Set
$$
|s\rangle=\left(L_{-2}+\frac32L_{-1}^2\right)|\phi\rangle.
$$
Its $L_0$ eigenvalue is $-1+2=1$. The Virasoro algebra gives
$$
L_1L_{-2}|\phi\rangle=3L_{-1}|\phi\rangle,
\qquad
L_1L_{-1}^2|\phi\rangle=-2L_{-1}|\phi\rangle,
$$
so $L_1|s\rangle=0$. At critical central charge $c=26$,
$$
L_2L_{-2}|\phi\rangle=(4L_0+13)|\phi\rangle=9|\phi\rangle,
\qquad
L_2L_{-1}^2|\phi\rangle=-6|\phi\rangle,
$$
so $L_2|s\rangle=0$. All $L_n$ with $n\ge3$ annihilate it directly. Hence $|s\rangle$ is physical and, being a positive-level Virasoro descendant, spurious; it is therefore a null state.


Solution

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

 Solution

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