Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 46 / 4 / Solution

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.