric $G$, Kalb–Ramond field $b$, and dilaton $\Phi$, one consistent Lorentzian convention isThe orientation fixes the two-form sign; here the Lorentzian curvature convention is chosen so Wick rotation gives the positive Euclidean dilaton term $S_{E,\Phi }=\int \sqrt{\gamma }\Phi R^{(2)}/(4\pi )$. This avoids hiding the convention in the topology argument. For constant vacuum value ${\Phi }_0$, the Gauss-Bonnet theorem gives $S_{E,\Phi }={\Phi }_0\chi$. A connected closed oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$, soThe connected vacuum amplitude has a string genus expansion ${\sum }_{g\ge 0}{g}_s^{2g-2}{\mathcal{A}}_g$; disconnected vacuum diagrams exponentiate the connected sum. Each additional handle supplies a factor $g_s^2$. This is dilaton Euler-characteristic weighting.

The rotating circle. For the specified embedding, the squared spatial speed and tangent length are both $1/2$, and their spatial inner product is zero. Including $X^0=t$ therefore givesThe metric is constant. Its equation reduces to $\ddot{X}-X^{\prime \prime }=0$, satisfied by the left- and right-moving trigonometric components and the linear time component. On a constant-time slice with $0\le \sigma <2\pi$, the proper length of a string isIt is time-independent.

Let $R(\theta )$ be an ordinary two-dimensional rotation matrix. Rotate the first coordinate pair by $R(t)$ and the second by $R(-t)$. In these time-dependent Cartesian coordinates both pairs become $(\cos \sigma ,\sin \sigma )/2$. A further fixed orthogonal change of basis givesThus every spatial slice is a planar circle of radius $1/\sqrt{2}$; its plane rotates in the ambient four-space. This is a rigid circular string in four spatial dimensions, not a pulsating circle. The rotating axes establish its spatial shape, not a transformation to an inertial spacetime frame.

The momentum density obtained from the Nambu–Goto action is $P_m=-T\sqrt{-h}{h}^{ta}{\partial }_a{X}_m$. Here $P_0=-T$, hence the conserved target-space energy isThe material velocity is transverse to the tangent and has magnitude $1/\sqrt{2}$. Its Lorentz factor is $\sqrt{2}$, so the excess over $TL$ is kinetic energy. The spatial circle being stationary in rotating axes does not eliminate this energy.

Constraints and endpoints. Varying the multipliers in the Nambu-Goto phase-space action imposesThese are first-class constraints generating normal and tangential worldsheet diffeomorphisms. They remove the two longitudinal embedding degrees of freedom; they are not extra physical force laws. Hamilton's equations areFor an open string, integration by parts leaves the spatial boundary variationAllowed variational open-string boundary conditions must make this vanish for every permitted endpoint displacement, and be preserved by the evolution. The bracketed expression is the open-string endpoint momentum flux. For free displacements, set that flux to zero. In a boundary-preserving gauge with $u=0$ and finite nonzero $e$, this is the Neumann boundary condition $X^{\prime m}=0$ at each end. This shows that free-end string boundary conditions are consistent.

At such an end, $C=0$ gives $P^2=0$, and $\dot{X}=eP$ gives ${\dot{X}}^2=0$. For a nontrivial endpoint trajectory with nonzero time velocity, its spatial speed is therefore one. This null motion of a free string endpoint follows from the boundary condition together with the constraints, not merely from the bulk wave equation. The induced metric may degenerate right at a free end; the result is understood as the endpoint limit or in the Polyakov description.

For a fixed spatial $p$-plane, split coordinates into $X^A$ along its $(p+1)$-dimensional worldvolume, including time, and $X^i$ transverse to it. Set $X^{\prime A}=0$ and $X^i=y^i$ at the ends. Tangential variations are free and their momentum flux vanishes; normal variations vanish, so their flux need not vanish. Evolution preserves the fixed normal values with $P^i=0$ at the boundary in the same gauge. These mixed Neumann boundary conditions and Dirichlet boundary conditions consistently restrict the endpoint worldlines to the plane.

