The zero-mode condition is automatic:
$$
L_0|\psi\rangle=\left(\frac12\alpha_0^2+2\right)|\psi\rangle=|\psi\rangle.
$$
Using $[L_m,\alpha_n^\mu]=-n\alpha_{m+n}^\mu$ and oscillator annihilation of the vacuum gives
$$
L_1|\psi\rangle
=2\left(t_{\mu\nu}\alpha_0^\nu+v_\mu\right)
\alpha_{-1}^\mu|0,p\rangle,
$$
which vanishes when $t_{\mu\nu}\alpha_0^\nu=-v_\mu$. Similarly,
$$
L_2|\psi\rangle
=\left(t^\mu{}_{\mu}+2v\mathbin\cdot\alpha_0\right)|0,p\rangle,
$$
which vanishes when $v\mathbin\cdot\alpha_0=-t^\mu{}_{\mu}/2$. Higher positive Virasoro modes annihilate a level-two state. These are precisely the stated physical-state conditions.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 306 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-306/4/d/solution