Normal-ordering constant of a string ($a$)

The normal-ordering constant is the zero-point shift in the quantum Virasoro generator. Closure of the target-space Lorentz algebra fixes $a=1$ and the spacetime dimension $D=26$ for the bosonic string.

In the usual critical Lorentz-invariant bosonic string, level-one transverse vector polarizations fit a massless representation, whereas a massive vector would require an extra polarization. This fixes the intercept to one. Polarization counting alone does not establish that conclusion in every dimension: in three dimensions the one transverse polarization transforms trivially under the transverse rotation group and can be interpreted as a scalar. The quantum consistency assumptions must accompany an argument that the first level must be massless.

Each of $d$ physical transverse bosons with a Neumann-Dirichlet open-string boundary condition contributes $\frac{1}{2}{\sum }_{r>0}r=1/48$ by the half-integer zeta-regularized mode sum. Thus $L_0^{\perp }=N+d/48$ and the convention $L_0={\alpha }^{\prime }{p}^2+N-a$ has $a=-d/48$. For $d=24$, the vacuum shift is $+1/2$. The ordinary integer-moded open-string shift is instead $-d/24$; these endpoint sectors have different normal-ordering constants.

With 24 physical half-integer open-string oscillators, the first three string level operator eigenvalues are $0,1/2,1$. Their states are the vacuum, $a_{1/2}^{i†}|0⟩$, and $a_{1/2}^{i†}{a}_{1/2}^{j†}|0⟩$. The creation operators commute, so the third level is the symmetric square of the 24-dimensional transverse vector, of dimension $24\cdot 25/2=300$. Its trace is a scalar and its trace-free part has dimension 299. Including the vacuum shift gives ${\alpha }^{\prime }{M}^2=1/2,1,3/2$ when the two longitudinal directions are common directions with Neumann boundary conditions. The representation labels refer to the surviving ND rotation group, not full unbroken target Lorentz symmetry.