The massless open-string excitations then separate into a gauge vector along the worldvolume and transverse scalar fields. A scalar displacement changes the corresponding $y^i$; its interpretation is a fluctuation of the plane's embedding. These worldvolume fields from open-string massless states support the interpretation as a dynamical planar D-brane. In the bosonic theory the still lower ground state is a tachyon, indicating an unstable brane; the interpretation does not require pretending that this tachyon is a stable massless field. Appropriate superstring sectors can remove that instability.

First-class reduction. A regular set of independent constraints is first class when their Poisson brackets vanish on the constraint surface, equivalently locally $\{{\phi }_i,{\phi }_j\}=C_{ij}{}^k{\phi }_k$. The coefficients can be functions on phase space. With the generator $G={\sum }_i{\xi }^i{\phi }_i$, an infinitesimal canonical gauge transformation isEach independent first-class condition removes one phase-space dimension, and quotienting its independent gauge orbit removes another. Thus the regular physical phase space hasThis regular first-class phase-space reduction requires independent constraints and gauge directions. The count can fail at singular strata or for reducible constraints; first-class closure alone does not guarantee independence. For a concrete counterexample to counting redundant equations, take two canonical pairs and ${\phi }_1=p_1$, ${\phi }_2=2p_1$. Both equations are first class, but there is only one independent condition and one gauge direction. The surviving pair $(q^2,p_2)$ has dimension two, whereas blindly substituting $N=n=2$ gives zero.

Oscillator symplectic form. The center-of-mass pair $(x^m,p_m)$ describes translations and total momentum. The nonzero Fourier coefficients ${\alpha }_k^m$ describe the standing-wave modes compatible with free-end Neumann boundary conditions; their reality condition is ${\alpha }_{-k}={\alpha }_k^{\ast }$, and ${\alpha }_0^m=\sqrt{2{\alpha }^{\prime }}{p}^m$ in the standard normalization. For $k>0$, regarding ${\alpha }_k^m$ as a complex coordinate makes its conjugate momentum $i{\alpha }_{-k,m}/k$. Equivalently write ${\alpha }_k=(q_k+i{s}_k)/\sqrt{2}$. Up to a total derivative, its kinetic term is $-q_k\cdot {\dot{s}}_k/k$, a real canonical form. Inverting this oscillator symplectic form of an open string givesThe zero oscillator commutes with nonzero oscillators but is not independent of $p$: $\{x^m,{\alpha }_0^n\}=\sqrt{2{\alpha }^{\prime }}{\eta }^{mn}$. Brackets between the center pair and independent nonzero oscillators vanish.

The quadratic constraints obeyFor example, the two terms in the bracket with $L_n=\frac{1}{2}{\sum }_j{\alpha }_j\cdot {\alpha }_{n-j}$ give equal contributions after relabelling $j$. Applying this identity to both factors of $L_m$ givesThis classical Virasoro constraint algebra has no central term and closes on the constraints, hence is first class. For $G={\sum }_n{\xi }_{-n}{L}_n$, the oscillator gauge transformation isReality is respected when ${\xi }_{-n}={\xi }_n^{\ast }$.

Light-cone reduction and mass. Choose light-cone coordinates $X^{\pm }=(X^0\pm X^{D-1})/\sqrt{2}$; a vector square is $-2{\alpha }^{+}{\alpha }^{-}+{\mathbf{\alpha }}^2$, with $D-2$ transverse components. On the proposed gauge slice ${\alpha }_{k\ne 0}^{+}=0$, the variation becomesFor ${\alpha }_0^{+}\ne 0$, every nonzero-mode gauge parameter has an invertible coefficient, so the conditions locally fix the corresponding gauge freedom. Globally this is the usual patch in which $X^{+}$ is an admissible worldsheet clock; it is not a claim about strings for which that coordinate has turning points. The zero-mode reparameterization is left over.

On that slice the nonzero Virasoro constraints are linear in the longitudinal oscillators:No nonzero longitudinal oscillator remains independent. The residual action isThis residual mass-shell action in light-cone string gauge displays the remaining zero-mode constraint. At the quantum level normal ordering replaces its oscillator term by $N-a$. One can further set $x^{+}$ equal to time and solve for the light-cone Hamiltonian $p^{-}=[{\mathbf{p}}^2+(N-a)/{\alpha }^{\prime }]/(2p^{+})$.

