Spectrum of the Bosonic String

Closed Strings The spectrum of the closed string can be obtained from the combination of the left-moving states and the right-moving ones. The ground state ($N_L =N_R =0$) is given by $\alpha 'M^2=-4$. That means that the ground state includes a tachyon. The first excited state ($N_L=1=N_R$) is massless and it is given by ${\alpha}^{i}_{-1} \widetilde{\alpha}^{j}_{-1} \vert,P \rangle$. This state can be naturally decomposed into irreducible representations of the little group $SO(24)$ as follows

$${\alpha}^i_{-1}\widetilde{\alpha}^j_{-1}\mid 0,P \rangle= ... ...{\alpha}^k_{-1}\widetilde{\alpha}^k_{-1}\bigg)\mid 0,P \rangle$$

$$+~{1\over D-2}{\delta}^{ij}{\alpha}^k_{-1}\widetilde{\alpha}^k_{-1}\mid 0,P \rangle .$$ (9)

The first term of the rhs is interpreted as a spin 2 massless particle $G_{ij}$ (graviton). The second term is a range 2 anti-symmetric tensor $B_{ij}$. While the last term is an scalar field $\Phi$ (dilaton). Higher excited massive states are combinations of irreducible representations of the corresponding little group SO$(25).$ truecm Open Strings For the open string, the ground state includes once again a tachyon since $\alpha 'M^2=-1$. The first exited state $N=1$ is given by a massless vector field in 26 dimensions. The second excitation level is given by the massive states ${\alpha}^{i}_{-2}\mid 0,P \rangle$ and ${\alpha}^{i}_{-1}{\alpha}^{j}_{-1}\mid 0,P \rangle$ which are in irreducible representations of the little group SO$(25)$. truecm