Physics Asked on July 13, 2021
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 $min mathbb{Z}$ and where we have $alpha_0^mu = tilde{alpha}_0^mu = frac{ell_s}{2}p^mu$. Now the Virasoro generators are given by
$$L_m=frac{1}{2}sum_{n=-infty}^infty alpha_{m-n}cdot alpha_n,quad tilde{L}_m=frac{1}{2}sum_{n=-infty}^infty tildealpha_{m-n}cdot tildealpha_ntag{1}$$
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:
$$L_0 = frac{1}{2}sum_{n=-infty}^infty alpha_{-n}cdot alpha_n = frac{1}{2}alpha_0^2+frac{1}{2}sum_{n=-infty}^{-1}alpha_{-n}cdot alpha_n +frac{1}{2}sum_{n=1}^infty alpha_{-n}cdot alpha_n=-frac{ell_s^2}{8}M^2+sum_{n=1}^infty alpha_{-n}cdot alpha_ntag{2}$$
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 $$M^2=frac{8}{ell_s^2}sum_{n=1}^{infty}alpha_{-n}cdot alpha_ntag{3}.$$
Now the same derivation in (2) with $tilde{L}_0$ gives $$M^2=frac{8}{ell_s^2}sum_{n=1}^infty tilde{alpha}_{-n}cdot tilde{alpha}_ntag{4},$$
and clearly if we sum them up and divide by two we find $$M^2=frac{4}{ell_s^2}sum_{n=1}^infty left(alpha_{-n}cdot alpha_n+tilde{alpha}_{-n}cdot tilde{alpha}_nright)tag{5}.$$
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?
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: $$ M^2=frac4{alpha'}sum_{n=1}^inftyalpha_ncdotalpha_{-n}=frac4{alpha'}sum_{n=1}^inftytildealpha_ncdottildealpha_{-n} $$
$L_0=0$ 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 $hat L_0simeq hat{tilde{L}}_0$ 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.
Correct answer by Nihar Karve on July 13, 2021
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