Why for the closed string we need to take both left and right moving modes into account when deriving the mass?

Question

In the closed bosonic string we have both left and right moving sectors with modes $\alpha_m^\mu$ and $\tilde{\alpha}_m^\mu$ respectively where $m\in \mathbb{Z}$ and where we have $\alpha_0^\mu = \tilde{\alpha}_0^\mu = \frac{\ell_s}{2}p^\mu$. Now the Virasoro generators are given by

$$L_m=\frac{1}{2}\sum_{n=-\infty}^\infty \alpha_{m-n}\cdot \alpha_n,\quad \tilde{L}_m=\frac{1}{2}\sum_{n=-\infty}^\infty \tilde\alpha_{m-n}\cdot \tilde\alpha_n\tag{1}$$

and classically the vanishing of the energy-momentum tensor requires $L_m = \tilde{L}_m =0$. Now it is said in some references (for example Becker, Becker & Schwarz, page 40) that to get the mass formula $M^2$ for the classical closed string we need to take both left-moving and right-moving modes into account. I don't get this, my impression is that we can obtain $M^2$ from just $L_0=0$ or $\tilde{L}_0=0$. Indeed:

$$L_0 = \frac{1}{2}\sum_{n=-\infty}^\infty \alpha_{-n}\cdot \alpha_n = \frac{1}{2}\alpha_0^2+\frac{1}{2}\sum_{n=-\infty}^{-1}\alpha_{-n}\cdot \alpha_n +\frac{1}{2}\sum_{n=1}^\infty \alpha_{-n}\cdot \alpha_n=-\frac{\ell_s^2}{8}M^2+\sum_{n=1}^\infty \alpha_{-n}\cdot \alpha_n\tag{2}$$

where in the third equality we have reindexed $m =-n$ in the first sum, used that the classical variables commute and used the definition of $\alpha_0$. When we set $L_0 =0$ we are able to get $$M^2=\frac{8}{\ell_s^2}\sum_{n=1}^{\infty}\alpha_{-n}\cdot \alpha_n\tag{3}.$$

Now the same derivation in (2) with $\tilde{L}_0$ gives $$M^2=\frac{8}{\ell_s^2}\sum_{n=1}^\infty \tilde{\alpha}_{-n}\cdot \tilde{\alpha}_n\tag{4},$$

and clearly if we sum them up and divide by two we find $$M^2=\frac{4}{\ell_s^2}\sum_{n=1}^\infty \left(\alpha_{-n}\cdot \alpha_n+\tilde{\alpha}_{-n}\cdot \tilde{\alpha}_n\right)\tag{5}.$$

My impression is that we can express $M^2$ either in terms of just $\alpha_m$ as in (3), equivalently in terms of just $\tilde{\alpha}_n$ as in (4) or in terms of both as in (5). But still, as I said, some references say that for the closed string we need both $L_0=0$ and $\tilde{L}_0=0$ to derive the mass. What am I missing here? We do we need both $L_0=0$ and $\tilde{L}_0=0$ to derive the mass formula?

Answer 1

1. You don't need to take both left- and right-moving modes. OP's equations $(3)$ and $(4)$ are both correct - the mass can be expressed as a sum over one set of modes only: $$ M^2=\frac4{\alpha'}\sum_{n=1}^\infty\alpha_n\cdot\alpha_{-n}=\frac4{\alpha'}\sum_{n=1}^\infty\tilde\alpha_n\cdot\tilde\alpha_{-n} $$

2. $L_0=0$ and $\tilde L_0=0$ - and more generally, $L_n=0$, $\tilde L_n=0$ - due to the vanishing of the independent energy-momentum tensor components $T_{--}=0$ and $T_{++}=0$, respectively. The equivalence $L_0=\tilde L_0$ is needed in the classical formula only if one wishes to express the mass in terms of both sets of oscillators.

   In the quantum theory, however, assuming a non-compact background for simplicity, the fact that $\hat L_0\simeq \hat{\tilde{L}}_0$ when acting on physical states implies level matching: $N=\tilde N$, and also ensures that the mass-squared matches between left-movers and right-movers.