= Solution
The <old covariant quantization> of a free-ended bosonic string retains oscillators in all target directions. In mostly-plus <Minkowski spacetime>, set $\alpha_0^m=\sqrt{2\alpha'}p^m$ and
$$
[\alpha_k^m,\alpha_l^n]=k\,\eta^{mn}\delta_{k+l,0},\qquad
L_n=\frac12\sum_k:\alpha_{n-k}\cdot\alpha_k:,\qquad
L_0=\alpha'p^2+N.
$$
Time-coordinate excitations have negative norms in this covariant <Fock space>. The <string ghost states> in this discussion are unwanted negative-norm physical states, distinct from the anticommuting <Faddeev-Popov ghost fields> in the <path integral>. Constraints and the <null-state quotient of a string> remove unphysical polarizations while maintaining target-space <Lorentz covariance>.
\b[Why only positive-mode constraints annihilate states.] The quantum <Virasoro algebra> is
$$
[L_m,L_n]=(m-n)L_{m+n}+\frac D{12}m(m^2-1)\delta_{m+n,0}.
$$
The physical conditions are
$$
\boxed{(L_0-a)|\psi\rangle=0,\qquad L_n|\psi\rangle=0\quad(n>0),}
$$
where $a$ is the <string intercept>. If both positive and negative modes annihilated states, $[L_1,L_{-1}]=2L_0$ would force $a=0$. Then $[L_2,L_{-2}]=4L_0+D/2$ would force $D=0$. Thus the <Virasoro central extension> and shifted zero mode prevent imposing every classical constraint strongly in a nontrivial string. As in <Gupta-Bleuler quantization>, negative-mode conditions act on physical bras, rather than also annihilating physical kets. Physical <null string states> are quotiented because their inner products with all physical states vanish.
\b[The intercept bound at level one.] For $|\epsilon;p\rangle=\epsilon_m\alpha_{-1}^m|p\rangle$, the constraints and norm are
$$
p\cdot\epsilon=0,\qquad p^2=\frac{a-1}{\alpha'},\qquad
\langle\epsilon;p|\epsilon;p\rangle=\epsilon^*\cdot\epsilon.
$$
If $a>1$, momentum is spacelike and its orthogonal complement contains a timelike negative-norm polarization. The <level-one intercept bound in covariant string quantization> is therefore
$$
\boxed{a\leq1.}
$$
For $a<1$, momentum is timelike and the $D-1$ orthogonal polarizations are positive. This avoids level-one ghosts but gives a <massive vector> with one more polarization than the transverse <light-cone gauge in string theory> spectrum. Ghost absence alone is weaker than equivalence.
For $a=1$, momentum is null. Its orthogonal complement contains $D-2$ positive directions and the null direction $p$. The state $p\cdot\alpha_{-1}|p\rangle$, proportional to $L_{-1}|p\rangle$, is physical, spurious and null. Removing it gives
$$
\boxed{\epsilon\sim\epsilon+\lambda p.}
$$
The quotient has exactly the $D-2$ massless transverse vector polarizations. Thus \b[equivalence at level one selects $a=1$ and the <null-state quotient of a string>], not just the inequality.
\b[Level two and the dimension.] Set $a=1$. At level two, $p^2=-1/\alpha'$ and $k=\sqrt{2\alpha'}p$ has $k^2=-2$. A general state is
$$
|\psi\rangle=\left(\frac12\epsilon_{mn}\alpha_{-1}^m\alpha_{-1}^n
+\zeta_m\alpha_{-2}^m\right)|p\rangle,\qquad\epsilon_{mn}=\epsilon_{nm}.
$$
Using $[L_m,\alpha_n^a]=-n\alpha_{m+n}^a$, the nontrivial positive-mode conditions are
$$
k^m\epsilon_{mn}+2\zeta_n=0,\qquad
\epsilon^m{}_m+4k\cdot\zeta=0.
$$
There are $D-1$ vector <null string states> $L_{-1}(v\cdot\alpha_{-1}|p\rangle)$ with $k\cdot v=0$. The parent has $L_0=0$, so the descendant norm is zero. Quotienting these leaves the massive <symmetric traceless rank-two tensor> plus one additional scalar.
The <level-two scalar in covariant string quantization> can be chosen, for every $D$, as
$$
|S_D\rangle=\left[\alpha_{-1}\cdot\alpha_{-1}
+\frac{D+4}{10}(k\cdot\alpha_{-1})^2
+\frac{D-1}{5}k\cdot\alpha_{-2}\right]|p\rangle.
$$
On its three displayed structures, $L_1$ gives respectively $2,-4,2$ times $k\cdot\alpha_{-1}|p\rangle$, and $L_2$ gives $D,-2,-4$ times the vacuum. The coefficients make both combinations vanish. The first two structures have norms $2D,8$ and cross inner product $-4$; the mode-two structure has norm $-4$ and is orthogonal to them. For $B=(D+4)/10$, $C=(D-1)/5$, this yields
$$
\boxed{\langle S_D|S_D\rangle
=2D-8B+8B^2-4C^2
=\frac{2}{25}(D-1)(26-D).}
$$
Above 26 this is a physical <negative-norm string state>. Below 26 it is an extra positive-norm scalar, which cannot be discarded just to force the ordinary light-cone state count. At 26 it becomes null and can be removed. Thus level-two ghost absence gives $D\leq26$, whereas \b[equivalence to the ordinary transverse spectrum requires $D=26$].
The critical scalar is also a <Virasoro descendant>. For $L_0|p\rangle=-|p\rangle$, the <level-two scalar Virasoro null state> candidate
$$
|Z\rangle=\left(L_{-2}+\frac32L_{-1}^2\right)|p\rangle
$$
obeys
$$
L_1|Z\rangle=0,\qquad L_2|Z\rangle=\frac{D-26}{2}|p\rangle.
$$
At $D=26$ it is physical and null, with $|S_{26}\rangle=2|Z\rangle$. At other dimensions it is not physical, so its norm must not be treated as a physical ghost test; the already-physical $|S_D\rangle$ gives the correct test.
Finally, the covariant level-two oscillator space has dimension $D(D+3)/2$. The $D$ conditions from $L_1$ and one from $L_2$ leave $(D-1)(D+2)/2$ physical components. Quotienting $D-1$ vector null states and the critical scalar null state leaves
$$
\boxed{\frac{D(D-1)}2-1=\frac{(D-2)(D+1)}2,}
$$
the <massive spin-two field> count and the level-two <light-cone gauge in string theory> count. The vector null descendants change the components of $\zeta$ orthogonal to $k$, and the critical scalar null descendant changes its component along $k$. Thus $\zeta$ can be set to zero. A representative then has $\zeta=0$, $k^m\epsilon_{mn}=0$ and $\epsilon^m{}_m=0$. The two approaches consequently agree on the massless level-one vector and massive level-two spin-two tensor for $a=1,D=26$.
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