Boson in Superstring

Question

I'm confused about a point. Superstring sigma model is

$$ S=-\frac{T}{2}\int\mathrm{d}^2z \left[\eta^{ab}\partial_aX^\mu\partial_b X_\mu -i\bar\psi^\mu\rho^a\partial_a\psi_\mu \right], $$

of course, the first term is in common with bosonic string one.

Then in addition to the bosonic string spectrum (the one coming from $X$s), that I have, as usual, I will have also the spectrum coming form $\psi$s. My questions are:

1. What is the fate of the bosonic string spectrum in superstring? i.e. how should I interpret the dilaton $\Phi$, the graviton $g_{\mu\nu}$ and the 2-form $B_{\mu\nu}$ coming form bosonic string spectrum? Why all books refer to the dilaton, gravinton and 2-form as the ones coming from NS part of $\psi$s spectrum?
2. After GSO projection the tachyon is cancelled form $\psi$s spectrum and the number of bosonic d.o.f. equals the number of fermionic ones. But this is referred again to the $\psi$s spectrum. If I consider also the $X$s spectrum I still have tachyon and extra boson that unbalance the d.o.f. counting.

Probabilly I make a mistake in my reasonment.

Answer 1

Your confusion comes from thinking that going to superstrings simply means adding fermions in the spectrum. The spectrum is instead different. For bosonic string (let's focus on NN boundary conditions and open strings) you have something like:

$$\alpha' m^2=N-1$$

where N is the number operator of the transverse vibrational excitations of the bosonic string. In superstring you find:

$$\alpha' m^2=N_{bos}+N_{ferm}-a_{NS/R}$$

where $N_{bos}$ is the number operator of the string coordinates $X$, while $N_{ferm}$ is the one for $\psi$. The ordering constant and the integer/semi-interger nature of $N_{ferm}$ depends on whether you are in the Ramond of NS sector.

In summary, they are two different theories, for instance notice that one lives in 26 dimensions and the other one in 10.

A good suggested reading on string theory is "Basic Concepts of String Theory" by Blumenhagen, Lüst, Theisen.

Answer 2

The answer to your first question is that the superstring analog of the first excited state of the bosonic string turns out to be massive. Instead, our friends the graviton, 2-form, dilaton and the massless vector are indeed obtained by acting with $\psi$ mode operators, not by $X$ mode operators.

I will now explain the contents of the NS sector of the superstring spectrum in some more detail. Recall that we have $\alpha$ oscillators coming from $\partial X \partial X$, which raise the level by integer numbers, and $b$ oscillators from the worldsheet fermions $\psi$, of which we consider the half-integer modes belonging to NS boundary conditions.

In the NS sector, the normal ordering constant $a_{NS}$ turns out to be equal to $\frac{1}{2}$. This means that the mass formula becomes $\alpha ' m ^2 = N - \frac{1}{2}$. For any state to be massless, one therefore needs $N = \frac{1}{2}$. Since states obtained by acting with $\alpha$ operators on the ground state have integer levels, there are no massless states $\sim \alpha|0\rangle$. Instead, the first $\alpha$ excited state, $\alpha ^\mu _{-1} |0\rangle$, has level $N = 1$ and therefore is massive with mass squared $m^2 = \frac{1}{2 \alpha '}$.

Excited states coming from acting with the $b$s can have half-integer levels. The first excited state of the NS open string is $b_{-1/2} ^\mu |0\rangle$, which has level $N= \frac{1}{2}$, as needed for a state to be massless. This state is a space-time vector and takes the role played by $\alpha ^\mu _{-1} |0\rangle$ in the bosonic string theory.

The closed string NS-NS sector is basically equivalent to twice the open string sector, with the requirement of level matching imposed and the modified mass formula $\alpha' m^2 = 4 (N - a) = 4n - 2$. The state $b^{\mu} _{-1/2} \tilde{b}^{\mu} _{-1/2} |0\rangle$ has level $N=1/2$ (we count only one set of modes) and is therefore massless. This is the state that contains the graviton, dilaton and the 2-form. In contrast, the state $\alpha ^\mu _{-1}\tilde{\alpha} _{-1} ^\mu |0\rangle$ has level $N=1$ and is massive with $m^2 = 1/\alpha'$. This state can be decomposed in the same way as in the bosonic theory. However, it is more meaningful to combine it with other $N=1$ states coming from the $b$ oscillators to yield a nice Lorentz multiplet, since, as a massive representation, it actually needs more states than a massless representation.

To answer your second question, note that the $X$ spectrum does not form a separate Hilbert space from the $\psi$ spectrum. A superstring can be in one of several ground states, labeled by their fermion boundary conditions and whether the string open or closed. Some of these states are tachyonic and removed by the GSO projection. The full spectrum can be generated from the ground states by the action of creation operators coming from $\psi$ or from $X$. There are no separate $X$ and $\psi$ sectors; the creation operators can be applied to all of the ground states. In particular, there is no separate tachyon 'coming from the $X$ spectrum'.