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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?

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  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.

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    $\begingroup$ hi @NiharKarve; if you say (and i quote) `this "level matching" constraint serves to cancel the normal-ordering constant.', then it can lead one to conclude that one imposes level matching in order to cancel the normal ordering constant; but this is (as far as I understand) wrong, and therefore misleading for a novice reading your answer. So perhaps you meant to write something slightly different; I don't know why you are mentioning the normal ordering constant in the first place. $\endgroup$ Jun 18, 2021 at 11:36
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    $\begingroup$ @Wakabaloola you are correct, and I will edit the post; it was indeed poorly worded $\endgroup$ Jun 18, 2021 at 11:37
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    $\begingroup$ There are a few ways of seeing why imposing level matching is a good thing, two of which are the following: 1) if you impose level matching then this ensures all the tensor indices are properly contracted in order to get something reparametrisation-invariant (e.g., see Weinberg's 1985 article on vertex operators); 2) the phase generated by acting with $L_0-\tilde{L}_0$ on a local vertex operator is in any case not globally-defined.. $\endgroup$ Jun 18, 2021 at 11:44
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    $\begingroup$ .. To define a phase you need two non-zero (e.g., unit) tangent vectors. But Hopf's index theorem says that globally-defined tangent vector must vanish at a discrete set of points on a general Riemann surface, and therefore the corresponding phase cannot be defined everywhere on the surface either. (see, e.g, the discussion beginning on page 82 in arxiv.org/pdf/1912.01055.pdf) $\endgroup$ Jun 18, 2021 at 11:44
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    $\begingroup$ @Wakabaloola Thanks! (and edited) $\endgroup$ Jun 18, 2021 at 11:45

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