1

2π X l>0 b aµ(-l) b aµ (l)

3.82) and similarly for φ-(0). I just wrote the sum over the positive and negative l’s separately, and replaced aµ(0) by the total momentum. In this form, there will never be a solution to that constraint, simply because the sums will never converge both when acting on some state. If the first converges when applied to a state |Ψi, we must have b aµ(l) b aµ (-l) | Ψi → 0 for l → ∞. (3.83) Using the commutator of the a’s this implies b aµ(-l) b aµ(l) | Ψi - π¯ h D l | Ψi → 0 for l → ∞, (3.84) where D is the dimension of the target spacetime. The second term will never converge, and so the second sum above cannot converge either. We have to reorder the operators such that both sums can converge. The condition for this is that almost all terms appear in the same order in both sums. But which order shall we choose? To see that there is a canonical choice, consider the following commutators of complex conjugate operators [b aµ(-k), b aν(k)] = π¯ h k gµν, [b bµ(k),b bν(-k)] = π¯

4π¯ h X k |k| qµ(k) qµ (-k), (3.91) with some wave function Φ(Q) that depends on the center of mass variable Qµ = qµ(0)/2π. When acting with the b φ±(0) on this functional, all terms in the sums vanish, and we are left with the Klein Gordon equation ( b Pµ b Pµ + 8π¯ h N) Φ(Q) = 0. (3.92) Here we need that the two normal order constants are equal, as otherwise we would get a contradiction. The state given can be interpreted as the ground state of the string’s internal degrees of freedom. In this ground state the string behaves like a point particle decribed by the wave function Φ(Q), which is completely analogous to the relativistic point particle in the last section. However, its mass depends on the choice we have made for the normal order constant, so effectively it depends on the operator ordering in the constraints. The explicit appearance of Planck’s constant in m2 = 8π¯ hN means that