Canonical quantization gives the transverse relationsHere $\hslash =1$, ${\alpha }_{k>0}$ annihilate the oscillator vacuum, and ${\alpha }_{-k}$ create excitations. Define $a_k^I={\alpha }_k^I/\sqrt{k}$ for $k>0$. The string level operator isThus it counts oscillator number weighted by mode number, and has nonnegative integer eigenvalues. The residual mass-shell condition givesThe constant $a$ is the normal-ordering constant of a string, or intercept, not an extra classical tension. In the usual Lorentz-invariant critical bosonic string, level one carries the $D-2$ transverse polarizations of a massless vector. A massive vector would need $D-1$ polarizations; the longitudinal one is not present. Lorentz consistency therefore requires that this vector level be massless, giving $a=1$. Equivalently the regularized transverse zero-point value $a=(D-2)/24$ gives $D=26$. This critical-vector assumption in the string intercept argument concerns the usual critical quantum theory. In $D=3$ the lone transverse level-one polarization transforms trivially under the transverse rotation group; a massive scalar interpretation is not excluded by counting. Thus the stated masslessness conclusion is not a consequence of polarization counting for arbitrary $D$. An arbitrary intercept in a transverse oscillator model also need not define the usual covariant critical theory.

The full self-dual massless spectrum. In the closed-string sector, $N$ and $\tilde{N}$ are the independent nonnegative integer oscillator levels of the two chiral sectors. The integer $n$ quantizes center momentum around the circle, $p_{\mathrm{circle}}=n/R$, and $w$ counts how many times the string winds it. At the self-dual circle $R=\sqrt{{\alpha }^{\prime }}$, zero mass requiresBoth levels are nonnegative, so their sum is at most two. Exhausting these possibilities gives the massless spectrum at the bosonic self-dual circle:

The last family is easy to miss because the uncompactified ground state is a tachyon; its positive compact energy cancels that negative contribution at these charges. There are no other possibilities: at level sum one, $n^2+w^2=2$ forces both charges to be $\pm 1$; at sum zero, $n^2+w^2=4$ gives exactly the four listed pairs. With $d=D-2$ transverse oscillators the number of independent massless polarizations is $d^2+4d+4=D^2$, hence 676 in the critical $D=26$ bosonic theory. The circle oscillator is included among the $d$ components; it should not be discarded when interpreting the lower-dimensional scalar states.

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.

Spinning-particle mechanics. Keep the mostly-plus metric and place the odd multiplier $\chi$ to the left. Varying the original PDF's dotted-fermion kinetic term givesThe last two equations are multiplier constraints. In deriving the fermion equation, the variation of $i\zeta \cdot \dot{\zeta }/2$ is $i\delta \zeta \cdot \dot{\zeta }$ after integration by parts; moving $\delta \zeta$ past $\chi$ supplies the sign in the interaction variation. The TeX conversion omits the dot, which would erase the fermionic symplectic structure; the PDF fixes it.

Under an infinitesimal Lorentz rotation the target vectors $x,p,\zeta$ transform together. The scalar contractions in the action are invariant. Its antisymmetric Noether charge isThe spin bilinear is antisymmetric because the Grassmann variables anticommute. Direct differentiation makes conservation particularly transparent:while the fermion equation givesThey cancel, so ${\dot{J}}^{mn}=0$. The same charge generates Lorentz transformations through the canonical and fermionic Dirac brackets, up to the convention for the sign of the antisymmetric transformation parameter.

The pseudoclassical spinning particle uses odd classical variables rather than assigning an ordinary commuting spatial vector to spin. Eliminating the fermionic momentum constraints givesThese are symmetric graded Poisson brackets. The even bilinears $S^{mn}$ have the Lorentz transformation law of an internal angular-momentum tensor. This is spin from Grassmann bilinears.

Quantization replaces the bracket by an anticommutator,The Clifford algebra is represented on spinors. The constraint $p\cdot \zeta =0$ then becomesthe massless Dirac equation. Its square gives the massless wave equation, matching $p^2=0$. The quantized spin generator is $-i[{\gamma }^m,{\gamma }^n]/4$. This spinning-particle Dirac quantization explains why the wavefunction carries a spinor index even though the original $x$ variables are vector coordinates.

