Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 306 / 4

In old covariant quantization, an open-string physical string state obeyswith analogous left- and right-moving conditions for a closed string. A spurious string state is a Virasoro descendant orthogonal to every physical state. A null string state is both physical and spurious; it has zero norm and zero inner product with every physical state, so quotienting by null states removes gauge redundancy without removing observable states.

Let $|s⟩=L_{-1}|\chi ⟩$ with $L_n|\chi ⟩=0$ for every $n\ge 0$. Since $[L_0,L_{-1}]=L_{-1}$, $|s⟩$ has $L_0$ eigenvalue one. Moreover,and for $n>1$,It is therefore physical. It is a Virasoro descendant and hence spurious, while its norm isThus it is null.

SetIts $L_0$ eigenvalue is $-1+2=1$. The Virasoro algebra givesso $L_1|s⟩=0$. At critical central charge $c=26$,so $L_2|s⟩=0$. All $L_n$ with $n\ge 3$ annihilate it directly. Hence $|s⟩$ is physical and, being a positive-level Virasoro descendant, spurious; it is therefore a null state.

The zero-mode condition is automatic:Using $[L_m,{\alpha }_n^{\mu }]=-n{\alpha }_{m+n}^{\mu }$ and oscillator annihilation of the vacuum giveswhich vanishes when $t_{\mu \nu }{\alpha }_0^{\nu }=-v_{\mu }$. Similarly,which vanishes when $v\cdot {\alpha }_0=-t^{\mu }{}_{\mu }/2$. Higher positive Virasoro modes annihilate a level-two state. These are precisely the stated physical-state conditions.

Using ${\alpha }_0^2=-2$, tracelessness and transversality of ${ϵ}_{\mu \nu }$ givesand $v\cdot {\alpha }_0=-t^{\lambda }{}_{\lambda }/2$, so the constraints hold.

On the momentum vacuum,andThereforewhereThe vacuum has $L_0|0,p⟩={\alpha }_0^2|0,p⟩/2=-|0,p⟩$ and is annihilated by positive modes, so part c proves that $|n⟩$ is null. The physical content is consequently the transverse traceless tensor ${ϵ}_{\mu \nu }$.