= Solution
\b[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 gives
$$
\boxed{\dot p_m=0,\qquad \dot x^m=e p^m+i\chi\zeta^m,
\qquad\dot\zeta^m=-\chi p^m,\qquad p^2=0,\qquad p\cdot\zeta=0.}
$$
The 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> is
$$
\boxed{J^{mn}=x^mp^n-x^np^m-i\zeta^m\zeta^n,
\qquad S^{mn}=-i\zeta^m\zeta^n.}
$$
The spin bilinear is antisymmetric because the <Grassmann variables> anticommute. Direct differentiation makes conservation particularly transparent:
$$
\frac d{dt}(x^mp^n-x^np^m)
=i\chi(\zeta^mp^n-\zeta^np^m),
$$
while the <fermion> equation gives
$$
\frac d{dt}S^{mn}=-i\chi(\zeta^mp^n-\zeta^np^m).
$$
They 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> gives
$$
\boxed{\{\zeta^m,\zeta^n\}_{D}=-i\eta^{mn}.}
$$
These 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,
$$
\{\widehat\zeta^m,\widehat\zeta^n\}_+=\eta^{mn},
\qquad\widehat\zeta^m=\frac1{\sqrt2}\gamma^m,
\quad\{\gamma^m,\gamma^n\}_+=2\eta^{mn}.
$$
The <Clifford algebra> is represented on spinors. The <constraint> $p\cdot\zeta=0$ then becomes
$$
\boxed{\gamma^m\partial_m\Psi(x)=0,}
$$
the <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.
\b[The NS phase-space action.] A Fourier action compatible with the displayed <constraints> is
$$
\boxed{S_{\mathrm{NS}}=\int dt\left[
\dot x\cdot p+\sum_{k>0}\frac{i}{k}\alpha_{-k}\cdot\dot\alpha_k
+i\sum_{r>0}b_{-r}\cdot\dot b_r
-\sum_{n\in\mathbb Z}\lambda_{-n}L_n
-i\sum_{r\in\mathbb Z+1/2}\chi_{-r}G_r\right].}
$$
The $\chi_r$ 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 $\alpha_0=\sqrt{2\alpha'}p$.
In <old covariant string quantization>, all target components are retained and obey
$$
\boxed{[\alpha_k^m,\alpha_l^n]=k\eta^{mn}\delta_{k+l,0},
\qquad\{b_r^m,b_s^n\}_+=\eta^{mn}\delta_{r+s,0}.}
$$
Other oscillator brackets vanish. The momentum-labelled <oscillator vacuum> satisfies $\alpha_{k>0}|0;p\rangle=b_{r>0}|0;p\rangle=0$, $\widehat p_m|0;p\rangle=p_m|0;p\rangle$, and $\alpha_0=\sqrt{2\alpha'}p$; take its matter norm positive. The zero-point convention stated in the question gives
$$
L_0=\alpha'p^2+N_{\mathrm{NS}},\qquad
N_{\mathrm{NS}}=\sum_{k>0}\alpha_{-k}\cdot\alpha_k
+\sum_{r>0}r\,b_{-r}\cdot b_r.
$$
A physical oscillator ground state has $\alpha'p^2=a$, so
$$
\boxed{M_0^2=-\frac a{\alpha'}.}
$$
It 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.
\b[The half-level vector and its norm.] Let $|A;p\rangle=A_m(p)b_{-1/2}^m|0;p\rangle$. The level is $N_{\mathrm{NS}}=1/2$. Its $L_0$ condition and the only potentially nonzero positive supercurrent condition give
$$
\boxed{\alpha'p^2+\frac12-a=0,\qquad p\cdot A=0.}
$$
Indeed $\{G_{1/2},b_{-1/2}^m\}=\alpha_0^m$ implies $G_{1/2}|A;p\rangle=\sqrt{2\alpha'}p\cdot A|0;p\rangle$. 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 $A^*\cdot A$. For a real, nonzero on-shell momentum in ordinary $D\ge2$ Minkowski space, three cases exhaust the possibilities:
* If $a<1/2$, then $p$ is timelike and its orthogonal complement is Euclidean. In its rest frame $p\cdot A=0$ sets $A^0=0$, so every nonzero physical polarization has positive norm.
* If $a=1/2$, then $p$ is null. Choose $p=E(1,0,\ldots,0,1)$, $E\ne0$. Transversality sets $A^{D-1}=A^0$, leaving $A^*\cdot A=\sum_{I=1}^{D-2}|A^I|^2\ge0$. Polarizations proportional to $p$ are null.
* If $a>1/2$, then $p$ is spacelike. A frame with $p=(0,k,0,\ldots)$ allows $A=(1,0,\ldots)$, which satisfies $p\cdot A=0$ but has norm $-1$. It is an explicit negative-norm physical polarization at this level.
Thus, with the usual nonzero particle-momentum hypothesis,
$$
\boxed{\text{all physical half-level vector polarizations have nonnegative norm}
\quad\Longleftrightarrow\quad a\le\frac12.}
$$
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 $G_{-1/2}|0;p\rangle=\sqrt{2\alpha'}p\cdot b_{-1/2}|0;p\rangle$. 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 $A=(1,0,\ldots)$ 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 $\alpha'p^2=a$ and $\alpha'p^2=a-1/2$ at the same momentum would incorrectly exclude all such excitations.
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