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After reading the chapter "Strings with world-sheet supersymmetry" in Becker, Becker, Schwarz book multiple times, I am still confused about the following things.

1) Open string doubt

They prove that, in order for the boundary term in the variation of the action to vanish, one has to impose EITHER Ramond condition OR Neveu-Schwarz condition, which lead to different mode expansions. After quantization and GSO projection, they then analyse the spectrum and find that the number of bosons in NS sector is the same as the number of fermions in the R sector, in both the ground state and first excited level. They then conclude that the theory is supersymmetric.

If one has to choose only one of the two mentioned condition, how can one have both spectra? How can the theory really be supersymmetric? I would say that it is just a curious feature of the theory, but I clearly miss their point.

2) Closed string doubt

As far as I understood, type II theories are closed string theories. At page 137, they again state what happens in each "sector" in both type IIA and type IIB. If I, for example, choose the NS-NS sector (which is the same for both theories), it would seem that there are no fermions in the theory.

What am I missing again?

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The ad-hoc quantization procedures for going from the classical theory of the superstring to the quantum superstring theory are just that - ad-hoc. We're not really looking at a single string with chosen boundary conditions and quantizing it in some clearly prescribed canonical fashion, we're playing around with the superstring and are trying to get to a well-defined quantum theory, one where the resulting theory on the worldsheet is conformal and which has no quantization-obstructing anomalies. In playing around with this, we find that there are these "sectors" corresponding to different boundary conditions, and we find that if we combine them in a particular fashion (which we then call grandiosely call "GSO projection"), then we get a consistent quantum theory.

If you are bound to the ontology that we are really thinking about "actual strings" in a 10-dimensional spacetime, then you must conclude that the states of the theory in which we have the sectors combined is the theory of more than one string, with different boundary conditions for different strings. But once we have arrived at the quantum theory, there is no way back - we cannot "unquantize" this theory to arrive back at some picture of a classical string corresponding to a particular state in the quantum theory.

If instead you view "string" theory as a theory of (supersymmetric) conformal field theories living on worldsheets, there is nothing strange about this.

For a little bit more about the two different ontologies, see this answer of mine.

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  • $\begingroup$ Thank you for the answer and for the link you cited. From what I understand then, the answer to my doubts is that, in the framework in which string theory is thought of as a theory of strings propagating in spacetime, we are considering a lot of different strings, which, depending on the boundary conditions we give, can appear as different particles. They can be fermions or bosons depending on these boundary conditions. Is this right? If this is true, then I would say that the spectrum of superstring theory is given by all the sectors combined, not just from one sector. Do you agree? $\endgroup$ – samario28 Mar 10 at 23:20
  • $\begingroup$ @samario28 It is not "the strings" that appear as different particles, as I say there is no "unquantization map" that would map specific quantum states to specific classical situations. It would be more common to say that the particle states are "excitations of the string(s)", but even that you shouldn't take too literally. And no, you won't get "the spectrum of superstring theory" by just taking all sectors, that's not a consistent quantum theory. As whatever text you're reading should point out, we get in fact five different quantum superstring theories for different choices of sectors. $\endgroup$ – ACuriousMind Mar 10 at 23:26
  • $\begingroup$ Damn, I thought I understood your first answer. Let me then rephrase what I said: type IIB spectrum is given by taking the states appearing IN ALL the sectors R-R, NS-NS, NS-R, R-NS (with the two R vacua having the same chirality). So you don't just take one sector, as there could be multiple strings... right? $\endgroup$ – samario28 Mar 11 at 8:14

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