(a) The potential for the one-dimensional harmonic oscillator is $V(x)=\frac{1}{2} m \omega^{2} x^{2}$. By considering the associated time-independent Schrödinger equation for the wavefunction $\psi(x)$ with substitutions

$$\xi=\left(\frac{m \omega}{\hbar}\right)^{1 / 2} x \quad \text { and } \quad \psi(x)=f(\xi) e^{-\xi^{2} / 2}$$

show that the allowed energy levels are given by $E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega$ for $n=0,1,2, \ldots$ [You may assume without proof that $f$ must be a polynomial for $\psi$ to be normalisable.]

(b) Consider a particle with charge $q$ and mass $m=1$ subject to the one-dimensional harmonic oscillator potential $U_{0}(x)=x^{2} / 2$. You may assume that the normalised ground state of this potential is

$$\psi_{0}(x)=\left(\frac{1}{\pi \hbar}\right)^{1 / 4} e^{-x^{2} /(2 \hbar)}$$

The particle is in the stationary state corresponding to $\psi_{0}(x)$ when at time $t=t_{0}$, an electric field of constant strength $E$ is turned on, adding an extra term $U_{1}(x)=-q E x$ to the harmonic potential.

(i) Using the result of part (a) or otherwise, find the energy levels of the new potential.

(ii) Show that the probability of finding the particle in the ground state immediately after $t_{0}$ is given by $e^{-q^{2} E^{2} /(2 \hbar)}$. [You may assume that $\int_{-\infty}^{\infty} e^{-x^{2}+2 A x} d x=$ $\sqrt{\pi} e^{A^{2}}$.]