= Solution
Using $\alpha_0^2=-2$, tracelessness and transversality of $\epsilon_{\mu\nu}$ gives
$$
t^\mu{}_{\mu}=\frac1{20}(3\alpha_0^2+26)t^\lambda{}_{\lambda}
=t^\lambda{}_{\lambda},
$$
$$
t_{\mu\nu}\alpha_0^\nu
=\frac1{20}(3\alpha_0^2+1)t^\lambda{}_{\lambda}\alpha_{0\mu}
=-\frac14t^\lambda{}_{\lambda}\alpha_{0\mu}=-v_\mu,
$$
and $v\mathbin\cdot\alpha_0=-t^\lambda{}_{\lambda}/2$, so the constraints hold.
On the momentum vacuum,
$$
L_{-1}|0,p\rangle=(\alpha_0\mathbin\cdot\alpha_{-1})|0,p\rangle,
$$
$$
L_{-2}|0,p\rangle
=\left(\alpha_0\mathbin\cdot\alpha_{-2}
+\frac12\alpha_{-1}\mathbin\cdot\alpha_{-1}\right)|0,p\rangle,
$$
and
$$
L_{-1}^2|0,p\rangle
=\left(\alpha_0\mathbin\cdot\alpha_{-2}
+(\alpha_0\mathbin\cdot\alpha_{-1})^2\right)|0,p\rangle.
$$
Therefore
$$
|\psi\rangle=\epsilon_{\mu\nu}\alpha_{-1}^\mu\alpha_{-1}^\nu|0,p\rangle+|n\rangle,
$$
where
$$
|n\rangle=\frac{t^\lambda{}_{\lambda}}{10}
\left(L_{-2}+\frac32L_{-1}^2\right)|0,p\rangle.
$$
The vacuum has $L_0|0,p\rangle=\alpha_0^2|0,p\rangle/2=-|0,p\rangle$ and is annihilated by positive modes, so part c proves that $|n\rangle$ is null. The physical content is consequently the transverse traceless tensor $\epsilon_{\mu\nu}$.
Back to article page
