Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 1 / A2 / d / Solution

The density $|{\Psi }_0{|}^2=\sqrt{\alpha /\pi }{e}^{-\alpha x^2}$ is even. Hence $⟨x⟩=0$; also ${\Psi }_0^{\prime }=-\alpha x{\Psi }_0$ gives $⟨p_x⟩=i\hslash \alpha ⟨x⟩=0$.

To evaluate the second position moment, differentiate the Gaussian integral with respect to its positive parameter:For the momentum second moment, apply the momentum operator twice:Both means vanish, soThe ground state attains the bound proved above and is therefore a minimum uncertainty state. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.