Past exam of the advanced quantum physics course of the University of Cambridge / 2014 / Paper 1 / 1

The normalized quantum harmonic oscillator ground state has $|A{|}^2=\sqrt{m\omega /(\pi \hslash )}$ and energy $E_0=\hslash \omega /2$. Its classical turning points are $x=\pm \sqrt{\hslash /(m\omega )}$. With $y=x\sqrt{m\omega /\hslash }$, the ground-state oscillator probability beyond the turning points isThe stated rounded error function gives $0.16$; the more precise Gaussian integral is $0.15730$.

An impulse integrated across its full support produces the impulsive quantum kick $K=e^{-i\hat{U}/\hslash }$. Thus the leading transition amplitude is $-i{U}_{21}/\hslash$ andHere $U_{21}$ has units of action. The off-diagonal two-state impulse is exactly $P={\sin }^2(|U_{21}|/\hslash )$ for a Hermitian two-state impulse with zero diagonal. No energy denominator occurs: the impulse is instantaneous.

For two equivalent $p$ electrons, antisymmetry permitsTheir $9+5+1=15$ states exhaust $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund's rules first maximize spin, then orbital angular momentum; the less-than-half-filled $p$ shell puts the smallest $J$ lowest. Hence the ground level is ${}^3{P}_0$. This is the allowed LS terms of a p-squared configuration.^[ref]