# NS sector and the Numberoperator

1. May 9, 2015

### moriheru

αιM2=(-1/2+N)
then, if N=1/2 αιM2=0, additionally if N=0 then αιM2=-1/2 and so on.
When N=1/2 then the fermions are all masless (as M^2 must be equivalent to 0 as α is not) and there are 8 states.
I don't know if this is to specific or not described well but my question is: how does one get to the 8 fermion states and how many states would one have when the N is ,say, equivalent to 3/2?

Please excuse, I Couldnt find a better header. Thanks for any help.

2. May 9, 2015

### fzero

I'm not sure what reference you're working from, but most texts (like Polchinski) do work the details out for the massless NS states. In light-cone gauge, we have the states $\psi^i_{-1/2} |0;k\rangle_\text{NS}$ and $i=1,\ldots 8$ because we only have the transverse degrees of freedom.

In the covariant description, we have $\psi^\mu_{-1/2} |0;k\rangle_\text{NS}$, but also the superconformal constraints

$$L_m |\Psi\rangle = G_r |\Psi\rangle =0, ~~~m,r\geq 0$$

to apply to any physical state. In addition, we must identify null states $|\chi\rangle$ satisfying $\langle \psi' | \chi\rangle=0$ for all physical states $|\psi'\rangle$. To outline this, we introduce a polarization vector $e_\mu$ so that we can write

$$|e;k\rangle_\text{NS} = e_\mu \psi^\mu_{-1/2} |0;k\rangle_\text{NS},$$

then requiring $L_0 |e;k\rangle_\text{NS}=0$ gives the mass shell condition $k^2=0$. Requiring $G_{1/2} |e;k\rangle_\text{NS}=0$ leads to $k^\mu e_\mu =0$. Finally, we find a null state $G_{-1/2}|0;k\rangle_\text{NS} \propto k_\mu \psi^\mu_{-1/2} |0;k\rangle_\text{NS}$ that requires us to identify $e^\mu \cong e^\mu + \lambda k^\mu$. These relations are completely parallel to the conditions found on the photon polarization 4-vector in EM: they remove the timelike and longitudinal components of $e_\mu$ leaving the, in this case, 8 transverse components that we compute in light-cone gauge.

At the first massless level, we find the states

$$\begin{split} \alpha_{-1}^i \psi_{-1/2}^j|0;k\rangle_\text{NS}, & ~~~8\cdot 8 = 64 ~\text{states}, \\ \psi_{-1/2}^i \psi_{-1/2}^j\psi_{-1/2}^k|0;k\rangle_\text{NS}, & ~~~8\cdot 7\cdot 6/(3\cdot 2\cdot 1) = 56 , \\ \psi_{-3/2}^i|0;k\rangle_\text{NS} , & ~~~8,\end{split}$$

leading to 128 states. Since these are massive, they must correspond to $SO(9)$ representations. We can fit them into $\mathbf{44}\oplus\mathbf{84}$, where the first is the symmetric, traceless 2-index tensor and the second is the completely antisymmetric 3-index tensor. The calculation in covariant formalism looks straightforward but quite tedious, so I won't attempt it.

3. May 10, 2015

### moriheru

Thanks, helped alot.

4. May 10, 2015

### moriheru

I was refering to Zwiebach chapter 14 on superstring theory. I assume that the ket with k,0 NS was the NS-vacuum state?! Not understadning the 8 fermion states was my own foolishness (one time 8 space) for 1/2, but I do not understand a) why you add the psi functions, b) why you change from the k super scripts to the i and the j. I mean if one uses the i and the j then why change to the k in one line then use the i in the next, the j in the next one and so on. I understand the
the second line of that but not why you divide by 6....
Thanks for any clarifications.

5. May 10, 2015

### fzero

Yes it's the NS ground state.

I'm not sure what you mean by "add," but I'll interpret it to mean why we have multiple factors of $\psi$ in order to get the first massive level? The physical mass of these states is related to the conformal weight which is in turn related to the weight of the oscillators used to construct them. To get $N=3/2$ we can either use a high-weight oscillator or some combination of lower weight ones.

$i,j,k=1,\ldots 8$ in the superscripts, I probably should not have used $k$ because I used it for the momentum. The particular choice of labeling doesn't really matter (as long as it's consistent with antisymmetry of the $\psi$s etc). You can relabel them some other way.

$6=3!$ accounts for the fact that permutations of the 3 indices are to be identified by antisymmetry.

6. May 10, 2015

### moriheru

I meant include the psi function sorry that was a bit unclear. Thanks I get it .