Past exam of the advanced quantum physics course of the University of Cambridge / 2012 / Paper 1 / 1 / b / Solution

For the linear rigid rotor, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ and the electric electric dipole moment interaction is $V=-E{\hat{d}}_z$. The isotropic $\ell =0$ state has no first-order shift. Its only coupled rotational state is $\ell =1,m_{\ell }=0$, at gap ${\Delta }_{\mathrm{rot}}={\hslash }^2/I$, giving the quadratic Stark shift of a polar linear rotorUsing $1\mathrm{kV}\mathrm{c}{\mathrm{m}}^{-1}=10^5\mathrm{V}{\mathrm{m}}^{-1}$ and the supplied I and d givesThe negative sign is the expected second-order Stark effect of a nondegenerate ground state. The mixing parameter $dE/(\sqrt{3}{\Delta }_{\mathrm{rot}})\simeq 0.074$ is small, supporting the weak-field estimate. The constants h and $\hslash =h/(2\pi )$ use their defined SI values; no molecular input beyond the supplied data is needed.