The NS phase-space action. A Fourier action compatible with the displayed constraints isThe ${\chi }_r$ are odd multipliers; an overall phase can instead be absorbed in their definition. The $b_r$ here are fermionic matter modes, not the $b$ antighost of Question3. This Neveu–Schwarz Fourier phase-space action gives the bosonic oscillator symplectic form and $\{b_r^m,b_s^n{\}}_{\mathrm{PB}}=-i{\eta }^{mn}{\delta }_{r+s,0}$.

For an open string, the two worldsheet fermion chiralities are related at each endpoint. Opposite relative signs at the two ends produce, after doubling the interval, $\psi (\sigma +2\pi )=-\psi (\sigma )$. The Neveu–Schwarz sector therefore has half-integer Fourier frequencies $r\in \mathbb{Z}+1/2$, with no matter-fermion zero mode. Equal endpoint signs give the periodic Ramond sector instead. The bosonic zero-mode normalization is again ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$.

In old covariant string quantization, all target components are retained and obeyOther oscillator brackets vanish. The momentum-labelled oscillator vacuum satisfies ${\alpha }_{k>0}|0;p⟩=b_{r>0}|0;p⟩=0$, ${\hat{p}}_m|0;p⟩=p_m|0;p⟩$, and ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$; take its matter norm positive. The zero-point convention stated in the question givesA physical oscillator ground state has ${\alpha }^{\prime }{p}^2=a$, soIt is a scalar, not a spinor, since the NS sector has no Clifford zero modes. It is a tachyon if $a>0$, massless if $a=0$, and massive if $a<0$. In the usual limiting $a=1/2$ theory it is the unprojected NS tachyon, removed by the standard GSO projection. Calling it tachyonic before specifying the sign of $a$ would be too strong.

The half-level vector and its norm. Let $|A;p⟩=A_m(p)b_{-1/2}^m|0;p⟩$. The level is $N_{\mathrm{NS}}=1/2$. Its $L_0$ condition and the only potentially nonzero positive supercurrent condition giveIndeed $\{G_{1/2},b_{-1/2}^m\}={\alpha }_0^m$ implies $G_{1/2}|A;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot A|0;p⟩$. Higher positive $G_r$ annihilate the state. Likewise $[L_n,b_{-1/2}^m]=(1/2-n/2)b_{n-1/2}^m$ is either zero or an annihilator, so all $L_{n>0}$ conditions hold. This is the half-level Neveu–Schwarz vector state.

Its matter norm is proportional to $A^{\ast }\cdot A$. For a real, nonzero on-shell momentum in ordinary $D\ge 2$ Minkowski space, three cases exhaust the possibilities:

Thus, with the usual nonzero particle-momentum hypothesis,This is a level-specific result, not a proof of the full NS no-ghost theorem in arbitrary dimension.

At the limiting intercept the vector becomes massless, and the longitudinal polarization is a null state. It is generated by $G_{-1/2}|0;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot b_{-1/2}|0;p⟩$. Factoring out that null direction identifies $A\sim A+\kappa p$ and leaves $D-2$ positive physical polarizations of a gauge vector. In the consistent critical NS string one additionally has $D=10$; the half-level positivity argument alone does not derive that dimension.

There is a genuine zero-momentum exception if the printed word “all” is read literally. At $a=1/2$ and $p=0$, the polarization $A=(1,0,\dots )$ obeys every displayed positive-mode constraint and the mass shell, but has norm $-1$. Transversality is vacuous, and $A\sim A+\kappa p$ removes nothing. This nonzero-momentum condition in the massless vector norm test is therefore necessary for the intended endpoint assertion. Including zero momentum changes the unrestricted elementary test to strict $a<1/2$; ordinary propagating massless particle states use nonzero null momentum.

Finally, an oscillator vacuum used to build the excited state is labelled by that state's momentum. It need not separately satisfy the scalar-ground-state mass shell: imposing both ${\alpha }^{\prime }{p}^2=a$ and ${\alpha }^{\prime }{p}^2=a-1/2$ at the same momentum would incorrectly exclude all such excitations.