Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 54 / 2 / Solution

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.