Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 46 / 3 / Solution

For the relativistic particle phase-space action, the first-class constraint $\phi =(p^2+m^2)/2$ generatesIndeed the integrand varies by $\frac{1}{2}d[ϵ(p^2-m^2)]/dt$. The canonical gauge transformation is an invariance when the gauge parameter vanishes at fixed temporal endpoints, or when all fields and the parameter are periodic. The boundary restriction matters for the proper-time modulus.

Normalize the worldline interval to $[0,1]$. Thenis invariant because $\delta s=ϵ(1)-ϵ(0)=0$. Every allowed $e$ in its orbit can be written $e(t)=s+\dot{ϵ}(t)$: set $ϵ(t)={\int }_0^t[e(u)-s]du$. Thus $s$ remains a gauge-invariant integration variable, not another removable nonconstant mode. For a worldline circle the constant gauge parameter is a residual zero mode. Without the endpoint restriction, the assertion that $s$ is invariant would not hold.

The worldline gauge-orbit determinant is the Jacobian from gauge-orbit coordinates $ϵ$ to the nonconstant part of $e$ is the Faddeev-Popov determinant of ${\partial }_t$. Equivalently, the gauge-fixing identity has the formThe determinant is taken between the appropriate boundary-condition spaces, with the modulus removed; on a circle the prime also removes the constant parameter. Gauge fixing therefore leaves a factor $\det {}^{\prime }{\partial }_t$ and a modulus measure, after division by any residual gauge volume. Even though this determinant is field independent in the present Abelian example, it is the required change-of-variables Jacobian. A Grassmann integral over the Faddeev-Popov ghosts exponentiates it:The overall determinant phase depends on the integration convention and can be absorbed into normalization. Zero modes and the same endpoint restrictions must be treated separately rather than included in an invertible determinant.

For the free-ended open string, take $0\le \sigma \le \pi$. A canonical cosine expansion at a fixed time isIt implements the Neumann boundary conditions and has ${\alpha }_n^{†}={\alpha }_{-n}$. With ${\alpha }_0^m=p^m/\sqrt{\pi T}$, its Nambu-Goto phase-space action, up to a total time derivative, isReality requires ${\lambda }_{-n}={\lambda }_n^{\ast }$. Numerical factors can be absorbed into these Lagrange multipliers. In this covariant quantization of the bosonic string the oscillators retain all $D$ spacetime components, in contrast to the transverse oscillators in the preceding solution. The canonical commutation relations areThe oscillator vacuum $|0;p⟩$ is annihilated by ${\alpha }_n^m$ for $n>0$. Its momentum label will sometimes be suppressed.

Define the matter Virasoro algebra generators using normal ordering:No additive intercept is included in this definition of $L_0$. For $n>0$ the indices of the two factors in each term add to $n$; they cannot both be negative. After normal ordering there is a positive-mode annihilation operator on the right, possibly accompanied by the zero mode. Hence $L_n|0;p⟩=0$ for every $n>0$. In $L_0$, commuting positive modes past negative modes formally adds $\frac{D}{2}{\sum }_{k>0}k$. This divergent constant needs a prescription, and a finite shift is an ordering ambiguity. Our convention instead puts the physical string intercept into the constraint $L_0-a$.

With this convention the matter Virasoro algebra isA different additive constant in $L_0$ would change the linear-in-$m$ central term, so stating the convention is essential.

For the worldsheet ghost fields, useand choose a ghost oscillator vacuum withThen $c_0|0{⟩}_{\mathrm{gh}}\ne 0$ since $\{b_0,c_0\}=1$. This choice specifies the ghost zero-mode doublet; it is not a claim that both zero modes annihilate one state. With the printed ghost Virasoro zero-mode convention, $L_0^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ and $L_m^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ for $m>0$. The latter follows by putting positive ghost modes on the right; a possible contraction requires $m=0$ and is absent here.

Apply the supplied BRST charge to the matter state times this ghost oscillator vacuum. Terms with a rightmost $c_m$, $m>0$, vanish, as do the positive-mode ghost generators. ThusThese one-ghost states are independent, as can also be seen by applying $b_0,b_m$. Therefore the BRST physical-state constraints of an open string areFor the matter oscillator vacuum, $L_0={\alpha }^{\prime }{p}^2$, so $p^2=1/{\alpha }^{\prime }$ and $M^2=-1/{\alpha }^{\prime }=-2\pi T$: the physical ground state is a tachyon. The momentum must satisfy this equation; the zero-momentum oscillator vacuum by itself would not be BRST-closed.

Matter and ghost generators commute with one another. Add their two algebras and write ${\mathcal{L}}_m=L_m+L_m^{\mathrm{gh}}$. The given ghost constant must be retained:Define the shifted generators ${\hat{\mathcal{L}}}_m={\mathcal{L}}_m-{\delta }_{m,0}$. ThenBRST nilpotence requires cancellation of the anomalous central term in this shifted constraint algebra, together with the intercept one already present in the charge. At $m=2,n=-2$ the remaining anomalous coefficient is $(D-26)/2$, so $D=26$. At this value the shifted total generators obey the Witt algebra. The unshifted ${\mathcal{L}}_m$ still have the displayed linear zero-mode shift; it must not be silently discarded.